6-1
Steam at 160 psia and 400oF enters a nozzle with a volumetric flow rate of 6615 cfm (cubic feet per minute). The inlet area is 14.5 in2. If the steam leaves at 1500 ft/s ft/s at a pressure of 40 psia, find the the exit temperature.
Approach: Use the first law for an open system, specialized for a nozzle. Find properties in the steam tables. P1 = 160psia
Assumptions:
P2 = 40 psia psia
T 1 = 400 F o
1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic.
V2
V 6615cfm 1 =
Solution: From the first law, specialized for a nozzle, h1 +
V1
2
= h2 +
V2
2
2 2 The inlet velocity may be found from
V1
V 1
( 6615)
ft 3 ⎛ 1 min ⎞
⎜
⎟
min ⎝ 60 s ⎠ ft = = = 1095 2 A s ⎛ 1ft ⎞ (14.5in.2 ) ⎜ 144in.2 ⎟ ⎝ ⎠
Taking the inlet enthalpy from Table B-12,
⎛ ⎞ ⎜ ⎟ ⎞ 1 l b f Btu h2 = h1 + − = 1218 + ⎡⎣10952 − 15002 ⎤⎦ 2 ⎜ = 1197 ⎜ ⎟ ⎟ lbm ⋅ ft 2 2 lbm 2 s ⎝ 778 ft ⋅ lbf ⎠ ⎜ lbm ⎜ 32.2 2 ⎟⎟ s ⎝ ⎠ V1
2
V2
2
At P2 = 40 ps psia, h2 =1197
Btu
Btu lbm
ft 2 ⎛
1
1 Btu
, fr from Ta Table BB-12,
T 2 = 320 o F
Answer
6- 1
= 1500
ft s
6-2
Oxygen at 220oF enters a well-insulated nozzle of inlet diameter 0.6 ft. The inlet velocity is 60 ft/s. The oxygen leaves at 75oF, 10 psia. The exit area is is 0.01767 ft2. Calculate the pressure at the inlet.
Approach: Use the first law for an open system, specialized for a nozzle. Apply conservation of mass and the ideal gas law to determine inlet pressure.
Assumptions: 1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic. 4. Specific heat is constant. 5. Oxygen behaves like an ideal gas under these conditions.
T 1 = 220 o F V1
= 60
T 2 = 75 F o
ft
P2 = 10 psia psia
s
Solution: From the first law, specialized for a nozzle, h1 +
Solving for V2
V1
2
= h2 +
2 V2
V2
2
2 and noting that, for an ideal gas with constant specific heat, ∆h = c p ∆T
= 2 ( h1 − h2 ) + V1 2 = 2c p ( T1 − T2 ) + V1 2
Using data for specific heat from Table B-8 2
V2
⎛ 778 lbf ⋅ ft Btu ⎞ ft ⎞ ⎛ 32.17 lbm ⋅ ft ⎞ ⎛ ft ⎞ ft ⎛ 6 0 1 2 6 9 = 2 ⎜ 0.2215 + = ( 220 − 75) R ⎜ ⎟⎜ ⎟ ⎟ ⎜ ⎟ 2 lbm ⋅ R ⎠ s ⎝ ⎝ 1 Btu ⎠ ⎝ 1lbf ⋅ s ⎠ ⎝ s⎠
From conservation of mass, =m m 1 2 ρ1V1 A1 = ρ 2V 2 A2 Using the ideal gas law, MP1 MP2 V1 A1 = V 2 A2 RT1 RT 2 Solving for inlet pressure, P1 = P2
V2 V1
⎛ 1269 ⎞ ⎛ 0.01767 ⎞ ⎛ 220 + 460 ⎞ ⎟⎜ = (10 psia ) ⎜ ⎟⎜ ⎟ = 16.8 psia 2 A1 T 2 ⎝ 60 ⎠ ⎜⎝ π ( 0.3) ⎟⎠ ⎝ 75 + 460 ⎠
A2 T 1
6- 2
Answer
6-2
Oxygen at 220oF enters a well-insulated nozzle of inlet diameter 0.6 ft. The inlet velocity is 60 ft/s. The oxygen leaves at 75oF, 10 psia. The exit area is is 0.01767 ft2. Calculate the pressure at the inlet.
Approach: Use the first law for an open system, specialized for a nozzle. Apply conservation of mass and the ideal gas law to determine inlet pressure.
Assumptions: 1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic. 4. Specific heat is constant. 5. Oxygen behaves like an ideal gas under these conditions.
T 1 = 220 o F V1
= 60
T 2 = 75 F o
ft
P2 = 10 psia psia
s
Solution: From the first law, specialized for a nozzle, h1 +
Solving for V2
V1
2
= h2 +
2 V2
V2
2
2 and noting that, for an ideal gas with constant specific heat, ∆h = c p ∆T
= 2 ( h1 − h2 ) + V1 2 = 2c p ( T1 − T2 ) + V1 2
Using data for specific heat from Table B-8 2
V2
⎛ 778 lbf ⋅ ft Btu ⎞ ft ⎞ ⎛ 32.17 lbm ⋅ ft ⎞ ⎛ ft ⎞ ft ⎛ 6 0 1 2 6 9 = 2 ⎜ 0.2215 + = ( 220 − 75) R ⎜ ⎟⎜ ⎟ ⎟ ⎜ ⎟ 2 lbm ⋅ R ⎠ s ⎝ ⎝ 1 Btu ⎠ ⎝ 1lbf ⋅ s ⎠ ⎝ s⎠
From conservation of mass, =m m 1 2 ρ1V1 A1 = ρ 2V 2 A2 Using the ideal gas law, MP1 MP2 V1 A1 = V 2 A2 RT1 RT 2 Solving for inlet pressure, P1 = P2
V2 V1
⎛ 1269 ⎞ ⎛ 0.01767 ⎞ ⎛ 220 + 460 ⎞ ⎟⎜ = (10 psia ) ⎜ ⎟⎜ ⎟ = 16.8 psia 2 A1 T 2 ⎝ 60 ⎠ ⎜⎝ π ( 0.3) ⎟⎠ ⎝ 75 + 460 ⎠
A2 T 1
6- 2
Answer
6-3
A well-insulated nozzle has an entrance area of 0.28 m2 and an exit area of 0.157 m 2. Air enters at a velocity velocity of 65 m/s and leaves at 274 m/s. The exit pressure is 101 kPa kPa and the exit temperature is is 12°C. What is the the entrance pressure?
Approach: Use the first law for an open system, specialized for a nozzle. Apply conservation of mass and the ideal gas law to determine inlet pressure. V1
Assumptions:
= 65
P2 = 101kPa
m
T 2 = 12 C o
s
1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic. 4. Specific heat is constant. 5. Air behaves like an ideal gas under these conditions.
V2
= 274
m s
Solution: From the first law, specialized for a nozzle, h1 +
V1
2
= h2 +
V2
2
2 2 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore c p ( T1 − T 2 ) =
V2
2
2
−
V1
2
2
Solving for T 1, T1 =
V2
2
− V1 2
2c p
+ T 2
Specific heat depends on temperature, and should be evaluated at the average of inlet and outlet temperatures, but the inlet temperature is unknown. As an approximation, evaluate the specific heat at the exit temperature, and correct later if necessary. From Table A-8, kJ At T2 = 273 + 12 = 285 K, c p ≈ 1.004 kg ⋅ K 2
⎡( 274 )2 − ( 65) 2 ⎤ m2 ⎣ ⎦s T 1 = + 285K = 320 320 K kJ ⎛ 1000 J ⎞ 2 (1.004 ) ⎜ ⎟ kg ⋅ K ⎝ 1 kJ ⎠ From conservation of mass, =m m 1 2 ρ1V1 A1 = ρ 2V 2 A2 Using the ideal gas law, MP1 MP2 V1 A1 = V 2 A2 RT1 RT 2 Solving for inlet pressure, V A T ⎛ 274 ⎞ ⎛ 0.157 ⎞ ⎛ 320 ⎞ P1 = P2 2 2 1 P1 = (101) ⎜ 268 kPa kPa ⎟⎜ ⎟⎜ ⎟ = 268 V1 A1 T 2 ⎝ 65 ⎠ ⎝ 0.28 ⎠ ⎝ 2 85 ⎠
Answer
Comment: The entrance temperature of 320 K is very close to the exit temperature of 285 K. Specific heat does not vary significantly between these two temperature and there is no need to iterate on specific heat.
6- 3
6-4
Carbon monoxide enters a nozzle at 520 kPa, 100oC, with a velocity of 10 m/s. m/s. The gas exits at 120 kPa and 500 m/s. Assuming no heat transfer transfer and ideal gas behavior, find find the exit temperature.
Approach: Use the first law for an open system, specialized for a nozzle. P1 = 520kPa
Assumptions: 1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic. 4. Specific heat is constant. 5. Carbon monoxide behaves like an ideal gas under these conditions.
