Thermals Fluids 1 Ch 12 Guide

12-1

An undergraduate heat transfer lab has an experiment to illustrate the effects of different boundary conditions on heat transfer from from a flat plate. A test section is installed installed into a wind tunnel. The test section consists of 100 thin strip heaters placed on a 2-m long flat plate. Each heater, 20-mm long and 250-mm wide, is located so that there is no space between adjoining heaters and is electrically and thermally insulated from the adjacent heaters; the backside of the plate is heavily insulated. insulated. The power to each heater can be individually controlled. controlled. The free-stream air temperature temperature is 25 ºC and has a velocity velocity of 4 m/s. By controlling the power to each strip, two different boundary conditions conditions can be modeled. Ignoring radiation, determine for strips 1, 5, 25, 100, and 200: a. the heat transfer rate when the power is adjusted in each heater to maintain a uniform plate temperature of 50 ºC (in W) b. the wall temperature temperature on strip strip number 25 when the power is adjusted in each heater to maintain maintain a uniform heat flux (equal to that on strip 25 from the first part) over the entire plate (in ºC).

Approach: Each heater is individually controlled so that the temperature or heat flux distribution distribution can be set. To find the total power dissipated by the heaters, we need to calculate the heat transfer coefficient for the two different boundary conditions.

Assumptions: 1. Air is at at one atmosphere. atmosphere. 2. Radiation is ignored. ignored.

Solution: Q = hA (Tw − T∞ )

a) The power from the plate plate can be calculated from

We ignore radiation. The heat transfer coefficient coefficient is determined from from the appropriate cor correlation. relation. We need the Reynolds number: Re = ρ V∞ L µ . The properties properties are evaluated (Appendix (Appendix A-7) A-7) at the film film temperature temperature T film = (T (T ∞ T w)/2 = (25 + 50)/2 = 37.5 oC, ∞ + 0.706 Re =

(1.137 kg

m3

= ρ =

) ( 4m s ) ( 2m) (1Ns

1.89 ×10 N ⋅ s/m -5

2

1.137 kg/m3, k = = 0.0270 W/mK, kg m )

2

= µ =

1.89 × 10-5 N⋅s/m2, Pr = =

= 481,000

This is laminar flow. The average heat transfer coefficient coefficient on a flat plate is obtained from 410 ( 0.02 0.0270W/ 70W/m mK ) Nu k 410 1/ 2 1/ 3 Nu L = 0.664 Re1L/ 2 Pr 1/ 3 = 0.664 ( 481000) ( 0.706) = 410 h = L = 5.54W W/m2 K = 5.54 L 2m →

Q = ( 5.54 W m2 K ) (100 ) ( 0.02m) ( 0.25m) ( 50 -25) o C =6 = 69.2 W

Answer

b) For the uniform heat flux situation, situation, we need to calculate the local heat transfer transfer coefficient coefficient on strip number number 25 in part (a) to calculate the local heat flux. Then, assuming that heat flux is uniform over the whole plate, we calculate the heat transfer coefficient with with the correlation for the uniform heat flux boundary condition. We assume that the heat transfer coefficient at the center of the strip is representative of the whole strip. Re x =

(1.137 ) ( 4) ⎡⎣ 24( 0.02) + 0.02 / 2⎤⎦ 1.89 ×10-5

=118,000

The local heat transfer coefficient with a uniform wall temperature is: h x (102 ) ( 0.0270) 1/ 2 1/ 3 118, 000) ( 0.706) = 102 = 5.59 Nu x = x = 0.332 Re1x / 2 Pr 1 / 3 = 0.332 (11 h x = 5.59W W/m2 K k 24 ( 0.02) + 0.02 / 2 →

So the heat flux on strip number 25 is

q ′′ = h (Tw − T∞ ) = ( 5.59 W m2 K )( 50 -2 -25) o C = =1 140 W/ W/m2

The heat transfer coefficient for a laminar flow with a uniform heat flux boundary condition is (139 ) ( 0.0270) 1/ 2 1/ 3 Nu x = 0.453 Re1x / 2 Pr 1 / 3 = 0.453 ( 11 118, 000) ( 0.706) = 139 h x = 3W/m2 K = 7.63W/m 24 ( 0.02 ) + 0.02 / 2 →

So the surface temperature is from q ′′ = h ( Tw − T∞ ) Tw = T∞ + q′′/h = 25 o C + (140 W m2 ) / ( 7.63W/m2 K ) = 43.3 o C

12- 1

Answer

12-2

Many schemes have been proposed to supply arid regions with fresh water. One plan involves towing icebergs from the polar regions to dry regions that need fresh water. Consider an iceberg, 1000-m long and 500-m wide, that is towed towed through 10 ºC water at a velocity velocity of 1km/hr. The density of ice is 917 kg/m3 and the heat of fusion is 333.4 kJ/kg. Determine: a. the average rate at which the flat bottom of the iceberg will melt (in mm/hr) b. how much ice will melt if the voyage is 1500 km long (in kg).

Approach: We define a control volume to encompass the iceberg. Heat transfer from the sea water melts the ice. An energy balance will equate the heat transfer with the change in the iceberg’s internal energy. The heat transfer rate is calculated once a heat transfer coefficient is estimated.

Assumptions: 1. The system is closed with with no work or potential or kinetic energy effects. 2. The bottom of the iceberg acts acts as a flat plate.

Solution: a) The average average melting rate dH /dt is is determined from an energy balance on a closed system that is defined as the iceberg. Assuming the bottom of the iceberg acts as a flat plate, there is no work, no mass flow across the control volume boundary, and negligible potential and kinetic energy changes. Conservation of energy gives us dU d d Q = = ( mu sf ) = ( ρ LWH us f ) dt dt dt  h L ( LW ) ( T∞ − Tice ) hL ( T∞ − Tice ) dH Q

=

dt

ρ u sf

=

LW

ρ usg

=

LW

ρ usf

The average heat transfer coefficient requires the Reynolds number, so for water from Appendix A-6 at the T film = o (T ∞ T ice = 999.9 kg/m3, k = = 0.578 W/mK, µ = = 15.0 × 10-4 N⋅s/m2, Pr = = 10.91 ∞ + ice)/2 = (10 + 0)/2 = 5 C, ρ = Re L =

ρ V∞ L

=

kg ( 999.9 kg

m3 ) (1 km hr ) (1000 m 1km) ( 1hr 3600s) ( 1000m)

(15.0 ×10

-4

µ

N⋅s m

2

1N ⋅ s ) ) ( kg m 1N 2

= 1.85×108

This is highly turbulent, so ignoring the very short laminar contribution, the average heat transfer coefficient is Nu L = 0.037 ReL0.8 Pr 1 / 3 = 0.037 (1. 1 .85× 108 ) h L = dH dt

Nu L k L

=

=

000 ) ( 0.578 W ( 337, 00

(195 W kg ( 999.9 kg

= 5.85 ×10-6

mK )

1000m

0.8

1/ 3 (10.91) = 337, 000

5W/m2 K =195W/

m2 K ) (10 − 0) o C (1J/s 1W 1W ) m3 ) ( 333.4 kJ kJ kg ) ( 1000J 1kJ) m

= 5.85×10-3

mm

s s b) The total ice melted for the the voyage

5 = 21.1

mm

Answer

hr

⎛ dH ⎞ ⎛ dH ⎞ ⎛ Lv ⎞ ∆m = ρ LW ⎜ ⎟ ⎟t = ρ LW ⎜ ⎟⎜ ⎝ dt ⎠ ⎝ dt ⎠ ⎝ V∞ ⎠ kg ⎞ m ⎞ ⎛ 1,500,000 m ⎞ ⎛ 1km/1000 m ⎞ ⎛ ⎛ = ⎜ 999.9 3 ⎟ (1000m ) ( 500m) ⎜ 5.85×10-6 ⎟ ⎜ ⎟⎜ ⎟ m ⎠ s ⎠ ⎝ 1km/hr ⎠ ⎝ 1hr/3600s ⎠ ⎝ ⎝ =1.58×1010 kg

Answer

12- 2

12-3

Two brothers have rooms side by side in a flat-roofed mobile home. The older brother continually complains that his room is colder than that of the younger brother. The older brother decides to add more heating to his room. As shown below, the first first room is 4-m long and the second second one is 3-m long; each is 4m deep into the plane of the page. The roof thickness is 0.25 m with a thermal conductivity of 1.2 W/m·K. The outside wind is parallel (out of the plane of the page) to the roof at a velocity of 20 km/hr at -10 ºC, the inside temperature is to be maintained at 21 ºC, and the inside heat transfer coefficient is 7.5 W/m 2·K. Determine the heat loss from the roof of each of the two rooms (in W).

Approach: The basic basic heat heat transf transfer er rate rate equat equation ion ( Q = ∆T / R Rtot ) can be used to calculate calculate the heat transfer transfer rate. The average heat transfer coefficient for each of the two rooms must be determined.

Assumptions: 1. The heat transfer transfer is one dimensional. dimensional. 2. Air is at at one atmosphere. atmosphere. 3. The roof can be treated as a flat flat plate.

Solution: Q =

The heat transfer can be calculated with

∆T

T IN − T f

=

Rtot

1

+

hi Ai

t kA

+

1 ho Ao

We need the average heat transfer coefficients for the two rooms, which requires the Reynolds number with air properties from Appendix A-7 evaluated evaluated at T film = (-10 + 21)/2 = 5.5 oC = 278.5 K, ρ = = 1.267 kg/m3, k = = 0.0245 W/mK, = 1.69 × 10-5 N⋅s/m2, Pr = = 0.714 µ = Re L1 =

(1.267kg/m ) ( 5.56m/s) ( 4m) ( N ⋅ sm 3

ρ V L1

=

1.69 ×10 N ⋅ s/m -5

µ

2

kg m)

2

=1.667 ×106

Similarly, Re L 2 = 2.918 ×10

6

This is turbulent flow, and ignoring the minor laminar contribution for Room A Nu L = 0.037 ReL0.8 Pr 1 / 3 = 0.037 (1. 1.667 ×106 ) h A =

Nuk L1

=

3140) ( 0.02 0.0245 45W W/mK /mK ) ( 3140 4m

0.8

1/ 3 ( 0.714) = 3140

19.2W/m 2W/m2 K =19.

For Room B, we need to evaluate the average h from L from L1 = 4m to L to L2 = 7 m h B =

1

h dA = A ∫ W (L

1

x

2

h B =

=

h Wd = L −L ) ∫ x

1

1

L2

k

∫ x ( 0.0296 Re

0 .8 x

2 − L1 L

Pr 1 / 3 )dx

1

0.037k

⎡⎣ ReL0.8 − ReL0.8 ⎤⎦ Pr 1 / 3 L2 − L1 2

1

0.03 0.037 7 ( 0.02 0.0245 45) ⎡

0.8

( 2.918 ×10 ) ⎣⎢

( 7 − 4)

6

0.8 1/ 3 − (1.667 × 106 ) ⎤⎥ ( 0.714) = 14.5 W m2 K ⎦

⎡⎣ 21 − ( −10 ) ⎤⎦ K

Q A =

1

( 7.5 W Q B =

m2 K ) ( 4 m ) ( 4m 4m )

+

0.25m

(1.2 W

m K ) ( 4 m) ( 4m)

⎡⎣ 21 − ( −10 ) ⎤⎦ 1

+

0.25

+

+

1

(19.2W

= 906W

1

( 7.5) ( 3) ( 4) (1.2) ( 3) ( 4) (14.5) ( 3) ( 4)

12- 3

=1260W

Answer

m2 K ) ( 4m 4 m) ( 4m 4 m)

Answer

12-4

Rolling mills are used to reduce the thickness of steel plates to create thin steel strips. The metal must be at a high temperature so that the power (force) required to reduce the metal thickness is not excessive and so that the desired material properties are obtained. Consider a 304 stainless steel strip 3-mm thick leaving a rolling mill at 1000 ºC at a speed of 20 m/s. A length of 50 m is exposed to air at 35 ºC. Convective heat transfer occurs on both the top and bottom surfaces of the strip. Ignoring radiation and axial conduction in the steel, estimate the temperature of the strip when it reaches 50 m from the roller (in ºC).

Approach: We will use conservation of energy on the strip, similar to what was done in Section 12.5 for a heat exchanger with a constant temperature fluid. Note that this “flow” of 304 stainless steel is analogous to the flow of a fluid. When we analyze this metal working process, we will obtain the same solution as we obtained for the constant temperature heat exchanger.

Assumptions: 1. The heat transfer coefficient is constant. 2. Radiation is ignored. 3. Air is at one atmosphere. 4. The flow can be approximated as flow over a flat plate.

Solution:  p) The equation for the outlet temperature is Tout = T f + (Tin − T f )exp ( − hA mc We have assumed that h is uniform over the length of the metal strip and that radiation can be ignored We assume that flow over a flat plate is a reasonable model of the flow. The average heat transfer coefficient requires the Reynolds number with air properties from Appendix A-7 evaluated at (approximately) T film = (1000 + 35)/2 = 517.5 oC = 790.5 K, so ρ = 0.4465 kg/m3, µ = 3.60 × 10-5 N⋅s/m2, k = 0.0573 W/mK, Pr = 0.688 Re =

ρ V L µ

( 0.4465 kg/m ) ( 20m/s)( 50m) 3

=

3.60 ×10-5 N ⋅ s/m 2

= 1.24 ×107

This is turbulent flow, so ignoring the minor laminar contribution Nu = 0.037 Re L0.8 Pr 1/ 3 = 0.037 (1.24× 107 ) h=

Nuk L

=

(15450)( 0.0573W/mK ) 50m

0.8

1/ 3 ( 0.688) = 15, 450

= 17.7W/m2 K

From Appendix A-2 for 304 SS, at 1200 K,

c p = 640 J/kg K,

ρ =

7900 kg/m3

⎡ ⎤ - (17.7W/m 2 K ) ( 50m ) W ⎥ = 997 o C T out = 35 C + (1000 -35) K exp ⎢ 3 ⎢⎣ ( 7900kg/m ) ( 0.003m) W ( 20 m/s) 640 J/kg k ⎥⎦ o

Comments: No radiation is not a good assumption because of the high metal temperature. The cooling rate with air is very low. Water is more typically used.

12- 4

Answer

12-5

The failure rate of computer chips increases with increasing operating temperature. Consider a 15-mm by 15-mm chip that is cooled on its top surface by a 5 m/s flow of 25 ºC air. Any heat transfer from its bottom surface to the circuit board is ignored. Because of the chip construction, the electrical power dissipated in the chip results in a uniform heat flux over the surface of the chip. The maximum temperature that any part of the chip can experience is 80 ºC. Determine: a. the maximum allowable chip power (in W) b. the maximum allowable chip power if this chip is the fifth in a column of identical chips all mounted flush to the surface with no space between the chips (in W).

Approach: With uniform heat flux on each chip, the maximum temperature will occur at the down steam end. The local heat transfer coefficient should be used in the rate equation ( Q = hA ∆T ) to determine the maximum permissible power.

Assumptions: 1. Air is at one atmosphere. 2. The flow can be treated as flow over a flat plate.

Solution: The maximum allowable power to keep all parts of the chip less than 80 oC can be calculated with Q = h x A ( (Ts − T f ) where h x is the local heat transfer coefficient. We assume flow over a flat plate with a uniform heat flux is a reasonable model of the flow. The Reynolds number must be evaluated at T film ~ (25 + 80)/2 = 52.5 o C = 325.5 K. So from Appendix A-7: ρ = 1.084 kg/m3, k = 0.0281 W/mK, µ = 1.96 × 10-5 N⋅s/m2, Pr = 0.703 a) Re =

ρ V L

(1.084kg/m ) ( 5m/s)( 0.015m) ( N 3

=

⋅

s2 kgm )

1.96 ×10-5 N ⋅ s/m2

µ

=

4150

This is laminar flow, so 1/ 2

Nu x = 0.453 Re1x / 2 Pr 1/ 3 = 0.453 ( 4150) h x =

Nuk x

=

( 25.9)( 0.0281W/mK ) 0.015m

=

1/ 3

( 0.703)

=

25.9

48.6W/m2 K

Q = ( 48.6W/m 2 K ) ( 0.015m )( 0.015m) [ 80 − 25] K = 0.60W

Answer

b) For the fifth chip, L = 5(0.015 m) = 0.075 m Re x = 20750 Nu x = 58.0 ( 58.0 )( 0.0281) 2 h x = = 21.7W/m K 0.075 Q = (21.7) (0.015) (0.015) (80 – 25) = 0.27 W

Answer

Comments: Because the heat transfer coefficient decreases with length, there is a dramatic decrease in allowable power if the chip is in the fifth place rather than the first. This illustrates well why you should be aware of whether a local or an average heat transfer coefficient is needed in a problem.

12- 5

12-6

The walls of a house are constructed of an exterior sheathing, insulation, framing timber, and drywall, and their composite resistance is estimated to be 4.15 m2·K/W. A winter wind blows parallel to the 3-m high, 18m long wall. The wind velocity is 30 km/hr, and its temperature is -5 ºC. The heat transfer coefficient at the interior of the house is 5 W/m2·K. For an inside air temperature of 21 ºC, determine the heat transfer rate through the wall (in W).

Approach: The basic heat transfer rate equation ( Q = ∆T / Rtot ) is used to calculate the heat transfer rate. The total thermal resistance is composed of the inside convective resistance, the wall resistance, and the external convective resistance, which must be determined from appropriate heat transfer coefficient correlations.

Assumptions: 1. The heat transfer is one dimensional. 2. The external flow can be treated as flow over a flat plate. 3. The air is at one atmosphere.

Solution: Tin − T f ∆T The heat transfer rate equation is Q = = Rtot Rconv ,in + Rwall + Rconv, out where

Rconv,in = Rwall =

′′ Rwall A

Rconv,out =

1 hi A

=

=

1

(5 W

m K ) ( 3m )(18m) 2

4.15m2 K/W

( 3m )(18m)

=0.00370K/W

= 0.0769K/W

1 ho A

We have forced convection over an assumed flat plate and need the Reynolds number. For T film we assume the temperature of the outside surface is T s ~ 0 oC, T film = (0-5)/2 = -2.5 oC = 270.5 K and the air properties (from Appendix A-7 by interpolation) are: ρ = 1.305 kg/m3, µ = 1.63 × 10-5 N⋅s/m2, k = 0.0238 W/mK, Pr = 0.716 Re L =

ρ V L

=

(1.305 kg

m3 ) ( 8.33m/s )(18m) ( Ns2 kgm )

(1.63×10

−5

µ

N⋅s m

2

)

= 1.20 × 107

This is turbulent flow. Ignoring the minor laminar contribution and using the correlation for the average heat transfer coefficient over the total length: Nu L = 0.037 ReL0.8 Pr 1/ 3 = 0.037 (1.20 × 107 ) h=

Nuk L

=

Rconv,out = Q =

(15300)( 0.0238W/mK ) 18m

1/ 3 ( 0.716) = 15,300

= 20.2W/m2 K

1

( 20.2W/m K )( 3m )(18m) 2

0.8

= 0.00092K W

⎡⎣ 21 − ( −5) ⎤⎦ K = 319W [0.00370 + 0.0769 + 0.00092] K/W

12- 6

Answer

12-7

One wall of an older office building (6-m high and 30-m long) is all glass 7-mm thick. Wind blows parallel to it at 20 km/hr and 5 ºC. The inside surface temperature of the glass is 20 ºC. Determine: a. the heat transfer rate from the glass (in W) b. the heat transfer rate if the wind velocity is tripled (in W).

Approach: The basic heat transfer rate equation ( Q = ∆T / Rtot ) coefficient over the glass.

Assumptions: 1. The heat transfer is one dimensional. 2. Air is at one atmosphere. 3. Ignore the laminar contribution to the heat transfer coefficient.

Solution: a) The heat transfer rate equation is: Tw − T f Tw − T f ∆T Q = = = Rtot Rglass + Rconv t kA + 1 h A From appendix A , for plate glass k = 1.4 W/mK. The heat transfer coefficient requires the Reynolds number with air properties from Appendix A-7 evaluated at the film temperature. Assuming T s ~ 10oC, T film = (10 + 5)/2 = 7.5 oC = 280.5K, so that ρ = 1.258 kg/m3, µ = 1.70 × 10-5 N⋅s/m2, k = 0.02465 W/mK, Pr = 0.714. ρ V L

Re =

=

(1.258kg

µ

m3 ) ( 5.56 m/s )( 30 m) ( Ns2 kg m) 1.70 × 10-5 N ⋅ s/m2

7

= 1.23 × 10

This is turbulent flow, and ignoring the minor laminar contribution, the average heat transfer coefficient for flow over a flat plate is: Nu L = 0.037 Re L0.8 Pr 1/3 = 0.037 (1.23 × 107)0.8 (0.714)1/3 = 15,600 h=

Nuk

Q =

b) If

L

=

(15600)( 0.02465W/mK ) 30 m

= 12.8

( 20 − 5) K 0.007 m (1.4 W mK ) ( 6m )( 30 m)

V2 =

3 V1 ,

+

W m2 K

=

1

(12.8 W

m2 K

32,500 W

Answer

) ( 6 m )( 30 m)

ReL,2 = 3 Re L,1

Nu L,2 = ( Re L,2 / Re L,1)0.8 Nu L,1 = 37,600 h2 = 30.9 W/m2K Q = 72,300 W

Answer

Comments: This is a substantial energy loss. The use of a double pane window would probably more than halve this energy loss.

12- 7

12-8

The roof of a minivan can be approximated as a flat plate, 2-m wide and 3.5-m long. The sun beats down on the roof such that the net solar radiation absorbed is 350 W/m 2. If the ambient air is 32 ºC, the car is moving at 100 km/hr, and the inside surface of the roof is heavily insulated, determine the steady-state temperature of the roof (in ºC). Ignore radiation emitted from the roof.

Approach: An energy balance on the roof will equate the incoming solar energy with the outgoing convective heat transfer. The convective heat transfer coefficient must be evaluated to calculate the convective heat transfer.

Assumptions: 1. 2. 3. 4. 5.

The control volume is steady. Potential and kinetic energy effects are negligible. No work is done on or by the control volume. The flow is over a flat plate. Laminar flow can be ignored.

Solution: We define a control volume around the roof and assume steady, negligible potential and kinetic energy effects, and no work. Applying conservation of energy, we obtain: Q = 0 Recognizing that this heat transfer is the net heat transfer: Q solar

 −Q conv =

0

′′ A − hA (Ts q solar ′′ = T f + qsolar

T s

)

− T f =

0

h

The convective heat transfer coefficient requires the Reynolds number with properties evaluated at T film

=

(T

− T f

s

)

2 . Assuming T s

Appendix A-7 are: Re L

=

ρ V L

ρ =

=

≈

45 o C , T film

=

(32 + 45) / 2 ≈ 38.5 o C , by interpolation the air properties from

1.133 kg/m3; µ = 1.90×10-5 N·s/m2; k = 0.0271 W/mK; Pr = 0.705

(1.133kg

m 3 ) ( 27.8 m s )( 3.5 m ) ( N is 2 kg im ) 2

-5

=

5.80 ×106

1.90×10 N is m This is turbulent flow. Assuming flow over a flat plate and ignoring the minor laminar contribution, the average heat transfer coefficient is:

Nu h= T s

µ

=

0.037 Re L0.8 Pr 1 3

Nu ik

=

L

=

32 o C+

=

0.037 (5.80 ×10 6 )

( 8480 )( 0.0271W

m iK )

3.5m

0.8

13

( 0.705 )

= 8480

=65.7 W m 2 i K

350 W m 2

=37.3 o C 65.7 W m iK

Answer

2

Comments: Our assumed surface temperature is close enough, or we could iterate one time for an improved answer.

12- 8

12-9

Power transformers change the voltage of electricity, but the devices are not 100% efficient. Dissipated heat must be removed from transformers so that they do not reach a temperature that could damage them. Consider a transformer that dissipates 30 W. It is 10-cm wide and 20-cm long with 8 fins 2-cm tall, 2-mm thick, and 20-cm long evenly distributed across the surface of the transformer. Air at 25 ºC is blown parallel along the length of the fins. Assume the fins have a fin efficiency of 100%. Ignoring radiation, and for a base temperature less than 65 ºC, determine the minimum air velocity required (in m/s).

Approach: The heat transfer rate equation ( Q = hA ∆T ) can be used to determine the required heat transfer coefficient, h. Once that is known, the appropriate correlation can be used to calculate the velocity.

Assumptions: 1. 2. 3. 4.

Air is at one atmosphere. Radiation is ignored. The fin efficiency is 100%. The flow can be approximated as that over a flat plate.

Solution: Q = hA (T s − T f ) ⇒

The heat transfer rate equation is:

h=

Q A (T s − T f )

where A = LW + 2 NHL = (0.2m)(0.1m) + 2(8)(0.02m)(0.2m) = 0.084 m2 30W h= = 8.93 W /m 2 K 2 ( 0.084m ) ( 65 - 25) K This is the minimum h needed to ensure T s < 65oC. We assume flow over a flat plate is applicable to this situation, and evaluate air properties from Appendix A-7 at T film = (65 + 25)/2 = 45 ºC, ρ = 1.110 kg/m2, µ = 1.93 × 10-5 N⋅s/m2, k = 0.0276 W/mK, Pr = 0.704. An average heat transfer coefficient is needed. We assume the flow is laminar. This will need to be checked for an isothermal flat plate. Nu L = hk L = 0.664 Re Pr 1/ 2 L

1/ 3

⎡ ⎤ hL ReL = ⎢ 1/ 3 ⎥ ⎣ 0.664 k Pr ⎦

→

2

2

⎡ ⎤ 8.93W/m2 K ) ( 0.2m ) ( ⎥ = 12,000 Re L = ⎢ 13 ⎢⎣ ( 0.664 )( 0.0276 W/mK )( 0.704) ⎥⎦ Assuming transition to turbulence occurs at a Reynolds number of about 500,000, this flow is laminar, so our assumption is valid. Therefore, Re L =

ρ V L µ

→

V

=

µ Re L ρ L

=

(1.93×10−

5

N ⋅ s/m2 ) (12, 000)

(1.110 kg/m ) ( 0.2 m) ( Ns 2

12- 9

2

kgm )

=1.04 m/s

Answer

12-10 Solar-powered planes have been designed to be able to stay aloft for very long times. Proposed uses include meteorology and surveillance. Photovoltaic (PV) cells are mounted on the top surface of the wing. The PV panel is 1.5 m wide and 8-m long. The solar energy absorbed but not converted to electricity equals about 850 W/m2, and the PV cell conversion efficiency decreases with increasing temperature. Determine the temperature of the trailing edge of the panel when the plane flies at 110 km/hr at 5000 m where the pressure is 54 kPa and temperature is 256 K.

Approach: The heat transfer rate equation ( Q =hA ∆T ) can be used. The heat transfer coefficient must be determined.

Assumptions: 1. Laminar effects at the leading edge are ignored. 2. Flow behaves as flow over a plate plate. 3. Air is an ideal gas.

Solution: An energy balance on the PV cells gives: Q solar T PV

= Tf +

 =Q conv

′′ qsolar A = hA (TPV −T f )

⇒

′′ /h qsolar

We assume that flow over a flat plate with a uniform heat flux wall condition is appropriate. Because we want the PV cell temperature at the trailing edge of the panel, the local heat transfer coefficient is used. The fluid properties from Appendix A-7 are evaluated at T film. We assume T PV ~ 274 K, so T film = (274 + 256)/2 = 265 K, ρ = 0.710 kg/m3, µ = 1.60 × 10-5 N⋅s/m2, k = 0.0235 W/mK, Pr = 0.718 Re =

L ρ V

=

( 0.710 kg

m3

) ( 30.6 m s )(1.5m) ( N ⋅ s

2

-5

1.60 ×10 N ⋅ s/m

µ

kgm)

2

6

= 2.04 ×10

This is turbulent flow, so for the local heat transfer coefficient Nu x = 0.0308 Rex0.8 Pr 1 / 3 = 0.0308 ( 2.04 × 106 ) h=

Nuk

=

( 3070)( 0.0235 W

L

T PV = 256K +

1.5m 850W/m 2 48.2W/m 2 K

mK )

0.8

2

= 48.2W/m

= 273.6K

1/ 3

( 0.718)

= 3070

K

Answer

12-10

12-11 For a quick solution to an overheating problem, brass rods, 6-mm in diameter and 5-cm long, are attached to a surface of a power supply. Air at 20 ºC and 5 m/s is blown perpendicular to the tubes. If the base temperature must not exceed 75 ºC, how much power can be dissipated by one rod (in W)?

