heat transger

January 11, 2008

ChE 314: Assignment 1

Due :

January 21 (Monday) before 16:00, in assignment box outside CME 567.

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

The first two problems are from Incropera et al. (5th ed.) 1.15

An electrical resistance heater is embedded in a long cylinder of diameter 30 mm. When water with a temperature of 25°C and velocity 1 m/s flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of 90°C is 28 kW/m. When air, also at 25°C, but with a velocity of 10 m/s is flowing, the power per unit length required to maintain the same surface temperature is 400 W/m. Calculate the convection coefficients for the flows of water and air. 2 2 [ Ans Ans: 4570 W/m K; 65 W/m K] ⋅

1.65

⋅

During its manufacture, plate glass at 600°C is cooled by passing air over its surface such 2 that the convection heat transfer coefficient is h = 5 W/m K. To prevent cracking, it is known that the temperature gradient must not exceed 15°C/mm at any point in the glass during the cooling process. If the thermal conductivity of the glass is 1.4 W/m K and its surface emissivity is 0.8, what is the lowest temperature of the air that can initially be used for cooling? Assume that the air temperature equals that of the surroundings, which functions effectively as an “enclosing wall.” [ Ans: 345°C] ⋅

⋅

Additional problem (to be handed in also):

A square silicon chip ( k 15 W/m K) has width w 6 mm on each side and thickness d 1 mm . The chip is mounted in a substrate such that its four edges and bottom surface are insulated, while the top surface is exposed to a flowing coolant that is at 20°C. For this problem, you can assume negligible heat transfer by radiation. =

⋅

=

=

coolant

d w

(a) With the chip dissipating 2 W of power, its bottom surface temperature is found to be 45°C at steady state. What is the effective heat transfer coefficient of the coolant flow? (b) What is the temperature at the chip’s top surface? (c) Justify the neglect of radiation in this problem.

Page 1 of 1

PROBLEM 1.15 KNOWN:

Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows.

FIND:

Convection coefficients for the water and air flow convection processes, h w and ha, respectively. SCHEMATIC:

ASSUMPTIONS:

(1) Flow is cross-wise over cylinder which is very long in the direction

normal to flow. ANALYSIS:

The convection heat rate from the cylinder per unit length of the cylinder has

the form q′ = h (π D )

( Ts − T∞ )

and solving for the heat transfer convection coefficient, find h=

q′ D ( Ts − T∞ )

.

π

Substituting numerical values for the water and air situations: Water

28

hw = π

Air

ha

103 W/ W/m D

0.030m ( 90-25 ) C 400 W/m

= π

COMMENTS:

×

×

×

D

= 4,570 W/m 2 ⋅ K

= 65 W/m

2 ⋅ K.

<

<

0.030m ( 90-25 ) C

Note that the air velocity is 10 times that of the water flow, yet

hw ≈ 70 × ha. These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation translation of this work beyond that permitted permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

PROBLEM 1.15 KNOWN:

Long, 30mm-diameter cylinder with embedded electrical heater; power required to maintain a specified surface temperature for water and air flows.

FIND:

Convection coefficients for the water and air flow convection processes, h w and ha, respectively. SCHEMATIC:

ASSUMPTIONS:

(1) Flow is cross-wise over cylinder which is very long in the direction

normal to flow. ANALYSIS:

The convection heat rate from the cylinder per unit length of the cylinder has

the form q′ = h (π D )

( Ts − T∞ )

and solving for the heat transfer convection coefficient, find h=

q′ D ( Ts − T∞ )

.

π

Substituting numerical values for the water and air situations: Water

28

hw = π

Air

ha

103 W/ W/m D

0.030m ( 90-25 ) C 400 W/m

= π

COMMENTS:

×

×

×

D

= 4,570 W/m 2 ⋅ K

= 65 W/m

2 ⋅ K.

<

<

0.030m ( 90-25 ) C

Note that the air velocity is 10 times that of the water flow, yet

hw ≈ 70 × ha. These values for the convection coefficient are typical for forced convection heat transfer with liquids and gases. See Table 1.1.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation translation of this work beyond that permitted permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

PROBLEM 1.65

Conditions associated with surface cooling of plate glass which is initially at 600 °C. Maximum allowable temperature gradient in the glass.

KNOWN:

FIND:

Lowest allowable air temperature, T ∞

SCHEMATIC:

(1) Surface of glass exchanges radiation with large surroundings at T sur = T∞, (2) One-dimensional conduction conduction in the x-direction. ASSUMPTIONS:

ANALYSIS:

The maximum temperature temperature gradient will exist at the surface of the glass and at the instant that cooling cooling is initiated. From the surface energy balance, balance, Eq. 1.12, and the rate equations, Eqs. 1.1, 1.3a and 1.7, it follows that

-k

dT dx

)

(

− h ( Ts − T∞ ) − εσ Ts4 − Ts4ur = 0

or, with (dT/dx) max = -15°C/mm = -15,000 °C/m and Tsur = T∞,

−1.4

W ⎡

C⎤

D

W

⎢ −15, 000 ⎥ = 5 (873 − T∞ ) K 2 m⋅K ⎢ m⎥ m ⋅K ⎣ ⎦ 5 .67 × 10 10−8 +0.8 × 5.

W

⎡8734 − T 4 ⎤ K 4 . ∞ ⎥⎦ 2 4⎢ m ⋅K ⎣

T∞ may be obtained from a trial-and-error solution, from which it follows that, for T ∞ = 618K,

21,000

W 2

≈ 1275

m

W 2

m

+ 19 ,730

W 2

.

m

Hence the lowest allowable air temperature is

T∞ ≈ 618K = 345 C. D

COMMENTS:

<

(1) Initially, cooling is determined primarily by radiation effects.

(2) For fixed T∞, the surface temperature gradient would decrease with increasing time into the cooling process. Accordingly, T∞ could be decreasing with increasing time and still keep within the maximum allowable temperature gradient.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit not-for-profit basis for testing or instructional purposes only to students enrolled in Sections 107 or 108 of the 1976 1976 courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections United States Copyright Act without the permission of the copyright owner is unlawful.

Assignment 1: Additional Problem

(a)

For conduction-convection composite system,

⎛ L q ′′ = ⎜ + ⎝ k

1 ⎞

⎟

h ⎠

−1

(T

bottom

− T ∞ ) −1

⎛ 0.001 1 ⎞ = + ⎟ ⎜ h ⎠ (0.006) 2 ⎝ 15 2

( 45 − 20 )

⇒

h = 2609 W/m ⋅ K

Also, the heat flux is q ′′ = 2 / (0.006) 2 = 55 556 W/m

(b)

q ′′ = − k

55 556 =

(c)

dT d x

= k

2

2

T bottom − T top d

(15) ( 45 − T top )

⇒

0.001

T top = 41.3 °C

Maximum radiative heat flux is from blackbody w/o enclosing walls: −

′ = σ T top4 = (5.67 ×10 8 ) (41.3 + 273) = 553 W/m q ′rad which is only 1% of the total heat flux.

4

2

January 20, 2008


Due: January 28 (Monday) before 16:00, in assignment box outside CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯

Reminder:

ChE 314 Midterm is on Feb 13 (Wednesday) Time: 13:00 − 13:50 Location TBA

1. Problem 2.25 from text. •

Part (c) is reworded as follows: Determine the surface heat fluxes q x′′ (− L) and q x′′ (+ L) . Show that these values, together with your answer in part (b), satisfy the condition of energy balance.

•

In part (f ), (f ), the “rate of change of energy stored” is on a per unit volume basis (i.e. answer 3 should have units W/m ). [Note: You should be able to write down the answer to part (f ) (f ) by inspection.]

2. Problem 2.38 from text.

3. Additional problem (to be handed in) in)

A cylindrical material of length L = 1 m and radius R = 1 cm carries a steady flow of electric current. The material has a thermal conductivity of k = 15 W/m ⋅ K , and its electrical resistance creates a power dissipation rate of 100 W (the dissipation of energy occurs uniformly throughout the body). The cylinder is cooled by a stream of fluid at temperature T ∞ , with an effective heat transfer coefficient h = 200 W/m 2 ⋅ K . (a) (b)

Determine, at steady state, the location where the temperature is highest in the cylinder. How much is this temperature higher than T ∞ ? Suppose, instead of convective cooling, the surface temperature of the cylinder is maintained at T = T ∞ . What would be your answers now to part (a)?

Page 1 of 1

PROBLEM 2.25 KNOWN: Analytical expression for the steady-state temperature distribution of a plane wall experiencing uniform volumetric heat generation q while convection occurs at both of its surfaces. FIND: (a) Sketch the temperature distribution, T(x), and identify significant physical features, (b)

Determine q , (c) Determine the surface heat fluxes, q′′x ( − L ) and q′′x ( + L ) ; how are these fluxes related to the generation rate; (d) Calculate the convection coefficients at the surfaces x = L and x = +L, (e) Obtain an expression for the heat flux distribution, q′′x ( x ) ; explain significant features of the distribution; (f) If the source of heat generation is suddenly deactivated ( q = 0), what is the rate of change of energy stored at this instant; (g) Determine the temperature that the wall will reach eventually with q = 0; determine the energy that must be removed by the fluid per unit area of the wall to reach this state. SCHEMATIC:

ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform volumetric heat generation, (3) Constant

properties. ANALYSIS: (a) Using the analytical expression in the Workspace of IHT, the temperature

distribution appears as shown below. The significant features include (1) parabolic shape, (2) maximum does not occur at the mid-plane, T(-5.25 mm) = 83.3°C, (3) the gradient at the x = +L surface is greater than at x = -L. Find also that T(-L) = 78.2°C and T(+L) = 69.8°C for use in part (d). Temperature distribution 90

85 ) C ( ) x ( T , e r u t a r e p m e T

80

75

70 -20

-10

0

10

20

x-coordinate, x (mm)

(b) Substituting the temperature distribution expression into the appropriate form of the heat diffusion equation, Eq. 2.19, the rate of volumetric heat generation can be determined.

d ⎛ dT ⎞ q ⎜ ⎟+ = 0 dx ⎝ dx ⎠ k d dx

T ( x ) = a + bx + cx

where q

q

k

k

( 0 + b + 2cx ) + = ( 0 + 2c ) + = 0

2

Continued …..