P2 = 120 120 kPa kPa
T 1 = 100 C o
V1
= 10
V2
m
= 500
m s
s
Solution: From the first law, specialized for a nozzle, h1 +
V1
2
V2
2
= h2 + 2 2 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore c p ( T1 − T 2 ) =
V2
2
−
V1
2
2 2 Specific heat depends on temperature, and should be evaluated at the average of inlet and outlet temperatures, but the outlet temperature is unknown. As an approximation, evaluate the specific heat at the inlet temperature, and kJ correct later if necessary. From Table A-8, at T 1 =100 + 273 = 373 K, K, c p ≈ 1.045 . kg ⋅ K 2
2
m⎞ ⎛ m⎞ ⎛ 10 ⎟ − ⎜ 500 ⎟ 2 2 ⎜ V − V2 s ⎠ ⎝ s ⎠ T2 = 1 + T 1 = ⎝ + 100 o C 2c p ⎛ kJ ⎞ ⎛ 1000 J ⎞ 2 ⎜1.045 ⎟⎜ ⎟ kg ⋅ K ⎠ ⎝ 1 kJ ⎠ ⎝ T 2 = − 19.6 C o
Answer
6- 4
6-5
Low-velocity steam with negligible kinetic energy enters a nozzle at 320 oC, 3 MPa. The steam leaves leaves the nozzle at 2 MPa with a velocity of 410 m/s. The mass flow flow rate is 0.37 0.37 kg/s. Determine a. the exit state b. the exit area area
Approach: Use the first law for an open system, specialized for a nozzle. Find properties in the steam tables. T 1 = 320 C o
Assumptions: 1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The nozzle is adiabatic.
P2 = 2 MPa
P1 = 3MPa V1
V2
0
Solution: a) From the the first law, specialized specialized for a nozzle, nozzle, h1 +
V1
2
V2
2
= h2 + 2 2 Determining the inlet enthalpy by interpolation in Table A-12, 2
⎛ ⎛ ⎛ kJ ⎞ kJ ⎞ ⎛ 1000 J ⎞ ⎛ 410 m/s ⎞ J ⎞ h2 = h1 + − = ⎜ 3042 ⎟ ⎜ = ⎜ 2.96 × 106 ⎟ = 2960 ⎜ ⎟ ⎟−⎜ ⎟ 2 2 kg ⎠ ⎝ 1 kJ ⎠ ⎝ 2 kg ⎠ ⎠ ⎝ ⎝ ⎝ kg ⎠ V1
2
V2
2
Interpolate in Table A-12 at 2 MPa and h = 2960 kJ/kg to find the final temperature of o T 2 = 274 C o
The exit state is superheated vapor at 2 MPa and 274 C. b) The mass flow rate is given by = ρ V A m 2 2 2 To determine the exit density, find the specific volume at the exit state by interpolation in Table A-12, v2 = 0.11 0.118m 8m 3 /kg /kg ρ 2 =
1 v2
=
1 3
0.118 m /k /kg
= 8.47
kg m3
Solving the mass flow rate equation for exit area, kg 0.37 m s A2 = = = 1.07 ×10−4 m2 kg ⎞ ⎛ m⎞ ρ 2 V 2 ⎛ ⎜ 8.47 m3 ⎟ ⎜ 410 s ⎟ ⎝ ⎠⎝ ⎠
6- 5
Answer
= 410
m s
6-6
Steam enters a diffuser at 250 oC and 50 kPa and exits at 300 oC and 150 kPa. The diameter at the entrance is 0.25 m and the diameter at the exit is 0.5m. If the mass flow rate is 9.4 kg/s, find the heat transfer to the surroundings.
Approach: Use the first law for an open system, eliminating work and potential energy. Find properties in the steam tables. T 1 = 250 C
T 2 = 300 C
P1 = 50kPa
P2 = 150 kPa
o
Assumptions: 1. Potential energy change is negligible. 2. The system operates in steady-state.
o
Solution: a) From the first law, 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Assuming steady conditions, one stream in and one stream out, no work, and no change in potential energy, the first law becomes
⎛
0 = Q cv + m ⎜ h1 +
2 ⎞ ⎛ V2 ⎞ m h − + ⎟ ⎜ 2 ⎟ 2 ⎠ 2 ⎠ ⎝
V1
⎝
2
The area at the inlet is 2
2
⎛ D ⎞ ⎛ 0.25 ⎞ A1 = π r = π ⎜ 1 ⎟ = π ⎜ = 0.0491m 2 ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ 2 1
The velocity at the inlet is, using data from Table A-12,
V1
=
mv 1
A1
=
m3 ⎞ ⎛ kg ⎞ ⎛ ⎜ 9.4 s ⎟ ⎜ 4.82 kg ⎟ ⎝ ⎠⎝ ⎠ 0.0491m 2
= 646 m s
The exit area is 2
⎛ 0.5 ⎞ = 0.196m2 ⎟ ⎝ 2 ⎠
A2 = π ⎜
Obtaining specific volume by interpolating in Table A-12, ( 9.4 )(1.98) mv m 2 V2 = = = 95.0 A2 0.196 s Rearranging the first law,
⎛
V2
⎝
2
Q cv = m ⎜ h2 +
2
− h1 −
⎞ ⎟ 2 ⎠
V1
2
2 2 ⎡ 2 m 2 m ⎤ 95 646 ) 2 ⎥ kg ⎢⎛ kJ ⎞ ⎛ 1000 J ⎞ ( ) s 2 ⎛ kJ ⎞ ⎛ 1000 J ⎞ ( s Q cv = 9.4 2976 + − − ⎢⎜ 3073 ⎟ ⎜ ⎥ ⎟ ⎟ ⎜ ⎟⎜ s ⎢⎝ kg ⎠ ⎝ 1 kJ ⎠ 2 kg 1 kJ 2 ⎥ ⎝ ⎠⎝ ⎠ ⎢⎣ ⎥⎦ Q cv = −107,146 W = −107.1 kW Answer
6- 6
6-7
Air enters a diffuser at 50 kPa, 85 oC with a velocity of 250 m/s. The exit pressure is atmospheric at 101 kPa. The exit temperature is 110oC. If the diameter at the inlet is 8 cm, a. Find the exit velocity. b. Find the diameter at the exit. Assume constant specific heats.
Approach: Use the first law for an open system, eliminating work, heat, and potential energy. Use ideal gas relations to evaluate properties.
T 1 = 85 o C
T 2 = 110 o C
P1 = 50kPa
Assumptions:
V1
1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The diffuser is adiabatic. 4. Air behaves like an ideal gas under these conditions. 5. Specific heats are constant.
P2 = 101kPa
= 250m/s
Solution: a)
From the first law, specialized for a diffuser h1 +
V1
2
= h2 +
V2
2
2 2 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore c p ( T1 − T 2 ) =
V2
2
−
V1
2
2 2 Evaluating specific heat at the average temperature of T ave = 197 o C = 370 K using data in Table A-8,
b)
V2
2 ⎛ kJ ⎞ ⎛ 1000 J ⎞ 2 m 85 110 250 = 2c p (T1 − T 2 ) + V1 2 = 2 ⎜ 1.01 − + ( ) ( ) ⎟ ⎟⎜ kg ⋅ K ⎠⎝ 1kJ ⎠ s2 ⎝
V2
= 110
m
s By conservation of mass = m m 1 2
Answer
ρ1V1 A1 = ρ 2V 2 A2 Using the ideal gas law, 2
⎛ D1 ⎞ P2 M V1π ⎜ ⎟ = RT RT1 ⎝ 2 ⎠ 2 P1M P1 T1
2
V1 D1
=
P2 T 2
⎛ D2 ⎞ V 2π ⎜ ⎟ ⎝ 2 ⎠
2
2
V 2 D2
Solving for exit diameter,
⎛ P1 ⎞⎛ T 2 ⎞ V1 ⎛ 50 ⎞⎛ 110 + 273 ⎞⎛ 250 ⎞ =8 ⎜ ⎟⎜ ⎟ ⎟⎜ 85 + 273 ⎟⎜ 110 ⎟ = 8.78 cm P T V 101 ⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ 2 ⎠⎝ 1 ⎠ 2
D2 = D1 ⎜
6- 7
Answer
6-8
Superheated steam enters a well-insulated diffuser at 14.7 psia, 320 oF and 400 ft/s. The steam exits as saturated vapor at a very low speed. Find the exit pressure and temperature.
Approach: Use the first law for an open system, eliminating work, heat, and potential energy. Find properties in the steam tables. T 1 = 320 o F
Assumptions:
P1 = 14.7psia
1. Potential energy change is negligible. 2. The system operates in steady-state. 3. The diffuser is adiabatic.
V1
V2
= 400ft/s
Solution: a)
From the first law, specialized for a diffuser h1 +
V1
2
V2
2
= h2 + 2 2 Using values of enthalpy at the inlet from Table B-12 at 14.7 psia, 320 oF, 2
2 ⎛ ft ⎞ ( 400) ⎜ ⎟ ⎧⎪ Btu ⎞ ⎛ ⎝s⎠ ⎪ = 1202.1 +
Btu ⎫ ⎪ lbm ⎪ =1205.3 Btu h2 ⎜ ⎨ ⎟ 2 ⎬ lbm ⎠ 2 lbm ⎝ ⎪ 25,037 ft ⎪ 2 ⎪ s ⎭ ⎩⎪ In Table B-10, for hg = 1205.3Btu/lbm, there are two possible solutions – one at 440oF, and 381.2 psia and the 1
other at about 470oF, and 514.1 psia. Either solution could occur, depending on the exit pressure imposed. Note also that the exit enthalpy is rather insensitive to pressure between these two values. Changing the exit pressure has little effect on the exit state. Answer
6-9
Steam enters an adiabatic turbine at 0.8 MPa and 500oC. It exits at 0.05 MPa and 150oC. If the turbine develops 24.5 MW of power, what is the mass flow rate?
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties in the steam tables.
T 1 = 500 o C P1 = 0.8MPa
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic.
Solution: Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
T 2 = 150 o C P2 = 0.05MPa
With values of enthalpy from Table A-12 m=
W
h1 − h2
=
24.5 ×103 kW
( 3480.6 − 2780.1kJ
kg )
= 34.98kg s
6- 8
Answer
0
6-10
Air enters an adiabatic turbine at 900 K and 1000 kPa. The air exits at 400 K and 100 kPa with a velocity of 30 m/s. Kinetic and potential energy changes are negligible. If the power delivered by the turbine is 1000 kW, a. find the mass flow rate. b. find the diameter of the duct at the exit. T 1 = 900K
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties using ideal gas relations.