Approach: This is a problem involving conduction heat transfer in a pin fin. The convective heat transfer must be evaluated first, and then the fin heat transfer evaluated.

Assumptions: 1. Air is at one atmosphere. 2. The fin is one dimensional. 3. The flow is as over a circular tube.

Solution: A pin fin was evaluated in Chapter 11. Heat transfer from it can be calculated with: Q = ( Tb − T f where m =

)

hPk Ax tanh ( mL )

hp/k A x . This is for an adiabatic tip. To account for convection from the tip, we use a corrected

length. L* = L + A x/ p = L + D/4. We assume the forced convection heat transfer coefficient for flow over a cylinder is applicable and is uniform over the fin. The Reynolds number is determined with air properties from Appendix A-7 evaluated at T film = (T b + T f )/2 = (75 + 20)/2 = 47.5 oC = 320.5 K, ρ = 1.101 kg/m3, µ = 1.94 × 10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.704. (This temperature is assumed to be a reasonable “average” over the length of the fin.) Re =

ρ V D

=

(1.101kg

) ( 5 m s )( 0.006m) (1Ns (1.94 ×10 N ⋅ s/m )

m3

2

-5

µ

kgm)

2

= 1700

From Table 12-1, Nu = 0.683 Re0.466 Pr 1/3 Nu = 0.683 (1700)0.466 (0.704)1/3 = 19.5

→

h=

Nuk D

=

(19.5)( 0.0278W/mK ) 0.006m

= 90.2W/m2 K

From Appendix A-2 for brass, k ≈110 W/mK

( 90.2W/m K )π ( 0.006m) 2

mL * =

(110W/mK )(π

0.006 ⎞ ⎛ ⎜ 0.05 + ⎟ m = 1.204 4 ⎠ 4 )( 0.006m) ⎝ 2

W ⎞ W ⎞⎛ π ⎞ ⎡⎛ 2⎤ ⎛ Q = ( 75 − 20) K ⎢⎜ 90.2 2 ⎟π ( 0.006m) ⎜ 110 ( 0.006m) ⎥ ⎟⎜ ⎟ m K ⎠ mK ⎠⎝ 4 ⎠ ⎝ ⎣⎝ ⎦

= 3.34W

Answer

12-11

0.5

tanh ( 1.204)

12-12 If the brass rods in P 12-11 are replaced by rectangular aluminum alloy fins, 10-cm wide (in the direction of air flow), 2-mm thick, and 5-cm long, determine how much power can be dissipated by one fin (in W).

Approach: This is a problem involving conduction heat transfer in a rectangular fin. The convective heat transfer coefficient must be evaluated first, and then the fin heat transfer evaluated.

Assumptions: 1. Air is at one atmosphere. 2. The fin is one dimensional. 3. The flow is as over a flat plate.

Solution: A fin with a constant cross-sectional area was evaluated in Chapter 11. Heat transfer from it can be calculated with Q = (Tb − T f where m =

)

h pkAx tanh ( mL )

h p/k A x . This is for an adiabatic tip. To account for convection from the tip, we use a corrected

length L* = L + A x/ p = L + t /2. (Heat transfer from the two end edges is ignored.) We assume forced convection heat transfer over a flat plate is applicable and is uniform over the fin. The Reynolds number is determined with air properties from Appendix A-7 evaluated at a film temperature that is reasonable over the length of the fin: T film = (75 + 20)/2 = 47.5 oC = 320.5K: ρ = 1.101 kg/m3, µ = 1.94 × 10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.704. Rew =

ρ VW

=

(1.101kg

m 3 ) ( 5 m s )( 0.1m ) ( N ⋅ s 2 kgm ) 1.94 ×10-5 N ⋅ s/m 2

µ

= 28,400

This is laminar flow, so for the length average heat transfer coefficient: hW Nuw = 0.664 Re1w/ 2 Pr 1/ 3 = 0.664 (28400)1/2 (0.704)1/3 = 99.5 k Nuk ( 99.5 )( 0.0278W/mK ) h= = = 27.7W/m2 K W 0.10m From Appendix A-2 for aluminum, k ≈ 177 W/mK mL * =

( 27.7 W (177W

m2 K ) ( 2 )( 0.10m )

mK )( 0.002m )( 0.10m)

( 0.05 + 0.002/2) m = 0.638

⎡⎛ ⎤ ⎛ W ⎞ W ⎞ Q = ( 75 − 20 ) K ⎢⎜ 27.7 2 ⎟ 2 ( 0.10m) ⎜ 177 ⎟ ( 0.002m)( 0.10m) ⎥ mK ⎠ m K ⎠ ⎝ ⎣⎝ ⎦ =13.7W Answer

12-12

0.5

tanh( 0.638)

12-13 In an electric hair drier, air at 25 ºC flows with a velocity of 5 m/s perpendicular to a nichrome heating element. The heating element is 1-mm in diameter and 40-cm long with a resistance of 1.38 Ω/m. The wire temperature cannot exceed 430 ºC so that the wire will not lose strength and sag. Determine: a. the total power dissipated (in W) b. the electric current in the wire (in A).

Approach: This is forced convection perpendicular to a cylinder. With the heat transfer coefficient, geometry and temperatures, we can calculate the maximum power to not exceed the temperature limit. Once the power is known, Ohm’s Law can be u sed to calculate current.

Assumptions: 1. Air is at one atmosphere. 2. Radiation is ignored. 3. The flow is perpendicular to a cylinder.

Solution: a) We assume radiation can be ignored, so that the heat transfer from the wire is

Q = hA (T s − T∞ )

To obtain h, we need the Reynolds number which requires the fluid properties. From Appendix A-7 at T film = (T ∞ + T s)/2 = (25 + 430)/2 = 227.5 oC = 500.5K, ρ = 0.7053 kg/m3, k = 0.04038 W/mK, µ = 2.671 × 10-5 N⋅s/m2, Pr = 0.680 Re =

ρ V∞ D

=

( 0.7053kg

2.671×10-5 N ⋅ s m 2

µ

Nu = 0.193 Re0.618 Pr 1/ 3 = 0.193 (132)

From Table 12-1 h=

Nuk D

=

m3 ) ( 5 m s )( 0.001m) ( 1Ns2 kgm)

( 8.44)( 0.04038W/mK ) 0.001m

= 341W/m

2

0.8

= 132 1/ 3

( 0.680)

= 8.44

K

Q = ( 341W m2 K ) π ( 0.001m)( 0.4m) ( 430 − 25) C =173W o

Answer

b) Electrical power is Q = I 2 R where I = current and R is resistance I = Q /R =

173W

(1.38 Ω

m )( 0.4m )

Answer

=17.7A

Comments: Because of the high wall temperature, the assumption that radiation can be ignored should be checked.

12-13

12-14 An existing electric powerline is being examined to determine if a higher current can be used. You are asked to calculate the maximum power dissipation per meter of length that is permissible by Joulean heating such that the inside surface of the cable insulation does not exceed 77 ºC. The copper wire in the cable is 2-cm in diameter and the insulation is 0.1-cm thick with a thermal conductivity of 0.08 W/m·K. The wind velocity perpendicular to the wire is 5 km/hr and the air temperature is 27 ºC. Neglecting radiation, determine the allowable power (heat generation rate) per unit length (W/m). [J12-001]

Approach: We must take into account two thermal resistances (conduction and convection) to determine the heat transfer rate from the inner surface of the insulation to the air. The heat transfer coefficient is calculated with the appropriate correlation.

Assumptions: 1. The heat transfer is one dimensional. 2. Air is at one atmosphere.

Solution: Q =

The heat transfer rate is calculated with where Rtot =

1 hA

+

1n( D2/ D1 ) 2π kL

=

1 hπ D2 L

+

∆T

Rtot

=

Ti − T ∞ Rtot

1n(D2/ D1 ) 2π kL

Combining these two equations and solving for heat transfer per unit length Ti − T ∞ Q L

=

1 hπ D2

+

1n( D2 − D1 ) 2π k

To obtain h, we need the Reynolds number. Properties should be evaluated at the film temperature, but we do not know it, so we estimate Ts ~ 67°C ° T film = (T s + T ∞ )/ 2 = (27 + 67)/2 = 47 C = 320 K From Appendix A-7 for air at 320 K (by interpolation) 3 -5 2 ρ = 1.110 kg/m , k = 0.0278 W/mK, µ = 1.94 × 10 N-s/m , Pr = 0.704 kg (1.110 3 ) (5 km ) ( 1000m ) ( 1hr ) (0.022m) ρ V∞ D 1km 3600s hr m Re = = = 1750 µ 1.94×10-5 Ν-s/m 2 From Table 12-1 for the heat transfer coefficient: Nu = 0.683 Re0.466 Pr 1/3 = 0.683(1750)0.466(0.704)1/3 = 19.7 Nuk (19.7 )( 0.0278 W mK ) h= = =24.9 W m 2 K D 0.022m ( 77 − 27 ) K W Q = = 64.9 1 1n (0.022 / 0.02) L m + 2 ( 24.9 W m K) π ( 0.022m) 2π ( 0.08 W/mK )

Answer

Check on T s Q = hA (T s - T ∞ ) = hπ D2 L(T s - T ∞ ) T s = T ∞ +

64.9W/m (Q / L) o o = 27 C + = 64.7 C 2 hπ D2 ( 24.9 W m K ) π (0.022m)

This assumed T s is close enough to the estimated temperature, so we do not need to reevaluate the film temperature.

12-14

12-15 The insulation on a 15-cm steam pipe deteriorates over time and is to be removed and replaced. The outer surface of the steam pipe is at 110 ºC. Air at -6 ºC blows perpendicular to the pipe at 40 km/hr. Determine: a. the heat transfer rate per unit length of pipe if it is left bare (in W/m) b. the heat transfer rate per unit length of pipe if 4-cm insulation (k = 0.04 W/m·K) is applied to the pipe (in W/m).

Approach: This is forced convection perpendicular to a circular cylinder, with and without insulation. The basic heat transfer rate equation can be used to find the heat transfer per unit length.

Assumptions: 1. The heat transfer is one dimensional. 2. The air is at one atmosphere.

Solution: The basic heat transfer rate equation is: T s − T f Ts − T f ∆T Q = = = Rtot Rconv + Rins 1 hπ DL + 1n (D2 / D1 ) 2π kL or

Q L

=

T s − T f 1 hπ D + 1n (D2 /D1 ) 2π k

a) With no insulation, D2 = D1 , D = D1, and Rins = 0. Evaluation of the forced convection heat transfer coefficient requires the Reynolds number with fluid properties evaluation at the film temperature, T film , (6+110)/2 = 52oC = 325 K. By interpolation in Appendix A-7: ρ = 1.086 kg/m3, k = 0.0281 W/mK, µ = 1.96 × 10-5 N⋅s/m2 , Pr = 0.703 ρ V D

Re =

(1.086 kg/m ) (11.1m/s)( 0.15 m) ( Ns 3

=

1.96 × 10 N ⋅ s/m -5

µ

2

kg m)

2

= 92,300

From Table 12-1, Nu = 0.193 Re0.618 Pr 1/3 Nu = 0.027 (92300)0.805 (0.703)1/3 = 238 Nuk ( 238)( 0.0281W/mK ) h= = = 44.7 W/m2 K D 0.15m Q L

=

⎡⎣(110 -(-6 ) ⎤⎦ o C ⎡1 ( 44.7 W m K ) π ( 0.15m ) ⎤ ⎣ ⎦ 2

= 2440

W

(without insulation)

m

Answer

b) With insulation, the film temperature will be much lower and the fluid properties different. We should iterate by guessing a temperature on the outside of the insulation, evaluating the properties, and proceeding as before. Let’s guess T s = 20oC, T film = (20 – 6)/2 = 7oC = 280 K ρ = 1.261 kg/m3, k = 0.02465 W/mK, µ = 1.70 × 10-5 , Pr = 0.714

(1.261) (11.1) ⎡⎣ 0.15 + 2 ( 0.04) ⎤⎦

Re =

1.70 × 10−5

= 189,400

Nu = 0.027 (189400)0.805 (0.714)1/3 = 427 Q L

h = (427)(0.02465) / (0.23) = 45.8 W/m2K

o

=

⎡⎣110 - ( -6 ) ⎤⎦ C

= 67.0

1n ( 23/15) 1 + (45.8 W m K ) π ( 0.23m ) 2 π ( 0.04 W/mK )

W m

2

We can calculate the outside wall temperature T s = T f +

( Q / L ) hπ D

= −6 +

Q = hA (T s – T f ) = hπ DL (T s – T f )

67.0W/m

( 45.8 W

m K ) π ( 0.23m ) 2

= −4 oC

Comments: We should recalculate with improved fluid properties if we want a more accurate result. 12-15

Answer

12-16 A very long cylinder 25-mm in diameter is placed in a large oven whose walls are maintained at 400 ºC. Air at 77 ºC flows perpendicular to the cylinder at a velocity of 2.5 m/s. The emissivity of the cylinder is 0.65 Determine the steady-state temperature of the cylinder (in ºC).

Approach: Two heat transfer modes – convection and radiation – must be taken into account. Radiation heats the cylinder, and convection cools it. At steady state these two heat transfer rates are equal. We must calculate the heat transfer coefficient to obtain the convective heat transfer.

Assumptions: 1. Air is at one atmosphere. 2. The cylinder is “small” relative to the larger oven.

Solution: At steady state, we apply conservation of energy to the cylinder, and assume no work and no potential or kinetic Q = 0

energy effects to obtain:

→

Q rad − Q conv = 0

4 4 Q rad = εσ A (Tw4 − Ts4 ) = εσπ DL (Tw −Ts )

Q conv = h A (Ts −T f ) = hπ DL (Ts −T f ) ε σ =

Therefore

(T

4 w

− Ts4 ) = h (Ts − Tf )

To evaluate h, we need Re with fluid properties from Appendix A-7 evaluated at T film. Assuming T s ≈227 oC = 500 K, T film = (350 + 500)/2 = 425 K, so by interpolation: ρ = 0.831 kg/m3, µ = 2.385×10-5 N⋅s/m2, k = 0.0354 W/mK, Pr = 0.686 Re =

D ρ V µ

( 0.831kg/m ) ( 2.5m/s)( 0.025m) ( N ⋅ s = ( 2.385×10 N ⋅ s/m ) 3

-5

2

kgm )

2

= 2180

From Table 12-1, Nu = 0.683 Re0.466 Pr 1 / 3 = 0.683 ( 2180) h=

Nuk D

=

( 21.6)( 0.0354W/mK ) 0.025m

0.466

1/ 3

( 0.686) = 21.6

= 30.6W/m2 K

Substituting into energy equation

⎛

( 0.65) ⎜ 5.67 ×10-8 ⎝

⎞⎡ ⎛ W ⎞ 4 4 ⎟ ⎣( 673K ) − T s ⎤⎦ = ⎜ 30.6 2 ⎟ (T s − 350K) m K ⎠ m K⎠ ⎝ W 2

4

Solving this equation, T s = 513.4 K = 240.4 oC This is close to the assumed T s, so we do not need to re-evaluate the fluid properties.

12-16

Answer

12-17 After the extrusion of a long solid plastic rod, it is cooled by a crossflow of 30 ºC air at a velocity of 20 m/s. The rod, whose diameter is 3.5-cm, initially has a uniform temperature of 200 ºC. The plastic’s properties are: ρ = 2300 kg/m3, c p = 850 J/kg·K, and k = 1.2 W/m·K. Determine: a. the time required for the surface temperature of the rod to drop to 100 ºC (in s) b. the centerline temperature at the same time (in ºC).

Approach: This is a transient conduction problem in an infinite cylinder. We ignore radiation, and must calculate the heat transfer coefficient.


Radiation is ignored. Air is at one atmosphere. The one-term approximation for transient conduction is valid. The heat transfer coefficient is constant during the cooling process.

Solution: a) Transient conduction in an infinite cylinder using the one-term approximation is governed by T ( ri t ) − T f = C1 exp ( − λ12 τ ) J 0 ( λ1 r/r0 ) τ > ∼ 0.2 Ti − T f where τ = α t/r 02 and C 1 and λ 1 require Bi = hr 0/k . τ = −

Solving the equation for τ :

⎡ T ( r1 t ) − T f ⎤ ⎢ ⎥ 1n λ 12 ⎢ C1 J 0 (Ti − T f ) ⎥ 1

⎣

⎦

To calculate h and Bi, we need the Reynolds number with fluid properties from Appendix A-7 evaluated at T film. Note that T film decreases with time, so we use T film ~ (150 + 30)/2 = 90 oC = 363 K so ρ = 0.972 kg/m3, µ = 2.13 × 10-5 N⋅s/m3, K = 0.0310 W/mK, Pr = 0.695. Re =

ρ V D µ

( 0.972kg/m ) ( 20m/s)( 0.035m) ( N ⋅ s 3

=

2.13 ×10 N ⋅ s/m -5

h=

D

=

kgm )

2

Nu = 0.193Re0.618 Pr 1/ 3 = 0.193 ( 31,900)

From Table 12-1, Nuk

2

(104 )( 0.0310W/mK ) 0.035m

= 31,900

0.618

1/ 3 ( 0.695) = 104

( 92.0 W/m K ) ( 0.035/2 m) 2

= 92.0W/m K 2

→

Bi =

1.2W/mk

= 1.342

From Table 11-2 by interpolation λ 1 = 1.373, C 1 = 1.252. From Table 11-3, J 0 (λ 1) = 0.5813 τ = −

⎡

1

(1.373)

2

⎤ (100-35) ⎥ = 0.326 ⎢⎣ (1.252 )( 200 − 35)( 0.5813) ⎥⎦

1n ⎢

Because τ > 0.2, the one-term approximation is okay. τ r 2 (1.2 W/mk )(1J/s 1W ) k t = 0 , α = = = 6.14 ×10-7 m2 /s 3 α ρ C p ( 2300 kg/m ) ( 850 J/kg k) t =

( 0.326)( 0.035/2m) 6.14 ×10-7 m2 /s

2

= 162 s

Answer

b) For the centerline temperature at the same time, J 0 (0) = 1 T = T f + (Ti − T f )C1 exp ( − λ12 τ ) = 35 oC + (200 – 35) K (1.252) exp [- (1.373) 2 (0.326)] = 146.7 oC

Answer

Comments: Probably a better way to solve this problem is to divide the process into several time intervals and evaluate the heat transfer coefficient and wall temperature during each interval. This would be done until the target wall temperature of 100 oC was obtained.

12-17

12-18 Lead shot is made by dropping molten lead ( ρ = 10,600 kg/m3) from a drop tower. Each pellet, a sphere 2mm in diameter, is cooled as it passes through air at 10 ºC. Assume the shot falls at its terminal velocity. The lead must be solidified from its molten state at 327 ºC to a solid state before it reaches the pool of water at the bottom of the drop tower. The enthalpy of fusion for lead is 24.5 kJ/kg. Ignoring radiation, determine the minimum required height of the drop tower (in m).

Approach: This is a transient cooling problem. The lead remains at a uniform temperature as it solidifies, and we will use an energy balance to develop the governing equation. This heat transfer coefficient is needed to calculate the convective heat transfer. That requires a velocity which will need to be calculated, too.

Assumptions: 1. The shot falls at its terminal velocity. 2. Air is at one atmosphere. 3. Radiation is ignore, and the heat transfer coefficient is constant. 4. Buoyancy force is ignored.

Solution: For a closed system defined as the lead sphere, conservation of energy is used assuming no work and no potential or kinetic energy effects: dU  = dU ⇒ Qt  = ∆U = m (u − u ) = −mu Q = Qdt ⇒ ⇒ hA (T f −Ts )t = −mu sf 2 1 sf dt We assume the convective heat transfer rate is constant; heat transfer from the control volume is negative. The time, t , required to completely solidify the lead is used with the terminal velocity to obtain the height, H , required: H = V t

∫

∫

We assume the sphere moves at the terminal velocity for the whole time, so that H =

− V mu sf hA (T f − Ts )

=

− V ρ L ( 4π R 3 3 )usf h 4π R 2 ( Tf − Ts )

=

−V

ρL

Ru sf

3h( Tf − Ts )

=

− Vρ L Dusf 6 h (Tf − Ts )

The terminal velocity is determined from a force balance between drag and gravity (buoyancy is negligible). Fdrag = Fgrav

⇒

⎛ π D 2 ⎞ ⎛ ρ a V 2 ⎞ ⎛ π D 3 ⎞ CD ⎜ ⎟ = ρ L ⎜ ⎟⎜ ⎟ g ⇒ ⎝ 4 ⎠⎝ 2 ⎠ ⎝ 6 ⎠

V

⎡ 4 ρ L gD ⎤ =⎢ ⎥ ⎣ 3 ρ a C D ⎦

0.5

The drag coefficient (given on Fig. 10-8) is a function of Reynolds number, so evaluating air properties from Appendix A-7 at 10 oC: ρ = 1.247 kg/m3, µ = 1.72×10-5 N⋅s/m2, k = 0.0249 W/mK, Pr = 0.712 (at 327 oC, -5 2 µ = 3.018×10 N⋅s/m ). The Reynolds number is Re = ρ V D µ , which requires the velocity that we do not know, so we iterate. We guess V , calculate Re, obtain C D, and then calculate matches the calculated value, then we have a solution. 1/ 2

V

⎡ 4 ⎛ 10600 ⎞ ⎛ ( 9.81m s2 ) ( 0.002m) ⎞ ⎤ ⎟⎥ =⎢ ⎜ ⎟⎜ ⎟⎥ C D ⎢ 3 ⎝ 1.247 ⎠ ⎜⎝ ⎠⎦ ⎣

=

14.91 C D1/ 2

[ m/s]

Estimate V = 25 m/s, Re = 3630, C D ≈ 0.4, V = 23.6 m/s m/s The heat transfer coefficient is determined from Nu = 2 + ⎡⎣0.4 Re

1/ 2

h=

Nuk D

=

+ 0.06 Re

2/3

1/ 4

⎤⎦ Pr [ µ /µ s ] 0.4

( 29.9 )( 0.0249W/mK ) 0.002m

Re = →

(1.247 kg

V

V

. When the guessed

m3 ) ( V m s )( 0.002m)

1.72 ×10-5 N ⋅ s/m 2

V

=145 V

= 23.6 m/s, Re = 3420, C D ≈ 0.4, V = 23.6

-5 ⎤ 1/ 2 2/3 0.4 ⎡ 1.72 ×10 = 2 + ⎡ 0.4 ( 3420) + 0.06 ( 3420) ⎤ ( 0.712) ⎢ ⎥ -5 ⎣ ⎦ ⎣ 3.018 ×10 ⎦

= 372W/m3 K

12-18

1/ 4

= 29.9

−

H

=

( 23.6m/s ) (10600kg/m3 ) ( 0.002m)( 24500 J 3 ( 372W/m2 K ) (10 327) K (1J/s 1W ) −

12-19

kg ) =

34.6m

Answer

12-19 False temperature readings can be obtained from temperature sensors if they are used incorrectly or if the effects of radiation are not taken into account. (For example, if a thermometer were used in direct sunlight or in shade to measure air temperature, significantly different readings would be obtained.) Consider a thermocouple that is a 1.5-mm sphere, which is used to measure the temperature of an air stream in a large duct. The air velocity is 4 m/s. The walls of the duct are at 150 ºC. The thermocouple indicates an air temperature of 300 ºC and has an emissivity of 0.5. Determine the actual air temperature (in ºC).

Approach: Two heat transfer modes-convection and radiation must be taken into account. Radiation cools the thermocouple, and convection heats it. At steady state, the two heat transfer rates are equal. We must calculate the heat transfer coefficient before we can obtain the convective heat transfer.

Assumptions: 1. Air is at one atmosphere. 2. The thermocouple is small relative to the duct size.

Solution: At steady state, we apply conservation of energy to the thermocouple, and assume no work and no potential or Q = 0

kinetic energy effects to obtain:

→

 =0. Q conv − Q rad

With Q conv = hA ( T f − Ts ) and

Q rad = εσ A ( Ts4 − Tw4 ) so that h (T f – T s) = εσ (T s4 – T w4). Solving for the fluid temperature: εσ

T f = Ts +

(T

4 s

− Tw4 )

h To evaluate h for a sphere, we use Eq. 12-11, which uses air properties from Appendix A-7 evaluated at the bulk fluid temperature. We assume T bulk ~ 300 oC, so that ρ = 0.616 kg/m3, µ = 2.926 × 10-5 N⋅s/m2, k = 0.0450 W/mK, Pr = 0.680. We assume the thermocouple is a sphere, so that Re =

D ρ V

=

( 0.616 kg

2.926 ×10-5 N ⋅ s/m2

µ

For a sphere

1/ 4

µ s ~

Nu = 2 + ⎡ 0.4 (126 )

1/ 2

Nuk D

= 126

2 0.06 ReD2 / 3 ⎤⎦ Pr 0.4 ( µ f /µ s ) Nu D = 2 + ⎡⎣ 0.4 Re1/ D +

Assume T surface ≈ 320 oC,

h=

m3 ) ( 4m/s)( 0.0015m) ( N ⋅ s2 kgm)

2.994 × 10-5 N⋅s/m2

⎣

+ 0.06 (126)

=

( 7.11)( 0.0450W/mK ) 0.0015m

2/3

⎤ ( 0.680) 0.4 ( 2.926× 10-5 2.994× 10-5 ) ⎦

1/ 4

= 7.11

= 213W/m2 K

Therefore: T f = 573K +

( 0.5) ( 5.67 x10-8 W 2

m2 K 4 )

213W/m K

⎡( 573K ) 4 − ( 423K ) 4 ⎤ = 583K = 310 o C ⎣ ⎦

Answer

Comments: For a better estimate, the fluid properties should be re-evaluated at this new fluid temperature and the calculations repeated.

12-20

12-20 Fluid velocities are often measured with hot wire anemometers. In such a device, the temperature of a small diameter cylinder is maintained constant by varying the electric current through it in response to varying fluid velocity; a wheatstone bridge is used to control the current. A typical hot wire is constructed of a 0.2-mm diameter polished platinum wire 10-mm long. Air at 23 ºC flows over the hot wire maintained at 200 ºC. The electrical resistivity of platinum is 17 µΩ-cm. Determine the electric current required for a velocity of: a) 1 and m/s b) 10 m/s.

Approach: This is forced convection perpendicular to a cylinder. With the heat transfer coefficient, geometry, and temperatures, we can calculate the power to reach the operating temperature. Once the power it known, Ohm’s Law can be used to calculate the current.

Assumptions: 1. Radiation is ignored. 2. Air is at one atmosphere.

Solution: a) We assume radiation can be ignored so that the heat transfer from the wire is

Q = hA (T s – T f ) where A

= π DL. From Ohm’s Law, Q = I 2 R where I is the current and R is the electrical resistance, R = ρ E L/ A x , where A x = π D2/4 and ρ E is the electrical resistivity. Substituting these expressions into the main equation and solving for I : 1/ 2

⎡ h π 2 D 3 (T s − T f ) ⎤ I = ⎢ ⎥ 4 ρ E ⎣⎢ ⎦⎥

To obtain h, we need the Reynolds number with air properties from Appendix A-7 evaluated at T film = (23 + 200)/2 = 111.5 oC : by interpolation of properties in the appendix ρ = 0.918 kg/m3, µ = 2.23 × 10-5 N⋅s/m2, k = 0.0327 W/mK, Pr = 0.691 Re =

ρ V D µ

( 0.918 kg/m ) (1 m/s)( 0.0002m) = 8.23 = ( 2.23×10 N×s/m )( kgm Ns ) 3

-5

2

2

From Table 12-1 Nu = 0.911 Re0.385 Pr 1/3 = 0.911(8.23)0.385(0.691)1/3 = 1.813 Nuk (1.813)( 0.0327 W/mK ) W h= = = 296 2 D 0.0002 m m K 1/2

⎡ ( 296W/m2 K ) π 2 ( 0.0002 m)3 ( 200 − 23) K ( 100cm m) ⎤ ⎥ =2.47A I = ⎢ ⎢ ⎥ 4(17×10-6 Ω-cm ) ⎣ ⎦ b) For V = 10 m s , Re = 82.3

Answer

Nu = 0.683 Re0.466 Pr 1/3 = 0.683(82.3)0.466(0.691)1/3 = 4.72 ( 4.72)( 0.0327) h= = 771W/m2 K 0.0002 1/ 2

⎡ ( 771) π 2 ( 0.0002)3 ( 200 − 23)( 100) ⎤ ⎥ I = ⎢ 4 (17 ×10−6 ) ⎢⎣ ⎥⎦

= 3.98A

12-21

Answer

12-21 After a heat treating process, a 2024-T6 aluminum sphere 20-mm in diameter is removed from an oven which is at 85 ºC. The sphere is placed in an air stream at 27 ºC that has a velocity of 10 m/s. Determine the time required for the sphere’s temperature to cool to 40 ºC (in s).