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

PROBLEM 2.25 (Cont.)

)

(

q = −2ck = −2 −2 ×104°C / m 2 5 W / m ⋅ K = 2 ×105 W / m3

<

(c) The heat fluxes at the two boundaries can be determined using Fourier’s law and the temperature distribution expression.

q′′x ( x ) = − k

dT

T ( x ) = a + bx + cx 2

where

dx

q′′x ( −L ) = −k [0 + b + 2cx ]

x =− L

= − [b − 2cL ] k

q′′x ( − L ) = − ⎡− 210° C / m − 2 − 2 × 10 ° C / m

(

⎢⎣

4

2

) 0.020m⎤⎥⎦ × 5 W / m⋅ K = − 2950 W / m

2

q′′x ( + L ) = − ( b + 2cL ) k = +5050 W / m2

< <

From an overall energy balance on the wall as shown in the sketch below, E in − E out + E gen = 0, ?

 =0 +q′′x ( −L ) − q′′x ( +L ) + 2qL 5

or

− 2950 W / m 2 − 5050 W / m 2 + 8000 W / m 2 = 0

3

2

 = 2 × 2 × 10 W / m × 0.020 m = 8000 W / m , so the equality is satisfied where 2qL

′′ qconv,l

′′ qconv,r

(d) The convection coefficients, h l and hr, for the left- and right-hand boundaries (x = -L and x= +L, respectively), can be determined from the convection heat fluxes that are equal to the conduction fluxes at the boundaries. See the surface energy balances in the sketch above. See also part (a) result for T(-L) and T(+L).

q′′conv,A = q′′x ( −L ) 2 h l ⎡⎣ T∞ − T ( −L ) ⎤⎦ = h l [20 − 78.2 ] K = −2950 W / m

2 h l = 51W / m ⋅ K

<

q′′conv,r = q′′x ( + L ) h r ⎡⎣T ( + L ) − T∞ ⎤⎦ = h r [69.8 − 20 ] K = +5050 W / m 2

h r = 101W / m 2 ⋅ K

<

(e) The expression for the heat flux distribution can be obtained from Fourier’s law with the temperature distribution

q′′x ( x ) = − k

dT dx

= − k [ 0 + b + 2cx ]

q′′x ( x ) = −5 W / m ⋅ K ⎡ −210°C / m + 2 −2 ×10 4°C / m 2 ⎤ x = 1050 + 2 ×10 5x

⎢⎣

(

)⎥⎦

<

Continued …..


PROBLEM 2.25 (Cont.) The distribution is linear with the x-coordinate. The maximum temperature will occur at the location where q′′x ( x max ) = 0,

x max = −

1050W/m2 5

2 ×10 W / m

= −5.25 ×10−3 m = −5.25 mm

3

<

(f) If the source of the heat generation is suddenly deactivated so that q = 0, the appropriate form of the heat diffusion equation for the ensuing transient conduction is

k

∂ ⎛ ∂T ⎞ ∂T = ρ c p ⎜ ⎟ ∂x ⎝ ∂x ⎠ ∂t 2

At the instant this occurs, the temperature distribution is still T(x) = a + bx + cx . The right-hand term represents the rate of energy storage per unit volume,  ′′ = k E st

∂ [0 + b + 2cx ] = k [0 + 2c] = 5 W / m ⋅ K × 2 − 2× 104° C / m2 = − 2× 105 W / m3 ∂x

(

)

<

(g) With no heat generation, the wall will eventually (t → ∞) come to equilibrium with the fluid, T(x,∞) = T∞ = 20°C. To determine the energy that must be removed from the wall to reach this state, apply the conservation of energy requirement over an interval basis, Eq. 1.11b. The “initial” state is that corresponding to the steady-state temperature distribution, T i, and the “final” state has Tf = 20°C. We’ve used T∞ as the reference condition for the energy terms.

′′ = Ef′′ − Ei′′ E′′in − E′′out = ∆Est

with

′′ = 0. E in

+L E′′out = c p ( T − T∞ ) dx −L i

∫

E′′out = ρ c p

∫

+L −L

+L

⎡a + bx + cx 2 − T ⎤ dx = ρ c ⎡ ax + bx 2 / 2 + cx 3 / 3 − T x ⎤ p ⎢⎣ ∞ ⎥⎦ ∞ ⎥⎦ ⎢⎣ −L

E′′out = ρ c p ⎡ 2aL + 0 + 2cL / 3 − 2T∞ L ⎤ 3

⎢⎣

⎥⎦

E′′out = 2600 kg / m3 × 800 J / kg ⋅ K ⎡⎢2 ×82 °C × 0.020m + 2 −2 ×10 4°C/ m 2

(

⎣

3

( 0.020m )

)

/ 3 − 2 (20 °C ) 0.020m ⎤⎥

⎦

E′′out = 4.94 ×106 J / m 2

<

COMMENTS: (1) In part (a), note that the temperature gradient is larger at x = + L than at x

= - L. This is consistent with the results of part (c) in which the conduction heat fluxes are evaluated. Continued …..


PROBLEM 2.25 (Cont.)

(2) In evaluating the conduction heat fluxes, q′′x ( x ) , it is important to recognize that this flux is in the positive x-direction. See how this convention is used in formulating the energy balance in part (c). (3) It is good practice to represent energy balances with a schematic, clearly defining the system or surface, showing the CV or CS with dashed lines, and labeling the processes. Review again the features in the schematics for the energy balances of parts (c & d). (4) Re-writing the heat diffusion equation introduced in part (b) as

−

d ⎛ dT ⎞ − + q = 0 k ⎜ ⎟ dx ⎝ dx ⎠

recognize that the term in parenthesis is the heat flux. From the differential equation, note that if the differential of this term is a constant ( q / k ) , then the term must be a linear function of the x-coordinate. This agrees with the analysis of part (e).  , the rate of energy change stored in the wall at the instant the (5) In part (f), we evaluated E st

5 3  volumetric heat generation was deactivated. Did you notice that E st = − 2 ×10 W / m is the same value of the deactivated q ? How do you explain this?


ChE 314: Solutions to Assignment 2 Problem 2 (Prob 2.38 from text):

For steady state radial heat flow, the heat diffusion eqn in cylindrical coordinates is: 1 d ⎛ dT ⎞ q& ⎜⎜ r ⎟⎟ = − r d r ⎝ d r ⎠ k If we insist on a linear radial temperature distribution, we must have dT

=

d r

T 2 − T 1 r 2 − r 1

= C (some constant)

The heat diffusion eqn now becomes

−

q& k

1 d

=

C d r

(r C ) =

r d r

r d r

=

C r

⇒

q& = −

k C r

i.e. to have a linear temperature distribution, the rate of heat generation q& cannot be uniform in the cylinder; it must instead vary as 1 / r .

Problem 3 (additional problem):

L = 1m k

R = 1cm

R

k = 15 W/m⋅K L

q& =

100 2

100 π

2

(0.01) (1)

q& 1 d ⎛ dT ⎞ ⎜⎜ r ⎟⎟ = − r d r ⎝ d r ⎠ k

Steady state:

dT

= −

r

Integrate again:

T = −

d r

q&

4k T (0) is finite

⇒

= 3.1831 × 10 5 W / m 3

⇒

2

Integrate and get

BC1 (at r = 0 ):

2

=

R L

π

h = 200 W/m ⋅K

q& r

+ c1

k 2

⇒

q& d ⎛ dT ⎞ ⎜⎜ r ⎟⎟ = − r k d r ⎝ d r ⎠

dT d r

= −

q& r k 2

+

c1 r

r 2 + c 1 ln r + c 2

⇒

T = −

c1 = 0

q&

4k

r 2 + c 2

(i)

dT

also (a)

d r

= −

q&

2k

r

From (i), it is clear that the highest temperature is at the centre line ( r = 0 ), and is equal to c 2 . This makes physical sense since the centre line is furthest away from convective cooling. Now need to find c 2 .

′ = q ′conv ′ q ′cond

BC2 (at r = R ):

⎛ dT ⎞ − k ⎜⎜ ⎟⎟ = h [T ( R) − T ∞ ] ⎝ d r ⎠ r = R ⎡ q& R 2 ⎤ ⎛ q& R ⎞ ⎟⎟ = h ⎢ − − k ⎜⎜ − + c 2 − T ∞ ⎥ ⎝ 2k ⎠ ⎣ 4k ⎦ Rearrange and get c 2 − T ∞ =

q& R ⎛ R h ⎞ ⎜⎜ 1 + ⎟⎟ h k 2 ⎝ 2 ⎠

Put in numbers: (3.1831 × 10 5 ) (0.01) ⎛ (0.01) (200) ⎞ ⎜⎜ 1 + ⎟⎟ = 8.49°C c 2 − T ∞ = (2) (200) ( 2 ) ( 15 ) ⎝ ⎠ (b)

New BC at r = R : From (i),

⇒

T ∞ = −

q&

R 2 + c 2 4k

c 2 − T ∞ =

q& R

2

4k

(3.1831 × 10 ) (0.01) 5

=

(4) (15)

2

= 0.53°C

January 25, 2008


Due: February 4 (Monday) before 16:00, in assignment box next to CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

1. Thermal resistance of cylindrical and spherical shells

(a) A hollow cylinder of length L has inner and outer radii r 1 and r 2 ( r 1

<

r 2 ), and the

corresponding surface temperatures are held constant at T ( r 1 ) = T H and T ( r 2 ) = T L . Show that, at steady state, the rate of radial heat flow is q r

=

( T H

−

T L )

⋅

2π k L ln (r 2 / r 1 )

,

where k is the thermal conductivity. Assume no internal heat generation. (b) A hollow sphere has inner and outer radii r 1 and r 2 ( r 1

<

r 2 ), and the corresponding

surface temperatures are held constant at T ( r 1 ) = T H and T ( r 2 ) = T L . Show that, at steady state, the rate of radial heat flow is q r

=

( T H

−

T L )

⋅

4π k 1 / r 1 − 1 / r 2

,

where k is the thermal conductivity. Assume no internal heat generation. Note: To calculate heat flux, you need only one of the two integration constants.