P1 = 1000kPa
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic. 5. Air behaves like an ideal gas under these conditions. 6. Specific heat is constant.
T 2 = 400K P2 = 100kPa
Solution: a) Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
V2
= 30m/s
For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore = m
W h1 − h2
=
W
c p ( T1 − T2 )
With values of specific heat from Table A-8 at the average temperature of 650 K, 1000 kW kg = m Answer = 1.88 kJ s 1.063 900 400 K − ( ) kg ⋅ K b) Exit area is related to velocity and mass flow rate through mv 2 A2 = V2
From the ideal gas law, kJ ⎛ ⎞ 8.314 ⎜ ⎟ ( 400 K ) RT 2 ⎝ m3 kmol ⋅ K ⎠ v2 = = = 1.15 kg ⎞ MP2 kg ⎛ ⎜ 28.97 kmol ⎟ (100 kPa ) ⎝ ⎠ Substituting values
A2 =
kg ⎞ ⎛ m3 ⎞ ⎛ 1.88 1.15 ⎟ ⎜ ⎟⎜ s ⎠⎝ kg ⎠ ⎝ 30
D2 =
4 A2 π
=
m
= 0.0721m 2
s
( 4 )( 0.0721) π
= 0.303m
Answer
6- 9
6-11
Saturated steam at 320oC enters a well-insulated turbine. The mass flow rate is 2 kg/s and the exit pressure is 50 kPa. Determine the final state if the power produced is: a. 100 kW b. 400 kW T 1 = 320 C o
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties in the steam tables.
saturated steam
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic.
Solution: a) Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
P2 = 50kPa
From Table A-10, the enthalpy of saturated steam at the initial temperature of 320 oC h1 = 2700kJ kg h2 = h1 −
⎛ kJ ⎞ ⎛ 1000 J ⎞ = ⎜ 2700 ⎟ ⎜ ⎟− m kg ⎠ ⎝ 1 kJ ⎠ ⎝
W
⎛ 1000 W ⎞ ⎟ ⎝ 1 kW ⎠ = 2.65 × 106 J = 2650 k J
(100 kW ) ⎜ 2
kg
kg
kg
s From Table A-11 at 50 kPa, we see that the exit enthalpy is greater than that of a saturated vapor; therefore, the exit state is in the superheated region and, by interpolation in Table A-12, o T 2 = 83.4 C Answer P2 = 50kPa
b) Recalculating the exit enthalpy for the higher power condition, ⎛ 1000 W ⎞ ( 400 kW ) ⎜ ⎟ ⎛ kJ ⎞ ⎛ 1000 J ⎞ ⎝ 1 kW ⎠ = 2.5 × 106 J = 2500 k J h2 = ⎜ 2700 − ⎟⎜ ⎟ kg ⎠ ⎝ 1 kJ ⎠ 2 kg s kg kg ⎝ From Table A-11 at 50 kPa, we see that the exit enthalpy is between that of a saturated liquid and a saturated vapor; therefore, the exit state is in the two-phase region and
(
h2 = h f + x hg − h f x =
h2 − h f hg − h f
=
)
2500 − 340.5 2646 − 340.5
= 0.937
Summarizing, the final state is two-phase with P2 = 50 kPa x = 0.937
Answer
6 - 10
6-12
Superheated steam at 1.6 MPa, 600 oC enters a well-insulated turbine. The exit pressure is 50 kPa. The turbine produces 10 MW of power. If the exit pipe is 1.6 m in diameter and carries 11 kg/s of flow, find the velocity at the exit. Neglect kinetic energy. T 1 = 600 o C
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties in the steam tables.
P1 = 1.6MPa
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic.
Solution: Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
P2 = 50kPa
With values of enthalpy from Table A-12 at 1.6 MPa, 600 oC,
h2 = h1 −
W m
= 3693
kJ kg
⎛ 1000kW ⎞ ⎟ ⎝ 1MW ⎠ = 2784 kJ
(10MW ) ⎜ −
11 kg/s
kg
From Table A-12, at P2 = 50kPa and h2 = 2784kJ/kg, T 2 = m
150 o C . To find the exit velocity,
V 2 A2
v2
Solving for velocity and using values of v2 from Table A-12 at T 2
V2
=
mv 2
A2
=
m3 ⎞ ⎛ kg ⎞ ⎛ ⎜11 s ⎟ ⎜ 3.89 kg ⎟ ⎝ ⎠⎝ ⎠ 2
⎛ 1.6 ⎞ 2 π ⎜ 2 ⎟ m ⎝ ⎠
= 21.3
m
150 o C and P2 = 50kPa,
Answer
s
6 - 11
6-13
Air at 550oC and 900 kPa is expanded through an adiabatic gas turbine to final conditions of 100 kPa and 300oC. The total power output desired is 1 MW. If the inlet velocity is 30 m/s, what should the inlet pipe diameter be? Neglect kinetic and potential energy. T 1 = 550 C
Approach:
o
Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties using ideal gas relations.
P1 = 900kPa V1
= 30m/s
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic. 5. Air behaves like an ideal gas under these conditions. 6. Specific heat is constant.
Solution: a) Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
T 2 = 300 o C P2 = 100kPa
For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore = m
W h1 − h2
=
W
c p ( T1 − T2 )
With values of specific heat from Table A-8 at the average temperature of 700 K, 1000 kW kg = m = 3.72 kJ s 1.075 ( 500 − 300) o C kg ⋅ K b) Exit area is related to velocity and mass flow rate through mv 1 A1 = V1
From the ideal gas law, kJ ⎛ ⎞ 8.314 ⎜ ⎟ ( 550 + 273 ) K RT 1 ⎝ m3 kmol ⋅ K ⎠ v1 = = = 0.262 kg ⎞ MP1 kg ⎛ ⎜ 28.97 kmol ⎟ ( 900 kPa ) ⎝ ⎠ Substituting values
A1 =
kg ⎞ ⎛ m3 ⎞ ⎛ 3.72 0.262 ⎟ ⎜ ⎟⎜ s ⎠⎝ kg ⎠ ⎝ 30
D1 =
4 A1 π
=
m
= 0.0325m2
s
( 4 )( 0.0325) π
= 0.203m
Answer
6 - 12
6-14
Air at 510°C and 450 kPa enters an ideal, adiabatic turbine. The exit pressure is 101 kPa. In steady state, the turbine produces 50 kW of power. Find a. the exit temperature. (Hint: use Eq. 2-56) b. the mass flow rate. T 1 = 510 o C
Approach: Use Eq. 2-56 to find the final temperature. Then apply the first law for an open system, eliminating heat, kinetic energy and potential energy to obtain mass flow rate.
P1 = 450kPa
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic. 5. Air behaves like an ideal gas under these conditions. 6. Specific heat is constant. 7. The process is quasi-static.
P2 = 101kPa
Solution: a) Since the turbine is ideal, the process is quasi-static. For an adiabatic, quasi-static, process of an ideal gas with constant specific heat (from Eq. 2-56),
⎛P ⎞ =⎜ 2 ⎟ T1 ⎝ P1 ⎠
T2
k −1 k
From Table A-8, for air k =1.4
⎛P ⎞ T2 = T 1 ⎜ 2 ⎟ ⎝ P1 ⎠ b)
k −1 k
⎛ 101 ⎞ = ( 510 + 273) ⎜ ⎟ ⎝ 450 ⎠
1.4 −1 1.4
= 511 K = 238 o C
Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore = m
W h1 − h2
=
W
c p ( T1 − T2 )
With values of specific heat from Table A-8, W 50 kW kg = m = = 0.173 c p ( T1 − T 2 ) ⎛ s kJ ⎞ ⎜1.06 kg ⋅ K ⎟ ( 510 − 238) K ⎝ ⎠
6 - 13
Answer
6-15
Saturated steam at 3 MPa enters a well-insulated turbine operating in steady state. The turbine produces 600 kW of power. The mass flow rate through the turbine is 84 kg/min and the exit quality is 0.93. Find the exit temperature. P1 = 3MPa
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. To find the final temperature, it is necessary to iterate with data from the steam tables.
saturated steam
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The turbine is adiabatic.
Solution: a) Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2
x2 = 0.93
From Table A-11, the enthalpy of saturated steam at the initial pressure of 3 MPa is hg = h1 = 2804 kJ kg h2 = −
W m
+ h1 = −
600 kW kJ kJ + 2804 = 2375 kg kg ⎛ kg ⎞⎛ 1min ⎞ ⎜ 84 min ⎟⎜ 60 s ⎟ ⎝ ⎠⎝ ⎠
The exit state is in the two-phase region, but neither temperature nor pressure is known. The exit temperature is found by trial and error. Begin by assuming a value for T 2 , then use x2 to compute h2 with data from Table A-10. If h2 = 2375kJ/kg , the iteration is complete, if not, select a new value of T 2 and recompute h2 . To begin, assume T 2 = 25°C. From Table A-10,
(
h2 = h f + x2 hg − hf
)
=104.9 + ( 0.93)( 2547 − 2442) = 2376 kJ kg This is very close to the calculated value of 2375, therefore T 2 = 25°C Answer
Comment: In this example, we selected the correct result immediately. In reality, it would be necessary to try several temperatures before zeroing in on the correct value.
6 - 14
6-16
In a 3-hp compressor, carbon dioxide flowing at 0.023 lbm/s is compressed to 120 psia. The gas enters at 60oF and 14.7 psia. The inlet and outlet pipes have the same diameter. Find the final temperature and the volumetric flow rate at the exit (in ft 3/min). Assume constant specific heat at 100oF. T 1 = 60 o F
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy. Find properties using ideal gas relations.