Approach: This is a transient conduction problem. We need to determine if lumped systems approach will work or if a one-dimensional transient solution is required. The heat transfer coefficient is determined from the appropriate correlation.

Assumptions: 1. All properties are constant.

Solution: To determine if the lumped systems approach is valid, we need to calculate Bi = h Lchar /k Lchar = V / A = r o / 3 = (0.01m/3) = 0.00333m For aluminum from Appendix A-2, k = 177 W/mK, c p = 875 J/kgK, ρ = 2770 kg/m3 The heat transfer coefficient requires the Reynolds number. The sphere heat transfer coefficient correlation uses properties at the bulk temperature: so from Appendix A-7 at 300K, ρ = 1.774 kg/m3, k = 0.02624 W/mK, µ = 1.846 × 10-5 N⋅s/m2, Pr = 0.708 (at 350 K, µ = 2.075 × 10-5 N⋅s/m2) Re =

ρ V D

(1.1774 kg/m ) (10m/s)( 0.02m) ( Ns 3

=

-5

1.846 ×10 N×s/m

µ

2

kgm )

2

= 12,800

The heat transfer coefficient is determined from: 1/ 4

Nu = 2 + ⎡⎣ 0.4 Re1 / 2 + 0.06 Re 2 / 3 ⎤⎦ Pr 0.4 [ µ /µ s ]

= 2 + ⎡0.4( 12800) ⎣

1/ 2

h=

Nuk D

=

+ 0.06 (12800)

( 67.9 )( 0.02624W/mK ) 0.02m

(89.1W/m K ) ( 0.00333m)

2/3

⎤ ( 0.708) 0.4 ⎡1.846× 10−5 2.075× 10−5 ⎤ ⎣ ⎦ ⎦

1/ 4

= 67.9

= 89.1W/m2 K

2

Bi =

177 W/mK

= 0.00168 < 0.1

so the lumped systems approach is valid:

⎡ −hA ⎤ ⎡ −ht ⎤ t ⎥ = exp ⎢ = exp ⎢ ⎥ Ti − T f ⎢⎣ mc p ⎥⎦ ⎣ ρ Lchar c P ⎦ T − T f

Solving for time t =

− ρ Lchar c p h

⎛ T − T f ⎞ − ( 2770kg m3 ) ( 0.00333 m)( 875 J/kgK) ⎡ 40 − 27 ⎤ ln ⎜ ln ⎢ = ⎥ 2 ⎜ Ti − T f ⎟⎟ 89.1 W/m K 1J/s 1 W ( ) ( ) ⎣ 85 − 27 ⎦ ⎝ ⎠

Answer

= 136 s

12-22

12-22 In an oil refinery, a steam pipe (k = 15 W/m·K) with an inside diameter of 10-cm and an outside diameter of 11-cm is covered with 3.5-cm of insulation (k = 0.03 W/m·K). The steam is at 300 ºC. Air and surrounding surfaces are at 17 ºC, and the air flows perpendicular to the pipe at a velocity of 2 m/s. The heat transfer coefficient of the steam is 100 W/m2·K. Determine the heat transfer rate per unit length of pipe (in W/m).

Approach: The heat transfer rate equation ( Q = ∆T / Rtot ) can be used directly. The heat transfer coefficient on the outside must be determined.

Assumptions: 1. The air is at one atmosphere. 2. The heat transfer is one dimensional.

Solution: The governing equation is Q =

∆T Rtot

T steam -T f

= 1

1n( D2 / D1 )

1n ( D3 /D2 )

1 h steam At 2π ksteel L 2π kins L ho Ao For flow perpendicular to the pipe, the Reynolds number with air properties from Appendix A-7 at T film is required to calculate h. Assuming T surface ≈ 27oC, T film ≈ (17 + 27)/2 = 22 oC = 295K: ρ = 1.197 kg/m2, k = 0.0258 W/mK, -5 2 µ = 1.81 × 10 N⋅s/m , Pr = 0.709 ρ V D

Re =

µ

+

+

+

(1.197kg/m ) ( 2m/s) ⎡⎣ 0.11+ 2 ( 0.035) ⎤⎦ m = = 23,800 (1.81×10 N ⋅ s/m )( ms Nkg ) 3

-5

2

2

From Table 12-1 Nu = 0.193 Re0.618 Pr 1/3 = (0.193)(23800)0.618(0.709)1/3 = 87.2 Nuk ( 87.2)( 0.0258W/mK ) h= = =12.5W/m2 K D ( 0.18m ) 1 h steam Asteam 1n ( D2 / D s ) 2π k steel L 1n ( D3 /D2 ) 2π kins L 1 ho Ao Q L

=

=

= = =

1

(100W/m K )π (1.10m) L 2

1n( 0.11/0.10) 2π (15W/mK ) L

=

1n ( 0.18/0.11) 2π ( 0.03W/mK ) L 1

=

=

0.00318 L

0.00101 L

=

2.613 L

0.1415

L (12.5)π ( 0.18) L ( 300 −17 ) K

( 0.00318 + 0.00101+ 2.613+ 0.1415) Km/W

=103W/m

Answer

Comments: The insulation dominants the total thermal resistance. Inaccuracies in the other resistances would have little effect.

12-23

12-23 If you have ever changed a hot incandescent light bulb, then you know that much of the p ower going into the bulb is converted to heat (about 90%) rather than to light (about 10%). (Fluorescent light bulbs are much more efficient.) All the heat is dissipated from the glass bulb. Consider an 8-cm diameter 100-W light bulb cooled by air at 30 ºC. Both convection and radiation (ε = 0.85) cool the glass. Assuming the surroundings are at 30 ºC for radiation purposes, determine the temperature of the glass bulb if air at 2 m/s flows across it (in ºC).

Approach: An energy balance on the light bulbs is needed. The power into the glass (90% of 100 W) is removed by convection and radiation. The heat transfer coefficient must be evaluated.

Assumptions: 1. Air is at one atmosphere. 2. Radiation is from a small body to a large area. 3. The light bulb is a sphere.

Solution: Q in = Q rad + Q conv

A steady energy balance on the glass gives Q in = εσ A (Ts4 − T f4 ) + hA (Ts − T f )

The light bulb is assumed to be a sphere, and the surroundings are very large. The heat transfer coefficient correlation for a sphere is based on air properties from Appendix A-7 evaluated at the bulk temperature, 300K 2 -5 2 -5 ρ = 1.1774 kg/m , µ = 1.846 × 10 N⋅s/m , k = 0.02624W/mK, Pr = 0.708 (assume T s = 450K, µ s = 2.484 × 10 N⋅s/m2). Re =

ρ V D

(1.1774kg/m ) ( 2m/s) ( 0.08m) = = 10,200 (1.846 ×10− N ⋅ s/m )( kgm Ns ) 3

5

µ

2

2

The heat transfer coefficient on a sphere is determined from 1/ 4

Nu = 2 + ⎡⎣ 0.4 Re1/ 2 + 0.06 Re 2 / 3 ⎤⎦ Pr 0.4 [ µ /µ s ]

= 2 + ⎡0.4 (10200) ⎣

1/ 2

h=

Nuk D

=

+ 0.06 (10200)

( 58.5)( 0.02624 W 0.08m

2/3

mK )

⎤ ( 0.708) 0.4 ( 1.846× 10−5 2.484× 10−5 ) ⎦

1/ 4

= 58.5

= 19.2 W/m2 K

2 4 2 0.9 (100W ) = ( 0.85) ( 5.67 × 10−8 W m2 K4 ) ( 4π )( 0.04m) ⎡T s4 − ( 300K) ⎤ + ( 22.6 W m2 K ) ( 4π )( 0.04m) (T s − 300) K

⎣

⎦

90=9.69 × 10-10 (T s4 − 8.1× 109 ) + 0.386 (T s − 300) Solving the equation for the surface temperature: T s = 450.3K = 177.3 oC

Answer

12-24

12-24 After preheating long 316 stainless steel rods 50-mm in diameter to a uniform temperature of 1000 ºC, they must be conveyed to another location in the plant for additional processing. The conveyor moves the rods perpendicular to the direction of travel at a velocity of 3 m/s. The rod emissivity is 0.5, and the air and surrounding temperatures are at 27 ºC. The centerline temperature of the rod must be greater than 900 ºC for the next processing step. Determine: a. the convective heat transfer coefficient at the start of the travel (in W/m 2·K) b. the radiation heat transfer coefficient at the start of the travel between one processing station and another (in W/m2·K) c. the allowable time for transit between the two stations assuming the total heat transfer coefficient (radiation and convection) is the sum of those calculated in parts (a) and (b) (in s).

Approach: The convective heat transfer coefficient for forced convection perpendicular to a cylinder and the radiative heat transfer coefficient both can be evaluated with the given information. Because the diameter is small relative to the cylinder length, we will analyze it as an infinite cylinder. The Biot number should be checked (with the total heat transfer coefficient) to determine if lumped systems approach is applicable. If not, the one-term approximation for one-dimensional transient conduction in an infinite cylinder will be used.

Assumptions: 1. Air is at one atmosphere.

Solution: a) To obtain the convective heat transfer coefficient, we need the Reynolds number with properties evaluated at T film = (T s + T f )/2. Because we want h at the start of travel, T s = 1000 oC, T film ~ 787 K. From Appendix A-7 by interpolation, ρ = 0.449 kg/m3, µ = 3.59 × 10-5 N⋅s/m2, k = 0.0571 W/mK, Pr = 0.688. Re =

D ρ V

=

( 0.449kg

m3 ) ( 3m s)( 0.050m)

3.59 ×10-5 N ⋅ s/m2

µ

= 1880

Nu = 0.683 Re0.466 Pr 1/3 = 0.683 (1880)0.466 (0.688)1/3 = 20.2 Nuk ( 20.2)( 0.0571W/mK ) h= = = 23.1W/m2 K D 0.050m

From Table 12-1,

b) Assuming the surroundings are very long and are at T f , the radiative heat transfer coefficient is defined as: hr = εσ (Ts + T f

) (T

s

2

+ T f 2 )

With T s = 1000 + 273 = 1273 K and T f = 27 + 273 = 300 K

⎛

hr = ( 0.5 ) ⎜ 5.67 ×10-8

⎝

⎞ 2 2 3 ⎟ (1273K + 300K) ⎡⎣( 1273K) + ( 300K) ⎤⎦ = 76.3W/m K m K ⎠ W 2

4

Bi =

c) The Biot number, with a combined coefficient, is:

( h+ hr ) Lchar

where LChar = V / A = D/4 k From Appendix A-2, for 316 stainless steel at 1000 K, ρ = 8238 kg/m3, k = 24.2 W/mK, c p = 602 J/kg K, Bi =

( 23.1+ 76.3) W/m2 K ( 0.05m/4) 24.2W/mK

= 0.051< 0.1

The lumped systems approach is valid, and the equation governing it is:

⎛ hA ⎞ ⎛ ht ⎞ t ⎟ = exp ⎜ − = exp ⎜ − ⎜ ρV c p ⎟ ⎜ ρ Lchar c p ⎟⎟ Ti − T f ⎝ ⎠ ⎝ ⎠ T − T f

t =

− ( 8238kg m3 ) ( 0.05 m 4)( 602 J kgK )

( 23.1+76.3) W/m

2

K

→

t =

− ρ Lchar c p

⎛ 900-27 ⎞ ⎟ = 67.6 s ⎝ 1000-27 ⎠

1n ⎜

12-25

h

⎛ T − T f ⎞ ⎜ T i − T f ⎟⎟ ⎝ ⎠

1n ⎜

Answer

12-25 Ventilation ducts are often uninsulated when they run through attics and other uninhabited spaces. Consider an air flow at 70 ºC and 15 m 3/min which enters a 20-m long 30-cm square duct. The duct runs through a space that is at 10 ºC. Ignoring the temperature drop across the metal duct, determine: a. the outlet temperature of the air (in ºC) b. the heat transfer rate from the hot air (in W).

Approach: This is a heat exchanger a constant wall temperature, so we can use the equation given in Section 12.5. The appropriate heat transfer coefficient correlation is used once the Reynolds number is determined.

Assumptions: 1. Air is at one atmosphere. 2. The duct wall temperature is the same as the air temperature outside of the duct. 3. The system is steady with negligible potential and kinetic energy effects, no work, and air an ideal gas with constant specific heat.

Solution: a) An equation for a constant wall temperature heat exchanger with only one convective thermal resistance is:

 p) Tout = T f + (Tin − T f ) exp ( −hA mc The mass flow rate is evaluated with the air density at the inlet ρ = 1.029 kg/m3

⎛ 15 m3 ⎞ ⎛ 1min ⎞ ⎟⎜ ⎟ = 0.257 kg/s ⎝ min ⎠ ⎝ 60 s ⎠

 = ρ V  = (1.029 kg/m3 ) ⎜ m

For the heat transfer coefficient, we evaluate the air properties from Appendix A-7 at the average air temperature. Because we do not know the exit temperature, assume T out ~ 60oC, T avg = (70 + 60)/2 ≅ 65oC, so ρ = 1.044 kg/m3, µ = 2.02 × 10-5 N⋅s/m2, k = 0.0291 W/mK, Pr = 0.700, c p = 1.009 kJ/kgK. This is a square duct, so the hydraulic diameter is needed 4 ( 0.30m )( 0.30m) 4 Ax D H = = = 0.30m P wetted 4 ( 0.30m ) Re =

ρ V D H

(1.044kg/m )(15 m min ) ( 0.30m)(1min 60s) = = 43,100 − 2.02 10 N s/m 0.30m kgm Ns × ⋅ ( ) ( ) ( ) 3

3

5

µ

2

2

2

This is turbulent flow, so using the Dittus-Boelter equation, with an exponent of 0.3 on Pr since the air is being cooled: Nu = 0.023 Re 0.8 Pr 0.3 = 0.023 ( 43,100) h=

NuK D H

=

(105)( 0.0291W/mK ) 0.30m

0.8

0.3

( 0.700) = 105

=10.2 W/m2 K

⎡ - (10.2W/m2 K ) ( 4 )( 0.30m )( 20m) ⎤ ⎥ = 39.4o C T out = 20 C + ( 70 - 20) C exp ⎢ ⎢⎣ ( 0.257kg/s )(1.009kJ/kgK )(1000J kJ) ⎥⎦ o

o

Answer

b) The heat transfer rate can be determined from conservation of energy. Assuming steady, negligible potential and kinetic energy effects, no work, and air an ideal gas with constant specific heat:

⎛ ⎝

 p ( Tin − Tout ) = ⎜ 0.257 Q = mc

kg ⎞ ⎛

kJ ⎞ ⎟ ( 70-39.4) K=7.93kW ⎟ ⎜ 1.009 s ⎠⎝ kgK ⎠

Answer

Comments: A better estimate can be obtained by re-evaluating the air properties at the average temperature.

12-26

12-26 Because of a shallow ocean floor and deep draft, an oil tanker must use an offshore oil depot to unload. The depot is connected to a shore installation by a 1100-m long 45-cm pipe. In the winter, the ocean water temperature is 5 ºC. The oil (properties equivalent to unused engine oil), initially at a temperature of 20 ºC, is pumped from the tanker at a flow rate of 0.08 m 3/min. Ignoring the water and pipe thermal resistances, determine: a. the outlet temperature of the oil (in ºC) b. the heat transfer rate (in W).

Approach: This is a heat exchanger with a constant wall temperature, so we can use the analysis given in Section 12.5. The appropriate heat transfer coefficient correlation is used once the Reynolds number is determined.

Assumptions: 1. The water and pipe thermal resistances are ignored. 2. The system is steady with no work, negligible potential and kinetic energy effects, and oil is an ideal liquid.

Solution: An equation for a constant wall temperature heat exchanger with one convective thermal resistance is:

⎛ −hA ⎞ ⎟⎟ ⎜ mc ⎝  p ⎠

Tout = T f + (Tin − T f ) exp ⎜

The heat transfer coefficient requires the Reynolds number with properties evaluated at the average oil temperature. Because we do not know T out , we assume T out ~ 15oC, so T avg = (15 + 20)/2 = 17.5 oC. For unused engine oil from Appendix A-6 3 -4 2 ρ = 890 kg/m , µ =9990×10 N⋅s/m , k = 0.145 W/mK, Pr = 12900, c p = 1.868 kJ/kgK.   ρ V D m 4m Re = but V = Re → = 2 µ ρ (π / 4 ) D πµ D 4 ( 890kg/m3 )( 0.08m3 / min ) (1min 60s)

Re = π

( 9990 × 10−

4

N ⋅ s/m2 ) ( 0.45m ) ( ms2 Nkg )

= 3.36

This is laminar flow, so check the entrance length: Lent = 0.037 Re Pr D = 0.037 (3.36) (12900) (0.45 m) = 722 m More than half of the 1100-m pipe is in the entrance region, so we take the entrance length into account, noting that Gz = RePrD/L 0.0668 ( 3.36 ) (12900)( 0.45 / 1100) 0.0668Gz Nu = 3.66 + = 3.66 + = 4.59 2/3 2/3 1 + 0.04 G z 1 + 0.04 ⎡⎣( 3.36 )(12900)( 0.45 /1100)⎤⎦ h=

Nuk D

=

( 4.59 )( 0.145W/mK ) 0.45m

⎡ T out = 5 C + ( 20 -5) C exp ⎢ ⎢⎣ ( 890 kg m3 o

o

= 10.3o C

= 1.48W/m2 K ⎤ ⎥ )( 0.08m3 min ) (1.868k J kgK )(1min 60s)( 1000 J kJ) ⎥⎦ - (1.48 W/m 2 K ) π ( 0.45m )(1100 m)

Answer

b) For the heat transfer rate, using the energy equation:

 (T − T )  P (Tout − Tin ) = ρ Vc Q = mc P out in

= ( 890 kg m3 )( 0.08 m3 min ) (1min 60s)( 1.868 kJ kgK)( 10.3-20) K =-21,500 W

Answer

12-27

12-27 A steam condenser downstream of a turbine in a Rankine cycle power plant, has 5,000 tubes, each with an internal diameter of 0.75-in. The steam condenses at 120 ºF on the outside of the tubes. The total cooling water flow rate is 3,500 lbm/s, enters the tubes at 54 ºF, and leaves at 85 ºF. Because the condensation heat transfer coefficient is very high, ignore the steam (and tube wall) thermal resistances. Determine: a. the heat transfer rate (in Btu/hr) b. the tube length required (in ft).

Approach: With the flow rating and inlet and outlet temperatures given for the cooling water, the heat transfer rate can be calculated with conservation of energy. Because this is a constant fluid temperature heat exchanger, the analysis given in Section 12.5 is used to calculate the tube length. This requires the heat transfer coefficient.

Assumptions: 1. The system is steady with no work and negligible potential and kinetic energy effects. 2. The liquid water is an ideal liquid with constant properties.

Solution: a) Applying the conservation of energy to the water, and assuming steady, no work, negligible potential and kinetic energy effects, and an ideal liquid with constant specific heat:

 P (Tout − Tin ) Q = mc For water at an average temperature of (54 + 85)/2 = 69.5 ºF, c p = 1.00 Btu/lbmR Q = ( 3500 lbm s )(1.00 Btu lbmR )( 85-54) R ( 3600 s hr) =3.906×108 Btu hr b) For a heat exchanger with one constant temperature fluid and Rtot = 1/hA (we ignore the condensation and wall  cp) where A = N π DL and  = m tot . Solving for L: thermal resistances): T out = T in + (T in – T f ) exp (-hA/ m m L =

 p − mc N π Dh

⎡ Tout − Tf ⎤ ⎥ ⎣⎢ Tin − T f ⎦⎥

ln ⎢

We need h, which requires the Reynolds number for flow in a single tube with the cooling water properties from Appendix B-6 evaluated at T avg = 69.5oF, µ = 65.8 × 10 -5 lbm/fts, k = 0.347 Btu/hrft oF, Pr = 6.82  tot  m ρ V D 4m Re = but V = ⇒ Re = 2 µ N ρ (π / 4 ) D N πµ D Re =

4 ( 3500lbm/s)

( 5000)π ( 65.8×10−5 lbm/fts ) ( 0.0625 ft )

= 21,700

This is turbulent flow so using the Dittus-Boelter correlation: Nu = 0.023 Re0.8 Pr 0.4 = 0.023(26700)0.8(6.82)0.4 = 146 h= L =

Nuk D

=

(146) ( 0.347 B tu/hr ft 0 F) 0.0625ft

= 810 Btu/hrft 0 F

− ( 3500 lbm/s ) (1.00 Btu lbm o F ) ( 3600s 1hr )

( 5000 )π ( 0.0625 ft ) ( 810 Btu/hrft 2 o F)

⎡ 85 − 120 ⎤ ⎥ = 10.0 ft 54 120 − ⎣ ⎦

1n ⎢

12-28

Answer

12-28 An air compressor used in a large car body shop is located in an inside equipment room. Fresh air at 5 ºC, 96 kPa is conveyed to the compressor from outside through a 30-cm circular duct that is 15-m long. The duct runs along the ceiling of the facility where the temperature is 34 ºC. If the air flow rate is 0.35 m 3/s, determine: a. the temperature of the air when it reaches the compressor (in ºC) b. the heat transfer rate (in W).

Approach: This is a heat exchanger with a constant wall temperature. We can use the analysis given in Section 12.5 (with Rtot = 1/hA) to calculate the outlet temperature. To do so, we need to determine the convective heat transfer coefficient. Once T OUT is determined, the heat transfer rate is calculated with conservation of energy.

Assumptions: 1. Air is at one atmosphere. 2. The system is steady with no work and negligible potential and kinetic energy effects. 3. Air is an ideal gas with constant specific heat.

Solution: a) An equation for a constant wall temperature heat exchanger, with Rtot = 1/hA, is  cp) T out = T f + (T in – T f ) exp (- hA/ m

 = ρ V  . Assuming air is an ideal gas The mass flow rate is m ρ =

PM RT

( 96 kN m ) ( 28.97 kg 2

=

( 8.314kJ

kmol )(1kJ kN ⋅ m )

kmolK )( 5 + 273) K

= 1.20kg/m3

 = (1.20kg/m3 )( 0.35m3 /s ) = 0.42kg/s m The heat transfer coefficient requires the Reynolds number with properties from Appendix A-7evaluated at T avg . Assume, T out ≈10 oC, T avg ≈ 7.5 oC ≈ 280.5 K, µ = 1.70×10-5 N⋅s/m2, k = 0.0246 W/mK, Pr = 0.714, c p = 1.005 kJ/kgK   ρ V D m 4m Re = but V = Re ⇒ = 2 µ ρ (π / 4 ) D πµ D 4( 0.42kg/s )

Re = π

(1.70 ×10−

5

N ⋅ s/m 2 )( 0.30m )

= 105,000

This is turbulent flow, so using the Dittus-Boelter equation Nu = 0.023Re0.8 Pr 0.4 = 0.023(105, 000) h=

Nuk D

=

( 209 )( 0.0246W/mK ) 0.30m

0.8

0.4

( 0.714) = 209

=17.1W/m2 K

⎡ - (17.1 W/m2 K ) π( 0.30m )(15m)(1J 1Ws) ⎤ ⎥ = 17.6 o C T out = 34 C + ( 5-34) K exp ⎢ ⎢⎣ ( 0.42kg/s )(1.005kJ/kgK)(1000J kJ) ⎥⎦ o

Answer

b) The heat transfer rate is determined from the energy equation. Assuming steady, no work, negligible potential and kinetic energy effects, and ideal gas with constant specific heat so that ∆h = c p ∆T :

 p (Tout − Tin ) = ( 0.42kg/s )(1.005k J kgK )(17.6 − 5) K (1000Ws 1kJ) = 5320 W Q = mc

12-29

Answer

12-29 A proposed cooling technique for high power computer chips is to machine micro-channels into the backside of the silicon chip. Because Nusselt number is defined as Nu = hk/Lchar , for a given value of Nu, as the characteristic length decreases the heat transfer coefficient increases. Consider a 1.5-cm by 1.5-cm computer chip that dissipates 50 W. Water at 20 ºC is used as the coolant, and its outlet temperature is limited to 25 ºC. Heat transfer is primarily from the base of the channel (that is, ignore heat transfer to the water from the channel sides). Assume the distance between the sides of two channels is 0.1-mm. Determine: a. the number of 0.25-mm deep and 0.25-mm wide micro-channels that are on a chip and the average surface temperature of the base of the micro-channels (in ºC) b. the number of 1-mm deep and 1-mm wide micro-channels that are on a chip and the average surface temperature of the base of the micro-channels (in ºC).

Approach: If we assume the chip is at a uniform temperature, T s, we can use the analysis given in Section 12.5 for a heat exchanger with a constant temperature wall. We need to determine the water flow rate and the heat transfer coefficient. The water flow rate can be determined from conservation of energy.

Assumptions: 1. The system is steady with no work and negligible potential and kinetic energy effects. 2. Water is an ideal liquid with constant specific heat.

Solution: a) For a constant wall temperature heat exchanger (with Rtot = 1/hA), the water exit temperature can be calculated from  cp) T out = T s + (T in – T s) exp (-hA/ m Solving for T s =

Tout −Tin exp ( − hA/m c p )

 c p ) 1- exp ( − hA/m

. Therefore, we must evaluate A, h, and m The heat transfer area depends on the number of channels. The number of channels, N , on a chip is based solely on geometry. W−D 0.015m − 0.0001m = = 42.5 W = ( NLc ) + ( N + 1) D ⇒ N = Lc + D 0.00025m + 0.0001m This must be an integer, so N = 42. Area for heat transfer is only the base of the channel A = NLc L = (42) (0.00025 m) (0.015 m) = 1.575 × 10 -4 m2 The water flow rate is determined from conservation of energy heat plus the heat transfer rate and temperatures. Assuming steady, no work, no potential or kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c p ∆T : Q  p (Tout − Tin ) ⇒ Q = mc m = c (Tout − T in ) For evaluating c p and eventually the heat transfer coefficient, we need water properties from Appendix A-6 at T avg = (20 + 25)/2 = 22.5 oC, so by interpolation: ρ = 997.6 kg/m3, c p = 4.180 kJ/kgK, k = 0.607 W/mK, µ = 9.29 × 10-4 N⋅s/m2, Pr = 6.40 ( 50W) (1J Ws) = m = 0.00239 kg/s ( 4.180k J/kgK )(1000 J kJ)( 25 - 20) K 12-30

The Reynolds number is Re =

D H ρ V µ

4( 0.00025m)( 0.000025m) 4 A x 4 HLc where D H = = = = 0.00025m Pwetted 2( H + Lc ) 2( 0.00025m + 0.00025m)

The velocity is for the flow in one channel  /N ( 0.00239kg/s ) m V

=

ρ A x

=

( 42 ) ( 997.6 kg/m3 ) ( 0.00025m)

( 997.6 kg/m ) ( 0.91 m/s )( 0.00025m)

= 0.91m/s

2

3

Re =

9.25 ×10-4 N ⋅ s/m2

= 245

This is laminar flow, so check the entrance length. Lent = 0.037 Re Pr D = (0.037) (245) (6.40) (0.00025 m) = 0.0145 m Because the entrance length is about the same length as the channel, we will take into account entrance conditions. Assuming T s ~ 25oC, µ s = 8.72 × 10-4 N⋅s/m2 1/ 3

Nu =1.86 Gz

1/ 3

h=

Nuk D H

hA

=

(µ

µ s

⎡ ( 245)( 6.40)( 0.00025) ⎤ =1.86 ⎢ ⎥ 0.015 ⎣ ⎦

( 5.57)( 0.607W/mK ) 0.00025m

⎛ 9.29 ×10−4 ⎞ ⎜ -4 ⎟ ⎝ 8.76×10 ⎠

0.14

4

m2 ) ( 1J Ws)

= 0.213

( 0.00239 kg/s )( 4.180k J/kgK )(1000 J kJ) ( 25+273) - ( 20+273) exp ( -0.213) =319.0K=46.0o C T s = 1-exp ( -0.213)  p mc

= 5.57

= 13,500W/m2 K

(13500W/m K )(1.575 x10− 2

=

)

0.14

b) For a channel with Lc = H = 0.001 m N = 13, A = 1.95 × 10-4 m2, D H = 0.001 m V = 0.184 m/s, Re = 199 Lent = 0.047 m, Nu = 8.24, h = 5,000 W/m2K T s = 346.7 K = 73.7 oC

Answer

Answer

Comments:

Increasing the size of the channels has a detrimental effect on the wall temperature. Even though the heat transfer area increased, the decrease in the heat transfer coefficient was larger, so that the hA product decreased, and it’s the hA product that is most important.