2. Problem 3.9 from text.

3. Problem 3.91 from text — part (a) only. •

Assume core region ( 0 < r < r 1 ) to be hollow

4. Problem 3.96 from text — part (a) only. •

Ans: 5.09°C, 5.07°C (book’s answers are wrong)

Page 1 of 1

ChE 314: Solutions to Assignment 3 Problem 1: Steady state and q& = 0

⇒ solve ∇ 2T = 0 subject to specified BCs.

(a) Hollow cylinder:

∇ 2T =

1 d ⎛ dT ⎞ ⎜ r ⎟ = 0 r d r ⎜⎝ d r ⎠⎟

⇒

r

dT

= c1

d r

dT

=

d r

c1

(∗)

r

T = c 1 ln r + c 2

Note that the c 1 ln r term survives b/c we don’t have r = 0 (cylinder is hollow). BCs at r = r 1 and r = r 2 : T H = c 1 ln r 1 + c 2 T L = c 1 ln r 2 + c 2

Subtract to eliminate c 2 :

⇒

T H − T L = c 1 ln r 1 / r 2

T H − T L

c1 =

ln ( r 1 / r 2 )

Put back into (∗) to calculate flux: q ′r ′ = − k

dT d r

= −

q r = q r ′′ ⋅ 2π r L =

k T H − T L

(

r ln r 1 / r 2

)

2π k L ln (r 2 / r 1 )

=

k T H − T L

(

r ln r 2 / r 1

(T H − T L )

Page 1 of 6

)

(b) Hollow spherical shell:

∇ 2T =

1 d ⎛ 2 dT ⎞ ⎜⎜ r ⎟⎟ = 0 2 d d r r r ⎝ ⎠

⇒

2

r

dT d r

dT

c1

=

d r

= c1 (∗)

2

r

T = −

c1 r

+ c2

BCs at r = r 1 and r = r 2 : T H = − c 1 / r 1 + c 2 T L = − c 1 / r 2 + c 2

Subtract to eliminate c 2 :

⎛ 1 1 ⎞ − ⎟ ⎜ r r ⎟ 2 ⎠ ⎝ 1

T H − T L = − c 1 ⎜

⇒

c1 = −

Put back into (∗) to calculate flux: q ′r ′ = − k

dT d r

=

k

T H − T L

r 2 1 / r 1 − 1 / r 2

q r = q ′r ′ ⋅ 4π r 2 =

4π k

(1 / r − 1 / r ) 1

(T H − T L )

2

Page 2 of 6

T H − T L

1 / r 1 − 1 / r 2

Page of 6 Excerpts from this work may be reproduced by instructors for distribution on a3not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Problem 3

(Prob 3.91 from text) T 1 T 2

2 r 2 2 r 1

2 r 3

h , T

∞

T 3

graphite thorium, q&

(

)

(

q = q& π r 2 − r 1 L = (10 ) π 11 − 8 2

2

8

Graphite:

R1 =

Convective cooling:

R 2 =

2

2

ln (r 3 / r 2 ) 2π k L 1 h ⋅ 2π r 3 L

)(10 − ) L 6

= =

= 17 907 L ( L in m; q in W)

ln (14 / 11)

=

2π (3) L

0.012 794 L

1 ( 2000) 2π (0.014) L

T 3

T 2

0.005 6841 L

T ∞

R 1

(

=

R2

)

T 2 − T ∞ = R 1 + R 2 q

⎛ 0.012 794 0.005 6841 ⎞ + ⎟ (17 907 L ) L L ⎝ ⎠

T 2 − 600 = ⎜

⇒

T 2 = 930.9 K

For T 1 , go back to heat diffusion equation: q& 1 d ⎛ dT ⎞ ⎜⎜ r ⎟⎟ = − r d r ⎝ d r ⎠ k

Integrate twice and get dT d r

= −

T = −

BC 1 (at r = r 1 ): From (1),

q&

2k

q&

4k

r +

c1

(1)

r

2 r + c1 ln r + c 2

(2)

dT / d r = 0 (zero heat flux)

−

q&

2k

r 1 +

c1 r 1

= 0

⇒

c1 =

q& r 1 2

2k

Page 4 of 6

Combining with (2): q&

T = −

4k

BC 2 (at r = r 2 ):

⇒

q& r 1 2

r + 2

2k

ln r + c 2

(3)

T ( r 2 ) = T 2

T 2 = −

q&

4k

r 2 + 2

q& r 1 2

2k

ln r 2 + c 2

(4)

Eliminate c2 by combining (3) and (4): T (r ) − T 2 = −

q&

(r

2

4k

− r 2

2

)+

q& r 1

2

2k

⎛ r ⎞ ⎟⎟ r ⎝ 2 ⎠

ln ⎜⎜

(5)

Using (5): T 1 − 930.9 = −

⇒

10

8

(4) (57)

(8

2

− 11 )(10 2

−6

) + (10 )(8

T 1 = 938.0 K

Page 5 of 6

8

2

× 10 −6 )

(2)(57)

⎛ 8 ⎞ ⎟ ⎝ 11 ⎠

ln ⎜

Problem 4

(Prob 3.96 a from text)

Approximate apple as sphere: r o

T ∞ = 5°C 2 h = 7.5 W / m ⋅K

3

= 840 kg / m k = 0.5 W / m⋅K r o = 4 cm

ρ

k

⎛ ⎞ J ⎞ ⎛ kg ⎞ ⎛ 1 day ⎟⎟ ⎜ 840 3 ⎟ ⎜⎜ ⎟⎟ = 38.89 W / m 3 q& = ⎜⎜ 4000 kg ⋅ day ⎠ ⎝ m ⎠ ⎝ (24) (3600) s ⎠ ⎝ Start with 1 d ⎛ 2 dT ⎞ q& ⎜ ⎟ = − r k r 2 d r ⎜⎝ d r ⎠⎟ Integrate twice and get T = −

q&

6k

r − 2

c1 r

+ c2

BC1 (at r = 0 ): T (0) finite

⇒

c1 = 0

⇒

T = −

q&

6k

(∗)

r 2 + c 2

BC2 (at r = r o ):

′ = q ′conv ′ q ′cond ⎛ dT ⎞ ⎟⎟ − k ⎜⎜ = h [T ( r o ) − T ∞ ] d r ⎝ ⎠ r = r o q& r o

3

⎡ q& 2 ⎤ = h ⎢− r o + c 2 − T ∞ ⎥ ⎣ 6k ⎦

(38.89) (0.04) 3 From (∗),

⎡ (38.89) (0.04) 2 ⎤ = (7.5) ⎢ − + c 2 − 5⎥ (6) (0.5) ⎣ ⎦

⇒

c 2 = 5.09°C

T (0) = c 2 = 5.09°C

T (r o ) = −

q&

6k

r o

2

+ c2 = −

(38.89) (0.04) (6) (0.5)

Page 6 of 6

2

+ 5.09 = 5.07°C

February 2, 2008


Due: February 11 (Monday) before 16:00, in assignment box next to CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Midterm Reminder Date: Feb 13, 2008 (Wednesday) Time: 13:00 − 13:50 Location: A − Liang

Low − Z

ETLC 2-001 ETLC 2-002

1. Problem 3.108 in text. •

Part (a): Make qualitative sketch only; do not include actual numbers.

•

Part (b): Omit.

•

Part (c):

Calculate

T 2

assuming

R ′t′, c → ∞

. [Note:

R ′t′, c

was finite in Part (a).]

Omit plot of temperature distribution.


Features of straight fins are summarized in Figure 3.18 (p. 150) and Table 3.5 (p. 152).

•

Error in Figure 3.18: Profile area for rectangular fin should be

•

Compare only rectangular and triangular fins; disregard parabolic profile.

•

Volume of a straight fin is in general

A p = L t

(not L c t ).

V = w A p .

3. Problem 3.129 in text.


See Figs. 3.14 (c) and 3.19 for sketches of annular fins of rectangular profiles.

•

For thermal conductivity of Al, assume an average temperature of 400 K.

Note: Values read from efficiency charts may vary by a few %.