P1 = 14.7psia
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The compressor is adiabatic. 5. Carbon dioxide behaves like an ideal gas under these conditions. 6. Specific heat is constant.
P2 = 120psia
Solution: Assuming an adiabatic compressor with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore c ( T − T ) W = m p 1 2
T2 = T 1 −
W c m p
With data from Table B-8 Btu ⎞ ⎛ ⎜ 2544 h ⎟ ( −3hp ) ⎜ ⎟ ⎜ 1hp ⎟ ⎝ ⎠ T 2 = 60 o F − = 533 o F lbm ⎞⎛ Btu ⎞ ⎛ 3600 s ⎞ ⎛ ⎜ 0.023 s ⎟⎜ 0.195 lbm ⋅ R ⎟ ⎜ 1h ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠
Answer
From conservation of mass = m = m 1 2
V2 v2
=
V2 P2 M RT 2
lbm ⎞ ⎛ psia ⋅ ft 3 ⎞ ⎛ ⎛ 60 s ⎞ 0.023 ⎜ 10.73 ⎟ ( 533 + 460) R ⎜ ⎜ ⎟ ⎟ s ⎠⎝ lbmol ⋅ R ⎠ RT m ft 3 ⎝ ⎝ 1min ⎠ 1 2 V 2.78 = = = 2 lbm ⎞ P2 M min ⎛ (120 psia ) ⎜ 44.01 ⎟ lbmol ⎠ ⎝ Comments:
Answer
If more accuracy is desired, the calculation should be repeated with specific heat evaluated at the average of the inlet and outlet temperatures, i.e. at (60+533)/2 or 296.5 F. The outlet temperature was not known at the beginning of the calculation, so specific heat at a temperature near the inlet temperature was used.
6 - 15
6-17
A well-insulated compressor is used to raise saturated R-134a vapor at a pressure of 360 kPa to a final pressure of 900 kPa. The compressor operates in steady state with a power input of 850 W. If the flow rate is 0.038 kg/s, what is the final temperature? P1 = 360kPa
Approach: Use the first law for an open system, eliminating heat, kinetic energy and potential energy.
saturated vapor
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The compressor is adiabatic.
Solution: Assuming an adiabatic compressor with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 2 1 h2 =
P2 = 900kPa
W
+ h1 m From Table A-15 at 360 kPa, h1 = hg = 250.6 kJ/kg 850 W kJ kJ + 250.6 = 272 kg kg kg 0.039 s From Table A-16, at P2 = 900 kPa and h2 = 271.6, T 2 ≈ 40 °C h2 =
6 - 16
Answer
6-18
Air flowing at 0.5 m3/min enters a compressor at 101 kPa and 25 oC. The air exits at 600 kPa and 300 oC. During this process, 250 W of heat are lost to the environment. What is the required power input? T 1 = 25 C
Approach:
o
Use the first law for an open system, eliminating kinetic energy and potential energy. Find properties using ideal gas relations.
P1 = 101kPa V = 0.5m 3 /min
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. Air behaves like an ideal gas under these conditions. 5. Specific heat is constant.
Solution: From the first law for an open system
⎛ ⎞ ⎛ ⎞ V Ve = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + gze ⎟ e ⎜ he + dt 2 2 ⎝ ⎠ ⎝ ⎠ 2
dE cv
P2 = 120psia
2
Assuming steady conditions, one stream in and one stream out, no change in kinetic or potential energy, the first law becomes (h − h ) 0 = Q − W + m 1 2
For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore c ( T − T ) W = Q + m p 1 2
Using data in Table A-1,
⎛ m3 ⎞ kg ⎞ ⎛ 1min ⎞ ⎛ 0.5 101kPa ) ⎜ 28.97 ( ⎜ ⎟ ⎟ ⎟⎜ min ⎠ kmol ⎠ ⎝ 60 s ⎠ V1 V1 PM kg ⎝ = m = 1 =⎝ = 9.84 ×10−3 kJ v1 RT 1 s ⎛ ⎞ ⎜ 8.314 kmol ⋅ K ⎟ ( 25 + 273) K ⎝ ⎠ The average temperature of the air is 25 + 300 T ave = = 162 o C = 435 K 2 Using specific heat values from Table A-8 interpolated at T ave
⎛ ⎝
W = −250 W + ⎜ 9.84 × 10−3
kg ⎞ ⎛
J ⎞ o ⎜ 1018 ⎟ ( 25 − 300) C = − 3000 W = − 3 kW ⎟ s ⎠⎝ kg ⋅ K ⎠
6 - 17
Answer
6-19
Refrigerant-134a enters a compressor at 0oF and 10 psia with a volumetric flow rate of 15 ft 3/min. The refrigerant exits at 70 psia and 140oF. If the power input is 2 hp, find the rate of heat transfer in Btu/h. T 1 = 0 F
Approach:
o
Use the first law for an open system, eliminating kinetic and potential energy.
P1 = 10psia V 15ft 3 /min 1 =
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state.
Solution: From the first law, neglecting kinetic and potential energy (h − h ) 0 = Q − W + m 1 2
T 2 = 140 F o
The mass flow rate is = m
P2 = 70psia
V
v1
Taking the inlet specific volume, v1 , from TableB-16, = m
15
ft 3
min = 3.19 lbm ft 3 min 4.703 lbm
(h − h ) Q = W + m 2 1
Again using Table B-16, Btu ⎞ ⎛ 3.412 ⎜ ⎛ 746 W ⎞ h ⎟ + ⎛ 3.19 lbm ⎞⎛ 129.1 − 102.9 Btu ⎞⎛ 60 min ⎞ Q = ( −2 hp) ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ min ⎠⎝ lbm ⎠⎝ 1 h ⎠ ⎝ 1 hp ⎠ ⎜⎜ 1W ⎟⎟ ⎝ ⎝ ⎠ Q = −76.0
Btu
Answer
h
6 - 18
6-20
A pump is used to raise the pressure of a stream of water from 10 kPa to 0.7 MPa. The temperature of the water is the same at inlet and outlet and equal to 20 oC. The velocity also does not change across the pump. If the mass flow rate is 14 kg/s, what power is needed to drive the pump? Assume frictionless flow and no significant elevation change.
Approach: ( P − P ). Use the equation for pump work in the form W = mv 1 2
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is ideal. 5. Water is incompressible.
Solution: For a frictionless pump with no elevation changes: (P − P ) W = mv 1 2 so, using the density of water at 20 oC from Table A-6, P1 − P2 W = m ρ
⎛ kg ⎞ (10 kPa − 700 kPa ) ⎛ 1000 Pa ⎞ = ⎜ 14 ⎟ ⎜ ⎟ = −9677 W = −9.68 kW kg 1kPa ⎠ ⎝ s ⎠ ⎝ 998.2
Answer
m3
6-21
A 2-hp pump is used to raise the pressure of saturated liquid water at 5 psia to a higher value. Assume the velocity is constant, the water is incompressible, and the flow is frictionless. If the mass flow rate is 6 lbm/sec, find the final pressure.
Approach: ( P − P ). Use the equation for pump work in the form W = mv 1 2
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is ideal. 5. Water is incompressible.
Solution: For a frictionless pump with no elevation changes: (P − P ) W = mv 1 2 P2 = P1 −
W
mv From Table B-11, v f = 0.0164 ft3/lbm at P = 5 psia. Substituting values ft ⋅ lbf ⎞ ⎛ ⎜ 550 s ⎟ ⎛ 1ft 2 ⎞ ( −2 hp ) ⎜ ⎟⎜ 2 ⎟ ⎜ 1 hp ⎟ ⎝ 144 in. ⎠ lbf ⎝ ⎠ P2 = 5 2 − in. ft 3 ⎞ ⎛ lbm ⎞ ⎛ 6 0.0164 ⎟ ⎜ s ⎟⎜ lbm ⎠ ⎝ ⎠⎝
P2 = 82.6 psia
Answer 6 - 19
6-22
Water is pumped at 12 m/s through a pipe of diameter 1.2 cm. The inlet pressure is 30 kPa. If the pump delivers 6 kW, find the final pressure. Assume frictionless, incompressible flow with no elevation or velocity changes.
Approach: ( P − P ). Use the equation for pump work in the form W = mv 1 2
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is ideal. 5. Water is incompressible.
Solution: For a frictionless pump with no elevation changes: (P − P ) m 1 2 (P − P ) = W = mv 1 2 ρ The mass flow rate is
⎛ ⎝
= ρ V A = 998 m ⎜
2
kg ⎞⎛
⎟⎜ 12
m3 ⎠⎝
m ⎞ ⎛ 0.012 m ⎞ kg π ⎜ = 1.35 ⎟ ⎟ s ⎠ ⎝ 2 s ⎠
1000 W ⎞⎛ kg ⎞ ⎛ 6 kW 998 − ⎜ ⎟⎜ ⎟ W ρ 1 kW ⎠⎝ m3 ⎠ ⎛ 1000Pa ⎞ ⎝ P2 = P1 − 30kPa =( − = 4.45 × 106 Pa = 4.45 MPa )⎜ ⎟ kg m ⎝ 1 kPa ⎠ 1.35 s
P2 = 4.45MPa
6-23
Answer
A 1-hp pump delivers oil at a rate of 10 lbm/s through a pipe 0.75 in. in diameter. There is no elevation change between inlet and exit, no velocity change, and no oil temperature change. The oil density is 56 lbm/ft3. Find the pressure rise across the pump.
Approach: ( P − P ). Use the equation for pump work in the form W = mv 1 2
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is ideal. 5. The oil is incompressible.