12-31

12-30 Consider turbulent flow of a fluid through a tube maintained at constant temperature. The mass flow rate is 0.32 kg/sec, and the heat transfer coefficient is 250 W/m2·K. Now the free-stream velocity of the fluid is doubled. Assume the flow regime remains unchanged. Estimate: a. the percent change in the pressure drop of the fluid between the old and new flow rates b. the percent change in the local heat flux between the fluid and the walls of the channel c. the percent change in the total heat transfer rate over the length of the channel if the heat transfer area is 3.7 m 2, specific heat is 2200 J/kg·K, inlet temperature is 25 C, and wall temperature is 75 C.

Approach: This is a comparison study of changes in performance ( ∆ P, h, Q ) when a turbulent flow is varied inside a tube. Basic expressions for ∆ P, f, h, and Q are used.

Assumptions: 1. The system is steady with no work or potential or kinetic energy effects.

Solution: a) Let

V2

= 2 V1 ⇒

L V 2  2 = 2m 1 . An expression for pressure drop is ∆ P = f ρ m . The tube geometric and D 2

2 ∆ P2 f 2 ( L / D ) ( ρ V2 2 ) f 2 ⎛ fluid properties remain fixed. Therefore, = = ⎜ ∆ P1 f1 ( L / D ) ( ρ V1 2 2 ) f 1 ⎝

For turbulent flow in a tube

f =

0.184 Re

0.2

=

⎞ ⎟ ⎠

V2 V1

2

0.184

( ρ D / µ )

0.2

V

0.2

Substituting this into the ∆ P expression and canceling like terms

∆ P 2 ⎛ =⎜ ∆ P 1 ⎝

V2 V1

1.8

⎞ 1.8 ⎟ = (2) = 3.48 ⎠

b) The change in the local heat flux

Using the Dittus-Boelter equation

Answer q2′′

=

h2 (Tw −T ) f

q1′′ h1 (Tw − T ) f

=

h2 h1

⎛ ρ VD ⎞ h = ( 0.023 ) ⎜ ⎟ D ⎝ µ ⎠ k

0.8

Pr 0.4 = C V 0.8

Substituting this into the local heat flux comparison and canceling like terms q2′′ ⎛

0.8

⎞ 0.8 =⎜ ⎟ = (2) =1.74 q1′′ ⎝ V1 ⎠ V2

Answer

c) For the total transfer rate, we can use conservation of energy. Assuming steady, no work, negligible potential energy and kinetic energy effects, and ideal fluid so that ∆h = c p ∆T  2 c p (Tout ,2 −Tin , 2 ) Q 2 m Q1

=

 1 c p ( Tout ,1 −Tin ,1 ) m

We can obtain the outlet temperatures for this constant wall temperature heat transfer situation from

 p) Tout = Tw − (Tw − Tin ) exp ( −hA mc

→

⎡ -(250 W m2 K ) (3.7m 2 ) ⎤ o ⎥ = 61.6 C ⎣ (0.32kg/s)(2200 J/kgK) ⎦

T out ,1 = 75o C − (75 − 25)o C exp ⎢

⎡ −(1.74)(250)(3.7) ⎤ = 59.1 o C ⎥ ⎣ 2(0.32)(2200) ⎦  2 c p (Tout ,2 −Tin ,2 ) ⎛ 59.1− 25 ⎞ m = = 2⎜ ⎟ = 1.86  1 c p (Tout ,1 −Tin ,1 ) m ⎝ 61.6− 25 ⎠

T out ,2 = 75 − (75 − 25)exp ⎢ Therefore,

Q 2 Q1

12-32

Answer

12-31 Glycerin is pumped through a 1.5 cm diameter tube that is 5 m long. The inlet temperature is 32 °C, the required outlet temperature is 22 °C, and the flow rate is 100 kg/hr. Determine the wall temperature required to obtain this outlet temperature (in °C).

Approach: This is a constant wall temperature heat exchanter, so we can use Eq. 12- 33. The appropriate heat transfer coefficient correlation is used once the Reynolds number is determined.

Assumptions: 1. The heat transfer is one dimensional.

Solution: An equation for a constant wall temperature heat exchanger is

 p ⎤⎦ Tout = Tw − (Tw − Tin ) exp ⎡⎣− hA mc Solving for T w  p ⎤⎦ Tout − Tin exp ⎡⎣ − hA mc  p ⎤⎦ 1 − exp ⎡⎣ − hA mc

T w =

with T in K

To calculate h, we need Re. Fluid properties are obtained from Appendix A-6 at the average bulk temperature (32+22)/2 = 27 ºC: c p = 2.427 kJ/kgK, k = 0.286 W/mK, µ = 7990 × 10-4 N-s/m2, Pr = 6780   ρ V D 4m 4m Re = but V = ⇒ Re = 2 µ ρπ D πµ D Re =

4(100kg/hr)(1hr 3600s) (7990×10-4 Ns/m2 )(0.015m)

= 2.95

π

Because this is laminar flow, we must check entrance length Lent  ,t = 0.037 Re Pr D = 0.037 (2.95) (6780) (0.015m) = 11.1m This is much longer than the tube length, so we use a correlation that takes into account the entrance length. Using Eq, 12- 42 and evaluating the viscosity at an assumed wall temperature of 7 ºC, µw= 53400 × 10−4 Ns/m2 1/ 3

Nu =1.86 Gz1/ 3 ( µ / µ w ) 0.14 =1.86( Re Pr D L ) 1/ 3

⎡ (2.95)(6780)(0.015) ⎤ =1.86 ⎢ ⎥⎦ 5 ⎣ h=

( µ / µw )

⎛ 7990 × 10−4 ⎞ ⎜ −4 ⎟ ⎝ 53400 × 10 ⎠

0.14

0.14

= 5.58

Nuk ⎛ 5.58)(0.286W/mk ⎞ =⎜ = 106 W /m2 K ⎟ D ⎝ 0.015m ⎠

⎡

⎤ -(106W/m 2 K)(π) (0.015m)(5m) ⎥ ⎣ (100 kg hr )(1hr 3600s)( 2.427 kJ kgK)( 1000 J kJ) ⎦

(273+ 22) − (273+ 32) exp ⎢ T w =

⎡

⎤ ⎥ ⎣ (100 kg hr )(1hr 3600s)( 2.427 kJ kgK)( 1000 J kJ) ⎦

1-exp ⎢

-(145W/m2 K)(π) (0.015m)(5m)

Answer

= 272.7K = -0.3 ºC

Comments: Because the viscosity changes so dramatically with temperature, we will iterate one time. At 273 K, , µw= 106, 000 × 10−4 Ns/m2, Nu = 5.07, h = 96.7 W/m2K, T w = 270.1 K = -2.9 ºC

12-33

12-32 In a pharmaceutical application, the product is subjected to a final sterilization by heating it from 32 ºC to 80 ºC. A flow of 60 cm3/s is passed through a 10-mm tube that is heated with a uniform heat flux produced by wrapping the tube with an electric resistance heater. If product properties can be approximated by those of ethylene glycol, and the tube is 25-m long, determine: a. the required power (in W) b. the wall temperature at the tube exit (in ºC).

Approach: The required heat transfer rate can be obtained from conservation of energy and the given information. The wall temperature at the tube outlet can be calculated using the basic rate equation ( q ′′ = h ∆T ).

Assumptions: 1. The system is steady with no work or potential or kinetic energy effects. 2. The liquid is ideal with constant specific heat.

Solution: a) For conservation of energy on the liquid, assume steady, no work, no potential or kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c p ∆T  Q = mc p ( Tout − Tin )

The properties required to calculate the heat transfer coefficient are evaluated from Appendix A-6 at T avg = (32 + 80)/2 = 56 oC ~ 330 K, ρ = 1089.5 kg/m3, c p = 2.549 kJ/kgK, µ = 56.1×10-4 N⋅s/m2, k = 0.260 W/mK, Pr = 55.0 Q = (1089.5kg m3 )( 60cm3 s ) (1m 100cm)

3

q′′ = h (T s −Tout )

b) For the wall temperature at the tube outlet T s = Tout + (Q/A ) h = Tout + Q

kgK)( 80 − 32) K( 1000W 1k J/s) = 8000W

( 2.549kJ

⇒

Ts = Tout + q ′′ h

π DLh

To calculate the heat transfer coefficient, we need the Reynolds number:   ρ V D m 4m Re = ⇒ V = ⇒ Re = 2 µ ρ (π /4 ) D πµ D 3

Re =

4 (1089.5 kg m3 )( 60 cm3 s ) (1m 100cm) π

( 56.1×10

-4

Ns/m 2 ) ( 0.01m )( kgm Ns2 )

= 1480

This is laminar flow, so we will check the entrance length Lent,t ~ 0.053 Re Pr D = 0.053 (1480) (55.0) (0.01 m) = 43.2 m This is significantly longer than the tube length, so entrance effects must be taken into account. We do not have a correlation that gives the heat transfer coefficient for laminar, developing flow with a constant wall heat flux boundary condition. So we will use the Seider and Tate correlation. Assuming T s ~ 350 K, -4 µ s = 34.2×10 N⋅s/m2 Nu =1.86 Gz

1/ 3

h=

Nuk D

=

( µ /µ s )

0.14

⎡ (1480 )( 55.0)( 0.01m) ⎤ = 1.86 ⎢ ⎥ 25m ⎣ ⎦

( 6.37)( 0.260W/mK ) 0.01m

π

⎛ 56.1× 10−4 ⎞ ⎜ −4 ⎟ ⎝ 34.2 × 10 ⎠

0.14

= 6.37

= 166W/m2 K

8000W

T s = 80 o C +

1/ 3

( 0.01m )( 25m ) (166W/m

2

K )

=141.3 o C

Re-evaluating µ s = at 373 K (highest temperatures for which we have properties µ s = 21.5 ×10−4 N⋅s/m2 Nu  6.80 , h  177W/m2 K , T s = 137.5 o C

Answer

Comments: The wall temperature will be lower than this, since

µ s should

12-34

be smaller than with what we have available.

12-33 Air enters a compressor operating at steady state with a volumetric flow rate of 37 m3/min at 105 kPa and 30 °C and exits with a pressure of 690 kPa and temperature of 240 °C. The compressor is cooled with 40 kg/min of water that circulates in a water jacket enclosing the compressor. The water jacket can be approximated as 25 3-cm diameter, 2-m long pipes. The water enters at 20 °C and the wall temperature of the pipes can be approximated as being constant at 135 °C. Assume fully developed flow. Determine: a. heat transfer rate from the compressor to the water (in kW) b. the mass flowrate of air (in kg/sec) c. the power input to the compressor (in kW).

Approach: The heat transfer rate from the compressor to the water jacket can be determined from a combination of an energy balance on the water and the use of the constant wall temperature heat exchanger equation given in Section 12.5. Assuming one-dimensional flow the air mass flow rate can be determined from its definition. Finally, an energy balance on the compressor is used to determine the required input power.

Assumptions: 1. The system is steady with negligible potential and kinetic energy effects. 2. Air and water are ideal fluids. 3. No work is done in the heat exchanger.

Solution: a) Using control volume CVII defined in the schematic, we apply conservation of energy to the water. Assume steady, no work, negligible potential and kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c p ∆T :

 c c p ( TD − TC ) Q II = m

The water outlet temperature, T D, is obtained with the analysis for a constant wall temperature heat exchanger  cp). The heat transfer coefficient is calculated with with Rtot = 1/hA: T D = T wall + (T c – T wall ) exp (- hA/ m the Reynolds number for flow in one tube evaluated at T avg = (T C + T D)/2. From Appendix A-7 and assuming T D ≈ o o -4 2 30 C, T avg = (20 + 30)/2 = 25 C, so µ = 8.72 × 10 N⋅s/m , k = 0.611 W/mK, c p = 4.178 kJ/kgK, Pr = 5.97. 4 ( 40kg/ min )(1min 60s )  /N  4m ρ V D m Re = ⇒ V= → Re = = = 1300 2 -4 2 µ ρ (π /4 ) D πµ DN π ( 8.72 ×10 N ⋅ s/m ) ( 0.03m )( 25) This is laminar flow, so assuming fully developed flow for constant wall temperature boundary condition Nuk ( 3.66)( 0.611W/mK ) Nu = 3.66 → h = = = 74.5W/m2 K D 0.03m

⎡ ⎤ - ( 74.5W m2 K ) π( 0.03m )( 2m)( 25) ⎥ = 33.6 o C T D = 135 C + ( 20 -135) K exp ⎢ ⎢⎣ ( 40 kg min )(1min 60s)( 4.178k J kgK)( 1000 J kJ) ⎥⎦ Q = ( 40 kg min )(1min 60s)( 4.178 kJ kgK )( 33.6 − 20) K = 37.9kW o

II

Answer

b) For the air mass flow rate 105 kN m2 ) ( 28.97 kg kmol ) ( 37 m3 min ) ( 1min 60s) ( PM kg  = ρ VA = ρ V = m V = 0.745 RT s ( 8.314 kg kmolK )( 30 + 273) K

Answer

c) Applying the energy equation to CVI , and assuming steady, negligible potential and kinetic energy effects, ideal gas with constant specific heat so ∆h = c p ∆T :

 ( hA − hB ) = 0 Q I − W + m Note that Q I = − Q II

→

→

 c p (TA − TB ) W = Q I + m

 p (TA − TB ) W = − Q II + mc

At the average air temperature, c p = 1013 J/kgK

 = − 37.9k W + ( 0.745 kg s )(1013J kgK )( 30 − 240) K( 1kJ 1000J) =− 196k W W

12-35

Answer

12-34 The condenser downstream of the turbine in a large Rankine cycle power plant is constructed of 30,000 25mm tubes. The steam condenses at 50 ºC with a heat transfer coefficient of 9,000 W/m2·K on the outside the tubes. The cooling water enters the tube side of the condenser at 20 ºC at a flow rate of 17,000 kg/s. For a 1000-MW (net) power output and a cycle thermal efficiency of 42%, determine: a. the cooling rate required (in MW) b. the outlet temperature of the cooling water (in ºC) c. the length of tubing required (in m).

Approach: The Rankine cycle thermal efficiency definition is used to calculate the cooling rate. The energy equation applied to the cooling water is used to calculate the outlet temperature. Condensing steam is a constant fluid temperature, so the analysis in Section 12.5 is used to calculate the required length.

Assumptions: 1. The system is steady, with no work or potential or kinetic energy effects. 2. Water is an ideal liquid with constant specific heat.

Solution: a) The definition of Rankine cycle thermal efficiency is Also Wnet = Q in − Q out

→

Q out = ⎡(1−ηcycle )

⎣

Q in =W net + Q out

η cycle

= Wnet Qin

Substituting this into the definition of η cycle:

 =⎡ η cycle ⎤ W net

⎣(1− 0.42) 0.42⎤⎦ (1000 MW) =1381MW

⎦

Answer

b) The outlet cooling water temperature is calculated with conservation of energy applied to a control volume around the cooling water. Assuming steady, no work, no potential or kinetic energy effects, and an ideal liquid

 p (Tout − Tin ) → with constant specific heat so ∆h = c p ∆T : Q = mc

 p Tout = Tin + Q /mc

Evaluating water properties from Appendix A-6 at the average temperature, assume T out ≈ 40 oC, T avg ≈ 30 oC c p = 4.176 kJ/kgK, k = 0.618 W/mK, µ = 7.79 × 10 -4 N⋅s/m2, Pr = 5.26 (1,381,000kW )(1kJ/s 1kW) T out = 20 o C + = 39.5 o C Answer (17000 kg s )( 4.176 kJ kgK )

 p Rtot ) Tout = T f + (Tin − T f )exp ( − 1/mc

c) For a heat exchange with one constant fluid temperature where Rtot =

1 hi A1

+

1n ( D2 /D1 ) 2π kL

+

1 h0 A0

. Because no information is given about the tube material or wall thickness,

wall resistance must be ignored, and Ai = A0 = N π DL.

Solving for the tube length:

L =

→

Rtot =

hi N π DL

+

1 ho N π DL

=

⎡1 1⎤ ⎢ + ⎥ N π DL ⎣ hi ho ⎦ 1

 p ⎡ 1 1 ⎤ ⎛ Tout − T f ⎞ − mc ⎟ ⎢ + ⎥1n ⎜ N π D ⎣ hi ho ⎦ ⎜⎝ Tin − T f ⎟⎠

The Reynolds number in one tube is

Re =

ρ V D

⇒

V

µ

Re =

1

4 (17000 kg s ) ( Ns2 kgm )

( 30000)( π ) ( 7.79×10-4 Ns/m2 ) ( 0.025m)

=

 /N m

(

ρ π /4

) D

2

⇒

 4m Re = N π µ D

=37050

Turbulent flow, so using Dittus-Boelter equation Nu = 0.023 Re0.8 Pr 0.4 = (0.023) (37050)0.8 (5.26)0.4 = 202 Nuk ( 202 )( 0.618W/mK ) h= = = 4990W/m2 K D 0.025m L = −

(17000 kg s )( 4.176 kJ kgK)( 1000 J ( 30000)π ( 0.025m)

kJ) ⎡

⎤ ⎡ 39.5- 50 ⎤ 1 1 + ⎢ ⎥1n ⎢ ⎥ = 9.85m Answer 2 2 4990W m K 9000W m K 20 50 ⎣ ⎦ ⎣ ⎦ 12-36

12-35 Air at a mass flow rate of 0.0015 lbm/s and an inlet temperature of 80 ºF enters a rectangular duct 3.5-ft long, 0.15-in. high, and 0.60-in. wide. A uniform heat flux of 50 Btu/hr·ft2 is imposed on the duct surface. Determine: a. the outlet temperature of the air (in ºF) b. the highest wall temperature and its location (in ºF and ft).

Approach: The outlet temperature can be determined from an energy balance on the air flow combined with the given information. The highest wall temperature will occur at the end of the duct where the fluid temperature is highest and can be found by using the rate equation ( Q = q″ A = hA ∆T )

Assumptions: 1. Air is an ideal gas at one atmosphere. 2. The system is steady, with no work or potential or kinetic energy effects.

Solution: a) For an energy balance on the air flow, assume steady, no work, no potential or kinetic energy effects, and an ideal gas with constant specific heat so that ∆h = c p ∆T :

 p (Tout − Tin ) Q = mc

 p =Tin + q ′′A /mc  p Tout =Tin + Q /mc

→

The fluid properties from Appendix B-7 are evaluated at T avg , so assuming T out ≈ 100 oF, T avg ≈ 90 oF; 3 o o -5 ρ = 0.00725 lbm/ft , c p = 0.240 Btu/lbm F, k = 0.0152 Btu/hr ft F, µ = 1.267 × 10 lbm/ft⋅s, Pr = 0.72 T out = 80 F +

( - 50Btu

o

hrft 2 ) ( 2 )( 0.0125 + 0.05) ft ( 3.5ft )

( 0.0015lbm s )( 3600s 1hr ) ( 0.240Btu Q = q ′′ A =hA (Tw − Tout )

b) Using the rate equation

lbm F) o

→

= 96.9 o F

Answer

Tw = Tout + q ′′/h

The Reynolds number is Re = ρ VD H µ . For the non-circular duct, we need the hydraulic diameter D H =

4 A x P wetted

The velocity is

Re =

V

4( 0.0125ft )( 0.05ft )

=

2( 0.0125+0.05) ft

=

 m ρ A x

=

( 0.0725lbm ft

3

= 0.020ft

0.0015lbm/s

( 0.0725lbm ft )( 0.0125ft )( 0.05ft ) )( 33.1ft s )( 0.02ft) 3

1.267 ×10-5 lbm/ft s

= 33.1

ft s

= 3790

Turbulent flow, so using the Gnielinski correlation Nu =

( f 8)( Re −1000) Pr 1/ 2 1 + 12.7( f 8) ( Pr 2 / 3 − 1)

−2

−2

f = [ 0.7901n Re − 1.64] = ⎡⎣ 0.791n ( 3790) − 1.64⎤⎦ = 0.0422 Nu =

h=

( 0.0422 8)( 3790 − 1000)( 0.72) = 13.0 1/ 2 2/3 1 + 12.7 ( 0.0422 8) ⎡( 0.72) − 1⎤ ⎣ ⎦

Nuk D H

=

(13.0 ) ( 0.0152Btu/hrft o F)

T w = 96.9 o F +

0.02ft 50Btu/hrft 2 9.89 Btu/hr ft

2 o

F

= 9.89

Btu hrft o F

=102.0 o F

Answer

12-37

12-36 Water enters at 40 ºF and flows at a rate of 0.25 ft 3/s inside a 20-ft long annulus, whose inner and outer radii at 1 in. and 2 in., respectively. The inner surface is maintained at 150 ºF and the outer surface is heavily insulated. Determine: a. the outlet temperature (in ºF) b. the heat transfer rate (in Btu/hr).

Approach: This is a constant wall temperature heat exchanger. Using the analysis from section 12.5, we can determine the outlet temperature. Once that is known, conservation of energy is used to calculate the heat transfer rate.

Assumptions: 1. The system is steady with no work and negligible potential and kinetic energy effects. 2. The water properties are constant.

Solution: a) For a constant wall temperature heat exchanger with Rtot = 1/hA, the exit temperature can be obtained from:  cp) T out = T s + (T in – T s) exp (-hA/ m The heat transfer coefficient requires the Reynolds number with fluid properties evaluation at the average temperature. Because we do not know the outlet temperature, we assume T out ~ 120oF, T avg = (40 + 120)/2 = 80oF, and the properties from Appendix B-6 are ρ = 62.2 lbm/ft3, c p = 1.00 Btu/lbmoF, k = 0.353 Btu/hr ft oF, µ = 57.8 × 10-5 lbm/ft⋅s, Pr = 5.89 Re =

ρ V D H µ

V

=

V A

where D H =

 V

=

(π 4 ) ( D22 − D12 )

4 A x Pwetted

=

4(π / 4 ) ( D22 − D12 ) π

( D2 + D1 )

4 ( 0.25 ft 3 /s )

= π

( 0.167

2

− 0.08332 ) ft 2

( 62.2 lbm/ft ) (15.3 ft/s )( 0.0833 ft )

= D2 − D1 = 0.0833ft

= 15.3ft/s

3

Re =

57.8×10-5 lbm/ft s

= 137,000

This is turbulent flow, so using the Dittus-Boelter correlation: Nu = 0.023 Re0.8 Pr 0.4 = (0.023) (137000)0.8 (5.89)0.4 = 601 h=

Nuk D H

=

( 601) ( 0.353Btu/hr ft o F) 0.0833ft

= 2550

Btu hr ft 2 o F

⎡ ⎤ - ( 2550Btu hrft 2 R ) π ( 0.0833 ft )( 20 ft ) ⎥ T out =150 F + (40 - 150) F exp ⎢ ⎢⎣ ( 62.2 lbm ft 3 )( 0.25 ft 3 s ) (1.00Btu lbmR )( 3600s 1hr ) ⎥⎦ o

o

= 63.4 o F

Answer

Because the calculated T out is significantly different than the assumed T out , the fluid properties should be re-evaluated and T out calculated again. b) The heat transfer rate can be determined from conservation of energy applied to the water. Assuming steady, no work, no potential or kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c p ∆T

 cp ∆T Q = m lbm ⎛ = ⎜ 62.2 3 ft ⎝

ft 3 ⎞ ⎛ Btu ⎞⎛ ⎟ ⎜ 0.25 s ⎟ ⎜1.00 lbmR ⎠⎝ ⎠⎝

o ⎞ ⎛ 3600s ⎞ × 106 Btu/hr ⎟ ( 63.4 - 40) F⎜ 1hr ⎟ =1.310 ⎠ ⎝ ⎠

12-38

Answer

12-37 Unused engine oil is to be heated from 20 ºC to 65 ºC using condensing steam at 100 ºC. The oil flows inside a 1-cm diameter tube at a flow rate of 0.1 kg/s. The resistance of the condensing steam and the tube wall can be ignored. Determine the length of tube required (in m).

Approach: This is a constant fluid temperature heat exchanger. Using the analysis given in Section 12.xx we can determine the required length. The heat transfer coefficient must be determined.

Assumptions: 1. The condensing and wall thermal resistances are ignored.

Solution: For a heat exchanger with one constant temperature fluid and ignoring the condensing coefficient and the wall

 p ) where A = π DL. Substituting this Tout = Tst + ( Tin − Tst ) exp ( − hA mc

resistance, then Rtot = 1/hA . Therefore, into the main equation and solving for L:

⎛ T − T ⎞  p 1n ⎜ out st ⎟ − mc ⎝ Tin − T st ⎠ L = π Dh

The heat transfer coefficient requires the Reynolds number with oil properties from Appendix A-6 evaluated at the average temperature (65 + 20)/2 = 42.5 oC; by interpolation µ = 1970×10-4 N⋅s/m2, k = 0.144 W/mK, c p = 1.972 kJ/kgK, Pr = 2683 Re =

ρ V D

→

V

=

µ

 m

(

ρ π /4

) D 2

→ Re =

 4m π µ D

4 ( 0.1kg/s ) ( Ns 2 kgm )

= π

(1970 ×10

-4

N ⋅ s/m2 ) ( 0.01m )

= 64.6

This is laminar flow, so check the entrance length Lent = 0.037 RePr D = 0.037 ( 64.6) ( 2683)( 0.01m) = 64.2m We assume the tube length will be shorter than this so we use the Seider-Tate correlation 1/ 3

Nu =1.86 ( Gr )

( µ /µ s )

0.14

=1.86 ( RePr D L )

1/ 3

( µ /µ s )

0.14

But this requires the tube length. Therefore, an iterative solution is required. The viscosity at the wall temperature of about 100 ºC is approximately 180 × 10-4 N⋅s/m2. Assume a tube length of 10 m. 1/ 3

⎡ ( 64.6) ( 2683)( 0.01m) ⎤ Nu =1.86 ⎢ ⎥ 10m ⎣ ⎦

⎛ 1970 × 10-4 ⎞ ⎜ -4 ⎟ ⎝ 180 × 10 ⎠

0.14

= 14.5

→

h=

Nuk D

=

(14.5)( 0.144W/mK ) 0.01m

= 209 W/m2 K

⎡ 65-100 ⎤ − ( 0.1kg/s )(1.972kJ/kgK ) 1n ⎢ ⎥ (1000J kJ ) 20-100 ⎦ ⎣ L = = 24.8m 2 π

( 0.01m ) ( 209W/m

K ) (1J/s 1W )

This is longer than our assumed length, so using this L 1/ 3

⎡ ( 64.6 )( 2683)( 0.01) ⎤ ⎛ 1970 × 10−4 ⎞ Nu =1.86 ⎢ ⎥ ⎜ −4 ⎟ 24.8 ⎣ ⎦ ⎝ 180 × 10 ⎠ ⎡ 65-100 ⎤ − ( 0.1)(1.972) 1n ⎢ ⎥ (1000) − 20 100 ⎣ ⎦ L = = 33.7m π

= 10.7

→

h=

(10.7 )( 0.144) 0.01

= 154W/m2 K

( 0.01)(154)

Additional iterations give Nu = 9.66, h =139W/m3 K Nu = 93.33,

0.14

h = 134 W/m2 K

L = 37.3m

Answer

L = 38.6m

12-39

12-38 In a small ship with limited space, water must be heated from 10 ºC to 50 ºC with condensing steam at 100 ºC. The water flow rate is 1.5 kg/s. Either one 4-cm diameter tube, two 3-cm tubes, or three 2-cm tubes can be used in parallel. Determine which configuration will yield the shortest length.