Page 1 of 1

ChE 314: Solutions to Assignment 4 Problem 1 [Based on Prob. 3.108 in text]

(a) T 1 T 3

•

Linear profile for − L < x < 0

•

dT / d x (i.e. heat flux) continuous at x = 0

•

T discontinuous at x = L (due to contact resistance)

T

∞

– L

L

(c) T 1

T 2

T ∞

R ins

Insulation: R ins =

L

0.050

=

k A

(100) ⋅π (0.005) 2 / 4

(

q f = h k P A c

Fin (with insulated tip): R f ≡

T 2 − T ∞

=

qf

R f

)

1 / 2

= 25.4648 K/W

tanh mL ⋅ (T 2 − T ∞ )

1

(hk P A )

1 / 2

c

(hk P A )

1 / 2

c

⎛ h P ⎞ ⎟ m= ⎜ ⎜ k A c ⎟ ⎝ ⎠

tanh mL

= [(500) (100) ⋅π (0.005) ⋅π (0.005) 2 / 4 ]

1 / 2

1 / 2

⎡ (500) ⋅ π (0.005) ⎤ = ⎢ ⎥ 2 ⎣ (100) ⋅ π (0.005) / 4 ⎦

= 0.12418 W/K

1 / 2

= 63.2456 m–1

tanh m L = tanh (63.2456 ) (0.050) = 0.99642

⇒

q f =

R f =

1 (0.12418) (0.99642)

T 1 − T ∞ Rins + R f

=

T 2 − T ∞ R f

⇒

= 8.0817 K/W 200 − 20 25.4648 + 8.0817 Page 1 of 5

=

T 2 − 20

8.0817

⇒

T 2 = 63.4 °C

Problem 2 [Prob. 3.123 in text]

Rectangular fin: L c = L + t / 2 = 0.015 + 0.003 / 2 = 0.0165 m

A p = L t = (0.015) (0.003) = 0.45 × 10

⎛ h ⎞ 3 / 2 ⎟ L c ⎜ ⎜ k A ⎟ ⎝ p ⎠

1 / 2

= (0.0165)

⇒

3 / 2

η f ≈ 0.97

−4

2

m

−4

⇒

⎛ ⎞ 50 ⎜⎜ ⎟ −4 ⎟ ( 185 ) ( 0 . 45 10 ) × ⎝ ⎠

V = w A p = 0.45 × 10

3

m

1 / 2

= 0.164

(from Chart) 2

A f = 2 w L c = 2 (1) (0.0165) = 0.033 m

q f = η f ⋅ h A f θ b = (0.97) ⋅ (50) (0.033) (100 − 20) =

128 W

Triangular fin: L c = L = 0.015 m

A p = L t / 2 = (0.015) (0.003) / 2 = 0.225 × 10 −4 m2

⎛ h ⎞ 3 / 2 ⎟ L c ⎜ ⎜ k A p ⎟ ⎝ ⎠

⇒

[

1 / 2

= (0.015) η f ≈ 0.98

Af = 2 w L 2 + (t / 2) 2

]

1 / 2

3 / 2

⎛ ⎞ 50 ⎜⎜ ⎟ −4 ⎟ × ( 185 ) ( 0 . 225 10 ) ⎝ ⎠

⇒

−4

V = w A p = 0.225 × 10

1 / 2

= 0.20

(from Chart)

= 2 (1) [(0.015) 2 + (0.003 / 2) 2 ]

1 / 2

= 0.03015 m2

q f = η f ⋅ h A f θ b = (0.98) ⋅ (50) (0.03015) (100 − 20) = 118 W

Note: Triangular fin uses half the material, but gives approximately the same rate of heat loss.

Page 2 of 5

3

m


k A = 200 W / m⋅K

A T ∞ = 25°C

T b = 100°C

B x 1

Long fins ⇒ case D in Table 3.4

⎛ θ ⎞ ⎜ ⎟ = exp (− mA x 1 ) ⎜ θ b ⎟ ⎝ ⎠ A

⇒

ln ⎜⎜

⎛ θ ⎞ ⎜ ⎟ = exp ( − mB x 1 ) ⎜ θ b ⎟ ⎝ ⎠ B

⇒

ln ⎜⎜

⎛ T A − T ∞ ⎞ ⎟⎟ = − mA x 1 − T T ∞ ⎠ ⎝ b

⎛ T B − T ∞ ⎞ ⎟⎟ = − mB x 1 − T T ∞ ⎠ ⎝ b

Divide the two equations: mA mB

⎛ 75 − 25 ⎞ ⎛ 60 − 25 ⎞ = ln ⎜ ⎟ ln ⎜ ⎟ = 0.532 100 25 100 25 − − ⎝ ⎠ ⎝ ⎠

⎛ h P ⎞ ⎟⎟ m ≡ ⎜⎜ k A ⎝ c ⎠

1 / 2

∴ (k B / 200)

⇒

1 / 2

= 0.532

mA mB

⎛ k ⎞ = ⎜⎜ B ⎟⎟ ⎝ k A ⎠

⇒

1 / 2

k B = 56.6 W / m ⋅ K

Page 3 of 5

Problem 4 (Prob 3.145 in text)

For single fin: CL

T b = 200°C

4 mm

T ∞ = 20°C 2 h = 40 W / m ⋅K

k = 240 W / m⋅K

15 mm 50 mm Referring to Fig 3.19 (p. 150): L = 15 mm ;

r 1 = 25 mm ;

r 2 = r 1 + L = 40 mm

r 2 c = r 2 + t / 2 = 42 mm L c = L + t / 2 = 17 mm

A p = L c t = 68 mm 2

(a)

L c

3 / 2

(h / k A ) p

1 / 2

= 0.11

r 2c / r 1 = 1.68

η f ≈ 0.97

(from Fig 3.19; value may vary by a few %)

Heat loss per fin: Recall η f ≡

q f h Af θ b

, where A f is the exposed fin area.

Here, Af = 2π r 2 − r 1 2

2

) + 2π r t = 2π [(40 2

2

− 25 2 ) + (40)(4) ] = 7131 mm2

∴ q f = η f ⋅ h A f θ b = (0.97)(40)(0.007131)(200 − 20) = 49.8 W ε f ≡

⇒

q f h A c θ b

ε f =

2

where A c = 2π r 1 t = 2π ( 25) ( 4) = 628.32 mm (“footprint” size) 49.8 ( 40) (0.00062832) ( 200 − 20)

=

11.0

Page 4 of 5

(b) For tube length of L tube = 1 m , unfinned area A b is A b = 2π r 1 L tube − N Ac = 2π (0.025) (1) − (125) (0.00062832) = 0.07854 m 2 A tot = A b + N A f = 0.07854 + (125)(0.007131) = 0.9699 m 2

Eqn 3.101:

⎡

(

q tot = h θ b A tot ⎢1 − 1 − η f

⎢⎣

⇒

N A f ⎤

)

A tot

(125)(0.007131) ⎤ ⎡ ⎥ = (40)(200 − 20)(0.9699) ⎢1 − (1 − 0.97) ⎥ 0.9699 ⎣ ⎦ ⎥⎦

q tot = 6790 W = 6.79 kW

Note: while covering 50% of the total surface area, the fins account for > 90% of the total heat loss!

Page 5 of 5

February 15, 2008


Due: March 3 (Monday) before 16:00, in assignment box next to CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

A rectangular fin, with dimensions L

= 12

cm , t = 2 cm , and w >> t , extends from a base

surface which is maintained at temperature T b

= 100 °C .

The fin is made of aluminum, which

has thermal conductivity k = 240 W / m ⋅ K . It is cooled by a stream of air at 20°C; the heat 2 transfer coefficient at the fin surface is h = 240 W / m ⋅ K .

h, T

T b

∞

y k

L

t

w

x

z

We are to determine the steady state temperature distribution T ( x, y ) far away from the side edges (i.e., far from z = 0 and z = w ); this is in effect a 2-D problem. Rather than solving the problem analytically, we will use here finite difference method to calculate temperatures at discrete nodal positions. To begin, a constant- z cross section (parallel to the x- y plane) is discretized into 1 cm × 1 cm squares. Set up the FDEs, as discussed in class, and determine the nodal temperatures. Once you have these temperatures, you should also calculate T ( x) , which is the average temperature across the fin thickness (from y = 0 to y = t ) for given values of x . Your submitted work should include: (a)

A brief description of your approach, as well as a listing of the computer source code.

(b)

A plot, in discrete symbols ( not joined by lines), of T (in °C) vs x (in cm).

(c)

On the same graph, a solid line representing the analytical 1-D solution (see Table 3.4).

Consider also a “fin” made of fireclay brick, with thermal conductivity k = 1. 2 W / m⋅K . All other parameters (geometric dimensions, h , T b , and T ∞ ) remain the same as above. (d)

Repeat parts (b) and (c) for the “clay fin.”

(e)

At x = 1 cm , plot T (in °C) vs y (in cm) for both the aluminum fin and the “clay fin.” Comment on the difference between the two curves and speculate reason. Page 1 of 1

ChE 314: Solutions to Assignment 5 (a)

Approach: See class notes

(b)

See below (black symbols)

(c)

Convection at tip ⇒ Case A in Table 3.4

(

m = h P / k A c

)

1 / 2

, where P / A c = 2(t + w) / t w ≈ 2 w / t w = 2 / t

⎛ h 2 ⎞ m = ⎜ ⋅ ⎟ ⎝ k t ⎠

⇒

h m k

=

1 / 2

⎡ (240) (2) ⎤ = ⎢ ⎥ ⎣ (240) (0.02) ⎦

240 (10) ( 240)

= 0.1 ;

1 / 2

= 10 m−1

m L = (10) (0.12) = 1.2

From Table 3.4, T − 20

100 − 20

=

cosh [10 (0.12 − x )] + 0.1 sinh [10 (0.12 − x)] cosh 1.2 + 0.1 sinh 1.2

T = 20 + 40.7830 [ cosh(1.2 − 10 x) + (0.1) sinh (1.2 − 10 x) ]

; x in m , T in °C

Plotted as solid line below.