Solution: For a frictionless pump with no elevation changes: (P − P ) W = mv 1 2 P1 − P2 =
W mv
=
ρ W m
⎛
⎜ ⎛ lbm ⎞ ⎜ 56 ft 3 ⎟ (1hp ) ⎜ ⎝ ⎠ ⎜ P1 − P2 =
550
ft ⋅ lbf ⎞
s 1 hp
⎝ 10
lbm
⎟ ⎛ 1ft 2 ⎞ ⎟⎜ 2 ⎟ ⎟ ⎝ 144 in. ⎠ ⎠ = 21.4psia
s 6 - 20
Answer
6-24
An architect needs to pump 2.3 lbm/s of water to the top of the Empire State building, which is about 1000 ft high. Assume water at 45 psia is available at the base of the building. What is the power of the pump needed, in hp, if the flow is assumed frictionless? The velocity of the water is constant.
Approach: Apply the first law for an open system, dropping terms for transients, heat, and kinetic energy. Replace the enthalpy difference with h2 − h1 = v ( P2 − P1 ) .
Assumptions: 1. The system is adiabatic and isothermal. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is ideal. 5. Water is incompressible.
Solution: The first law for an open system is 2 2 ⎛ ⎞ ⎛ ⎞ V Ve ⎜ h + = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + gze ⎟ e e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Apply this between station 1 and station 2 on the figure above, neglecting kinetic energy and assuming steady, adiabatic operation, to get ( h − h ) + mg 0 = −W + m ( z1 − z2 ) 1 2 For an incompressible liquid undergoing an isothermal process h2 − h1 = v ( P2 − P1 ) Substituting this into the first law and noting that v = 1 ρ
⎛ P1 − P2 ⎞ ( z1 − z2 ) ⎟ + mg ⎝ ρ ⎠
W = m ⎜
Using values for the density of water from Table B-6 and being very careful with units,
⎛ lbm ⎞ W = ⎜ 2.3 ⎟ s ⎠ ⎝
( 45 − 14.7 )
lbf ⎛ 144 in.2 ⎞
⎜
in.2 ⎝ 1 ft 2 lbm 62.1 3 ft
⎟ ⎠
1 hp lbf ⋅ ft 550 s
⎛ ⎞ ⎛ ⎞ ⎟ ⎜ 1 hp ⎟ ft ⎞ ⎜ 1 lbf ⎛ lbm ⎞⎛ + ⎜ 2.3 ⎟( −1000ft )⎜ ⎟ ⎟⎜ 32.17 s2 ⎟ ⎜ lbf ⋅ ft ⎟ s ⎠⎝ ⎝ ⎠ ⎜ 32.17 lbm ⋅ft ⎟ ⎜ 550 ⎜ ⎟ ⎜ ⎟ s2 ⎠ s ⎠ ⎝ ⎝ W Answer = −3.89 hp
Comment: The actual pump chosen should have more horsepower than this because, in reality, there are some frictional effects.
6 - 21
6-25
Air at 150oC, 40 kPa is throttled to 100 kPa. The inlet velocity is 3.6 m/s. Find the exit velocity.
Approach:
P1 = 40kPa
Use the first law specialized for a throttle.
T 1 = 150 C
P2 = 100kPa
o
Assumptions:
V1
= 3.6m/s
1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. Air may be considered an ideal gas under these conditions. 5. The throttle is adiabatic. 6. The inlet and exit pipe have the same diameter. 7. Specific heat is constant.
Solution: For a throttle with no heat transfer, no change in kinetic or potential energy, and no work, the first law reduces to 0 = h2 − h1 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore 0 = c p ( T2 − T1 )
→ T1 = T2
From conservation of mass = m m 1 2 ρ V1 A1 = ρ 2 V 2 A2 Assuming the diameter of the inlet and exit pipes are the same and using the ideal gas law, RT1 RT 2 P1M
V1
=
V2
P2 M
Solving for exit velocity, V2
⎛ P ⎞ ⎛ m ⎞⎛ 400 ⎞ m = V1 ⎜ 2 ⎟ = ⎜ 3.6 ⎟⎜ = 14.4 ⎟ s ⎠⎝ 100 ⎠ s ⎝ P1 ⎠ ⎝
Answer
6 - 22
6-26
Saturated liquid R-134a at 24 oC is throttled until the final quality is 0.116. Find the final temperature and pressure.
Approach: T 1 = 24 o C
Use the first law specialized for a throttle.
saturated liquid
Assumptions:
x2 = 0.116
1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The throttle is adiabatic.
Solution: From Table A-14 at 24 oC h1 = h f =82.9 kJ/kg From the first law applied to a throttle, h1 = h2 This problem must be solved iteratively, since neither the final temperature nor the final pressure is known. First assume a final temperature and determine the corresponding quality, x2. For example, assume T 2 = 0oC. Then, from Table A-14 h2 − h f 82.9 − 50.0 x2 = = = 0.167 hg − h f 247 − 50.0 This value of x2 is too high, so a different value of T 2 is chosen. By trial and error, the final value of T 2 is found to be o T 2 ≈ 8 C To verify this result, calculate the final quality as h2 − h f 82.9 − 60.7 x2 = = = 0.116 hg − h f 252 − 60.7 The final pressure is the saturation pressure at 8oC, which is given in Table A-14 as P2 = 0.388MPa
Answer
6 - 23
6-27
Saturated liquid R-134a at 80 oF undergoes a throttling process. The pressure decreases to ¼ of its original value. Find the exit quality.
Approach: T 1 = 80 o F
Use the first law specialized for a throttle.
P2 = P1 / 4
saturated liquid
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The throttle is adiabatic.
Solution: From the first law applied to a throttle, h1 = h2 From Table B-14, If T sat = 80 o F
At P1 = 101.4, h1 = h f = 37.3 P2 =
P1
=
Psat = 101.4psia
then Btu lbm
101
= 25.2psia 4 4 To find the quality at the exit, we need to interpolate in Table B-14. From Table B-14 T
P
5 10
23.8 26.7
h f
13.14 14.66
By interpolation at 25.2 psia, h f =13.9 and hg =102.8. The quality is h2 − h f 25.2 − 13.9 x2 = = = 0.127 hg − h f 102.8 − 13.9
6 - 24
hg
102.5 103.2
Answer
6-28
A supply line contains a two-phase mixture of steam and water at 240 °C. To determine the quality of the mixture, a throttling calorimeter is used. In this device, a small sample of the two-phase mixture is bled off from the line and expanded through a throttling valve to atmospheric pressure. If the temperature on the downstream side of the throttling valve is measured to be 125°C, what is the quality of the mixture in the main steam line?
Approach: Use the first law specialized for a throttle.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The throttle is adiabatic.
Solution: From the first law applied to a throttle, h1 = h2 From Table A-12 (superheated vapor), kJ h2 = 2726 kg From Table A-10 at T 1 = 240°C kJ kJ h f = 1037 hg = 2804 kg kg
(
)
h1 = h f + x hg − hf = h2 x =
h1 − h f hg − h f
=
2726 − 1037 2804 − 1037
= 0.956
Answer
6 - 25
6-29
In a heat pump, R-134a is throttled through an expansion coil, which is a long copper tube of small diameter. The tube is bent in a coil both to fit in a compact space and to provide a large pressure drop. The refrigerant enters as saturated liquid at 5 oC with a flow rate of 0.025 kg/s and exits as a two-phase mixture at a pressure of 200 kPa. The wall of the coil may be assumed to be at the average temperature of the inlet and outlet. Heat is exchanged by natural convection and radiation from the outer surface of the coil with a combined heat transfer coefficient of 6 W/m 2·oC to the surroundings at 20oC. The expansion coil has an outside diameter of 8mm and a length of 2.2 m. Calculate the quality at the exit state.
Approach: Use the first law specialized for an open system, eliminating kinetic and potential energy changes, work, and the transient term. Calculate the heat transfer from the convection rate equation.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The heat transfer coefficient is uniform over the surface of the coil and independent of temperature.
Solution: From the first law for an open system, 2 2 ⎛ ⎞ ⎛ ⎞ Vi Ve = Qcv − W + ∑ m i ⎜ hi + + gzi ⎟ − ∑ m + gze ⎟ e ⎜ he + dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
cv
Assuming no change in kinetic or potential energy and steady conditions with no work, the first law becomes (h − h ) 0 = Q + m 1 2
Interpolating in Table A-14 gives the enthalpy of saturated liquid at 5oC as h1 = 56.7 kJ/kg. The heat lost from the surface of the coil is Q = hA ( Tave − Tsurr )
The saturation temperature at the exit condition of 200 kPa is, from Table A-15, T 2 = −10.1o C. The average temperature of the outside of the coil is T + T 2 5 + ( −10.1) T ave = 1 = = −2.55 o C 2 2 The surface area of the coil is
⎛
⎞ 2 ⎟ ( 2.2 m ) = 0.0553 m 1000mm ⎝ ⎠
A = π DL = π ( 8 mm ) ⎜
1m
Therefore, the rate of heat loss by convection/radiation is W ⎞ ⎛ Q = hA ( Tave − Tsurr ) = ⎜ 6 2 o ⎟ ( 0.0553 m2 ) ( −2.55 − 20) o C = −7.48 W ⎝ m ⋅ C⎠ Solving the first law for exit enthalpy, h2 =
Q m
+ h1 =
⎛ kJ ⎞ ⎛ 1000 J ⎞ + ⎜ 56.7 ⎟ ⎜ ⎟ = 56, 396 J = 56.4 kJ kg ⎝ kg ⎠⎝ 1kJ ⎠ 0.025 −7.48 W
s With values at the exit pressure of 200 kPa from Table A-15, the exit quality is h2 − h f 56.4 − 36.84 x = Answer = = 0.0956 hg − h f 241.3 − 36.84
Comment: The exit enthalpy was very close to the inlet enthalpy, with heat loss having little influence on the final enthalpy. This is often the case; therefore, throttles are usually assumed to be adiabatic.