Approach: Whether there is 1, 2 or 3 tubes, the flow is evenly distributed among the tubes. This is a constant fluid temperature heat exchanger. Using the analysis given in Section 12.5 and the appropriate flow in each tube (which depends on the diameter), the required length can be determined. The heat transfer coefficient for each configuration must be evaluated.

Assumptions: 1. The condensing and wall thermal resistances are ignored.

Solution: For a heat exchanger with one constant temperature fluid and ignoring the condensing heat transfer coefficient and

 p ) where A = π DL for one tube. Tout = Tin + (Tin −T st ) exp ( − hA mc

the wall resistance, then Rtot = 1/hA:

 /N: Substituting this into the equation, recognizing that the flow in one tube is m

⎛ T − T st ⎞  p 1n ⎜ out − mc ⎟ Tin − T st ⎠ ⎝ L = N π DH

The heat transfer coefficient requires the Reynolds number (for flow in one tube) with fluid properties from Appendix A-6 evaluated at the average temperature, (10 + 50)/2 = 30 0C, so c p = 4.176 kJ/kgK, µ = 7.79 × 10-4 N⋅s/m2, k = 0.618 W/mK, Pr = 5.26  /N  ρ V D 4m m Re = → V= → Re = µ πµ D N πµ D For D = 4 cm and N = 1 Re =

4 (1.5kg/s ) ( Ns 2 kgm )

(1) π ( 0.04m ) ( 7.79 ×10-4 N ⋅ s/m2 )

= 61,300

Similarly, for D = 3 cm and N = 2, Re = 40860 D = 2 cm and N = 3, Re = 40860 These are all turbulent flow, so using the Dittus-Boelter correlation: For D = 4 cm Nu = 0.023 Re0.8 Pr 0.4 = 0.023(61300)0.8 (5.26)0.4 = 302 Nuk ( 302 )( 0.618W/mK ) h= = = 4,670W/m2 K 0.04m D Similarly for D = 3 cm D = 2 cm For D = 4 cm

Nu = 218 Nu = 218

h = 4500 W/m2 K h = 6750 W/m2 K

⎡ 50 − 100 ⎤ − (1.5kg/s )( 4176J/kgK ) ln ⎢ ⎥ (1000 J kJ) 10 − 100 ⎦ ⎣ L = = 6.27m 2

(1)π ( 0.04m ) ( 4670W/m

Similarly for D = 3 cm and N = 2, D = 2 cm and N = 3,

K )

L = 8.68 m L = 8.68 m

Answer

The single 4 cm tube yields the shortest length.

12-40

12-39 A thick, stainless steel (AISI 316) pipe with inside and outside diameters of 20 mm and 40 mm, respectively, is heated electrically to provide a uniform heat generation rate of 107 W/m3. This pipe is encased within a larger concentric tube with an inside diameter of 50 mm, whose outer surface is heavily insulated. Pressurized water flows through the annular region between the two tubes with a flow rate of 0.6 kg/s. The water inlet temperature is 20 ºC. Determine: a. the required pipe length if the desired outlet temperature is 40 ºC (in m) b. highest surface temperature and its location (in ºC and m).

Approach: This is an internal forced convection problem. With the known heat generation rate and water flow, we can determine the length using conservation of energy. The highest surface temperature can be determined from the heat transfer rate and the heat transfer coefficient.

Assumptions: 1. The heat transfer is one dimensional. 2. The system is steady with no work or potential or kinetic energy effects. 3. Water is an ideal liquid with constant specific heat.

Solution: a) Assume a steady flow with no work, negligible potential and kinetic energy effects, and water is an ideal liquid with constant specific heat so that ∆h = c p∆T . Applying these to the conservation of energy and mass:

 (h in – hout ) = 0 Q + m q '''

→

 c p (T in – T out ) = 0 q''' V wall + m

π

⎡⎣ Do2 − Di2 ⎤⎦ L + m c p (Tin − Tout ) = 0 4

 c p (Tin − Tout ) m

L =

→

q ''' (π 4 ) ⎡⎣ Do2 − Di2 ⎤⎦ Evaluating the water specific heat at the average temperature T avg = (T in – T out )/2 = (20 + 40)/2 = 30 °C. From Appendix A-6. c p = 4.176 kJ/kgK (0.6kg/s)(4.176 kJ/kgK)(40− 20)K (1000J 1kJ) =5.32m Answer L = (107 W m3 )(π 4) ⎡⎣ (0.04m)2 − (0.02m)2 ⎤⎦ (1J Ws) b) The location of the highest surface temperature is at the tube outlet. There the bulk water temperature is 40ºC, 3 -4 2 so from Appendix A-6: ρ = 992.2 kg/m , k = 0.631 W/mK, µ = 6.34 × 10 N-s/m , Pr = 4.19 The heat flux of that location is q′′ = h(T s- T out )

T s = Tout +

→

q′′ h

= Tout +

q ′′′V wall

= Tout +

hA s

2 2 q ''' ⎣⎡ Do − Di ⎦⎤

h

4 Do L

Because this is flow in an annulus, we need the hydraulic diameter for Re and Nu 4 A x Dh = = Do − Di = 50 mm – 40 mm = 10mm = 0.01m pwetted V

=

 m ρ A x

Re =

 m

=

ρ V Dh

4 ) ⎡⎣ D − D ⎤⎦

(

2

ρ π

=

( 992.2 kg

2 o

0.6kg/s

=

( 992.2 kg

2 m3 ) (π 4) ⎡( 0.05m) − (0.04m)2 ⎤

⎣⎢

⎦⎥

m3 ) ( 0.855 m s) ( 0.01m) (1Ns2 kgm) 6.34×10-4 N-s/m2

µ

= 0.855

m s

= 13,400

This is turbulent flow, so using the Dittus-Boelter equation Nu = 0.023 Re0.8 Pr 0.4 = 0.023 (13400)0.8 (4.19)0.4 = 81.7

→

h=

Nuk Dh

=

( 81.7 ) (0.631W/mK)

(10 W m ) ⎡⎣ (0.04m) − (0.02m) ⎤⎦ T = 40 C + = 40 C+2.73 C=42.7 C ( 5160 W/m K ) 4 ( 0.04m)( 5.32m) 7

3

2

W m2 K

2

o

s

0.01m

= 5160

o

2

12-41

o

o

Answer

12-40 Pasteurization is the sterilization of milk to ensure no diseases are transmitted with the milk. Consider a flow of 1.5 kg/s of milk whose temperature must be raised from 35 ºC to 75 ºC in a 2-cm tube. The wall temperature is 100 ºC. Milk properties are: ρ = 1030 kg/m3, µ = 2.12 × 10 -3 Ns/m2, c p = 3850 J/kgK, and k = 0.6 W/mK. Determine the required tube length (in m).

Approach: With the given information, the analysis in Section 12.5 for a constant wall temperature heat exchanger can be used to calculate the required length. The heat transfer coefficient must be evaluated.

Assumptions: 1. Only the milk thermal resistance is taken into account.

Solution: For a constant wall temperature heat exchanger, the governing equation from Section 12.5 is:

⎛ hA ⎞ ⎟⎟ ⎜ mc ⎝  p ⎠

Tout = Tin + (Tin − Tw ) exp ⎜ −

where A = π DL . Substituting this in and solving for L: L = −

 mc p

hπ D

⎛ Tout − T w ⎞ ⎟ ⎝ Tin −T w ⎠

ln ⎜

To calculate the heat transfer coefficient, we need the Reynolds number:   ρ V D m 4m → V= → Re = Re = µ ρ A x π µ D Using the given fluid properties 4 (1.5 kg s ) ( Nis 2 kgim )

Re = π

( 2.12 ×10

-3

Nis m 2 ) ( 0.02 m)

= 45,040

This is turbulent flow, so using the Dittus-Boelter correlation: Pr =

µ c p

k

( 2.12×10

-3

=

Nis m 2 ) ( 3850 J kgi K ) 0.6 W mi K

Nu = 0.023Re0.8 Pr 0.4 = 0.023 ( 45040) h=

Nu ik

L = −

L

=

( 345)( 0.6 W 0.02m

mi K )

0.8

= 13.6

0.4 (13.6) = 345

=10,360 W m2 i K

(1.5 kg s )( 3850 J kgi K ) ⎛ 75 − 100 ⎞ ln = 8.48m (10360 W m2 i K )π ( 0.02 m) ⎜⎝ 35 − 100 ⎟⎠

12-42

Answer

12-41 Pressurized liquid water enters a 2-cm diameter, 6-m long tube at 20 ºC at a flow rate of 0.5 kg/s. The tube surface temperature is constant, and the total power transferred to the water is 150 kW. Determine the surface temperature (in ºC).

Approach: This is a constant wall temperature heat exchanger. Using the analysis given in Section 12.5, we can determine the required surface temperature.

Assumptions: 1. The system is steady with no work or potential or kinetic energy effects. 2. Liquid water is an ideal liquid with constant specific heat.

Solution: For a constant wall temperature heat exchanger, (with Rtot = 1 hA ) the exit temperature can be calculated from: Tout = Ts Solving for

+

 p) (Tin − Ts ) exp ( − hA mc

= T s

 p) Tout − Tin exp ( − hA mc  p ) 1 − exp ( − hA mc

The outlet temperature can be obtained from an energy balance on the water. Assuming steady, no work, no potential or kinetic energy effects, and an ideal liquid with constant specific heat so that ∆h = c p ∆T :


→

Tout = Tin

+

 p Q /mc

All the properties should be evaluated at the average water temperature. From Appendix A-6 and assuming T out o o 3 -4 ≈100 C, T avg = (20 + 100)/2 = 60 C, c p = 4.181 kJ/kg K, ρ = 983.2 kg/m , k = 0.653 W/mK, µ = 4.52×10 N⋅s/m2, Pr = 2.89 150kW (1000J/s 1kW) T out = 20 o C + = 91.8 o C ( 0.5kg/s )( 4.181kJ/kg K)(1000 J kJ) To calculate h, we need the Reynolds number   ρ V D m m ⇒ V = = Re = 2 µ ρ A ρ (π /4 ) D 4 ( 0.5kg/s ) ( Ns 2 kgm )

Re = π

( 4.52

10-4 N ⋅ s/m 2 ) ( 0.02m)

=

Re =

⇒

 4m π µ D

70,400

×

This is turbulent flow, so using the Dittus-Boelter equation Nu = 0.023 Re0.8 Pr 0.4 = 0.023(70400)0.8(2.89)0.4 = 266 Nuk ( 266 )( 0.653W/mk ) 2 = = 8670W/m K h= D 0.02m hA

(8670W/m K )π ( 0.02m)( 6m)(1J/s 1W) 2

=

=

1.564

( 0.5kg/s )( 4.181k J/kg K)(1000 J kJ) ( 91.8 + 273) − ( 20 + 273) exp( − 1.564) T = 383.8K s = 1 − exp ( −1.564)

 p mc

12-43

= 110.8 o C

Answer

12-42 Parts of the Alaskan oil pipeline (1-m in diameter) are buried 3-m below the surface of the earth (k = 0.65 W/m·K) and covered with 20-cm of insulation (k = 0.05 W/m·K). Pumping stations are 60-km apart. To decrease pumping power the oil is heated to about 100 ºC (to reduce its viscosity) before it enters the pipeline at a pumping station. In the winter the surface temperature of the earth is -30 ºC. Assume the oil properties can be approximated with those of unused oil given in the appendix. For a flow rate of 0.5 m3/s, and using properties evaluated at the average temperature of the oil, determine: a. the oil temperature when it reaches the next pumping station (in ºC) b. the heat transfer required at the pumping station to raise the oil temperature back to 100 ºC (in W) c. the pumping power required (in W) d. the pumping power if the inlet oil temperature is 50 ºC instead of 100 ºC (in W).

Approach: This is a heat exchanger with one fluid with constant temperature, so that analysis given in Section 12.5 can be used to calculate the outlet temperature. Once that temperature is known, conservation of energy is used to calculate the heat transfer rate. The pumping power also is calculated with conservation of energy, but first the pressure drop is needed.

Assumptions: 1. The heat transfer is one dimensional. 2. The heat transfer coefficient is uniform. 3. The system is steady with no work or potential or kinetic energy effects. 4. Oil is an ideal liquid with constant specific heat.

Solution: a) For a constant wall temperature heat exchanger, the exit temperature is calculated with

 p Rtot ) Tout = Ts + ( Tin − Ts ) exp ( − 1 mc

Three resistances contribute to Rtot. ln ( D2 D1 ) 1 1 Rtot = + + k ground S 2π kins L hA First evaluating the ground resistance, we assume two-dimensional conduction in the ground with constant k ground , and from Table 11-1 for this configuration z ~D S = 2π L/cosh −1 ( 2 z /D ) = 2π ( 60000m) /cosh-1 ⎡⎣ 2( 3m) 1m⎤⎦ = 152,100m Evaluating the oil properties from Appendix A-6 at T avg ≈ 370 K, ρ = 841.8 kg/m3, c p = 2.206 kJ/kgK, 10-4 N⋅s/m2, k = 0.137 W/mK, Pr = 300

µ =

186 ×

 = ρ VA = ( 841.8 kg m3 )( 0.5m3 /s ) = 421kg/s m Re =

For a circular tube

 4m π µ D

4( 421kg/s)

= π

(186 × 10

-4

N ⋅ s/m 2 ) (1m )

= 28,800

This is turbulent flow, so using the Dittus-Boelter correlation: Nu = 0.023 Re0.8 Pr 0.3 = 0.023 ( 28,800) h= Rtot =

Nuk D

=

( 470 )( 0.137W/mK ) 1m 1

( 0.65 W mK )(152,100m)

+

0.8

0.3

( 300) = 470

= 64.5W/m2 K 1n (1.4/1)

1

= 2.80 ×10-5 K/W

( 64.5 W m K )π (1m)( 60000m) s )( 2.206 kJ kgK ) ( 2.80× 10 K W ) ( 1000 J kJ) ⎤ ⎦

2π ( 0.05W mK)( 60000m)

T out = −30 0 C + ⎡⎣100 − ( − 30) ⎤⎦ K exp ⎡− 1/ ( 421kg ⎣

+

2

-5

= 95.1 oC = 368.1 K Answer Using this temperature, T avg = 370.5 K. This is close enough to our assumed value so no iteration is required.

12-44

b) Applying conservation of energy to the oil, and assuming steady, no work, negligible potential or kinetic energy effects, and ideal liquid with constant specific heat so that ∆h = c p ∆T 6  p (Tin − Tout ) = ( 421kg s )( 2.206 kJ kgK)( 100 − 95.1 ) K ( 1000 J kJ) = 4.55× 10 W = 4550kW Q = mc

Answer

c) For pumping power, we define a control volume around the pump. Apply conservation of energy, and assume steady, adiabatic, negligible potential and kinetic energy effects, isothermal, and incompressible: W = V ∆P

V

=

V A x

=

L

ρ V

D

2

∆ P = f

Pressure drop is

3

4V  π D

4 ( 0.5 m s )

2

2

=

π

(1m)

2

= 0.637m/s

For the friction factor, we need the Reynolds nu mber, which is the same as we used above. −2

f = ⎡⎣0.790 1n ( 28,800) − 1.64⎤⎦ = 0.0239 2

⎛ 60000m ⎞ ⎛ kg ⎞ ( 0.637m s ) ⎛ Ns2 ⎞ N 841.8 ∆ P = ( 0.0239 ) ⎜ ⎜ ⎟ = 244, 700 2 ⎟⎜ 3 ⎟ m ⎠ 2 m ⎝ 1m ⎠ ⎝ ⎝ kgm ⎠ Finally, pumping power is

⎛ m3  W = − ⎜ 0.5 s ⎝

⎞⎛ N ⎞ ⎛ 1J ⎞⎛ 1W ⎞ ⎟ ⎜ 244700 2 ⎟ ⎜ ⎟⎜ ⎟ = − 122,300W = − 122.3kW m ⎠ ⎝ 1N ⋅ m ⎠⎝ 1J/s ⎠ ⎠⎝

d) Using oil properties from Appendix A-6 for T in ≈ 50 oC 1410 × 10-4 N⋅s/m2 4 ( 421kg/s) Re = = 3800 -4 2 π (1410 × 10 N ⋅ s/m ) (1m )

→

use T avg ≈ 320 K,

Answer ρ =

871.8 kg/m3

µ =

f = 0.0421 2

⎛ 60000m ⎞ ⎛ 0.637 ⎞ ∆ P = ( 0.0421) ⎜ = 447, 000N/m2 ( 871.8) ⎜ ⎟ ⎟ 1 ⎝ ⎠ ⎝ 2 ⎠  = - (0.5) (447,000) = - 223,500 W = - 223.5 kW W

Answer

Comments:

Because of the steep change in viscosity with changing temperature, the pumping power increases dramatically when the oil temperature is decreased.

12-45

12-43 Pressurized liquid water flowing inside a tube at a rate of 1 kg/s is to be heated from 25 ºC to 90 ºC using condensing steam. The 304 stainless steel tube has an inside diameter of 25-mm, a wall thickness of 1-mm, and a length of 6-m. The condensing coefficient on the outside of the tube is 6,500 W/m2·K. Determine: a. the steam temperature and pressure required (in ºC and kPa) b. the condensation rate of the steam assuming the steam enters as a saturated vapor and exits as a saturated liquid (in kg/s).

Approach: This is a constant fluid temperature heat exchanger. Using the analysis given in Section 12.5, we can determine the required steam temperature.


Solution: For a heat exchanger with one constant temperature fluid, T st =

Solving for

 p Rtot ) Tout − Tin exp ( −1/mc  p Rtot ) 1 − exp ( −1/mc

 p Rtot Tout = Tst + (Tin − Tst ) exp ( − 1 mc

where Rtot =

1 hi Ai

+

1n ( Do /Di ) 2π kL

+

)

1 ho Ao

From Appendix A-2, for 304 stainless steel, k = 14.9 W/mK. The inside heat transfer coefficient requires Re with properties from Appendix A-6 evaluated at the average temperature, T avg = (90 + 25)/2 = 56.5 oC, so by -4 2 interpolation µ = 4.70 × 10 N⋅s/m , k = 0.650 W/mK, Pr = 3.02, c p = 4.179 kJ/kg K Re =

ρ V Di

→

=

V

µ

 m

(

ρ π /4

) Di2

→

Re =

 4m π µ Di

4 (1 kg s ) ( N ⋅ s2 kgm )

= π

( 4.70 ×10

-4

N ⋅ s/m2 ) ( 0.025m)

= 108,400

This is turbulent flow, so using the Gnielinski correction ( f 8)( Re −1000) Pr Nu = 1/ 2 1 + 12.7( f 8) ( Pr 2 / 3 − 1) −2

−2

f = [ 0.7901n Re − 1.64] = ⎡⎣ 0.7901n (108400) − 1.64⎤⎦ = 0.0177 Nu =

Rtot =

( 0.0177 /8)(108360 − 1000)( 3.02) = 435 1/ 2 2/3 1 + 12.7 ( 0.0177/8) ⎡( 3.02) − 1⎤ ⎣ ⎦ 1

(11, 300 W m K )π ( 0.025m)( 6m) 2

+

→

h=

Nuk Di

=

1n ( 27/25) 2π (14.9W mK )( 6m )

= 0.000188 + 0.000137 + 0.000302 = 0.000627 K/W 1 (1J/s 1W ) 1

=

(1kg s )( 4.179kJ kgK )( 0.000627 K W )( 1000 J kJ) ( 90 + 273) - ( 25 + 273) exp ( - 0.382) T st = =502.9K= 230 o C 1-exp ( - 0.382)  p Rtot mc

( 435)( 0.650W/mk ) 0.025m

+

=11,300W/m2 K

1

( 6500 W

m2 K )π ( 0.027m)( 6m)

= 0.382

Answer

From the saturated water table, Appendix A-10, at 230 oC, P = 2795 kPa Answer b) For the condensation rate, we apply conservation of energy to a control volume around the steam and liquid water. Assume steady, adiabatic, no work, no potential or kinetic energy effects, and an ideal liquid water:  st h fg = m w c p ,w (Tout − Tin ) m

 st = m

 w cp , w (Tout − Tin ) m h fg

=

(1kg s )( 4.179 kJ

kgK )( 90 − 25) K

1813.8 kJ kg

with h fg from the steam table.

12-46

= 0.150

kg s

Answer

12-44 The oil from a large Diesel engine flows through an oil cooler before it is returned to the engine. Consider a flow rate of 0.1 kg/s that must be cooled from 90 ºC to 40 ºC by passing through a thin-walled tube with a diameter of 12.7 mm. Air at 30 ºC is in crossflow outside the tubes with a velocity of 10 m/s. Determine the required tube length (in m).

Approach: This is a constant fluid temperature heat exchanger. Using the analysis given in Section 12.5, we can determine the required length. The outside heat transfer coefficient must be determined.

Assumptions: 1. The heat transfer is one dimensional. 2. Air is at one atmosphere. 3. Ignore wall resistance.

Solution:  p Rtot ⎤⎦ Tout = T f + (Tin − Tout ) exp ⎡⎣ − 1/mc

For a heat exchanger with one constant temperature fluid:

where Rtot =

⎛ 1 1 ⎞ + ⎜ ⎟ Solving for L hoil π DL hair π DL π DL ⎝ hoil hair ⎠ 1

+

1

=

1

⎛ T − T f ⎞ ⎛ 1 1 ⎞  p 1n ⎜ out − mc + ⎟ ⎜ ⎜ Tin − T f ⎟ hoil hair ⎟ ⎠ ⎝ ⎠⎝ L = π D

The oil heat transfer coefficient requires the Reynolds number with fluid properties at T avg = (90 + 40)/2 = 65 o C=338 K. From Appendix A-6 at 340 K µ = 531×10-4 N⋅s/m3, k = 0.139 W/mK, Pr = 793, c p = 2.076 kJ/kg K Re =

ρ V D

→

V

µ

=

 m

(

ρ π /4

→

) D 2

Re =

 4m π µ D

4 ( 0.1kg s ) ( N ⋅ s2 kgm )

= π

( 531×10−

4

N ⋅ s/m 2 ) ( 0.0127m)

= 189

This is laminar flow, so the entrance length is Lent,t ≈ 0.037 Re Pr D = 0.037 (189) (793) (0.0127 m) = 70.4 m We assume the tube length will be shorter than this so we will use the Seider-Tate equation: Nu = 1.86 (Gz )1/3 (µ /µ s)0.14 = 1.86 ( Re Pr D/ L)1/3 (µ /µ s)0.14 This requires the tube length, so an iterative solution is needed. Assuming T s ≈ 330 K, µ s = 836×10-4 N⋅s/m2 and a tube length of 40 m.

⎡ (189 )( 793)( 0.0127) ⎤ Nu =1.86 ⎢ ⎥ 40 ⎣ ⎦

1/ 3

⎛ 531× 10−4 ⎞ ⎜ −4 ⎟ ⎝ 836 × 10 ⎠

0.14

= 6.33

→

hoil =

Nuk D

=

( 6.33)( 0.139W/mK ) 0.0127

= 69.2W/m2 K

The air side heat transfer coefficient requires the Reynolds number with properties from Appendix A-7 at T film = (T s + T f )/2= (340 + 273 + 30)/2 = 322 K: ρ = 1.096 kg/m3, µ = 1.95 × 10-5 N⋅s/m2, k = 0.0279 W/mK, Pr = 0.703 Re =

ρ V D

=

(1.096kg

m3 ) (10 m s )( 0.0127m) ( N ⋅ s2 kgm)

µ

(1.95 ×10

-5

N ⋅ s/m 2 )

= 7140

From Table 12-1 for crossflow over a cylinder Nu = 0.193 Re0.618 Pr 1/3 = 0.193 (7140)0.618 (0.703)1/3 = 41.3 hair = Nuk D = ( 41.3)( 0.0279) 0.0127 = 90.8W/m2 K

⎛ ⎝

- ⎜ 0.1 L =

kg ⎞ ⎛ kJ ⎜ 2.076 ⎟ s ⎠⎝ kgK

⎞⎛ J ⎞ ⎡ 40 -30 ⎤ ⎛ 1 1 ⎞ m2 K 1000 1n + ⎟⎜ ⎟ ⎢ ⎟ ⎥⎜ kJ ⎠ ⎣ 90 -30 ⎦ ⎝ 69.2 90.8 ⎠ W ⎠⎝ π

( 0.0127m )

= 237m

This is much longer than the calculated entrance length, so we assume fully developed laminar flow with constant properties for the constant wall temperature boundary condition. ( 3.66) ( 0.139) Nu = 3.66 → h = = 40.1W/m2 K 0.0127 The air side h is the same as before, so calculating the new length: L = 335 m Answer

Comments: This is an impractical length. Multiple tubes in parallel should be used. 12-47

12-45 For some applications, enhanced cooling capabilities are obtained by attaching a heat generating system to a cold plate, which is maintained at a cold temperature by passing water through it. Consider the copper cold plate (shown below) that has heat generating equipment attached to its top and bottom surfaces. Each of the six channels is 6-mm square and 100-mm long, and the walls of each channel is 4-mm thick. If chilled water at 10 ºC is pumped through the channels at a velocity of 0.5 m/s, and the surfaces of the cold plate must stay below 45 ºC, determine: a. the maximum allowable power (top and bottom surfaces) to the cold plate (in W) b. the water outlet temperature (in ºC).

Approach:

Assumptions:

Heat enters the cold plate from the top and bottom surfaces. An energy balance on the water can give the total heat transfer but the exit temperature is unknown. With the top and bottom surfaces at a constant temperature, we have a heat exchanger with one constant temperature fluid, so the analysis from Section 12.5 is applicable. Examining the figure, a unit cell (shown) can be analyzed that is representative of all 12 unit cells. We have heat transfer from a fin. The heat transfer coefficient must be determined once one unit cell is analyzed, we multiply by 12 to determine the total performance of the cold plate.

1. The heat transfer is one dimensional. 2. The system is steady with no work and no potential or kinetic energy effects. 3. Water is an ideal liquid with constant specific heat.

Solution: An energy balance is applied to the water flow. Defining a control volume that encompasses only the water, and assuming steady, no work, negligible potential and kinetic energy effects, and ideal liquid with constant specific heat so that ∆h = c p

∆T :


The outlet temperature can be determined by applying the analysis from Section 12.5 for a heat exchanger with a

 p Rtot ⎤⎦ Tout = Tb + (Tin − Tb ) exp ⎡⎣ − 1/mc

constant wall temperature.