Aluminum Fin 100

FDE result Analytical 1-D solution

90 ) C g e d (

80

e v a

T

70

60

0

2

4

6 x (cm)

Page 1 of 3

8

10

12

(d)

Clay “fin”: k = 1.2 W/m ⋅K 2

⇒

−1

h ∆ / k = 2 ;

m = 141.42 m

;

h / m k = 1.4142

Clay Fin 100

FDE result Analytical 1-D solution

80 ) C g e d (

60

T

40

e v a

20

0 0

2

4

6 x (cm)

Page 2 of 3

8

10

12

(e)

Temperature across fin at x = 1 cm (based on FDE result):

2.0

aluminum clay

1.5

) m c 1.0 ( y

0.5

0.0 20

40

60 T (deg

80

100

C)

Consider Biot number in the y direction: Bi = (240)(0.02)/240 = 0.02 for aluminum Bi = (240)(0.02)/1.2 = 4.0

for clay

⇒ expect negligible T variations ⇒ expect significant T variations

Page 3 of 3

March 3, 2008



1. Two identical aluminum cylinders of radius R and length 2 R are left overnight in an oven. The cylinders are removed from the oven the next day and placed on an insulating surface, with cylinder 1 standing on its end and cylinder 2 lying on its side. The air flow in the room 2 gives rise to an effective heat transfer coefficient of h = 50 W / m K. The thermal properties of aluminum are assumed to be those at 300 K. ⋅

(a)

You are told that the lumped capacitance method is applicable in this situation. What can you say about the dimension R ?

(b)

Which cylinder will first reach room temperature? What is the ratio of the two equilibration times?


Note: one of the online answers is wrong,


Note that the initial temperature is not given (and not needed).

•

Omit Part (b).


5. (a) (b)

Problem 5.69 in text. In Problem 5.69, what is the significance of an “extremely large” heat transfer coefficient h ?

Page 1 of 1

ChE 314: Solutions to Assignment 6 Problem 1

(a)

Aluminum at 300K: ρ = 2702 kg / m 3 ; c = 903 J / kg⋅K ; k = 237 W / m⋅K Also, h = 50 W / m 2 ⋅ K If lumped capacitance method were applicable, then must have Bi = h ( 2 R ) / k < 0.1

Note: Here, chose 2 R as characteristic length L c . If instead one chooses R as L c , it would be correct also (i.e. there is no sharp dividing line between “large” and “small” Biot numbers). Based on above expression, one gets (0.1) ( 237) 0.1 k m R < < 2 h ( 2) (50)

⇒ For LC method to work, should have R < 24 cm (b)

Obviously, cylinder 2 will reach room temperature first b/c it has more exposed area. Mathematically, τ =

ρ c V h A

⇒ larger A will give rise to shorter time constant τ

With all else equal, τ 1 / τ 2 = A 2 / A1 , where A 2 = 2 π R

2

+ ( 2π R )( 2 R ) = 6π R 2

A1 = π R + ( 2π R )( 2 R ) = 5π R 2

⇒

2

τ 1 / τ 2 = 6 / 5 = 1.2

Note: Cooling time does not depend on oven temperature or room temperarture; this is characteristic of exponential decay.

Page 1 of 4


Treat problem as having 2 coatings:

2 mm

Bi =

h L

=

k

( 200)(0.002) 0.25

= 1 .6

ζ 1 = 0.9903

⇒

C 1 = 1.1547

(by interpolation)

Eqn 5.40a: 42 − 25 200 − 25

= 1.1547 exp (− 0.9903 2 Fo ) cos (0.9903)

(

0.1534 = exp − 0.9903 Fo Fo ≡ t α / L

2

2

⇒

)

⇒

1.9116 =

Fo = 1.9116

(1.2 × 10 − 7 ) t −3 2

(2 × 10 )

⇒

t = 63.7 s

⇒

T o = 56.0°C

Temperature at interface ( x ∗ = 0 ) : T o − 25

200 − 25

= 1.1547 exp [− (0.99032 ) (1.9116) ]

-------------------------------------------------------------------------------------------------------Problem 3 [Prob. 5.48 in text]

Bi =

h r o k

T s − T ∞ T i − T ∞ T o − T ∞ T i − T ∞

=

(1000)(0.030) 50

= 0.6

ζ 1 = 1.0184

⇒

C 1 = 1.1345

= C 1 exp (−ζ 1 2 Fo ) J 0 (ζ 1 )

T s − T ∞ T o − T ∞

= C 1 exp (−ζ 1 2 Fo )

J 0 (1.0184 ) = 0.7568 =

550 − 750 T o − 750

⇒

Page 2 of 4

= J 0 (ζ 1 )

T o = 486 K

Problem 4 [Prob. 5.62 in text] r o = 0.020 mm ; T i = 400 °C ;

Bi = α =

h r o

=

k k

ρ c

=

h = 300 W / m ⋅ K ;

k = 15 W / m ⋅ K

2

T ∞ = 25 °C

(300)(0.020) 15 15 (3000) (850)

= 0.4

ζ 1 = 1.0528

⇒

C 1 = 1.1164

= 5.8823 × 10 − 6 m2 / s

(a) Eqn 5.52: sin ζ 1 − ζ 1 cos ζ 1

ζ 1 3

= 0.2978

0.80 = 1 − 3θ o∗ (0.2978)

θ o∗ = 0.22386

⇒

Eqn 5.50c:

(

0.22386 = 1.1164 exp −1.0528 2 ⋅ Fo Fo = 1.4497 =

(

t ⋅ 5.8823 × 10

(20 × 10

−3

)

−6

)

2

) ⇒

t = 98.6 s

(b) Q

uniform temperature distribution in both cases T i

T

before

after

where Q = 0.8 Qo = 0.8 ρ c V (T i − T ∞ ) Energy balance:

Q = ρ c V (T i − T )

Combining with above relation, we get 0.8 (T i − T ∞ ) = T i − T

0.8 ( 400 − 25) = 400 − T

Page 3 of 4

⇒

T = 100 °C


(a) steel slab T s = 25°C x

Steel:

T i = 300°C

ρ = 7800 kg / m 3 ; c = 480 J / kg⋅K ; k = 50 W / m⋅K α =

k

ρ c

=

50 (7800) ( 480)

= 1.335 × 10 −5 m 2 / s

Set T = 50°C in eqn 5.57:

⎛ x ⎞ ⎟ = 50 − 25 = 0.0909 ⎜ 2 α t ⎟ 300 − 25 ⎝ ⎠

erf ⎜

⇒

x

2 α t

= 0.08073

x = 25 mm :

0.025 −5

2 (1.335 × 10 ) t

(b)

= 0.08073

⇒

t = 1796 s =

29.9 min

As h → ∞ , the surface temperature will approach that of the free stream, with T (0 , t ) = T ∞ = constant

i.e. the boundary condition is Case 1 in Table 2.2 (p. 78). For moderate (i.e. finite) h , the boundary condition is like before, i.e. back to Case 3 in Table 2.2.

Page 4 of 4

March 9, 2008






Note:

h x

denotes

h ( x) ,

i.e. the local heat transfer coefficient at

(Yet another unnecessary confusion in notation.)


Page 1 of 1

x.

PROBLEM 6.6 KNOWN:

Expression for the local heat transfer coefficient of a circular, hot gas jet at T∞

directed normal to a circular plate at Ts of radius ro. FIND:

Heat transfer rate to the plate by convection.

SCHEMATIC:

ASSUMPTIONS:

(1) Steady-state conditions, (2) Flow is axisymmetric about the plate, (3) For h(r), a and b are constants and n ≠ -2. ANALYSIS:

The convective heat transfer rate to the plate follows from Newton’s law of

cooling q conv =

∫A dqconv = ∫A h ( r ) ⋅ dA ⋅ ( T∞ − Ts ) .

The local heat transfer coefficient is known to have the form,

h (r ) = a + br n and the differential area on the plate surface is

dA = 2π r dr. Hence, the heat rate is q conv =

ro

∫0

( a + brn ) ⋅ 2

π

r dr ⋅ ( T∞ − Ts ) r

b n+2 ⎤ o ⎡a q conv = 2π ( T∞ − Ts ) ⎢ r 2 + r ⎥⎦ 2 n 2 + ⎣ 0 b n+2 ⎤ ⎡a q conv = 2π ⎢ ro2 + ro ⎥ n+2 ⎣2 ⎦

( T∞ − Ts ) .

<

Note the importance of the requirement, n ≠ -2. Typically, the radius of the jet is much smaller than that of the plate. COMMENTS:


PROBLEM 6.17 KNOWN: Pressure dependence of the dynamic viscosity, thermal conductivity and specific heat. FIND: (a) Variation of the kinematic viscosity and thermal diffusivity with pressure for an

incompressible liquid and an ideal gas, (b) Value of the thermal diffusivity of air at 350 K for pressures of 1, 5 and 10 atm, (c) Location where transition occurs for air flow over a flat plate with T∞ = 350 K, p = 1, 5 and 10 atm, and u∞ = 2 m/s. ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Transition at Re x,c = 5 ×

105, (4) Ideal gas behavior. -7

2

PROPERTIES: Table A.4, air (350 K): µ = 208.2 × 10 N⋅s/m , k = 0.030 W/m⋅K, cp = 1009

J/kg⋅K, ρ = 0.995 kg/m3. ANALYSIS:

(a) For an ideal gas p = ρRT or ρ = p/RT

(1)

while for an incompressible liquid,

ρ

= constant

The kinematic viscosity is ν = µ/ ρ

(2) (3)

Therefore, for an ideal gas ν

= µRT/p or

ν ∝

p-1

(4)

<

and for an incompressible liquid ν

<

= µ/ ρ or ν is independent of pressure.

The thermal diffusivity is α = k / ρc

Therefore, for an ideal gas, α

= kRT/pc or

α∝

For an incompressible liquid

p -1 α

(6)

= k/ρc or α is independent of pressure

< <

(b) For T = 350 K, p = 1 atm, the thermal diffusivity of air is

α

=

0.030 W/m ⋅ K 3

0.995 kg/m × 1009 J/kg ⋅ K

= 29.9 × 10-6 m2 /s

<

Using Equation 6, at p = 5 atm, Continued…


PROBLEM 6.17 ( Cont.)