6 - 26
6-30
One way to produce saturated liquid water is to mix subcooled liquid water with steam. In the tank below, 40 kg/s of subcooled liquid water enters at 15oC and 50 kPa. Superheated steam enters at 200oC and 50 kPa. What mass flow rate is required so that the exit stream is saturated liquid water at 50 kPa? Assume the tank is well-insulated.
Approach: Apply conservation of mass and the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The tank is adiabatic.
Solution: From conservation of mass + m = m m 1 2 3 From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic tank and steady operation, the first law becomes h + m h − m h 0=m 1 1 2 2 3 3 For state 1, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 15 oC, which is, from Table A-10, kJ h1 = 63 kg From Table A-12 at 200 oC and 50 kPa, kJ h2 = 2877 kg From Table A-11, the enthalpy of saturated liquid at 50 kPa is kJ h3 = 340.5 kg in the first law by using conservation of mass, Eliminating m 3 h + m h −(m + m ) h 0=m 1 1 2 2 1 2 3 (h − h ) + m (h − h ) 0=m 1 1 3 2 2 3 Solving for m 2
⎛ kJ ⎞ ⎛ kg ⎞ − ⎜ 40 ⎟ ( 63 − 340.5) ⎜ ⎟ −m 1 ( h1 − h3 ) s ⎠ ⎝ ⎝ kg ⎠ = 4.38 kg s = m = 2 h2 − h3 ⎛ kJ ⎞ ( 2877 − 340.5) ⎜ ⎟ ⎝ kg ⎠
6 - 27
Answer
6-31
In a desuperheater, superheated steam is converted to saturated steam by spraying liquid water into the steam. Using data on the figure, calculate the mass flow rate of liquid water.
Approach: Apply conservation of mass and the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The desuperheater is adiabatic. 5. Pressure is constant during the process.
Solution: From conservation of mass + m = m m 1 2 3 From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic tank and steady operation, the first law becomes h + m h − m h 0=m 1 1 2 2 3 3 From Table A-12, the enthalpy of superheated steam at 250oC and 2 MPa is h1 = 2902kJ kg For state 2, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 30 oC, which is, from Table A-10, h2 =125.8 kJ kg The desuperheater is effectively a mixing chamber, and the pressure is the same for all three streams. The enthalpy of saturated vapor at 2 MPa, is, from Table A-11, h3 = 2799 kJ kg in the first law by using conservation of mass, Eliminating m 3 h + m h −(m + m ) h 0=m 1 1 2 2 1 2 3 (h − h ) + m (h − h ) 0=m 1 1 3 2 2 3 Solving for m 2
⎛ kJ ⎞ ⎛ kg ⎞ − ⎜ 0.3 ⎟ ( 2902 − 2799) ⎜ ⎟ −m 1 ( h1 − h3 ) s ⎠ ⎝ ⎝ kg ⎠ = 0.0116 kg s = m = 2 h2 − h3 ⎛ kJ ⎞ (125.8 − 2799) ⎜ ⎟ ⎝ kg ⎠
6 - 28
Answer
6-32
A laundry requires a stream of 8 kg/sec of hot water at 40 oC. To obtain this supply, liquid water at 20oC is mixed in an adiabatic chamber with saturated steam. All three process streams are at 100 kPa. What are the required mass flow rates of the two inlet streams?
Approach: Apply conservation of mass and the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The mixing chamber is adiabatic. 5. Pressure is constant during the process.
Solution: From conservation of mass + m = m m 1 2 3 From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic tank and steady operation, the first law becomes h + m h − m h 0=m 1 1 2 2 3 3 From Table A-11, for saturated steam at 100 kPa, h1 = 2675.5kJ kg For state 2, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 20 oC, which is, from Table A-10, h2 = 83.96kJ kg For state 3, the enthalpy of the subcooled liquid is approximated by the enthalpy of saturated liquid at 40 oC, which is, from Table A-10, h3 = 167.57kJ kg Using conservation of mass to eliminate m in the first law 1 −m )h + m h − m h 0 = (m 3 2 1 2 2 3 3 h −m h + m h − m h 0=m 3 1 2 1 2 2 3 3
Gathering terms (h − h ) = m (h − h ) m 2 1 2 3 1 3 = m 2
(h − h ) m 3 1 3
( h1 − h2 )
8
=
kg s
( 2675.5 − 167.57)
( 2675.5 − 83.96)
=8−m = 0.26 kg s m 1 2
kJ
kJ kg
= 7.74 kg s
kg
Answer
6 - 29
Answer
6-33
Steam with a quality of 0.88 and a pressure of 20 kPa enters a condenser. The steam flow is divided equally among 20 tubes 2.1-cm in diameter which run in parallel through the condenser. The same amount of heat is removed from each tube. Liquid water exits each tube with a velocity of 1.5 m/s and a temperature of 55oC. Find the total amount of heat removed from the entire condenser.
Approach: Apply the first law for an open system, eliminating all terms except heat and enthalpy change.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. Pressure is constant in each tube.
Solution: From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ Vi Ve = Qcv − W + ∑ mi ⎜ hi + + gzi ⎟ − ∑ me ⎜ he + + gze ⎟ dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
cv
Neglecting changes in kinetic and potential energy, and assuming steady operation with no work, the first law becomes 0 = Q + mh mh 1 − 2
Using data from Table A-11 for saturated steam at 20 kPa,
(
)
h1 = h f + x hg − hf = 251.4 + 0.88 ( 2610 − 251.4) = 2327 kJ kg
For state 2, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 55 oC, which is, from Table A-10, h2 = 230 k J kg The mass flow rate in each tube is (with values of density from Table A-6) 2
m tube
kg ⎞⎛ m ⎞ ⎡⎛ 2.1cm ⎞ ⎛ 1m ⎞ ⎤ kg ⎛ = ρ 2 V 2 A2 = ⎜ 985.7 3 ⎟⎜ 1.5 ⎟ π ⎢⎜ ⎜ ⎟ ⎥ = 0.512 ⎟ m ⎠⎝ s ⎠ ⎣⎝ 2 ⎠ ⎝ 100 cm ⎠ ⎦ s ⎝
The total mass flow rate for all 20 tubes is kg = 20 m m tube = 20 ( 0.512 ) = 10.24 s From the first law kg kJ ( h − h ) = 10.24 Q = m ( 230 − 2327) = −21, 478 kW 2 1 s kg
6 - 30
Answer
6-34
Saturated steam at 120oF is condensed in a tube, as shown. Cooling water at 50 oF flows in crossflow over the exterior of the pipe, giving a heat transfer coefficient of 200 Btu/h·ft2·oF. Find the exit quality.
Approach: Use the thermal resistance analogy to determine the rate of heat transfer. Apply the first law for an open system, eliminating all terms except heat and enthalpy change.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. Pressure is constant in each tube. 5. The interior of the tube wall is at the condensing steam temperature.
Solution: Apply the thermal resistance analogy to find the heat removed from the pipe. Assuming that the condensing heat transfer coefficient is very large, the interior wall of the tube is at the temperature of the saturated steam. The resistance to conduction in the tube wall is ⎛ 3.25 ⎞ ln ⎜ ⎟ Btu ⎝ 3 ⎠ R1 = = 0.000212 o Btu ⎞ h⋅ F ⎛ 2π ( 2 ft ) ⎜ 30 ⎟ o ⎝ h ⋅ ft ⋅ F ⎠ The resistance to convection on the exterior of the tube is 1 Btu R2 = = 0.00147 o h⋅ F ⎛ 1ft ⎞ Btu ⎞ ⎛ ⎜ 200 h ⋅ ft 2 ⋅o F ⎟ 2π ( 3.25in.) ⎜ 12 in. ⎟ ( 2 ft ) ⎝ ⎠ ⎝ ⎠ The two resistances add in series. The total heat transferred is Q=
∆T
Rtot
(120 − 50) o F
=
( 0.000212 + 0.00147)
Btu
= 41,640
Btu
⎞
⎛
h
h ⋅o F
From the first law for an open system dE cv dt
= Q cv − Wcv +
∑
⎛
Vi
⎝
2
m i ⎜ hi +
2
+ gzi ⎟ − ⎠
∑
m e ⎜ he +
⎝
⎞ + gze ⎟ 2 ⎠
Ve
2
Neglecting changes in kinetic and potential energy, and assuming steady operation with no work, the first law becomes 0 = Q + mh mh 1 − 2
Solving for exit enthalpy and using data in Table A-10, Btu −41,640 Q h + 1113.5 Btu = 735 Btu h2 = + h1 = lbm m lbm lbm 110 h Using data in Table A-10, h2 − h f 735 − 88 x = = = 0.63 hg − h f 1113.5 − 88
6 - 31
Answer
6-35
Superheated R-134a enters a well-insulated heat exchanger at 0.7 MPa, 70 oC. It exits as saturated liquid at 0.7 MPa with a volumetric flow rate of 6000 cm3/min. The R-134a exchanges heat with an air flow, which enters at 18oC at a mass flow rate of 195 kg/min. Find the exit air temperature.
Approach: Apply the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The heat exchanger is adiabatic. 5. Pressure is constant during the process. 6. Air may be considered an ideal gas under these conditions. 7. The specific heat of air is constant.