The resistances that must be taken into account are the conduction through the plane wall with thickness δ and through the fin, as described by the unit cell. Therefore, the total resistance is: NA f δ 1 Rtot = where η0 =1 − + (1−η f ) kcu Asur η o h Atot Atot For a unit cell, N = 1 A sur = (S + W)L = (0.004 m + 0.006 m) (0.1 m) = 0.001 m 2 A f = 2(W/2)L = (0.006 m) (0.1 m) = 0.0006 m 2 Ab = WL = 0.0006 m2 Atot = Ab + NA f = 0.0006 m2 + (1)(0.0006 m2) = 0.0012 m2 The convective resistance and the fin efficiency require the heat transfer coefficient. From Appendix A-6 assuming T out ≈ 30 oC, T avg ≈ (10 + 30)/2 = 20 oC, ρ = 998.2 kg/m3, c p = 4.182 kJ/kg K, k = 0.603 W/mK, µ = 9.85 × 10-4 N⋅s/m2, Pr = 6.83. In addition for copper, k = 401 W/mK. For the heat transfer coefficient, we Re = ρ VD H µ with

need the Reynolds number Re =

( 998.2 kg

m3 ) ( 0.5m s )( 0.006m)

9.85 ×10-4 N ⋅ s/m2

= 3040

This is turbulent flow, so using the Gnielinski correlation −2

−2

f = [ 0.7901n Re − 1.64] = ⎡⎣ 0.791n ( 3040) − 1.64⎤⎦ = 0.0454

12-48

2

D H = 4 Ax /P wetted = 4( W ) / ⎡⎣2 ( W + W ) ⎤⎦ = W = 0.006m

Nu =

h=

( f 8)( Re −1000) Pr ( 0.0454/8)( 3040 − 1000)( 6.83) = = 22.7 1/ 2 1/ 2 1 + 12.7 ( f 8) ( Pr 2 / 3 − 1) 1+ 12.7 ( 0.0454/8) ( 6.832 / 3 − 1)

Nuk D H

=

( 22.7 ) ( 0.603 W/m3 K ) 0.006m

= 2280W/m2 K

The fin is a straight rectangular fin with an adiabatic tip, so fin efficiency is η f

=

tanh ( mW 2 )

=

mW 2 0.006m ⎡ 2

mW

where

2

1/2

=

W ⎡ h2L ⎤

2 ⎣ k A x ⎦

2 ⎣kS L⎦

⎢

m2 K ) 2

η f

= tanh ( 0.160) /0.160= 0.992

η o

=1−

(1)( 0.0006) 0.0012

⎥ =

⎢

⎥

1/ 2

⎤ ⎢ ⎥ ⎢⎣ ( 401W mK )( 0.004m) ⎥⎦

( 2280 W

1/ 2

W ⎡ hp ⎤

= 0.160

(1− 0.992) = 0.996

For one unit cell Rtot =

0.004m

( 401W/mK ) ( 0.001m

2

+

1

) ( 0.996) ( 2280W/m K)( 0.0012m ) 2

2

= 0.377m2 K/W

 = ρ V A x = ( 998.2 kg m3 ) ( 0.5m/s )( 0.006m)( 0.006m) = 0.0180kg/s m T out = 45o C + (10 - 45) K exp ⎡ -1/ ( 0.018kg/s)( 4.182 kJ/kgK)( 1000 J kJ) ( 0.377m2 K /W)⎤ =11.2o C

⎣

⎦

Answer

We have a total of 12 unit cells, so the total heat transfer rate is Q = (12 )( 0.018 kg s )( 4.182 kJ kgK )( 11.2 − 10) K ( 1000 J kJ) =1095W

Answer

Comments:

Because the assumed T out was higher than the calculated T out , the fluid properties probably should be reevaluated and the calculations repeated.

12-49

12-46 The total thermal resistance between the outside and inside of a home consists of the external convective resistance, the wall resistance, and the internal convective resistance. Adding additional insulation reduces the heat transfer but it also changes the inside wall temperature. Compare the average natural convection heat transfer coefficient on a 2.5-m tall wall for two situations: a. inside air temperature of 22 ºC and wall temperature of 10 ºC b. inside air temperature of 22 ºC and wall temperature of 17 ºC.

Approach: This is a natural convection problem. We assume that the wall can be approximated as a vertical flat plate.

Assumptions: 1. Air is at one atmosphere.

Solution: The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A-7 evaluated at T film = (T w + T f )/2. 1) T film = (10 + 22)/2 = 16 0C = 289 K, so ρ = 1.221 kg/m3 ; µ = 1.77×10-5 N⋅s/m2 ; k = 0.0254 W/mk, Pr = 0.711 2) T film = (17 + 22)/2 = 19.5 0C = 292.5 K, so ρ = 1.207 kg/m3, µ = 1.79×105 N⋅s/m2; k = 0.0256 W/mK; Pr = 0.710 1) Ra =

g β ρ 2 (Tw −

)δ

3

f

=

2

µ

2

( 9.81m s ) (1 289K ) (1.221kg m ) ( 22 (1.77 10 N s/m ) 2

Pr

3

×

-5

−

⋅

2

10) K ( 2.5m)

3

( 0.711) =

2

2.15 ×1010

From Table 12-3, for a vertical wall 1/ 3

Nu = 0.1 Ra1/ 3 = 0.1( 2.15 ×1010 ) h=

Nuk H

=

278

=

( 278)( 0.0254W/mK )

=

2.5m

2.83W/m2 K

2

2) Ra =

( 9.81)(1/292.5)(1.207) ( 22

(1.79

-5

10

×

−

)

17)( 2.5)

3

Answer

( 0.710)

2

Nu = 0.1 (8.46 × 19 9)1/3 = 204 ( 204 )( 0.0257) 2 h= = 2.09W/m K (~ 26% smaller) 2.5

=

8.46 ×109

Answer

Comments: As can be seen, a small change in the driving temperature difference reduces the inside convective heat transfer coefficient. So adding insulation helps in two ways: increased conduction resistance and increased convection resistance.

12-50

12-47 Consider again the light bulb in Problem P 12-23. For all the same conditions, determine the temperature of the glass bulb if it is cooled by natural convection (in ºC).

Approach: An energy balance on the light bulbs is needed. The power into the glass (90% of 100W) is removed by convection and radiation. The natural convection heat transfer coefficient must be evaluated.

Assumptions: 1. Air is at one atmosphere. 2. Radiation is from a small body to a large area. 3. The light bulb is a sphere.

Solution: A steady energy balance on the glass gives Q in = Q rad + Q conv = εσ A (Ts4 − T f4 ) + hA (Ts −T f )

The light bulb is assumed to be a sphere and the surroundings are very large. The natural convection heat transfer coefficient correlation for a sphere is Nu = 2 +

where

Ra =

0.589 Ra1/ 4

⎡1 + ( 0.469/ Pr )9/16 ⎤ ⎣ ⎦

4/ 9

g βρ 2 (T s − T f ) D 3 Pr

µ 2 Evaluate the fluid properties from Appendix A-7 at T film = (T s + T f )/2; because we seek T s, we must first estimate T s = 200oC. T film = (200 + 27)/2 = 113.5 oC = 386.5K, ρ = 0.913kg/m3, µ = 2.23 × 10-5 N⋅s/m2, k = 0.0327 W/mK, Pr = 0.691 2

3

( 9.81m/s )(1 386.5K ) ( 0.913 kg/m ) ( 200 − 27) K ( 0.08m) ( 0.693) = 2.61×10 Ra = − 2.23 10 N s/m × ⋅ ( ) 0.589 ( 2.61×10 ) 2

3

6

5

6

Nu = 2 +

h=

Nuk D

⎡1 + ( 0.469 0.691) ⎣ =

1/ 4

9/16

( 20.2 )( 0.0327 W

2

2

⎤ ⎦

4/9

= 20.2

mK )

0.08m

= 8.26 W/m2 K

Q in = 0.9 (100 W) = 90 W Q rad = ( 0.85 ) ( 5.67 ×10−8 W m2 K4

) ( 4π )( 0.04 m)

2

⎡T s4 − ( 300K) 4 ⎤ = 9.690× 10−10 ⎡T s4 − 8.1× 109 ⎤ ⎣ ⎦ ⎣ ⎦

Q conv = (8.26W/m2K) (4π ) (0.04m)2 (T s – 300)K = 0.1661(T s – 300) Combining these expressions 90 = 9.69 × 10-10 ⎡⎣T s4 − 8.1×109 ⎤⎦ + 0.1661(T s − 300) Solving T s ~ 506.1 K = 233 oC This is reasonably closed to our assumed value of T s so we will not iterate.

12-51

Answer

12-48 In any design process decisions have to be made about placement of components. Cost, performance, and maintainability are some of the criteria used. Consider the placement of a 1.2 W, 60-mm by 60-mm electric component in a larger device. The component’s surface temperature must not exceed 85 ºC. The air is quiescent at 25 ºC. Neglecting radiation, determine if the component can be located facing downward or facing upward. (In other words, what is the component’s surface temperature if it is facing upward or downward?)

Approach: This is a natural convection heat transfer problem. The maximum power transferred from a 85 °C surface to 25°C air should be calculated and compared to the 1.2W specification. Appropriate heat transfer coefficient correlations for the two geometrics must be used.

Assumptions: 1. Radiation is neglected. 2. The plates match the geometry for the correlations in Table 12-3. 3. Air is at one atmosphere.

Solution: Q = hA (T w - T ∞ )

Neglecting radiation, the heat transfer rate is calculated with

We need the Rayleigh number to calculate h. Evaluating air properties from Appendix A-7 at T film = (85 + 25)/2 = 55oC = 328K, (by interpolation) ρ = 1.076 kg/m3, k = 0.0283 W/mK, µ = 1.99 × 10-5 N⋅s/m2, -6 2 ν = 18.5 × 10 m /s, Pr = 0.702 Ra =

g β (Tw − T )δ 3 ∞

Pr ν From flat plates, the characteristic length is δ = A/p = (0.06m) (0.06) / (4) (0.06m) = 0.015m 2

2

Ra =

3

( 9.81m/s ) (1/ 328K )( 85 25) K ( 0.015m) ( 0.702) (18.5 10 m /s ) −

6

2

−

×

2

=

12400

From Table 12-3 for a horizontal plate 1) Hot plate facing upward Nu = 0.54 Ra1/4 = 0.54(12400)1/4 = 5.70 Nuk ( 5.70)( 0.0283W/mK ) W h= = = 10.8 δ 0.015m m2 K Q

=

(10.8 W

m2 K

2

) ( 0.06m) ( 85

−

25) K = 2.33W

Answer

2) Hot plate facing downward Nu = 0.27 Ra1/4 = 0.27 (12400)1/4 = 2.85 h = (2.85) (0.0283)/0.015 = 5.38 W/m2K Q = (5.38) (0.06) (0.06) (85 – 25) = 1.16 W

Answer

The safe choice is to use the location with the component facing upward. However, because facing downward is close to satisfying the design specification, it might usable if other design considerations had to taken into account.

Comment: A second approach would be to estimate the wall temperature for the 1.2W power input. This would require an iterative solution.

12-52

12-49 To lower the viscosity of an oil before it is used in a process, an electric resistance heater 1.5-cm in diameter and 30-cm long is immersed horizontally in a vat of unused engine oil which is at 20 ºC. If the heater surface should not rise above 150 ºC so that the oil does not smoke, determine the maximum power that can be dissipated in the heater (in W).

Approach: This is a natural convection problem. Because of the large length compared to the diameter, we will ignore heat transfer from the two ends. The basic heat

(

)

transfer rate equation Q = hA∆T is used to calculate the maximum power. The heat transfer coefficient must be evaluated.

Assumptions: 1. The system is steady. 2. All properties are constant and are evaluated at the appropriate temperatures. 3. The ends of the heater are ignored.

Solution: Q = hA (T s − T f ) . Ignoring the ends, A = π DL .

The heat transfer rate is calculated with:

From Table 12-3, the heat transfer coefficient correlation is:

⎧ ⎫ 0.387 Ra1 6 ⎪ ⎪ Nu = ⎨0.6 + 8 27 ⎬ 9 16 ⎡1 + ( 0.559 Pr ) ⎤ ⎪ ⎪ ⎣ ⎦ ⎭ ⎩ where Ra =

g β (T s − T f ) D 3 Pr ν 2

2

. Evaluating the oil properties from Appendix A-6 at

T film = (150 + 20) / 2 = 85 o C=358 K , ν= 32.1×10-6 m2/s; k = 0.138 W/mK; Pr = 425, β = 0.7×10-3/K Ra =

( 9.81m s ) ( 0.7×10-3

K ) (150-20) K ( 0.015 m)

( 32.1×10

-6

m2 s )

3

2

( 425)

= 1.24 × 106

2

16 ⎧ ⎫ 0.387 (1.24 × 106 ) ⎪ ⎪ Nu = ⎨0.6 + = 21.0 8 27 ⎬ 9 16 ⎡ ⎤ ⎪ ⎪ 1 + 0.559 425) ⎣ ( ⎦ ⎭ ⎩

h=

Nu ik D

=

( 21.0 )( 0.138 W 0.015m

mi K )

=193 W m2 i K

Q = (193 W m2 K ) π ( 0.015 m )( 0.30 m)(150-20) K=355 W

Answer

Comments: A factor of safety might be used to derate the power to ensure the surface temperature remains below 150 ºC. In addition, as the oil temperature increases, lower power would be used to ensure the surface temperature remains below 150 ºC.

12-53

12-50 An electric resistance heater, 10-mm in diameter and 300-mm long, is rated at 550 W. If the heater is horizontally positioned in a large tank of water that is at 20 °C, estimate the surface temperature of the heater (in °C).

Approach: This is a natural convection heat transfer problem. We will ignore heat transfer from this ends of this cylinder and assume that the tank is large enough to not influence this heat transfer coefficient. Because the driving force is the temperature difference and we do not know it, an iterative solution is required.

Assumptions: 1. The ends of the heater are ignored. 2. Only natural convection occurs.

Solution: The surface temperature can be determined from the basic heat transfer rate equation: Q = hA (Tw − T∞ )

→

Tw = T∞ + Q hA

We ignore the cylinder ends, so that A = π DL. The natural convection heat transfer coefficient for a horizontal cylinder is used: 2

⎧ ⎫ 0.387 Ra1/ 6 ⎪ ⎪ Nu = ⎨0.6 + where 8/27 ⎬ ⎡⎣1 + (0.559 Pr )9/16 ⎤⎦ ⎪ ⎪⎩ ⎭

Ra =

g β (Tw − T∞ ) D 3 ν 2

Pr

The properties should be evaluated at the film temperature T film =( T w - T ∞)/2, so as first approximation, estimate o o T w ~ 40 C. Therefore T film = 30 C, and the properties from Appendix A-6 are β = 3.06 × 10 -4/K, µ = 7.79 × 10-4 N⋅s/m2, ρ = 995.6 kg/m3, Pr = 5.26 , ν = 0.782 × 10-6 m2/s , k = 0.618 W/mK 3

( 9.81m/s )( 3.06 ×10− /K ) ( 40 − 20) K ( 0.01m) ( 5.26) Ra = = 516,000 ( 0.782 ×10 m /s ) 2

4

-6

2

2

2

1/ 6 ⎧ ⎫ 0.387 ( 516,000) ⎪ ⎪ Nu = ⎨0.6 + = 14.6 8/27 ⎬ 9/16 ⎡ ⎤ 1 (0.559/ 5.26) ⎪⎩ ⎣ + ⎦ ⎪⎭

T w = 20o C +

550W

( 902W/m K )π ( 0.01m)( 0.3m) 2

→

h=

Nuk D

=

(14.6 )( 0.618W/mK ) 0.01m

= 902W/m2 K

= 84.7 o C

This temperature is significantly about what we first estimated, so we iterate T film = (84.7 + 20)/2 = 52.3 oC at 50 oC the fluid properties are: β = 4.95 × 10-4/k, µ = 4.88 × 15-4 N⋅s/m2, ρ = 985.7 kg/m3, Pr = 3.15, k = 0.648 W/mK, ν = 4.95 × 10-6 m2/s →

3

Ra =

( 9.81) ( 4.95×10−4 ) ( 84.7 − 20)( 0.01) ( 3.15) −6 2

( 0.495×10 )

= 4.04 × 106

2

1/ 6 ⎧ ⎫ 6 0.387 4.04 10 × ( ) ⎪ ⎪ Nu = ⎨0.6 + = 25.4 8/27 ⎬ 9/16 ⎡ ⎤ ⎪ ⎪ 1 + 0.559/ 3.15) ⎣ ( ⎦ ⎭ ⎩

T w = 20 +

550

(1650) π ( 0.01) ( 0.30)

h = (25.4) (0.648)/0.01 = 1650 W/m2K

= 55.4 o C

One more iteration, based on trends, let’s assume T w ~ 70oC ⇒ T film ~ 45oC, β = 4.23 × 10 -4/K, µ = 5.77 × 10-4 m2/s, ρ = 990.2 kg/m3, Pr = 4.23, k = 0.637 W/mK, ν = 0.583 × 10-6 m2/s Ra = 2.58×106, Nu = 22.8 , h = 1450 W/m2K T w = 60.2 oC Another iteration gives T w ~ 62oC. Answer →

→

12-54

12-51 A passive solar heating technique is to use a massive masonry wall (a Trombe wall) to absorb solar energy and then to release it slowly when the air temperature surrounding the wall is lower than that of the wall. Consider a long 3-m tall wall, well insulated on its backside, that has a net radiant solar energy flux into the wall of 150 W/m2. The air temperature is 21 ºC. Assuming that the temperature of the wall changes very slowly and the wall operation can be approximated as quasi-steady, determine the average surface temperature of the wall (in ºC).

Approach: This is a natural convection problem that can be solved with the basic rate equation ( Q = q″ A = hA ∆T ).

The heat transfer coefficient must be calculated.

Assumptions: 1. The system is quasi-steady. 2. Air is at one atmosphere.

Solution: Q = q ′′A = hA (T s − T f )

Assuming a steady state system, the convective rate equation is: T s = Tf + q ′′/h

Solving for T f :

Using an average heat transfer coefficient will give us the average wall temperature. We need to evaluate the Rayleigh number at T film = (T s + T f )/2. Because T s is what we seek, we must first estimate T s, evaluate h, and then calculate T s. If the estimate and calculated values are close, then we are done; if not, we should iterate. From Appendix A-7 and assuming T s ~ 73 oC T film = (21 + 73)/2 = 47 oC = 320 K, ρ = 1.103 kg/m3, µ = 1.94 × 10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.703 →

2

m ⎞⎛ 1 ⎞⎛ kg ⎞ 3 ⎛ 9.81 2 ⎟ ⎜ 1.103 3 ⎟ ( 73 − 21) K ( 3m) ( 0.703) 2 3 ⎟ ⎜ ⎜ g βρ (T s − T f ) H Pr ⎝ s ⎠ ⎝ 320K ⎠ ⎝ m ⎠ Ra = = = 9.78 ×1010 2 2 -5 2 µ (1.94 ×10 N ⋅ s/m ) 1/ 3

Nu = 0.1 Ra 1/ 3 = 0.1( 9.78× 1010 )

From Table 12-3, h=

Nuk H

=

( 461)( 0.0278W/mK )

T s = 21o C +

3m

= 461

= 4.27W/m2 K

150W/m 2

= 56.1o C 4.27W/m K 2

Assuming T s ~ 56.4 oC gives T film ~ 312 K so that ρ = 1.131 kg/m3, µ = 1.90 × 10-5 N⋅s/m2, k = 0.0272 W/mK, Pr = 0.704 2

Ra =

3

( 9.81)(1/312)(1.131) ( 56.4 − 21)( 3) ( 0.704)

(1.90 ×10 ) -5

2

= 7.50x1010

10 1/3

Nu = 0.1 (7.50 × 10 ) = 421 ( 421)( 0.0272) h= = 3.82W/m2 K 3 T s = 21 + 150 /3.82 = 6 0.2 0 C

Answer

This is close enough, and another iteration is not needed.

12-55

12-52 A steam-heated cooking vat in a food processing plant has a bottom that is 1.5 m by 1.5 m. The vat is filled with water, initially at 25 ºC, and the bottom is heated with condensing steam at 105 ºC. Determine: a. the initial heat transfer rate from the bottom of the vat to the water (in W) b. how long it would take for the water temperature to rise to 30 ºC if the water depth is 60-cm (in min).

Approach: This is a natural convection heat transfer problem. We will assume that the heat transfer coefficient from an upward facing heated plate is applicable. The time to heat the water from 25 0C to 30 0C can be determined with conservation of energy.

Assumptions: 1. The heat transfer coefficient correlation for an upward facing heated plate is applicable. 2. The system is closed with no work or potential or kinetic energy effects. 3. The condensing resistance is ignored. 4. The heat transfer rate is constant.

Solution: a) The initial heat transfer rate is calculated with

Q = hA (Tw − T∞ )

We assume the condensing heat transfer coefficient is very high and the conduction resistance through the vat bottom is very small, so that the wall temperature is the same as the condensing temperature. The heat transfer coefficient calculation requires the Rayleigh number with properties evaluated at T film = (T ∞ + T w)/2 = (25 + 105)/2 = 65 oC. From Appendix A-6 ρ = 980.5 kg/m3, k = 0.658 W/mK, µ = 4.21 × -4 2 10 N⋅s/m , ν = 0.429 × 10-6 m2/s, β = 5.53 × 10 -4/K, Pr = 2.67 The characteristic length is δ = A/ p = (1.5 m) (1.5 m) / (4 × 15 m) = 0.375 m g β (Tw − T∞ )δ 3 Pr

Ra =

ν 2

3

( 9.81m/s )( 5.53×10 /K ) (105 − 25) K ( 0.375m) ( 2.67) = 3.32 ×10 = ( 0.429 ×10 m /s ) 2

-4

11

-6

2

From Table 12-3, for an upward facing heated plate Nu = 0.15 Ra1/3 We are outside the range of the correlation’s applicability, but with nothing else available we are forced to use this. Nu = 0.15 (3.32 × 10 11)1/3 = 1040 Nuk (1040)( 0.658W/mK ) h= = =1820W/m2 K δ 0.375m Q = (1820 W/m2K) (1.5 m) (1.5 m) (105 – 25) K = 328,000 W

Answer

o

b) The time to reach 30 C can be determined by conservation of energy applied to a closed system. Assume no work and no potential or kinetic energy changes: dU d Q = = ( mu ) dt dt For an ideal liquid, du = c dT , c ~ c p ~ cv dT Q = mc p dt Assuming the heat transfer rate is approximately constant, we integrate this expression: mc p ∆T  = mc dT → t = Qdt p Q

∫

∫

At the average bulk temperature (25 + 30)/2 = 27.5 oC, c p = 4.177 kJ/kg K, ρ = 996 kg/m3 t =

( 996 kg

m3 ) (1.5m )(1.5m )( 0.6m)( 4.177 kJ kgK )( 1000 J kJ)( 5K)

( 328,000W)(1J/s 1W)

12-56

85.6s=1.43min

Answer

12-53 You devise a transient heat transfer experiment to measure the natural convection heat transfer coefficient on a 2024-T6 aluminum sphere. Initially, the 3-cm sphere is at a uniform temperature of 90 ºC as measured by a thermocouple inserted into the sphere’s center. You plunge the sphere into 10 ºC water and record the center temperature as it decreases with time. The center temperature reaches 80 ºC after 0.83 s, 50 ºC after 5.66 s, and 40 ºC after 9.78 s. Determine: a. the average heat transfer coefficient as the sphere changes temperature from 90 to 80 ºC (in W/m2·K) b. the average heat transfer coefficient as the sphere changes temperature from 50 to 40 ºC (in W/m2·K) c. whether or not this approach is valid.

Approach: This is a transient heat transfer problem. Because the sphere is aluminum and natural convection has a relatively low heat transfer coefficient h, we will assume a lumped systems analysis is applicable to this problem, and the recorded sphere temperatures are uniform. We also assume h is relatively constant over the time intervals.

Assumptions: 1. The lumped systems analysis is valid. 2. The heat transfer coefficient is constant over the time intervals.

Solution: ⎛ − hAt ⎞ = exp ⎜ ⎜ ρ Vc p ⎟⎟ Ti − T f ⎝ ⎠ − ρ V c p ⎛ T − T f ⎞ Solving for h: h = ln ⎜ ⎜ Ti − T f ⎟⎟ At ⎝ ⎠ 3 3 For a sphere V = 4π R 3 = π D 6 and A = 4π R 2 = π D 2 . Combining these with the main equation: The lumped systems equation is

h=

− ρ Dc p 6t

T − T f

⎛ T − T f ⎞ ⎜ Ti − T f ⎟⎟ ⎝ ⎠

ln ⎜

From Appendix A-2 for 2023-T6 aluminum, ρ= 2770 kg/m3, k = 177 W/mK, c p = 875 J/kgK a) For the average heat transfer coefficient as the sphere changes temperature from 90 to 80 ºC h=

−( 2770 kg m3 ) ( 0.03m)( 875J/kgK ) 6 ( 0.83s )

⎡ 80-10 ⎤ 2 ⎥ = 1950 W/m K − 90 10 ⎣ ⎦

ln ⎢

Answer

b) For this time interval, we must reset T i = 50 oC T = 40 oC, and the time interval is t = 9.785 – 5.66s = 4.125 − ( 2770)( 0.03)( 875) ⎡ 40-10 ⎤ h= ln ⎢ = 846W/m2 K Answer ⎥ 6 ( 4.12 ) ⎣ 50 -10 ⎦ c) To determine if this approach is valid, check the Blot number. Bi =

hLchar k

=

h( D/6 ) k

(1950 W/m K ) ( 0.03m) 2

=

6 (177W/mk )

= 0.055 < ∼ 0.1

So the approach is valid. The second condition does not need to be checked explicitly because that h is smaller than the one used.

12-57

12-54 Farmer Brown installs an electric resistance heater in the watering trough for his cows so that the water will not freeze during the long cold winter. He does not want a cow to burn its tongue if it accidentally touches the heater. He places a 25-mm diameter, 30-cm long, 100-W heater horizontally in the water which is maintained at 5 ºC. Determine: a. the surface temperature of the heater (in ºC) b. the surface temperature of the heater if the water trough develops a leak, all the water drains out, and the air temperature is -15 ºC (in ºC).

Approach: This is a natural convection problem. We ignore the heat transfer from the heater ends and use the rate equation ( Q = h A ∆T ) to calculate the surface temperature. The appropriate heat transfer coefficient must be evaluated.

Assumptions: 1. Ignore heat transfer from the ends. 2. Air is at one atmosphere. 3. There is no radiation in air.

Solution: a) The surface temperature can be determined from the basic convective heat transfer rate equation (ignoring heat Q = hπ DL (T s − Tf

transfer from the ends):

)

→

Ts = Tf + Q h π DL

The natural convection heat transfer coefficient for a horizontal cylinder is 2

⎧ ⎫ 0.387 Ra1/ 6 ⎪ ⎪ Nu = ⎨0.6 + 8/27 ⎬ 9/16 ⎡ ⎤ ⎪ ⎪ 1 + 0.559/ Pr ) ⎣ ( ⎦ ⎭ ⎩

where

Ra =

g βρ 2 (T s − T f ) D 3 Pr µ 2

Evaluating the properties of water from Appendix A-6 at T film = (T s – T f )/2, we first assume T s ≈ 30 oC → T film ≈ = 290 K, ρ = 999 kg/m3, µ = 10.8 × 10-4, N⋅s/m2, k = 0.598 W/mK, Pr = 7.56, β = 1.74 × 10-4/K

( 9.81m/s )(1.74 ×10 2

Ra =

-4

2

/K )( 999kg/m3 ) ( 30 − 5) K ( 0.025m)

(10.8×10

-4

N ⋅ s/m

2

)

3

( 7.56)

2

= 4.31×106

2

1/ 6 ⎧ ⎫ 0.387 ( 4.31×106 ) ⎪ ⎪ Nu = ⎨0.6 + = 27.5 8/27 ⎬ ⎡1 + ( 0.559 /7.56)9/16 ⎤ ⎪ ⎪ ⎣ ⎦ ⎭ ⎩

→

Nuk ( 27.5 )( 0.598W/mK ) h= = = 657 W/m2 K D 0.025m

T s = 5 o C +100W ⎡( 657 W m2 K ) π ( 0.025m )( 0.30m ) ⎤ =11.5 o C

⎣

⎦

The first estimated T s is significantly larger than the calculated temperature. The fluid properties should be reevaluated and a new T s calculated. Using T s = 11.5 oC, and properties evaluated at T film ~ 280 K. Ra = 2.33×105 → Nu = 12.1 → h = 282W/m2 K → Ts = 20.1o C Guessing T s ≈ 16 oC and T film ≈ 285K: Ra =1.13×106 → Nu = 18.8 → h = 443W/m2 K → T s = 14.6 0 C Answer This is close enough, so additional iterations are not required. b) Now, with air we also assume no radiation. Following the same procedure as above, guess T s ~ 500 oC, T film ~ 515K, ρ = 0.685 kg/m3, µ = 2.75 × 10-5 N⋅s/m2, k = 0.0417 W/mK, Pr = 0.680 Ra = ( 9.81) (1/515)( 0.685) Nu = 6.91 →

2

3

( 500 + 15)( 0.025) ( 0.680)

h = 11.6W/m2 K

( 2.75× 10 ) -5

2

= 64, 600

→ T s = −15 o C +100 ⎡⎣(11.6)π ( 0.025)( 0.30) ⎤⎦ = 353o C

Iterating a second time with T s ≈ 353 oC, and evaluating properties at T film ≈ 480 K Ra = 77650 → Nu = 7.25 → h = 11.4W/m2 K → T s = 357 o C

Answer

Comment: At this temperature, radiation is significant, so more calculations with only convection are not justified.