α

= 29.9 × 10-6 m 2 /s/5 = 5.98 × 10-6 m2 /s

<

At p = 10 atm, α

= 29.9 × 10-6 m 2 /s/10 = 2.99 × 10-6 m2 /s

<

(c) For transition over a flat plate, x u Rex,c = c ∞ = 5 × 105 ν

Therefore x c = 5 × 105 ( ν /u ∞ )

For T∞ = 350 K, p = 1 atm, ν

= µ/ ρ = 208.2 × 10-7 N ⋅ s/m2 0.995 kg/m3 = 20.92 × 10-6 m2 /s

Using Equation 4, at p = 5 atm ν

= 20.92 × 10-6 m2 /s 5 = 4.18 × 10-6 m2 /s

At p = 10 atm, ν

= 20.92 × 10-6 m2 /s 10 = 2.09 × 10-6 m2 /s

Therefore, at p = 1 atm x c = 5 × 105 × 20.92 × 10-6 m2 /s/(2m/s) = 5.23 m

<

At p = 5 atm, x c = 5 × 105 × 4.18 × 10-6 m2 /s/(2m/s) = 1.05 m

<

At p = 10 atm x c = 5 × 105 × 2.09 × 10-6 m2 /s/(2m/s) = 0.523 m

<

COMMENT: Note the strong dependence of the transition length upon the pressure for the gas

(the transition length is independent of pressure for the incompressible liquid).


PROBLEM 6.21 KNOWN: FIND:

Local Nusselt number correlation for flow over a roughened surface.

Ratio of average heat transfer coefficient to local coefficient.

SCHEMATIC:

ANALYSIS:

The local convection coefficient is obtained from the prescribed correlation, k

k

1/3 = 0.04 Re0.9 x Pr x x 0.9 0.9 ⎡V⎤ 1/3 x ≡ C1x-0.1. h x = 0.04 k ⎢ ⎥ Pr x ⎣ ν ⎦

h x = Nu x

To determine the average heat transfer coefficient for the length zero to x, hx ≡ hx =

1 x

1

x 0 C1 x 0.9

x

∫ h x dx =

x 0.9

x

C1 ∫ x-0.1dx 0

= 1.11 C1 x-0.1.

Hence, the ratio of the average to local coefficient is hx hx

=

1.11 C1 x-0.1

COMMENTS:

Nu x ≠

C1 x -0.1

= 1.11.

<

Note that Nu x / Nux is also equal to 1.11. Note, however, that

1 x

∫ Nu x dx.

x 0


PROBLEM 6.32 KNOWN:

Ambient, interior and dewpoint temperatures. Vehicle speed and dimensions of windshield. Heat transfer correlation for external flow. FIND:

Minimum value of convection coefficient needed to prevent condensation on interior surface of windshield. SCHEMATIC:

ASSUMPTIONS:

(1) Steady-state, (2) One-dimensional heat transfer, (3) Constant properties.

PROPERTIES: Table A-3,

10

-6

2

glass: k g = 1.4 W/m⋅K. Prescribed, air: k = 0.023 W/m⋅K, ν = 12.5 ×

m /s, Pr = 0.70.

ANALYSIS:

From the prescribed thermal circuit, conservation of energy yields

T∞,i − Ts,i Ts,i − T∞,o = 1/ hi t / k g + 1/ h o where h o may be obtained from the correlation

h L 0.8 1/ 3 Nu L = o = 0.030 Re L Pr k -6 2 With V = (70 mph × 1585 m/mile)/3600 s/h = 30.8 m/s, ReD = (30.8 m/s × 0.800 m)/12.5 × 10 m /s 6 = 1.97 × 10 and ho =

0.023 W / m ⋅ K 0.800 m

(

0.030 1.97 × 106

0.8

)

( 0.70)1/ 3 = 83.1W / m2 ⋅ K

From the energy balance, with Ts,i = Tdp = 10°C

− Ts,i − T∞,o ) ⎛ t ( 1 ⎞ hi = + ⎜ ⎟ ( T∞,i − Ts,i ) ⎜⎝ kg ho ⎟⎠

1

⎞ (10 + 15 ) °C ⎛ 0.006 m 1 hi = ⎜ + ⎟ ( 50 − 10 ) °C ⎜⎝ 1.4 W / m ⋅ K 83.1W / m2 ⋅ K ⎟⎠

−1

hi = 38.3 W / m 2 ⋅ K

<

COMMENTS:

The output of the fan in the automobile’s heater/defroster system must maintain a velocity for flow over the inner surface that is large enough to provide the foregoing value of hi . In addition, the output of the heater must be sufficient to maintain the prescribed value of T ∞,i at this velocity.


March 18, 2008


Due: March 28 (Friday) before 16:00, in assignment box next to CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

1. Go back the first problem in Assignment 1 (Problem 1.15 in text) and re-calculate the heat transfer coefficients without considering the given wattages (i.e. do not use the data 28 kW/m and 400 W/m in the problem). Use the Churchill-Bernstein correlation (eqn 7.46).

2. (Based on Problem 7.13 in text.)

Consider the parallel flow of a fluid over a flat plate of dimensions L

×

2 L ; the surface

temperature of the plate is maintained constant at T s . As shown below, the plate can be in one of two orientations relative to the flow: u∞ , T ∞

Case A

u∞ , T ∞

Case B

(a) Assuming the boundary layers remain laminar over the entire plate in both cases, is it possible to have q A = q B ? ( q is the total rate of heat transfer from the plate.) Explain.

If we ignored the transition region (see Figure 6.4) and assume Re x , c r = 5 × 10 5 , the critical position x c r at which the boundary layer becomes turbulent is given by x c r ∗ Introducing the dimensionless position x

laminar for 0 < x ∗

< 1,

≡ x / x c r ,

and turbulent for x ∗

=

(5 × 10 )ν / u 5

we can say that the boundary layer is

> 1.

1 / 3 in terms of x ∗ ; write separate (b) Express the dimensionless quantity h x c r / k Pr

expressions for 0 < x ∗ numerical values.

<1

and x ∗

> 1.

1 / 3 (c) Plot h x c r / k Pr vs x ∗ for 0.1 < x ∗

(d) Is it possible in general to have q A

=

Your answers should comprise only x ∗ and

<

3.

q B ? Explain.

Page 1 of 2

∞

.

March 18, 2008

3. Re-analyze Example 7.6 (p. 419) for a more compact tube bank, in which the longitudinal and transverse pitches are S L = S T = 20 mm. All other parameters (including N L = 7 and

N T

=

8 ) remain the same. Carry out the following iterations:

(a) Using the Zhukauskas correlation (eqns 7.56 & 7.57), and assuming equal inlet and outlet temperatures (i.e. T i = T o = 15°C ), determine the air-side heat transfer coefficient h , the

improved estimate of T o , and the rate of heat transfer per unit length (length is in 2

direction of cylinder axes). [ Ans: 250 W / m ⋅ K , 40.2°C, 29.7 kW/m.] (b) Repeat the above calculations based on the new outlet temperature. (Do not update the air density ρ in eqn 7.59; see Comment 3 on page 422.) When do you know you have “the final answers”?

Page 2 of 2

ChE 314: Solutions to Assignment 8 Problem 1 [Based on Prob. 1.15 in text] D = 0.030 m ; T f = (T s + T ∞ ) / 2 = (90 + 25) / 2 = 57.5°C = 330.6 K

(a)

Water at 330.6 K: k = 0.6507 W/m⋅K µ =

0.4847 × 10 −3 Pa⋅s ;

1 / ρ = 1.016 × 10 −3 m / kg 3

Pr = 3.118 Re D = V D / ν =

(1) (0.030) 0.4924 × 10 − 6

= 60926

Using eqn 7.46, Nu D = 0.3 +

223.6 1.058

h = Nu D ⋅ k / D =

(b)

(1.297) = 274.4

( 274.4) (0.6507 )

= 5950 W / m2⋅K

0.030

Air at 330.6 K: k = 28.56 × 10 −3 W/m⋅K ν

= 18.97 × 10 −6 m2 /s

Pr = 0.7027 Re D = V D / ν =

(10) (0.030) 18.97 × 10 − 6

= 15814

Using eqn 7.46, Nu D = 0.3 +

69.32 1.1396

h = Nu D ⋅ k / D =

(1.130) = 69.04

(69.04) (0.02856) 0.030

= 65.7 W / m2⋅K

Page 1 of 5

⇒

ν

= µ / ρ = 0.4924 × 10 −6 m2 /s

Problem 2 [Based on Prob. 7.13 in text]

(a)

From Newton’s law of cooling, q A = h 2 L ( T s − T ∞ ) ⋅ 2 L 2 q B = h L ( T s − T ∞ ) ⋅ 2 L

For q A = q B , must have h L = h 2 L

2

But for laminar boundary layer, h x ∝ 1 / x (eqn 7.25); i.e. it is a monotonically decreasing fcn.