Solution: Using data in Table A-15 for a saturated liquid at 0.7 MPa, the mass flow rate of the refrigerant is cm3 ⎛ 1m
3
⎞ 6000 ⎜ ⎟ min ⎝ 100 cm ⎠ Vr V kg r mr = ρ rV r = = = = 7.2 3 vr
v f
0.0008328
m
min
kg
From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ Vi Ve ⎜ h + = Qcv − W + ∑ m i ⎜ hi + + gzi ⎟ − ∑ m + gze ⎟ e e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic heat exchanger in steady operation, the first law becomes (h − h ) + m (h −h ) 0=m r a 1 2 3 4 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore (h − h ) + m c ( T − T ) 0=m r a p 1 2 3 4
Using data from Tables A-16, A-15, and A-8, kg ⎞ kJ ⎛ 7.2 307 − 86.78) ( ⎜ ⎟ (h − h ) m min ⎠ kg r 1 2 T4 = + T 3 = ⎝ + 18 = 26.1o C c m kg ⎞ ⎛ kJ ⎞ ⎛ a p ⎜195 min ⎟ ⎜ 1 kg ⋅ K ⎟ ⎝ ⎠⎝ ⎠
6 - 32
Answer
6-36
R-134a flows through the evaporator of a refrigeration cycle at a rate of 5 kg/s. The R-134a enters as saturated liquid and leaves as saturated vapor at 12oC. Air at 25oC enters the shell side of the heat exchanger. If the air leaves at 15oC, what mass flow rate of air is required?
Approach: Apply the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The evaporator is adiabatic. 5. Pressure is constant during the process. 6. Air may be considered an ideal gas under these conditions. 7. The specific heat of air is constant.
Solution: From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve ⎜ h + = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + gze ⎟ e e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic evaporator in steady operation, the first law becomes (h − h ) + m (h −h ) 0=m r a 1 2 3 4 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore (h − h ) + m c ( T − T ) 0=m r a p 1 2 3 4
Solving for the mass flow rate of the air, h2 − h1 m m a = r c p (T3 − T 4 ) With values of enthalpy from Table A-14 and specific heat from Table A-8, kJ ( 254.03 − 66.18) kg kg ⎛ kg ⎞ m = 93.5 a = ⎜5 ⎟ s kJ ⎞ ⎝ s ⎠⎛ o ⎜ 1.005 kg K ⎟ ( 25 − 15) C ⋅ ⎠ ⎝
6 - 33
Answer
6-37
Superheated steam at 5 psia and 200°F is condensed in a heat exchanger. The steam flows at 39 lbm/s and exits as saturated liquid. Cooling water at 45 °F is used to condense the steam. The water and steam are not mixed in the heat exchanger, but enter and leave as separate streams. If the maximum allowable water temperature rise is 15 °F and the maximum allowable water velocity is 11 ft/s, what is the diameter of the pipe which carries water to the heat exchanger?
Approach: Apply the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The condenser is adiabatic. 5. Pressure is constant during the process. 6. The cooling water is incompressible. 7. The specific heat of the water is constant.
Solution: From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic condenser in steady operation, the first law becomes h + m h = m h + m h m 1 1 3 3 2 2 4 4 = m and m = m ; therefore Since the streams do not mix, m 1 2 3 4 (h − h ) = m (h − h ) m 1 1 2 3 4 3
Assuming water is an ideal fluid, (h − h ) = m c ( T − T ) m 1 1 2 3 p 4 3 = m m 3 1
( h1 − h2 ) c p (T4 − T 3 )
With h1 from Table B-12, h2 from Table B-11, and c p from Table B-6, Btu 1148 − 130.2) ( lbm ⎛ ⎞ lbm = 2646 lbm = 39 m 3 ⎜ ⎟ Btu ⎞ s ⎠⎛ s o ⎝ ⎜ 1 lbm ⋅ o F ⎟ (15 F ) ⎝ ⎠ To find the diameter = ρ V A = ρ V m 3
D = 2
m3
V
ρπ
⎛ D ⎞ π ⎜ ⎟ ⎝2⎠
2
2646
=2
lbm
s = 2.21ft lbm ⎞ ⎛ ft ⎞⎛ ⎜11 s ⎟⎜ 62.4 ft 3 ⎟ π ⎝ ⎠⎝ ⎠
6 - 34
Answer
6-38
A two-phase mixture of steam and water with a quality of 0.93 and a pressure of 5 psia enters a condenser at 14.3 lbm/s. The mixture exits as saturated liquid. River water at 45oF is fed to the condenser through a large pipe. The exit temperature of the river water is 70oF less than the exit temperature of the other stream. If the maximum allowable average velocity in the pipe carrying river water is 15 ft/s, calculate the pipe diameter.
Approach: Apply the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The condenser is adiabatic. 5. Pressure is constant during the process. 6. The river water is incompressible. 7. The specific heat of the river water is constant.
Solution: From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic condenser in steady operation, the first law becomes h + m h = m h + m h m 1 1 3 3 2 2 4 4 = m and m = m ; therefore Since the streams do not mix, m 1 2 3 4 (h − h ) = m (h − h ) m 1 1 2 3 4 3
Assuming water is an ideal fluid, (h − h ) = m c (T − T ) m 1 1 2 3 p 4 3 = m m 3 1
( h1 − h2 ) c p (T4 − T 3 )
State 2 is a saturated liquid at 5 psia, since pressure is constant across the condenser. From Table B-11 T 2 = 162.2 o F T4 = T 2 − 70 = 92.2 F o
State 1 is a two-phase mixture at 5 psia. With data from Table B-11, Btu h1 = h f + x ( hg − hf ) = 130.2 + 0.93 (1131 − 130.2) = 1061 lbm The enthalpy of state 2 is h2 = h f = 130.2 Btu lbm Using the value of specific heat from Table B-6 at the average river water temperature of 69 oF lbm ⎞ Btu ⎛ 14.3 1061− 130.2) ( ⎜ ⎟ (h − h ) m s ⎠ lbm 1 1 4 = m =⎝ = 282 lbm s 3 ( h3 − h4 ) ⎛1.00 Btu ⎞ ( 92.2 − 45) o F ⎜ ⎟ lbm ⋅ o F ⎠ ⎝ = ρ V A = ρ V m 3
D =
4m 3
ρ V π
⎛ D ⎞ π ⎜ ⎟ ⎝2⎠
2
⎛ ⎝
4 ⎜ 282
D =
lbm ⎞
⎟
s ⎠ = 0.620 ft lbm ⎞⎛ ft ⎞ ⎛ ⎜ 62.2 ft 3 ⎟⎜ 15 s ⎟ π ⎝ ⎠⎝ ⎠ 6 - 35
Answer
6-39
A heat exchanger is used to cool engine oil. The specific heat of the oil is 0.6 Btu/lbm·oF. Using data on the figure below, find the exit temperature of the air.
Approach: Apply the first law for an open system.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The heat exchanger is adiabatic. 5. Pressure is constant during the process. 6. The oil is incompressible. 7. The specific heat is constant. 8. Air behaves like an ideal gas under these conditions.
Solution: From the first law for an open system 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic heat exchanger in steady operation, the first law becomes h + m h = m h + m h m 1 1 3 3 2 2 4 4 = m and m = m ; therefore Since the streams do not mix, m 1 2 3 4 (h − h ) = m (h − h ) m 1 1 2 3 4 3
Assuming the oil is an ideal liquid and the air is an ideal gas, c c m T T m T T 1 p ,oil ( 1 − 2 ) = 3 p ,air ( 4 − 3 ) Btu ⎞ ⎛ lbm ⎞⎛ 3.8 0.6 160 − 90) o F ( ⎜ ⎟⎜ ⎟ o m1c p ,oil ( T1 − T2 ) s ⎠⎝ lbm ⋅ F ⎠ T4 = T 3 + = 50 + ⎝ = 85 o F lbm Btu c m ⎛ ⎞⎛ ⎞ 3 p ,air ⎜19 s ⎟⎜ 0.24 lbm ⋅ R ⎟ ⎝ ⎠⎝ ⎠
6 - 36
Answer
6-40
Saturated liquid Refrigerant-134a at 36 oC is throttled to –8oC. The refrigerant then enters an evaporator and exits as saturated vapor. The evaporator is used to cool liquid water from 20oC to 10oC. If the mass flow rate of refrigerant is 0.013 kg/s, what is the mass flow rate o f the water?
Approach: Apply the first law for an open system to each component.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The throttle and evaporator are adiabatic. 5. Pressure is constant across the evaporator. 6. The water is incompressible. 7. The specific heat of the water is constant.
Solution: Taking a control volume around the throttle, and applying the first law, h1 = h2 Using data in Table A-14 kJ h1 = h f = 100.3 kg From Table A-14 at –8oC, h f = 39.54 kJ/kg and hg = 242.5 kJ/kg. Since h2 = 100.3 kJ/kg falls between these two values, state 2 is a two-phase mixture. Taking a control volume around the evaporator and applying the first law 2 2 ⎛ ⎞ ⎛ ⎞ V Ve h = Q cv − Wcv + ∑ m i ⎜ hi + i + gzi ⎟ − ∑ m + + gze ⎟ e ⎜ e dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
Neglecting changes in kinetic and potential energy, and assuming an adiabatic evaporator in steady operation, the first law becomes h + m h = m h + m h m 2 2 4 4 3 3 5 5 The water and R-134a do not mix, therefore = m = m = m = m m m R W 2 3 4 5 (h − h ) + m (h − h ) 0=m R w 2 3 4 5 m w =
(h − h ) m R 3 2
( h4 − h5 )
Assuming the water is an ideal liquid (h − h ) m R 3 2 m w = c p ( T4 − T 5 ) The saturated vapor at state 3 is at the same pressure as state 2. We have already concluded that state 2 is a two phase mixture, so its saturation temperature is –8oC. During a constant pressure process entirely in the two-phase region, temperature remains constant. Therefore, state 3 is a saturated vapor at –8oC. From Table A-14 h3 = hg = 242.5kJ/kg h2 = h1 = 100.3kJ/kg
Using values of specific heat at the average water temperature of 15 oC, m w =
( 0.013 kg s )( 242.5 − 100.3) kJ kg kg = 0.0441 o s ( 4.187 kJ kg ⋅ K)( 20 − 10) C
6 - 37
Answer
6-41
In a flash chamber, a pressurized liquid is throttled to a lower pressure where it becomes a two-phase mixture. The saturated liquid and vapor streams are removed in separate lines. In the figure below, liquid R-134a at 10°F and 30 psia is throttled to 12 psia. If 21.6 lbm/h of saturated vapor exits the flash chamber, what is the inlet flow rate? Assume the flash chamber is adiabatic.