12-58

12-55 The manufacturer of the electric resistance heater described in Problem P 12-54 wants to expand her sales and considers using the heater for fuel oil tanks, too. Concern about a possible fire hazard if the oil anywhere in an oil tank reaches too high a temperature makes her contact a consulting engineer for an analysis. If the oil has properties of unused oil and the oil is at 0 ºC, determine the surface temperature of the heater (in ºC).

Approach: This is a natural convection problem. We will ignore heat transfer from the two ends. The basic convective heat transfer equation ( (Q = hA ∆ T ) is used to calculate the surface temperature. The appropriate heat transfer coefficient must be evaluated.

Assumptions: 1. Ignore heat transfer from the ends.

Solution: Ignoring heat transfer from the two ends, the basic convective heat transfer rate equation is used to calculate the Q = hπ DL (T s − T f

surface temperature:

)

→

Ts =T f + Q hπ DL

The natural convection heat transfer coefficient for a horizontal cylinder is: 2

⎧ ⎫ 0.387 Ra1/ 6 ⎪ ⎪ Nu = ⎨0.6 + where 8/27 ⎬ ⎡1 + ( 0.559/ Pr )9/16 ⎤ ⎪ ⎪ ⎣ ⎦ ⎭ ⎩

Ra =

g β ρ 2 (T s − T f )D 3 Pr µ 2

Evaluating the oil properties from Appendix A-6 at T film = (T s + T f )/2, we first must estimate T s ≈ 35 oC → T film ≈ 290.5 K, ρ = 890 kg/m3, µ = 9990×10-4 N⋅s/m2, k = 0.145 W/mK, Pr = 12900, β = 0.7 × 10-3 /K

( 9.81m s )( 0.7 ×10 2

Ra =

-3

/K )( 890kg m3 )

( 9990 × 10−

4

2

3

( 35 − 0) K ( 0.025m) ( 12900)

N ⋅ s/m2 )

2

= 38,500

2

⎧ ⎫ 1/ 6 0.387( 38,500) ⎪ ⎪ Nu = ⎨0.6 + = 8.07 8/27 ⎬ 9/16 ⎡ ⎤ ⎪ ⎪ 1 + 0.559 /12900) ⎣ ( ⎦ ⎭ ⎩ h=

Nuk D

=

T s = 0o C +

( 8.07 )( 0.145W/mK ) 0.025m

= 46.8W/m2 K

100W

( 46.8W/m K )π ( 0.025m)( 0.30m) 2

= 90 o C

Because this calculated T s is significantly different than the estimated T s, we must iterate. Guessing T s = (35 + 90)/2 = 62.5 oC ⇒ T film ~ 305K, ρ = 881 kg/m3, µ = 3700 × 10-4 N⋅s/m2, k = 0.145 W/mK, Pr = 4900, β = 0.7 × 10-3/K 2

Ra =

3

( 9.81) ( 0.7 ×10-3 ) ( 881) ( 62.5 − 0)( 0.025) ( 4900)

Nu =12.4

( 3700×10 ) -4

→

h = 72.0W/m2 K

2

→

T s = 59 o C

Close enough so no more iterations are required.

12-59

= 1.86 ×105

Answer

12-56 The coils in electric power transformers mounted on telephone poles in every neighborhood are cooled by oil. If the coils reach too high a temperature, the transformer can fail. To prevent this problem, the transformer is externally cooled by air. The worst-case scenario occurs on hot, still summer days. Consider a transformer that is 55-cm in diameter and 1.5-m tall on a day when the temperature is 40 ºC. Assume the heat transfer coefficients on the ends are the same as on the cylinder sides. If 250 W must be dissipated, determine: a. the surface temperature of the transformer (in ºC) when ignoring radiation b. the surface temperature if radiation is included with an emissivity of 0.6 (in ºC).

Approach: This is a natural convection problem. Using the basic heat transfer rate equation ( Q = hA ∆T ), the surface temperature can be determined. An iterative solution must be used to find the natural convective heat transfer coefficient.

Assumptions: 1. Air is at one atmosphere. 2. The same heat transfer coefficient is used on sides and ends. 3. The transformer is “small” relative to its surroundings.

Solution: a) The rate equation is Q = hA∆T = hA (T s − T f ) . Solving for the surface temperature

T s = T f + Q hA

where A = π DH + 2π D2/4 = π (0.55 m) (1.5 m) + 2π (0.55 m2)/4 = 3.070 m2 We assume that h can be approximated by that on a vertical wall, and this also is assumed to be valid for the two ends. The Rayleigh number is evaluated with properties from Appendix A-7 at T film = (T s + T f )/2. We assume T s ≈ 65 oC, so T film ≈ (65 + 40)/2 = 52.5 oC ≈ 325 K, and ρ = 1.086 kg/m3, µ = 1.96 × 10-5 N⋅s/m2, k = 0.0282 W/mK, Pr = 0.703 Ra =

2

µ

From Table 12-3, h=

Nuk H

=

2

3

3

9

-5

1/ 3

(176)( 0.0281W/mK )

T s = 40 0 C +

2

( 9.81m s ) (1 325K ) (1.086 kg m ) ( 65 − 40)( 1.5m) ( 0.703) = = 5.50 ×10 (1.96 ×10 N ⋅ s/m ) Nu = 0.1 Ra = 0.1( 5.50 ×10 ) = 176

g β ρ 2 (T s − T f ) H 3 Pr

1.5m

9

2

2

1/ 3

= 3.31W/m2 K

250W

( 3.31W/m K )(3.070m ) 2

2

= 64.7 0 C

Answer

This is reasonably close to our assumed T s, so iteration is not required. b) If we include radiation, then the energy balance is Q = Q conv + Q rad = hA (Ts − T f ) + εσ A (Ts4 −T f4 ) The surface temperature will be lower, so h should be re-evaluated at a lower T s. Assume T s ≈ 55 oC = 328 K T film ≈ 47.5 oC ≈ 320 K, ρ = 1.103 kg/m3, µ = 1.94 × 10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.703 2

Ra =

3

( 9.81)(1 320)(1.103) ( 55 − 40)( 1.5) ( 0.703)

(1.94 ×10

-5

1/ 3

Nu = 0.1( 3.53× 109 )

N ⋅ s/m

= 152

⇒

2

)

2

h=

= 3.53×109

(152)( 0.0278 ) 1.5

= 2.82W/m2 K

Substituting into the energy balance

(

4

250W = ( 2.82 W m2 K )( 3.07m2 ) (T s -313) C + ( 0.6) 5.67×10-8 W m2 K o

Solving for T s ~ 324.3 K ~ 51.3 oC

) ( 3.07m ) ⎡⎣T -( 313K) ⎤⎦ 2

Answer 12-60

4

s

4

12-57 To improve the heat transfer from the transformer described in Problem P 12-56, 16 fins made of plain carbon steel are attached. Each fin has the same length as the transformer, is 4-mm thick, and extends from the surface 100-mm. Using only natural convection, determine the surface temperature of the transformer (in ºC).

Approach: This is a natural convection problem from fins. Using the convective heat transfer rate equation applied to the finned surface and to the ends, the surface temperature can be determined. An iterative solution may be required, since the calculation of the natural convection heat transfer coefficient depends on the surface temperature.


Solution: The rate equation for convective heat transfer from a finned surface and from the ends is: Q =η 0 h Atot (Ts − T f ) + 2h Aend (Ts − T f ) Solving for the surface temperature

⎡ ⎤ 1 T s = T f + Q ⎢ ⎥ ⎣η 0 h Atot + 2 h Aend ⎦

We assume that the same heat transfer coefficient is applicable to the ends as to the finned surface. Assuming a straight, rectangular profile fin with an adiabatic tip, and using a corrected length: N A f η0 =1 − (1−η f ) and Atot = Ab + N A f Atot N = 16 A f = 2 L*H = 2(L + t /2) H = (2) (0.10 + 0.004/2) m (1.5 m) = 0.31 m2 Ab = π DH - Nt H = π (0.55 m) (1.5 m) – 16 (0.004 m) (1.5 m) = 2.50 m2 Atot = 2.50 + 16 (0.31) = 7.40 m2 Aend = π D2/4 = π (0.55 m)2/4 = 0.24 m2 η f =

The fin efficiency is

tan h ( mL* ) mL*

1/ 2

⎛ hp ⎞ m =⎜ ⎟ ⎝ k A x ⎠

where

/2

⎛ h 2 H ⎞ =⎜ ⎟ ⎝ k t H ⎠

We assume that h can be approximated by that on a vertical wall, and this is also valid for the ends. The Rayleigh number is evaluated with properties from Appendix A-7 at T film = (T s + T f )/2. We assume T s ≈ 54 oC → T film ≈ 320 K, ρ = 1.103 kg/m3, µ = 1.94 × 10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.703 2 3 ⎛ 1 ⎞ 9.81m s 2 ) ⎜ 1.103 kg m3 ) ( 54 − 40) K (1.5m) ( 0.703) 2 3 ( ( ⎟ g β ρ (T s − T f ) H Pr ⎝ 320 ⎠ Ra = = = 3.29 ×109 2 2 µ (1.94 ×10-5 N ⋅ s/m2 ) From Table 12-3

1/ 3

Nu = 0.1 Ra1 / 3 = 0.1( 3.29 ×109 )

For plain carbon steel, k = 60.5 W/mK,

η f =

tanh ( 0.487) 0.487

T s = 40 o C +

= 0.928

= 149

→

h=

Nuk H

=

⎡ ( 2.76 W m2 K ) 2 ⎤ * ⎥ mL = ⎢ ⎢⎣ ( 60.5W m K)( 0.004m ) ⎥⎦ η 0 =1 −

16( 0.31m2 ) 7.40m 2

(149)( 0.0278W/mK ) 1.5m

= 2.76W/m3 K

1/ 2

( 0.10 + 0.004 /2) m = 0 .487

(1 − 0.928) = 0.951

⎡ ⎤ 1 ⎢ ⎥ = 52.0 o C 2 2 2 ( 2.76W m K ) ⎢⎣ ( 0.951) ( 7.40m ) + 2 ( 0.24m ) ⎥⎦ 250W

Answer

This is close enough to our estimate, so we do not need to iterate.

Comments: Without the fins or radiation, the surface temperature was 64.7 ºC. This problem illustrates well the effectiveness of fins in increasing heat transfer. 12-61

12-58 Two 1.5-m in diameter, 2-m tall tanks connected to a common piping header are used to store propane for use in an isolated cabin in the mountains. Unknown to the owner, the spring in the pressure relief valve on the system weakens and allows the pressure in the two tanks to drop to atmospheric. The temperature of the propane falls to -42 ºC when the pressure inside the two tanks reaches one atmosphere. The still ambient air is at 15 ºC. Heat transfer from the air to the propane causes it to vaporize, and the vapor is vented from the tank. Properties of propane are v f = 0.001755 m3/kg, v g = 0.4127 m3/kg, h fg = 425 kJ/kg, and h g = 493 kJ/kg. Ignoring the wall resistance of the tank and radiation, determine how long it will take for the tank to empty (in days).

Approach: The rate of the conservation of energy and mass are applied to the control volume shown. These equations are integrated with respect to time. The heat transfer coefficient is required to calculate the heat transfer rate.

Assumptions: 1. The air is at one atmosphere. 2. Radiation and wall resistance is ignored. 3. The system has no work, and potential and kinetic energy effects are negligible.

Solution: For the control volume shown, assume no work, constant pressure, and no potential or kinetic energy effects. Conservation of mass and energy give:

− m out = dm/dt

 Q − m h dU dt where hout is the o ut o ut =

and

enthalpy of the leaving vapor and is equal to h g (saturated vapor). Integrating both equations with respect to time:

− ∫ m out dt = ∫ dm

⇒

− ∫ m out dt = m2 − m1

With h g constant and assuming Q is constant

∫ Qdt − ∫ m 



out

∫

hg dt = dU

 + ( m2 − m1 )h g = m2 u 2 − m1u1 . Solving for time: Qt

t = ⎡⎣ m2 u 2 − m1u1 − ( m2 − m1 )hg ⎤⎦ Q The initial internal energy is saturated liquid, u1 = u f . The final internal energy is saturated vapor, u2 = u g . From the given information we can determine the internal energies and masses: m1 = V /ν 1 = V /ν f V = 2π D2 H /4 = 2π (1.5 m)2 (2 m)/4 = 7.07 m 3 m1 = (7.07 m3) / (0.001755 m3/kg) = 4030 kg and m2 = V /ν g = (7.07 m3) / (0.4127 m 3/kg) = 17.1 kg The final mass is negligible compared to the initial mass, so one will ignore m2; that is, use m2 ~ 0. The initial internal energy is u1 =u f = h f − Pv f =( hg − h fg ) − Pv f = ( 493− 424) kJ/kg −(101.3 kN m2

)( 0.001755 m

3

kg )( 1kJ 1kN ⋅ m) = 68.8kJ kg

This is a natural convection problem (ignore radiation). We assume the heat transfer coefficient can be calculated for a vertical wall, the convection from the top of the tanks can use the same heat transfer coefficient, and there is no heat transfer from the tank bottoms (they sit on the ground). Evaluating air properties from Appendix A-7 at T film = (T f + TP ) 2 260K , ρ = 1.359 kg/m3, µ = 1.56 × 10-5 N⋅s/m2, k = 0.0230 W/mK, Pr = 0.719 Ra =

g βρ 2 (T f − TP )H 3 Pr 2

µ

( 9.81m/s ) (1/260K ) (1.359 kg = (1.56 × 10 2

-5

From Table 12-3 Nu = 0.1 Ra1/3 = 0.1(9.39 × 10 10)1/3 = 454 The surface area is (insulated bottoms)

→

2

m3 ) ( 15 + 42) K( 2m) ( 0.719) N ⋅ s/m

h=

Nuk H

3

2

)

=

2

( 454 )( 0.0230W/mK ) 2m

= 9.39 × 1010 = 5.23W/m2 K

A = 2π DH + 2π D /4 = 2π ⎡⎣(1.5m )( 2m ) + (1.5m) 2/4⎤⎦ = 23.6m2 2

The heat transfer rate is Q = hA (T f − TP ) = ( 5.23W m2 K )( 23.6m2 ) ⎡⎣15 − ( − 42) ⎤⎦ K = 7040W The time requires for the tanks to empty −( 4030kg )( 68.8kJ kg) + ( 4030kg)( 493kJ/kg) t = = 2.43×105 s = 67.5 hrs = 2.81 days ( 7040 W)(1kW 1000 W)( 1kJ/s 1kW)

Comments:

Answer

Note that radiation heat transfer will have an influence and will accelerate vaporization.

12-62

12-59 An experiment is performed to determine the heat transfer coefficient on a horizontal circular cylinder. Radiation effects are minimized by polishing the cylinder’s surface. The 30-cm long, 2.5-cm diameter cylinder has well insulated ends. Measurements show that 30 W are dissipated when the cylinder surface temperature is 95 ºC and the surrounding air and surfaces are at 20 ºC. Determine: a. the natural convection heat transfer coefficient from the data b. the natural convection heat transfer coefficient if radiation is taken into account and the surface emissivity is estimated to be 0.07 (in W/m2·K) c. compare the calculated heat transfer coefficient to one calculated with the appropriate correlation (in W/m2·K). Comment on the accuracy of the experimental results.

Approach: The convective heat transfer rate equation ( Q = hA ∆T ) is used to determine the convective heat transfer coefficient. If radiation is included, the heat transfer rate due to convection is reduced, so h increases.

Assumptions: 1. The air is at one atmosphere.

Solution: a) The rate equation is: Q = hA (T s – T f ). Solving for h and noting that the ends are well insulated h=

Q

30W

=

Answer

A(T s − T f ) π ( 0.025m )( 0.30m )( 95 − 20) K

(

Q tot = Qc onv + Q rad = hA (T s - T f ) +εσ A T s4 −T f4

b) If radiation is included Solving for h

=

=17.0W/m2 K

h=

)

Q tot − εσ A(T − T ) 4 s

4 f

A(T s − T f )

4 4 30W − ( 0.07 ) ( 5.67 ×10−8 W m2 K 4 )π ( 0.025m)( 0.30m) ⎡( 95 + 273) − ( 20+ 273) ⎤ K4

⎣

⎦

π ( 0.025m )( 0.30m )( 95 − 20) K

=16.4W/m2 K

Answer

c) To calculate h from a correlation, use properties from Appendix A-7 T film = (95 + 20)/2 = 57.5oC = 330.5 K, ρ = 1.068 kg/m3, k = 0.0285 W/mK, µ = 1.98 × 10-5 N⋅s/m2, Pr = 0.701 g β ρ 2 (T s − T f ) D 3 Pr

Ra =

µ 2 2

3

( 9.81m/s ) (1/330.5K ) (1.068 kg/m ) ( 95 − 20) K ( 0.025m) ( 0.701) = = 70,900 (1.98×10− N ⋅ s/m ) 2

3

5

2

2

From Table 12-3 2 ⎧ ⎫ ⎧ ⎫ 1/ 6 1/ 6 ⎪⎪ ⎪⎪ ⎪ 0.387( 70900) 0.387 Ra ⎪ Nu = ⎨0.6 + 0.6 = + = 7.10 ⎬ ⎨ 8/27 8/27 ⎬ 9/16 ⎡1 + ( 0.559/ Pr ) ⎤ ⎪ ⎪ ⎡1 + ( 0.559/ 0.701) ⎤ ⎪ ⎪ ⎣ ⎦ ⎭ ⎢⎣ ⎥⎦ ⎭⎪ ⎩ ⎩⎪ 9/16

h=

( 7.10 ) ( 0.0285 W/m2 K ) 0.025m

= 8.1 W/m2 K

Answer

Comments: This is poor accuracy. Radiation plays a small role, reducing the experimental heat transfer coefficient by about 3.5%. However, that is minor compared to the factor of 2 differences between the experimental h and the h calculated with the correlation. The experimental is flawed and the measurement sensors must be checked as well as the experimental procedures. For example, the smallest air draft can raise the experimental h significantly, so care must be exercised to enclose the test section with panels to minimize drafts. 12-63

12-60 Arrays of vertical fins are often attached to equipment to aid passive (i.e., natural convection) cooling of the device. Consider the assembly shown below that is located in air at 20 ºC. Each fin has a length of 25 mm, a thickness of 1.5 mm, and a height of 100 mm. Assume the fin has a fin efficiency of 100%, the base temperature is 75 ºC, and each fin operates as if it is independent of all other surfaces nearby. For a fin spacing of 8 mm, determine the heat transfer rate from an array of fins that covers 150 mm of wall (in W).

Approach: This is a natural convection between fins. We assume that it is similar to natural convection in vertical parallel plate channels, and the heat transfer coefficient calculated from the appropriate correlation is applicable to all the surfaces. The number of fins must be determined, and then the basic rate equation is used to calculate the heat transfer rate.

Assumptions: 1. Air is at one atmosphere. 2. The heat transfer is one dimensional.

Solution: Q =η 0 h Atot (Tb − T f )

The rate equation for a finned surface is

Because η f = 1, the overall surface efficiency also is unity, that is, η 0 = 1. The total area is Atot = N 2 LH + WL W = N (S +t )

For the number of fins, N :

→ N =

W

=

0.150m

S + t 0.008m+0.0015m

= 15.8

Because there must be an integer number of fins, N = 15 Atot = (15) (2) (0.10 m) (0.025 m) + (0.15 m) (0.1 m) = 0.090 m2 For the average heat transfer coefficient, we assume isothermal surfaces and use the Bar-Cohen and Rohsenow correlation.

⎧ ⎫ ⎪ ⎪ ⎡ q ′′ ⎤ S ⎪ 576 2.87 ⎪ Nu s = ⎢ + ⎥ =⎨ 2 1/ 2 ⎬ ⎡ ⎛ S ⎞ ⎤ ⎪ ⎣⎢ Tb − T f ⎦⎥ k ⎪ ⎡ Ra ⎛ S ⎞ ⎤ ⎢ Ras ⎜ L ⎟ ⎥ ⎪ ⎪ ⎢ s ⎜⎝ L ⎟⎠ ⎥ ⎝ ⎠⎦ ⎭ ⎦ ⎣ ⎩⎣ Ra s =

−1/ 2

g β ρ 2 (T f − T f )S 3 Pr µ 2

From Appendix A-7 at T film = (74 + 20)/2 = 46.5 oC = 320.5 K, ρ = 1.101 kg/m3, µ = 1.94×10-5 N⋅s/m2, k = 0.0278 W/mK, Pr = 0.704 2

Ra s

2

3

(9.81m s ) (1/320.5K ) (1.101kg m ) ( 75 − 20) K ( 0.008m) ( 0.704) = = 1950 (1.94 ×10 N ⋅ s/m ) 3

-5

2

2

⎧ ⎫ 576 2.87 ⎪ ⎪ Nu s = ⎨ + 2 1/ 2 ⎬ ⎪⎩ ⎡⎣1950 ( 0.008 0.10) ⎤⎦ ⎡⎣1950 ( 0.008 0.10) ⎤⎦ ⎪⎭ q′′ =

Nuk (Tb − T f ) S

=

−1/ 2

=1.99

(1.99 )( 0.0278W/mK )( 75 − 20) K

Q = q ′′ Atot = ( 380 W m2

= 380W/m2

0.008m

)( 0.090m ) = 34.2W 2

12-64

Answer

12-61 A window 30-cm tall and 45-cm wide is centered in an oven door that is 50-cm tall and 75-cm wide. During operation when the room temperature is 24 ºC, the window reaches a temperature of 45 ºC, and the door surface reaches 33 ºC. Assume that both the door and window have an emissivity of 1.0 and the surroundings also are at 24 ºC. Estimate: a. the heat transfer from the door and window (in W) b. the heat transfer if the door did not have a window (in W).

Approach: We will need to use approximations for part A, because of the natural convection over two surfaces at different temperatures. Heat loss is due to radiation and convection. We will assume that the window is flush with the door, and we have the same natural convection heat transfer coefficient over both door and window.

Assumptions: 1. The heat transfer coefficient for he window and door is the same. 2. Air is at one atmosphere.

Solution: a) The total heat loss flow the door and window combination is: Q tot = Q DOOR + QWINDOW Q DOOR = εσ AD (T24 − T f4 ) + h AD (T2 −T f

)

QWINDOW = εσ Aw (T14 − T f4 ) + h Aw (T1 −T f

AD = L 2 H 2 − L1 N 1

where

)

Aw = L1 H 1

where

The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A-7 evaluated at T film = (T s + T f )/2. Because the door and window are at different temperatures, we assume T s ~ (T s + T 2)/2 = (45 + 33)/2 = 39 oC = 312 K, so that T film = (312 + 297)/2 = 304.5 K. The properties are: 3 -5 2 ρ= 1.159 kg/m , µ = 1.887 × 10 N⋅s/m , k = 0.0266 W/mK, Pr = 0.707 g βρ 2 (T s − T f ) H 23 Pr

Ra =

2

µ

2

( 9.81m s ) (1 304.5K ) (1.159 kg m ) ( 312 − 297) K( 0.5m) ( 0.707) = =1.61×10 1.887 10 N s/m × ⋅ ( ) 2

3

3

8

-5

2

2

From Table 12-3, the vertical plate correlation is: 1/ 4

Nu = 0.59 Ra / 4 = 0.59 (1.61×108 )

= 66.5

h=

→

Nuk H 2

=

( 66.5)( 0.0266W/mK ) 0.5m

= 3.54W/m2 K

2

Aw = (0.45 m) (0.30 m) = 0.135 m A D = (0.50 m) (0.75 m) = 0.135 m 2 = 0.240 m2

⎛

Q DOOR = (1) ⎜ 5.67 ×10-8

⎝

⎞ W⎞ 4 4 ⎛ 0.24m2 ) ⎡( 306K ) − ( 297K) ⎤ + ⎜ 3.54 2 ⎟ ( 0.24m2 ) ( 306− 297) K 2 4 ⎟( ⎣ ⎦ m K ⎠ m ⎠ ⎝ W

= 13.4 + 7.6 = 21.0 W 4 4 QWINDOW = (1) ( 5.67 x10-8 ) ( 0.135) ⎡( 318) − ( 297) ⎤ + ( 3.54)( 0.135)( 318− 297) = 18.7+ 10.0= 28.7W

⎣

⎦

Q tot = 21.0 + 28.7 = 49.7 W

Answer

b) For a door with no window, assuming the heat transfer coefficient is the same as in part (a) 4 4 Q = (1) ( 5.67 ×10-8 ) ( 0.5) ( 0.75) ⎡( 306) − ( 297) ⎤ + ( 3.54)( 0.5)( 0.75)( 306 − 297)

⎣

⎦

= 21.0 + 11.9 = 32.9 W

Answer

12-65

12-62 The heat loss situation described in Problem P 12-15 changes when the wind stops and the air is calm. For this new condition, determine: a) the heat transfer rate per unit length of pipe (in W/m) with no insulation, and b) the heat transfer rate per unit length of pipe if 4-cm insulation (k = 0.04 W/m·K) is applied to the pipe (in W/m).

Approach: This is a natural convection problem on a circular cylinder. The basic heat transfer rate equation ( Q = ∆T / Rtot )

can be used to find the heat transfer per unit length. The heat transfer coefficient must be evaluated.


Solution: The basic heat transfer rate equation is T s − T f Ts − T f ∆T Q = = = Rtot Rconv + Rcond 1 hπ DL + 1n ( D2 / D1 ) 2π k L

or

Q

=

L

T s − T f 1 hπ D + 1n( D2 / D1 ) 2π k

a) With no insulation D2 = D1, D = D1, and Rcond = 0. Evaluation of the natural convection h requires the Rayleigh number with fluid properties evaluated of T film = (-6 + 110)/2 = 52 oC = 325K. From Appendix A-7, ρ = 1.086 kg/m3, µ = 1.96 × 10-5 N⋅s/m2, k = 0.0281W/mK, Pr = 0.703 g βρ 2 (T s − T f ) D 3 Pr

Ra =

µ 2

2

3

( 9.81m/s ) (1 325 K ) (1.086 kg/m ) ⎡⎣110 −( − 6) ⎤⎦ ( 0.15 m) ( 0.703) = = 2.55 × 107 − (1.96 ×10 N ⋅ s/m ) 2

3

5

2

2

From Table 12-3 for a horizontal cylinder 2

2

1/ 6 ⎧ ⎫ ⎧ ⎫ 0.387( 2.55 ×107 ) 0.387 Ra1/ 6 ⎪ ⎪ ⎪ ⎪ Nu = ⎨0.6 + = ⎨0.6 + = 37.3 8 / 27 ⎬ 8 / 27 ⎬ ⎡1 + ( 0.559/ Pr )9 /16 ⎤ ⎡1+ ( 0.559 0.703)9 /16 ⎤ ⎪ ⎪ ⎪ ⎪ ⎣ ⎦ ⎣ ⎦ ⎭ ⎩ ⎭ ⎩

h= Q L

Nuk

=

D

=

( 37.3)( 0.0281W/mK ) 0.15m

= 6.99 W/m2 K

⎡⎣110 − ( −6 ) ⎤⎦ K 1/ ⎡( 6.99 W/m2 K ) π ( 0.15 m) ⎤

⎣

= 382

⎦

W m

(without insulation)

Answer

b) Once we add the insulation, the outside insulation temperature will be low. Assuming T s ~ 0oC ⇒ T film = 270K, ρ = 1.307 kg/m3, µ = 1.63 × 10-5 N⋅s/m2, k = 0.0239W/mK, Pr = 0.716 2

Ra =

3

( 9.81)(1/ 270)(1.307) ⎡⎣110−( − 6) ⎤⎦ ( 0.23) ( 0.716)

Nu = 73.8

−5 2

(1.63×10 ) ⇒

h=

( 73.8)( 0.0239) 0.23

= 7.67 W/m2 K

= 2.36 × 108

⎡⎣110 − ( −6 ) ⎤⎦ = 61.7 W/m L 1 ⎡⎣( 7.67 )π ( 0.23) ⎤⎦ + 1n ( 23/15) ⎡⎣ 2π ( 0.04) ⎤⎦ The outside wall temperature can be calculated with Q = hπ DL (T s – T f ) ⇒ T s = T f + T s = T f + ( Q / L ) ( hπ D ) Q

=

T s = − 60 C +

( 61.7W/m) = 5.1o C 2 ( 7.67W/m K ) π ( 0.23 m)

This is close to what we assumed, so no iteration is required.