⇒ not possible to have h L = h 2 L , and hence qA cannot be equal to qB for laminar BLs (b)

∗

Using eqn 7.25 for laminar BL (i.e. for 0 < x < 1 ),

⎛ u ⎞ = 0.664 ⎜ ∞ ⎟ k ⎝ ν ⎠

h x

h x c r k Pr

1 / 3

1 / 2

x1 / 2 Pr 1 / 3

⎛ u ⎞ = 0.664 ⎜ ∞ ⎟ ⎝ ν ⎠

1 / 2

x −1 / 2 x c r

⎛ u ⎞ = 0.664 ⎜ ∞ ⎟ ⎝ ν ⎠

1 / 2

1 / 2

⎛ x c r ⎞ ⎜⎜ ⎟⎟ x c r 1 / 2 ⎝ x ⎠

1 / 2

⎛ x c r ⎞ ⎜⎜ ⎟⎟ x ⎝ ⎠

⎛ u ⎞ = 0.664 ⎜ ∞ ⎟ ⎝ ν ⎠

= (0.664) (5 × 10

)

5 1 / 2

1 / 2

⎛ x c r ⎞ ⎜⎜ ⎟⎟ x ⎝ ⎠

−1 / 2

= 470 (x ∗ ) For turbulent BL (i.e. x ∗ > 1 ), use eqn 7.31: 4 / 5 ⎡ ⎤ ⎛ u ∞ ⎞ = ⎢ 0.037 ⎜ ⎟ x 4 / 5 − 871⎥ Pr 1 / 3 k ⎝ ν ⎠ ⎢⎣ ⎥⎦

h x

h x c r k Pr

1 / 3

⎛ u ⎞ = 0.037 ⎜ ∞ ⎟ ⎝ ν ⎠

4 / 5

⎛ u ⎞ = 0.037 ⎜ ∞ ⎟ ⎝ ν ⎠

4 / 5

x −

1 / 5

= 0.037 (5 × 10 −1 / 5

= 1341 ( x ∗ )

)

k Pr

1 / 3

=

∗ 470 ( x )

−1 / 5

1341 ( x ∗ )

1 / 5

⎛ x c r ⎞ ⎜ ⎟ ⎜ x ⎟ ⎝ ⎠

871 x c r x

⎛ (5 × 10 5 )ν ⎞ ⎜⎜ ⎟⎟ u ∞ ⎝ ⎠ 1 / 5

−

4 / 5

871 x c r x

871 x

−1 / 2

h x c r

⎛ x c r ⎞ ⎜⎜ ⎟⎟ x ⎝ ⎠

5 4 / 5

−

x c r −

∗

∗

; 0 < x < 1

−

871 x

∗

; x ∗ > 1 Page 2 of 5

−

⎛ (5 × 10 5 )ν ⎞ ⎜⎜ ⎟⎟ u ∞ ⎝ ⎠

871 x c r x

1 / 2

1 / 2

(c)

1500

1000

h xc r k Pr

1 / 3

500

0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

x ∗

(d)

Since h is no longer a monotonically decreasing function once we enter into the turbulent regime, it is possible to have a situation where h ( x ) = h ( 2 x) . From the above plot, it can be seen (if you look real close) that h is the same when x ∗ is 0.63 and 2 × 0.63. This means we can have q A = q B when L = 0.63 x c r .

Page 3 of 5

Problem 3

(a)

Air at T i = 15°C = 288 K :

= 1.217 kg / m 3 ; c p = 1007 J / kg ⋅ K ; k = 0.02534 W / m ⋅ K ; Pr = 0.710

ρ

ν

= 14.82 × 10 −6 m 2 /s

Air at T s = 70°C = 343 K : Pr s = 0.701 Estimate V max : given V = 6 m / s ; S L = S T = 20 mm ; D = 16.4 mm ; N L = 7 ; N T = 8

(

S D = 20 2 + 10 2 S T + D

=

2

Re D , max =

1 / 2

20 + 16.4 2

V max =

⇒

)

= 22.36 mm

S D >

= 18.2 mm

S T + D

2

⎛ 20 ⎞ ⋅ V = ⎜ ⎟ ⋅ (6) = 33.33 m / s S T − D ⎝ 20 − 16.4 ⎠ S T

(33.33) (0.0164) 14.82 × 10 − 6

Staggered ; S T / S L = 1

= 36887

C = 0.35 ;

Re D , max = 36887

m = 0.6 ; C 2 = 0.95

Nu D = (0.95) (0.35) (36887 ) 0.6 (0.710) 0.36 (0.710 / 0.701) 0.25 = 162.1 h =

Nu D ⋅ k D

=

(162.1) (0.02534) 0.0164

= 250.5 W / m 2 ⋅K

Eqn 7.59:

⎡ ⎤ π (0.0164) (56) (250.5) = exp ⎢− ⎥ 70 − 15 ⎣ (1.217) (6) (8) (0.020) (1007) ⎦

70 − T o

⇒

T o = 40.24°C

Now calculate LMTD (eqn 7.58): T s − T i = 70 − 15 = 55°C ; T s − T o = 70 − 40.24 = 29.76°C

∆T lm =

55 − 29.76 ln (55 / 29.76)

= 41.10 K

Eqn 7.60: q / L = h ⋅ N π D ⋅ ∆T lm = (250.5) ⋅ (56)(π )(0.0164) ⋅ (41.10) W / m = 29.7 kW / m

Page 4 of 5

(b)

New mean temperature: T i + T o

2

=

15 + 40.24 2

= 27.62°C ≈ 300 K

Air at 300 K : c p = 1007 J / kg ⋅ K ;

Re D , max =

(33.33) (0.0164) 15.89 × 10

−6

ν

= 15.89 × 10 −6 m 2 /s ; k = 0.0263 W / m ⋅ K ; Pr = 0.707

= 34400

⇒

C = 0.35 ;

m = 0.6 ; C 2 = 0.95

Nu D = (0.95) (0.35) (34400 ) 0.6 (0.707 ) 0.36 (0.707 / 0.701) 0.25 = 155.04 h =

Nu D ⋅ k D

=

(155.04) (0.0263) 0.0164

= 248.6 W / m 2 ⋅K

⎡ ⎤ π (0.0164) (56) ( 248.6) = exp ⎢− ⎥ 70 − 15 ⎣ (1.217) (6) (8) (0.020) (1007) ⎦

70 − T o

⇒

T o = 40.1°C

T s − T i = 55°C ; T s − T o = 70 − 40.10 = 29.90°C

∆T lm =

55 − 29.90 ln (55 / 29.90)

= 41.18 K

q / L = h ⋅ N π D ⋅ ∆T lm = (248.6) ⋅ (56)(π )(0.0164) ⋅ (41.18) W / m = 29.5 kW / m

Page 5 of 5

March 27, 2008


Due: April 4 (Friday) before 16:00, in assignment box next to CME 567. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Final Exam Reminder

Date and Time: April 18 (Friday) at 14:00 Location: TBA Format: Open book and notes; calculator needed

1. Engine oil flows through a tube at a rate of 0.5 kg/s; the tube is 80 m long and has a diameter of 25 mm. The oil enters the tube at 14°C, while the tube’s inner surface temperature is maintained at 100 °C.

What are the outlet temperature T m,o and total heat transfer rate q conv after two iterations? (i.e. after an initial trial assuming T m, o

=

T s , followed by one more iteration.)


Assume fully developed hydrodynamic and thermal conditions.

•

Comment on the validity of the above assumptions.


Omit part (b).

•

Solve problem assuming again T m, o

•

Use the Dittus-Boelter correlation (eqn 8.60) for tubulent flow in pipe.

•

Compared to the first problem, why is iterative refinement not so critical here?

= T s .

No need to iterate this time.


For annular pipe, Re D

=

& 4m µ ( Di

π

+ Do )

and laminar flow occurs when Re D < 2300. •

Properties of liquid water at 320 K: k = 0.64 W / m⋅K ; c p

=

4180 J / kg⋅K ; µ

Page 1 of 1

=

0.6 × 10 − 3 Pa⋅s

ChE 314: Solutions to Assignment 9 Problem 1: T s = 100°C T m, i = 14°C

D = 25 mm

engine oil

⋅

m = 0.5 kg/s

L = 80 m

Assume T m , o = T s = 100°C Engine oil at 330 K:

µ

⇒

T m = 57°C = 330 K

= 0.0836 Pa ⋅ s ; k = 0.141 W / m⋅K ; Pr = 1205 ; c p = 2035 J / kg⋅K

Eqn 8.6: Re D =

& 4m π D µ

( 4) (0.5)

=

π (0.025) (0.0836)

= 304.6

⇒ laminar flow (i.e. Re D < 2300 )

x fd , t = 0.05 Re D Pr ⋅ D = (0.05) (304.6) (1205) (0.025) = 459 m

⇒ entrance flow

L < x fd , t

Also, x fd , h = 0.05 Re D ⋅ D = (0.05) (304.6) (0.025) = 0.38 m L < xfd , t

thermal entry; use eqn 8.56

L >> xfd , h

( D / L) Re D Pr =

Nu D ≡

h D k

(0.025) (304.6) (1205) 80

h (0.025)

=

0.141

= 3.66 +

= 114.7

(0.0668) (114.7) 1 + (0.04) (114.7)

2 / 3

⇒

h = 42.87 W / m ⋅ K 2

Eqn 8.41b: 100 − T m, o

⎡ π (0.025) (80) (42.87) ⎤ = exp ⎢ − ⎥ 100 − 14 (0.5) (2035) ⎣ ⎦

⇒

T m , o = 34.00°C

Eqn 8.34:

(

)

& c p T m , o − T m , i = (0.5) ( 2035) (34.00 − 14) W = 20.35 kW q conv = m -----------------------------2nd iteration: T m =

14 + 34.0 2

Engine oil at 297 K:

= 24°C = 297 K µ

= 0.6399 Pa ⋅ s ; k = 0.145 W / m⋅K ; Pr = 8350 ; c p = 1897 J / kg⋅K

(note hugh changes in

µ

and Pr )

Page 1 of 5

Re D =

& 4m π D µ

( 4) (0.5)

=

π (0.025) (0.6399)

= 39.79

⇒ laminar flow (i.e. Re D < 2300 )

x fd , t = 0.05 Re D Pr ⋅ D = (0.05) (39.79) (8350) (0.025) = 415 m

thermal entry; use eqn 8.56

x fd , h = 0.05 Re D ⋅ D = (0.05) (39.79) (0.025) m = 5.0 cm ( D / L) Re D Pr =

Nu D ≡

h D k

=

(0.025) (39.79) (8350) 80

h (0.025)

0.145

= 3.66 +

= 103.83

(0.0668) (103.83) 1 + (0.04) (103.83)

2 / 3

⇒

h = 42.584 W / m 2 ⋅ K

Eqn 8.41b: 100 − T m , o

⎡ π (0.025) (80) (42.584) ⎤ = exp ⎢ − ⎥ 100 − 14 (0.5) (1897) ⎣ ⎦

⇒

T m , o = 35.1°C

Eqn 8.34:

& c p T m , o − T m , i = (0.5) (1897) (35.1 − 14) W = q conv = m

Page 2 of 5

20.0 kW

Problem 2: (Prob 8.26 in text) T s = 25°C T m, i = 85°C

⋅ = 0.01 kg/s m

T m =

35 + 85 2

D = 3 mm

ethylene glycol L = ?

T m, o = 35°C

= 60°C = 333 K

Ethylene glycol at 333 K:

= 0.522 × 10 −2 Pa ⋅ s ; k = 0.2603 W / m⋅K ; Pr = 51.34 ; c p = 2562 J / kg⋅K µ

(

)

& c p T m , o − T m , i = (0.01) (2562) (35 − 85) = q conv = m Re D =

& 4m π D µ

( 4) (0.01)

= π

−2

(0.003) (0.522 × 10 )

= 813

laminar + fully developed flow + const T s Nu D ≡

h D k

=

h (0.003)

0.2603

= 3.66

⇒

⇒

− 1280 W

⇒ laminar flow (i.e. Re D < 2300 ) eqn 8.55

h = 317.6 W / m ⋅ K 2

Eqn 8.41b:

⎡ π (0.003) L (317.6) ⎤ = exp ⎢ − ⎥ 25 − 85 (0.01) (2562) ⎦ ⎣

25 − 35

⇒

L = 15.3 m

Check assumptions: x fd , h = 0.05 Re D ⋅ D = (0.05) (813) (0.003) = 0.12 m

⇒

f d assumption OK

x fd , t = 0.05 Re D Pr ⋅ D = (0.05) (813) (51.34) (0.003) = 6.26 m

Page 3 of 5

⇒ f d assumption not OK

Problem 3: (Prob 8.46 in text) T s = 100°C T m, i = 20°C

D = 50 mm

air

⋅ = 0.01 kg/s m

L = 5 m

Assume T m , o = T s = 100°C

⇒

T m = 60°C = 333 K

= 2.002 × 10 −5 Pa ⋅ s ; k = 0.02874 W / m⋅K ; Pr = 0.702 ; c p = 1008 J / kg⋅K

Air at 333 K:

µ

Eqn 8.6: Re D =

& 4m π D µ

=

( 4) (0.01) π (0.050) ( 2.002 × 10

−5

)

= 12 720

⇒ turbulent flow (i.e. Re D > 2300 )

Eqn 8.60: Nu D = 0.023 Re D4 / 5 Pr 0.4 = (0.023) (12720) 0.8 (0.702) 0.4 = 38.358 h =

Nu D ⋅ k D

=

(38.358) (0.02874) 0.050

= 22.05 W/m2 ⋅ K

Eqn 8.41b: 100 − T m , o

⎡ π (0.050) (5) (22.05) ⎤ = exp ⎢ − ⎥ 100 − 20 (0.01) (1008) ⎣ ⎦

⇒

T m , o = 85.6 °C

Eqn 8.34:

& c p T m , o − T m , i = (0.01) (1008) (85.6 − 20) = 661 W q conv = m

For air,

µ ,

k , Pr and c p do not vary strongly with temperature, ∴ iterations perhaps not needed.

Page 4 of 5

Problem 4 (Prob 8.93 in text)

insulated liquid water T m , i = 20°C

T m , o = 75°C

steam at 100 °C liquid water

T m =

20 + 75 2

= 47.5°C ≈ 320 K

D o = 100 mm ; D i = 25 mm

⇒ D h = D o − D i = 0.075 m

& = 0.02 kg / s m Re D =

& 4m π µ ( D i + D o )

Assume fd conditions D i / D o = 0.25

=

(4) (0.02) π (0.6 × 10

−3

) (0.100 + 0.025)

= 339.5

⇒ use Table 8.2 Nu i = 7.37

⇒

⇒

hi =

Nu i ⋅ k D h

=

(7.37) (0.64)

Eqn 8.41b (for const T s ) :

⎡ π (0.025) (62.89) L ⎤ = exp ⎢ − ⎥ 100 − 20 (0.02) (4180) ⎦ ⎣ 100 − 75

⇒ L = 19.7 m

At exit,

(

⇒ laminar flow

)

q ′i′ = h i T s − T m , o = (62.89) (100 − 75) = 1570 W / m 2

Page 5 of 5

0.075

= 62.89 W / m 2 ⋅ K

April 4, 2008


For practice only; not to be handed in. ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Final Exam Reminder

Date and Time: April 18 (Friday) 14:00

−

16:30

Location: Main Gym (not Butterdome), Rows 12, 14, 16 Format: Open book and notes; bring calculator


Parts (a), (b) and (d) only.

•

Note: To even start the problem, you will need to know the heat capacities. Here, we will take the following crude approach: we assume the exit temperatures to be T h , o ≈ 100°C and T c , o ≈ 85°C. For the heat capacities, just use the closest tabulated values from the steam table; it would be silly to interpolate.


Complete part (b) using the LMTD method.

•

Additional parts to problem: (c)

Determine the required heat exchanger length using the NTU method.

(d)

Based on the tabulated average properties, calculate the overall heat transfer 2 coefficient U and compare to the given value of 60 W / m ⋅ K. ¾

For this problem, assume pipe flow is turbulent when Re D exceeds 7000, in which case the Dittus-Boelter correlation (eqn 8.60) should be used.

¾

For annular flow of oil, assume fully developed conditions and uniform inner surface temperature.

Page 1 of 1

ChE 314: Solutions to Assignment 10 Problem 1 (Prob 11.64 in text) (a)

Assume T h , o ≈ 100°C Assume T c, o ≈ 85°C

⇒ ⇒

200 + 100

T h = T c =

Hot stream:

& h cp , h = C h = m

Cold stream:

& c cp, c = C c = m

2 35 + 85 2

= 150°C ≈ 420 K

= 60°C ≈ 335 K

( 42) (4302) 3600 (84) ( 4186) 3600

⇒

⇒

c p , h = 4302 J / kg⋅K

c p , h = 4186 J / kg⋅K

= 50.19 W/K = C min = 97.67 W/K = C max

⇒ q max = C min T h , i − T c , i = (50.19) ( 200 − 35) = 8280 W (b)

Use NTU method: C r = C min / C max = 50.19 / 97.67 = 0.5139

NTU = U A / C min = (180) (0.33) / 50.19 = 1.183 ε

=

q =

1 − exp [ − 1.183 (1 − 0.0.5139)] 1 − (0.5139) exp [ − 1.183 (1 − 0.5139)] ε ⋅ q max

(

)

q = C c T c , o − T c , i

5094 = 50.19 ( 200 − T h , o )

⇒

5094 = 97.67 ( T c , o − 35 )

)

⇒

(

C h T h , i + C c T c , i C c + C h

=

)

⇒

⇒ ⇒

T h , o = 98.5°C T c , o = 87.2°C

C h T h , i + C c T c , i = T o ( C c + C h )

(50.19) (200) + (97.67)(35) 50.19 + 97.67

q = C h T h , i − T o = (50.19) ( 200 − 91.00) = ε

= 0.6152

T h , o = T c , o ≡ T o

q = C h T h , i − T o = C c T o − T c , i

T o =

1 − (0.5139) (0.5627)

⇒

Parallel flow with L → ∞

(

1 − 0.5627

= (0.6152) (8280) = 5094 W

q = C h T h , i − T h , o

(d)

=

= 91.00°C

5470 W

= q / q max = 5470 / 8280 = 0.66

Page 1 of 3

[all T ’s in °C or K]

Problem 2: (Prob 11.20 in text)

oil D i = 25 mm

water

D o = 45 mm

oil L U = 60 W / m 2 ⋅ K

Hot stream (oil):

Cold stream (water):

(a)

& h = 0.1 kg / s ; m

c p , h = 1900 J / kg ⋅ K

T h , i = 100°C ;

T h , o = 60°C

& c = 0.1 kg / s ; m

c p , c = 4200 J / kg ⋅ K ;

T c , i = 30°C

& c p ) h = (0.1) (1900) = 190 W / K C h = ( m

& cp C c = m

c

= (0.1) (4200) = 420 W / K

q = C h T h , i − T h , o = 190 (100 − 60) = 7600 W

Also,

(

q = C c T c , o − T c , i

)

7600 = 420 ( T c , o − 30 ) (b)

⇒

T c , o = 48.1°C

∆T 1 = T h , i − T c , o = 100 − 48.1 = 51.9°C ∆T 2 = T h , o − T c , i = 60 − 30 = 30°C

∆T lm =

∆T 2 − ∆T 1 30 − 51.9 = = 39.95°C ln (∆T 2 / ∆T 1 ) ln (30 / 51.9)

q = U A ∆T lm (c)

⇒

⇒

7600 = (60) ⋅ π (0.025) L ⋅ (39.95)

L = 40.4 m

C min = C h = 190 W / K ; C max = C c = 420 W / K C r = C min / C max = 190 / 420 = 0.4524 q max = C min T h , i − T c , i = 190 (100 − 30) = 13 300 W ε

= q / q max = 7600 / 13300 = 0.5714

NTU = NTU =

⎡

1 (0.4524 − 1) U A C min

⇒

⎤ ⎥ = 1.001 ( 0 . 5714 ) ( 0 . 4524 ) 1 − ⎣ ⎦

ln ⎢

1.001 =

0.5714 − 1

(60) ⋅ π (0.025) L 190

⇒

Page 2 of 3

L = 40.4 m

heat transger

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