12-66

Answer

12-63 A power amplifier is mounted vertically in air that is at 27 ºC. The case is made of anodized aluminum with a surface area of 3800 mm2 and a height of 40 mm. If the amplifier operates at 127 ºC, estimate the total power dissipation (natural convection and radiation) from the unit (in W). Assume a surface emissivity of 0.76.

Approach: This is a combined natural convection plus radiation heat transfer problem. With the surface temperature known, this is a straight forward calculation. The appropriate heat transfer coefficient correlation must be chosen, and we assume the surface is small relative to its surroundings.

Assumptions: 1. Air is at one atmosphere. 2. The power amp is “small” relative to the large surrounding area.

Solution: The total heat transfer from the surface by convection and radiation is: Q = Q conv + Q rad = hA (Tw − T∞ ) + εσ A (Tw4 − T∞4 ) To evaluate h, we need the Rayleigh number with properties from Appendix A-7 evaluated at T film = (T w + T ∞)/2 = (400 + 300)/2 = 350 K, ρ = 0.998 kg/m3, ν = 20.76 × 10-6 m2/s, k = 0.03003 W/mK, Pr = 0.697 g β (Tw − T∞ ) H 3 Pr

Ra =

ν 2

3

( 9.81m/s ) (1 350K )(127 − 27) K( 0.04m) ( 0.70) = ( 20.76×10 m /s) 2

-6

2

2

= 2.90 × 105

From Table 12-3, Nu = 0.59 Ra1/4 = 0.59 (2.90 × 10 5)1/4 = 13.7 Nuk (13.7)( 0.03003W/mK ) h= = = 10.3W/m2 K H 0.04m

⎛

Q = ⎜10.3

⎝

W ⎞

⎛ W ⎞ 4 4 2 -8 3 ⎟ ( 0.0038m ) ( 400 − 300) K + ( 0.76) ⎜ 5.67× 10 2 4 ⎟ ( 0.0038m ) ⎡⎣( 400K) − ( 300K) ⎤⎦ m K⎠ m K ⎠ ⎝ 2

Q = 3.91W + 2.87W = 6.78W

Answer

12-67

12-64 In car paint shops and other drying applications, radiant heaters are often used because the radiant thermal energy heats the surface directly with minimal heating of the surrounding air. Consider a vertical flat panel 1-m tall and 4-m long with an emissivity of 0.85 mounted on the wall of a large room. The panel is maintained at a uniform temperature of 330 ºC and the walls and air in the room are maintained at 25 ºC. Determine the heat transfer rate from the panel to the room (in W).

Approach: Power input to the heater must equal radiation plus convection from panel. We assume the panel is located in a very large room for radiation purposes, and we use Q = hA ∆T for the convection. The natural convection heat transfer coefficient must be evaluated.

Assumptions: 1. Air is at one atmosphere. 2. The heater is small relative to the large room.

Solution: An energy balance on the heater gives

 + Q Q in = Q εσ A (Ts4 − T f4 ) + hA (Ts −T f ) rad conv =

where A = HL. The natural convection heat transfer coefficient requires the Rayleigh number with properties from Appendix A-7 evaluated at T film = (330 + 25)/2 = 177.5 oC = 450.5 K, ρ = 0.784 kg/m3, µ = 2.484 × 10-5 N⋅s/m2, k = 0.03707 W/mK, Pr = 0.683 2

Ra =

g βρ 2 (T s − T f ) H 3 Pr 2

µ

=

⎞⎛ m ⎞⎛ 1 kg ⎞ 3 ⎛ ⎜ 9.81 2 ⎟ ⎜ 450.5K ⎟ ⎜ 0.784 3 ⎟ ( 603 − 298) K ( 1m) ( 0.683) s ⎠⎝ m ⎠ ⎝ ⎠⎝

( 2.484 ×10

-5

N ⋅ s/m

2

)

2

= 4.52 ×109

From Table 12-3 for a vertical wall Nu = 0.1 Ra1/3 = 0.1 (4.52 × 10 9)1/3 = 165 Nuk (165)( 0.03707W/mK ) h= = = 6.13W/m2 K H 1m

⎛

Q rad = ( 0.85 ) ⎜ 5.67 ×10-8

⎝

⎛

W ⎞

⎝

m2 K ⎠

Q conv = ⎜ 6.13

⎞ 4 4 ⎟ (1m)( 4m) ⎡⎣( 603K) − ( 298K) ⎤⎦ = 23,970W m K ⎠ W 2

4

⎟ (1m )( 4m )( 603 − 298) K = 7,480W

Q in = 23,970W+7,480W=31,450W

Answer

12-68

12-65 A home hobbyist builds a kiln to fire her ceramic pots. Plans obtained from the Internet state that because of the thick fireclay bricks used to construct the kiln, insulation on the outside surface are not required. When she uses the 1-m by 1-m by 1-m kiln for the first time in a room at 30 ºC, the kiln’s outside wall temperature is 90 ºC. Assuming that there is heat loss from the four sides and the top only, and that these surfaces have an emissivity of 0.8, determine: a. the total heat loss from the kiln (in W) b. the total heat loss from the kiln i f 4-cm thick insulation with k = 0.04 W/m·K and ε = 0.1 (assume the outside brick temperature remains at 90 ºC c. the simple payback time for the insulation if the insulation costs $400, the cost of natural gas is $0.50/105 kJ, the furnace has an efficiency of 84%, and the furnace operates 2000 hr/yr.

Approach: This is a combined convection and radiation problem in part (a), and conduction, convection, and radiation in part (b). The natural convection heat transfer coefficient must be evaluated for the two different situations.

Assumptions: 1. Air is at one atmosphere. 2. The kiln is “small” relative to the large room size.

Solution: a) The heat transfer rate is calculated with the convective rate equation for the sides and top and the radiation rate equation assuming the kiln is in a very large room: Q = ( h s As + hT AT

)(Ts − T f ) + εσ Atot (Ts4 −T f 4 )

The heat transfer coefficient on the sides is assumed to be that on a vertical wall; for the top, we use a correlation for a heated plate facing upward. From Appendix A-7 at T film ≈ (90 + 30)/2 = 60oC = 333 K, ρ = 1.060 kg/m3, -5 2 µ = 2.00 × 10 N⋅s/m , k = 0.0288 W/mK, Pr = 0.702. For vertical wall (sides) 2

3

( 9.81m s ) (1/333K ) (1.060kg/m ) ( 90 − 30) K(1m) ( 0.702) Ra = = 3.49 ×10 ( 2.00 ×10− N ⋅ s/m ) 2

3

9

5

Nu = 0.1 Ra1/3 = 0.1 (3.49 × 109)1/3 = 152

From Table 12-3 h s =

Nuk H

=

2

2

(152 ) ( 0.0288W/m2 K ) 1m

4.38W/m2 K

2

A s = 4 HW = 4m For the top, Lchar = A/ P = HD/2( H + D) = 0.25m RaT = 5.45 × 107 Nu = 0.15 Ra1/3 = 0.15 (5.45 × 107)1/3 = 56.9 hT = (56.9) (0.0288)/0.25 = 6.55 W/m2K AT = 1m2

⎡⎛

Q = ⎢⎜ 4.38

⎣⎝

W ⎞ W ⎞ W ⎞ ⎤ 4 4 ⎛ ⎛ 4m 2 ) + ⎜ 6.55 2 ⎟ (1m 2 )⎥ [ 90 − 30]K + ( 0.8) ⎜ 5.67x10−8 2 4 ⎟ ( 5m 2 ) ⎡( 90 + 273K ) − ( 30 + 273K ) ⎤ ( ⎟ 2 ⎣ ⎦ m K⎠ m K⎠ m K ⎠ ⎝ ⎝ ⎦

= [1051 + 393] + 2026 W = 3470 W b) Draw a circuit diagram to evaluate the heat loss when insulation is added.

Answer

From the diagram: Q tot = Q conv + Q rad Tins − T f T s − T ins = + εσ Atot (Tins4 − T f4 ) 1 t kins Atot hs As + hT AT

We want the outside temperature of the insulation but the heat transfer coefficient depends on that temperature, so assuming T ins ≈ 44oC, T film ≈ 310 K, ρ = 1.139 kg/m3, µ = 1.89 × 10-5 N⋅s/m2, k = 0.0270 W/mK, Pr = 0.706 12-69

2

Ra s =

3

( 9.81)(1/310 )(1.139 ) ( 44 − 30)(1) ( 0.706)

(1.89 ×10 ) −5

=1.14 × 109

Nu s = 0.1 (1.14 × 109)1/3 = 104 h s = 2.82 W/m2K RaT = 1.78 x 107 NuT = 0.15 (1.78 × 107)1/3 = 39.2 hT = 4.23 W/m2K Now substituting these into the main energy balance ( 363 − Tins ) K (T ins − 303 ) K = + ( 0.1) ( 5.67 × 10−8 ) ( 5 ) ⎡⎣T ins4 − 3034 ⎤⎦ 0.04m 1 →

→

( 0.04 W/mK ) ( 5m2 ) ( 2.82 )( 4 ) + ( 4.23)(1) Solving for T ins = 315.3 K = 42.3oC

and

Q tot = 239 W

Answer

Because the calculated T ins is close to the assumed value, no iteration is required. c) The energy savings with the insulation is 3470 – 239 = 3231 W For 2000 hrs/yr in a 84% efficient furnace, this is results in an energy savings of ( 2000hr/yr )( 3231W )(1kW 1000W )(1kJ/s 1kW )( 3600s hr ) = 2.77 ×107 kJ/yr 0.84 Gas costs $0.50/105 kJ, so total fuel expense savings is ( 2.77 × 107 kJ yr )( $0.50 105 kJ ) = $138/yr If the insulation costs $400, the simple payback is $400 = 2.9 yrs. $138/yr

Answer

12-70

12-66 In oil refineries and chemical process plants, insulation on pipes is wrapped in thin aluminum metal sheaths to protect the insulation from the weather. After weathering and exposure to harsh air borne chemicals around the plants, the surface of the metal sheath corrodes and the emissivity is about 0.4. Consider a 30cm I.D., 40-cm O.D. carbon steel pipe (k = 60 W/m·K) carrying saturated steam at 350 ºC covered with 7.5-cm of fiberglass insulation (k = 0.036 W/m·K). The steam convective heat transfer coefficient is 600 W/m2·K. With an air temperature of 0 ºC, determine the heat transfer rate per unit meter of pipe length when: a. a crossflow at 30 km/hr is on the outside of the pipe (in W/m) b. natural convection is on the outside of the pipe (in W/m) (use an approximation based on the results of part (a) and justify).

Approach: Assuming one-dimensional heat transfer, the base heat transfer rate equation ( Q = ∆T / Rtot ) can be used to calculate the heat transfer rate. The heat transfer coefficient on the outside must be evaluated, and the outside surface temperature is needed to determine the radiation heat transfer.


Solution: For steady one-dimensional heat transfer, the total heat transfer from steam to outer surface is: T st − T 3 ∆T Q = = 1n ( D2 /D1 ) 1n( D3 /D2 ) Rtot 1 h st π D1 L

+

+

2π k s L

2π kins L

where the outside surface temperatures, T 3, is unknown. This heat transfer rate must equal that from the surface to Q = Q conv + Q rad

the air

= h π D3 L

(T

−T f

3

) + εσ π D L (T 3

4 3

4

−T f

)

The length, L, can be factored out of both equations. We have assumed that the surroundings are very large and are at the same temperature as the air. We have two equations and two unknowns ( Q /L and T 3) that can be solved simultaneously. We can recast the radiation contribution in terms of a radiative heat transfer coefficient T3 − T f Q = where hr = εσ (T3 + T f ) (T32 + T f 2 ) 1 1 + hπ D3 L hr π D3 L For the forced convection heat transfer coefficient, from Appendix A-7 and assuming T 3 → 7 oC, T film = (0 + 7)/2 = 3.5 oC ≈ 275 K, ρ = 1.284 kg/m3, µ = 1.67 × 10-5 N⋅s/m2, k = 0.0243 W/mK, Pr = 0.715 Re =

ρ V D

=

(1.284kg

m3 ) ( 30000m hr )( 1hr 3600s)( 0.55m) 1.67 ×10-5 N ⋅ s/m2

µ

From Table 12-1 Nu Nu =

Nuk D

=

=

0.027 Re0.806 Pr 1 3

( 706 )( 0.0243W/mK ) 0.55m

The resistances are 1 h st π D1 L

=

1n ( D2 /D1 ) 2π k s L

=

2

= 31.2W/m

1

( 600W/m K )π ( 0.30m) L 2

=

1n ( 0.40/0.30) 2π ( 60W/mK ) L

=

0.806

0.027 ( 3.48 × 105 )

=

K

0.00177 L

0.000763 L

12-71

5

= 3.48 ×10 13

( 0.713)

=

706

1n ( D3 /D2 ) 2π kins L 1 hπ D3 L

=

=

1n ( 0.55/0.40) 2π ( 0.036 W/mK ) L

=

1.408

1

( 31.2W

m K )π ( 0.55m) L 2

L

=

0.0185 L

With a first approximation of T 3 = 7 oC = 280 K 2 2 hr = ( 0.4 ) ( 5.67 ×10-8 W m2 K 4 ) ( 280K + 273K ) ⎡( 280K ) + ( 273K ) ⎤ = 1.9W/m2 K

⎣

1 hr π D3 L

=

1

(1.9 W m K ) ( 0.55m) L 2

π

=

⎦

0.302 L

Solving the two rate equations, we obtain T 3 = 337.9 K. This does not match our assumed value. Iterating on T 3, we obtain T 3 = 326.4 K hr = 2.46 W/m2K 1 0.2351 hr π D3 L

=

L

Q /L = 210 W/m

Answer

b) For the natural convection case, if we assume hnat



0.1h forced as a first approximation, then solving for the

conv

conv

heat loss as above: T 3 = 350 K = 77 oC hr = 2.79 W/m2K 1 0.2079 hr π D3 L

=

L

Q /L = 194 W/m

Answer

Comments:

Because the heat loss per unit length does not change much with a factor of ten change in the air side heat transfer coefficient, h, a more accurate estimate of h is not needed.

12-72

12-67 The power cables for an electric welding rig are suspended above the floor of a factory so that a tripping hazard is not created. The cables are 25-m long; the the copper is 10-mm in diameter diameter and is covered by a 2mm thick rubberized cover (k ( k = = 0.26 W/m·K), which is black, has an emissivity of 0.9 and cannot have a temperature greater than than 65 ºC. The cable resistance resistance is 4 × 10-4 ohms. If the cable is suspended in calm air at 27 ºC, determine the maximum allowable current (in A).

Approach: Because the cable is long, we assume one-dimensional heat transfer. This is a combined natural convection convection and radiation problem. We will need to estimate the surface surface temperature to obtain the natural convection heat transfer coefficient and the radiation. radiation. An iterative solution solution may be required. Once the heat transfer rate rate is determined, the maximum allowable current can be calculated.

Assumptions: 1. The air is at one atmosphere. atmosphere. 2. The heat transfer transfer is one dimensional. dimensional.

Solution: Q tot = Q cond = Q conv + Q rad

The total heat transfer rate is calculated with

Assuming the surroundings are large and at the same temperature as the air, and all properties are constant 2π kL(Tmax − Ts ) = hA (T s - T f ) +εσ A (Ts4 − T f 4 ) 1n( D2 /D1 ) The convective heat transfer coefficient must be determined with fluid properties from Appendix A-7 evaluated at the film temperature, T film = (T (T s + T f )/2. We assume T s ≈ 45oC = 318K, T film ≈ (27 + 45)/2 = 36 0C ≈ 310 K, so 3 ρ = = 1.143 kg/m , k = = 0.0268 W/mK, µ = = 1.90 × 10-5 N⋅s/m2, Pr = = 0.711. When calculating the Rayleigh number, number, we will keep these properties constant and only change T s during the iteration. This is for simplicity simplicity only: Ra =

g βρ 2 (T s − T f ) D23 Pr µ 2 2

=

⎛

2

⎞

( 9.81m s ) ⎜ T +300 ⎟ (1.143 kg ⎝

s

m3

⎠

(1.90 ×10−

5

2

2

) (T −300) ( 0.014m) ( 0.711) s

N ⋅ s/m 2 )

2

⎛ T − 300 ⎞ = (1.385×10−5 ) ⎜ s ⎟ ⎝ T s + 300 ⎠

For T s = 318 K, Ra = Ra = 4035, so from Table 12-3

⎧ ⎫ ⎧ 1/ 6 0.38 0.387 7 ( 4035 4035) 0.387 Ra1/ 6 ⎪ ⎪ ⎪ Nu = ⎨0.6 + = ⎨0.6 + 8/27 ⎬ 9/16 ⎡1 + ( 0.559/ Pr )9/16 ⎤ ⎪ ⎪ ⎪ ⎡1 + ( 0.559 /0 /0.711) ⎤ ⎣ ⎦ ⎭ ⎩ ⎩ ⎣ ⎦

2

2

h=

NuK D

=

3.54) ( 0.02 0.0268 68W W/mK /mK ) ( 3.54 0.014m

8/27

⎫ ⎪ ⎬ = 3.54 ⎪ ⎭

6.78W/ W/m m2 K = 6.78

Therefore, the right hand side of the heat transfer rate equation is: Q = (6.78 W/m2K) π (0.014 m) (25 m) (318 – 300) K 4 4 + ( 0.9 ) ( 5.67×10-8 W m2 K 4 ) π ( 0.014 m)( 25m) ⎡( 318K) − ( 300 K) ⎤ = 134.2+ 119.3= 253.5W ⎣ ⎦

The left hand side is

Q =

2π ( 0.26 W mK ) ( 25m ) ln ( 0.014 0.010)

338-31 318 8) K=24 K=2428W 28W ( 338-

These two heat transfer rates do not match, so an iteration is required. Doing so, we obtain: T s = 333.6 K =60.6o C

Ra = 7350

Nu = 4.06

h = 7.78 W m2 K

Q = 529 W

Electric power is Q = I 2 R , where I where I is is current. current. Solving for current: I = ( Q R )

0.5

0.5

= ( 529W 4 × 10−4 Ω )

= 1150 A

12-73 12- 73

Answer

12-68 A manufacturer of prefabricated buildings is considering using the same structure (shown below) for walls and roofs for quickly and cheaply assembled assembled buildings. The inner and outer surfaces surfaces are 1.27-cm thick plywood (k (k = = 0.115 W/m·K), W /m·K), and the air-filled gap is 10-cm wide; the panels are 2.5-m long and 1.25-m deep out of the plane of the page. For temperatures on the outside of the two plywood sheets sheets of -10 ºC and 15 ºC, determine: a. the heat transfer rate for both horizontal and vertical orientations (in W) b. the effect of inserting a baffle baffle at midheight for when the panel panel would be used in a vertical orientation (in W).

Approach: Assuming steady, one-dimensional heat transfer, the basic rate equation, equation, Q = ∆T / R Rtot , can be used to calculate the heat transfer rate. rate. The natural convection heat transfer coefficient inside the wall must be estimated.


The heat transfer transfer is one dimensional. dimensional. Air is at at one atmosphere. atmosphere. Radiation is ignored. ignored. Heat transfer transfer from the ends is ignored.

Solution: Assuming adiabatic ends, and that radiation can be ignored, the rate equation is: T s − T 1 ∆T Q = = Rtot RWALL ,1 + RWALL , 2 + Rconv The wall resistances are RWALL ,1 = RWALL , 2 =

t kA

=

0.0127m W/mK ) (1.25m ) ( 2.50 m) ( 0.115 W/

= 0.0353K/W

a) The convective resistance in a vertical enclosure requires requires the natural convective heat transfer coefficient, coefficient, with properties evaluated at at the average of the two wall temperatures. We first estimate estimate the heat transfer transfer rate by assuming the convective resistance is equal to twice that contributed by the plywood, and then use this estimate to calculate the inside wall temperatures Q est =

⎡⎣15 − ( − 10 ) ⎤⎦ K =177W 4( 0.03 0.035k 5k/W /W)

The temperature drop across both sheets of plywood is the same, so T s − T 2, in Q est = → T2 − T2, in = Q est (t kA ) = 177 W ( 0.0353 K W ) = 6.3K t kA o So T s,in ≈ = 10 + 6.3 = -3.7 oC and T 2,in Therefore, from Appendix A-7 at 2,in ≈15 – 6.3 = 8.7 C 3 -5 o T avg = ( −3.7 + 8.7 ) 2 = 2.5 C = 27 275K, ρ = = 1.284 kg/m , µ = = 1.67 × 10 N⋅s/m2, k = = 0.0243 W/mK, Pr = = 0.715

Ra s =

h= 1 hA

Nuk S

=

2

3

=

3

6

2

From Table 12-6

2

2 75K ) (1.284 kg m ) ( 8.7+ 3.7) ( 0.10 m) ( 0.715) ( 9.81m/s ) (1 27 = =1.87 ×10 µ (1.67 ×10 N ⋅ s/m ) − − Nu = 0.073 Ra ( H S ) = 0.073 (1.87× 10 ) ( 2.50 0.10) = 6.06

g βρ 2 (T2,in − T1,in )S3 Pr

-5

1/ 3 s

6.06) ( 0.02 0.0243 43W W/mK /mK ) ( 6.06 0.10m

6

1.47W//m3 K = 1.47W

1

(1.47 W m K ) (1.25m) ( 2.50m) 2

1/ 9

= 0.217K/W

Therefore, for vertical configuration:

12-74 12- 74

2

1/ 3

1/ 9

Q =

⎡⎣15 − ( −10 ) ⎤⎦ K = 86.8W 2 ( 0.0353K/W ) + 0.217 K/ K/W

o o Calculating a new T 1,in 1,in = - 6.9 C and T 2,in 2,in = 11.9 C

Ra s = 2.84 ×106 ,

Nu = 7.23 ,

h = 1.76 W/m2 K

Q = 98.9 W

Answer

We could iterate again for an improved estimate. For the horizontal configuration, configuration, from Table Table 12-6 and using the same Ra same Ra as as for the vertical configuration 1/ 3

Nu = 0.061 Ra1 / 3 = 0.061( 2.84× 106 ) h=

Nuk

1 hA

S

=

Q =

=

( 8.64 ) ( 0.0243) 0.10

= 8.64

2.10W W/m2 K = 2.10

1

( 2.10 W/m K ) (1.25m) ( 2.50m) 2

= 0.152K/W

⎡⎣15 − ( −10) ⎤⎦ =112W 2 ( 0.0353) + 0.152

Answer

b) If a baffle is is inserted in the the vertical wall, wall, the only change change in the calculation calculation of the heat transfer coefficient coefficient is in -1/9 the term ( H ( H /S ) hbaffle h

−1/ 9

( H/S )baffle ⎛ 1.25 ⎞−1/ 9 = =⎜ ⎟ = 1.08 −1/ 9 ⎝ 2.50 ⎠ ( H/S )

hbaffle = 1.08 (1.76 W/m 2K) = 1.90 W/m2 K Q = 105 W

Answer

12-75 12- 75

12-69 A thermal pane window often is constructed of two panes of plate glass separated by a short distance; this arrangement provides increased thermal resistance compared to a single pane of glass. Consider a 4-ft wide and 6-ft high window whose inside wall temperature is 75 ºF and outside wall temperature is 45 ºF. It is constructed of glass that is 0.25-in thick. Determine the heat transfer rate: a. for a single pane of glass (in Btu/hr) b. for two panes separated by a 1-in air gap (in Btu/hr) c. for two panes of glass with if a thin (0.1-in.) sheet of glass is inserted between the other two panes that are still separated by 1 in. (in Btu/hr).

Approach: Assuming one-dimensional heat transfer, the heat transfer rate is determined by the rate equation, Q = ∆T / Rtot . This thermal resistance can be found for combinations of t /kA and 1 /hA. Radiation is ignored.

Assumptions: 1. Air is at one atmosphere between the two panes of glass. 2. Radiation is ignored.

Solution: a) The thermal resistance of a single pane of glass is:

R glass

=t

kA

From Appendix B-3, for plate glass k = 0.8 Btu/hr ft oF, and ( 0.25/12ft) o R glass = = 0.001085hr F Btu o ( 0.8 Btu hrft F ) ( 6ft )( 4ft ) The heat transfer rate is Q = ∆T / R glass = (75 – 45)oF/(0.001085 hr oF/Btu) = 27,650 Btu/hr b) Resistance of two panes of glass and an air gap is:

Rtotal

=

Answer

2 Rglass + 1 hA

For air properties from Appendix B-7 at T avg = (75+45)/2= 60oF, ρ = 0.077 lbm/ft3, k = 0.0146 Btu/hrftoF, µ = 1.214×10-5 lbm/ft⋅s,, Pr = 0.72 g βρ 2 (T1,in − T2,in ) S 3 Pr

Ra =

µ 2 The temperatures to use are the inside wall temperatures. We could estimate the temperatures from the glass resistance. Let’s assume for the air gap, that resistance is approximation equal to that contribution by both panes of glass. Therefore: Q

estimate

=

T1 − T 2 4 R glass

=

( 75-45)

o

F

4 ( 0.001085hr o F/Btu)

=6912

Btu hr

Now calculate the ∆T across the glass:  Qt o o  ∆T = =Q R = ( 6912 Btu hr ) ( 0.001085 hr F Btu ) = 7.5 F estimate glass kA So T 1,in = 75 – 7.5 = 67.50F and T s,in = 45 + 7.5 = 52.5 oF 2

( 32.2ft/s ) (1 520R ) ( 0.077 lbm ft ) ( 67.5 − 52.5) R (1 12 ft) ( 0.72) Ra = =15570 1.214 × 10 lbm fts ( ) From Table 12-5 Nu = 0.197 Ra ( H S) = 0.197 ( 15570) ( 6 [1/12]) = 1.369 Nuk (1.369 ) ( 0.0146 Btu hr ft F ) 2

3

3

2

−5

−1/ 9

1/ 4

1/ 4

o

h= 1

hA

S =

=

1/12ft 1

( 0.24 Btu

hrft

2 o

F ) ( 6ft )( 4ft )

= 0.24 Btu

= 0.1738 hr

0

hrft 2 o F

F Btu

12-76

−1/

9

Therefore, Q = 

( 75 -45)

o

F

2 ( 0.001085 hr F Btu ) +0.1738 hr F Btu o

o

=170

Btu hr

We need to re-estimate h. Because of the symmetric system, the fluid properties are the same, so ∆T glass = (170) (0.001085) = 0.185 oF, T 1,in = 74.8 oF, T 2,in = 45.2 oF, Ra = 30720, Nu = 1.623, h = 0.28 Btu/hr ft 2 oF, 1 /hA = 0.147 hr oF/Btu, and Q = 201.7 Btu/hr.

Answer

c) Adding the thin glass between the two panes changes the Rayleigh number, because the characteristic length is changed as is the ∆T . We will keep the same fluid properties. Because of symmetry, half the temperature drop occurs on each side of the thin glass partition, so 3

2

Ra 

( 32.2 )(1 520)( 0.077) ( 75 − 60) ⎡⎣( 1− 0.1) ( 2i12) ⎤⎦ ( 0.72)

(1.214×10− ) 5

2

=1419

From the Table 12-5, Nu = 1 (1)( 0.0146) Btu h= = 0.39 hrft 0 F ⎡ 1− 0.1 ⎤

⎣ 2 (12) ⎦

1 hA Q =

=

1

( 0.39 Btu

hrft 0 F ) ( 6ft )( 4ft )

= 0.107

hr 0 F Btu

∆T 2 R glass + 2 Rconv + R partition

( 75 − 45)

=

2 ( 0.001085hr F Btu ) + 2 ( 0.107hr F 0

0

o

F

( 0.1 12) ft Btu ) + ( 0.8Btu hrft 0 F ) ( 6ft )( 4ft )

= 138

Btu

Answer

hr

Comments:

Using a double pane window reduces the heat loss substantially. Inserting a thin glass partition between the two other panes of glass further reduces the heat loss by 32%. Another technique to reduce heat loss is to lower the interpane pressure and to use a gas (such as argon) with a lower thermal conductivity.

12-77

Thermals Fluids 1 Ch 12 Guide

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