Approach: Apply conservation of mass and the first law for an open system to the combination of the two components.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The throttle and flash chamber are adiabatic. 5. Pressure is constant in the flash chamber.
Solution: In this problem, the easiest approach is to define a control volume around the combination of the two components. You could also analyze each component separately, but that would take more effort and produce the same final result. Applying conservation of mass to the control volume shown, = m + m m 1 2 3 From the first law with no kinetic or potential energy change, and no heat transfer or work, h = m h + m h m 1 1 2 2 3 3 State 1 is a compressed liquid. Approximate the enthalpy of the liquid as the enthalpy of the saturated liquid at the same temperature. From Table B-14 h1 = h f at 10°F = 14.66 Btu/lbm From Table B-15, the enthalpy of states 2 and 3 is h2 = hg at 5 psi = 93.79 Btu/lbm h3 = h f at 5 psi = –3.73 Btu/lbm Substituting the given value of mass flow rate into conservation of mass produces lbm = 21.6 m + m 3 1 h Substituting m 1 and the values of enthalpy into the first law,
( 21.6 + m 3 )
lbm ⎛
Btu ⎞ ⎛ lbm ⎞⎛ Btu ⎞ lbm ⎛ Btu ⎞ 14.66 93.79 = ⎜ 21.6 + m 3 ⎜ ⎟ ⎟⎜ ⎟ ⎜ − 3.73 ⎟ h ⎝ lbm ⎠ ⎝ h ⎠⎝ lbm ⎠ h ⎝ lbm ⎠
gives Solving for m 3 = 92.9 m 3
lbm
h From conservation of mass lbm lbm lbm = 21.6 m + 92.9 = 114.5 1 h h h
Answer
6 - 38
6-42
Saturated liquid water at 40 kPa enters a 140 kW pump. The output of the pump is fed into a boiler where heat is added at a rate of 302 MW. There is negligible pressure drop across the boiler. If the mass flow rate of water is 70 kg/s, determine the boiler pressure and the state at the exit of the boiler.
Approach: ( P − P ) to the pump to Apply W = mv 1 2 determine the exit pressure form the pump. This is equal to the pressure at the exit of the boiler, since there is negligible pressure drop across the boiler. Draw a control volume around both components and apply the first law to determine exit temperature from the boiler.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The system operates in steady-state. 4. The pump is adiabatic and ideal. 5. Pressure is constant across the boiler. 6. The water is incompressible.
Solution: For the pump, power is related to pressure rise by (P − P ) Wcv = mv 1 2 Solving for exit pressure and using the specific volume of saturated liquid water at 40kPa from Table A-11,
⎛ 1000W ⎞ ⎟ W ⎛ 1000Pa ⎞ ⎝ 1 kW ⎠ = 1.99 × 106 Pa ≈ 2.0 MPa P2 = P1 − cv = 40kPa ⎜ − ⎟ mv m3 ⎞ ⎝ 1 kPa ⎠ ⎛ kg ⎞ ⎛ 70 0.001027 ⎟ ⎜ ⎟⎜ s ⎠⎝ kg ⎠ ⎝
( −140kW ) ⎜
Pressure is constant across the boiler, so P3 = P2 = 2.0MPa
Answer
To find the state at the exit of the boiler, draw a control volume around both components. The first law for this control volume is (assuming steady-state, and no kinetic or potential energy changes) h − m h 0 = Q cv − Wcv + m i i e e (h − h ) 0 = Q cv − Wcv + m 1 3
h3 =
Q cv − W cv m
+ h1
302, 000 kW − (− 140)kW kJ + 318 kg kg 70 s kJ h3 = 4634 kg
=
The exit state of the boiler has a pressure of 2 MPa and an enthalpy of 4634 kJ/kg. From Table A-12, this is superheated vapor with a temperature of T = 1000 o C
Answer
6 - 39
6-43
Air at 2000 R enters the turbine of a turbojet engine. The turbine is well insulated and produces 100 Btu of work per pound mass of air flowing through the engine. Upon exiting the turbine, the air enters the inlet of an insulated nozzle at 20 ft/s. The air leaves the nozzle at 2800 ft/s through an exit flow area of 0.6 ft2. The pressure at the nozzle exit is 10 lbf/in 2. What is the mass flow rate of air through the engine in lbm/s?
Approach: Draw a control volume around each component and apply the first law.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change in the turbine is negligible. 3. The system operates in steady-state. 4. The nozzle and turbine are adiabatic. 5. Air behaves like an ideal gas under these conditions. 6. The specific heat of the air is constant.
Solution: Begin by constructing a control volume around the turbine. Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes (h − h ) W = m 1 2 For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore W = mc T T p ( 1 − 2 )
Specific heat depends on the average air temperature in the turbine, however, the exit temperature is unknown. To make further progress, use c p of air at the inlet temperature of 2000 R (about 1500 o F ) and correct the calculation later if necessary. Solving for T 2 and using values of c p from Table B-8 at 1500°F, W 100Btu lbm T2 = T 1 − cv = 2000 R − = 1638 R Btu c m p 0.276 lbm ⋅ R Apply the first law for an open system to a control volume around the nozzle: 2 2 ⎛ ⎞ ⎛ ⎞ V3 V2 = Qcv − W + m ⎜ h2 + + gz2 ⎟ − m ⎜ h3 + + gz3 ⎟ dt 2 2 ⎝ ⎠ ⎝ ⎠
dE cv
cv
Assuming an adiabatic nozzle, no work, no change in potential energy and steady conditions,
⎛
0=m ⎜ h2 +
⎝
2 ⎛ ⎞ V3 ⎞ m h − + ⎜ 3 ⎟ ⎟ 2 ⎠ 2 ⎠ ⎝
V2
2
For an ideal gas with constant specific heat, ∆h = c p ∆T , therefore 0 = c p (T2 − T 3 ) + T3 =
V2
2
− V3
2c p
V2
2
− V3 2 2
2
+ T 2
The specific heat should be evaluated at the average temperature in the nozzle; however, the exit temperature is unknown. The nozzle inlet temperature is 1638 R or 1178°F and we expect the outlet temperature to be lower. For simplicity, we assume an average temperature of 1000 °F and will correct the calculation later if necessary. Using data from Table B-8,
⎛ ⎞ ⎞⎜ 1lbf ⎟ T 3 = ⎟ + 1638 R = 1043 R ⎜ ⎟⎜ ⎛ Btu ⎞ ⎝ 778ft ⋅ lbf ⎠⎜ 32.2 ft ⋅ lbm ⎟ ⎜ ⎟ 2 ⎜ 0.263 ⎟ s2 ⎠ ⎝ lbmR ⎠ ⎝
( 20
2
- 28002 ) ft 2 s2 ⎛
1Btu
From the ideal gas law: 6 - 40
⎛ 1ft 2 ⎞ ft ⋅ lbf ⎞ ⎛ 1545 1043 R ( ) ⎜ ⎟ ⎜ ⎟ lbmol ⋅ R ⎠ 144 in.2 ⎠ RT 3 ⎝ ⎝ v3 = = = 38.6 ft 3 lbm lbm ⎞ MP3 ⎛ 2 ⎜ 28.97 lbmol ⎟ (10 lbf in ) ⎝ ⎠ The mass flow rate may now be calculated as m=
V 3 A3
v3
=
( 2800 ft s ) ( 0.6 ft 2 ) 38.6 ft 3 lbm
= 43.5lbm s
Answer
Comments: For greater accuracy, you could recalculate the mass flow rate with improved values of specific heat, based on the calculated temperatures. The specific heat is not a strong function of temperature over the range considered, so this may not be necessary.
6 - 41
6-44
A well-insulated, rigid tank of volume 0.7 m 3 is initially evacuated. The tank develops a leak and atmospheric air at 20oC, 100 kPa enters. Eventually the air in the tank reaches a pressure of 100 kPa. Find the final temperature.
Approach: Apply the first law for an open system and integrate over time. Note that the temperature of the air leaking into the tank is constant with time, so the enthalpy of the air is also a constant.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The tank is adiabatic. 4. Air may be considered to be an ideal gas. 5. The specific heat is constant.
Solution: From the first law for an open system with no change in kinetic or potential energy dU cv = Q cv − Wcv + m i hi − m e he dt The tank is adiabatic. No work is done and nothing leaves the tank, so the first law reduces to dU cv = m i hi dt Integrating over time
∑
∫ dU
cv
∑
= ∫ m i hi dt
For an ideal gas, enthalpy depends only on temperature. The air entering the tank is at the same temperature throughout the process, so enthalpy is constant and may be removed from the integral.
∫
dt = h m U 2 − U1 = hi m i i 2
where m2 is the mass in the tank at the end of the process. At the beginning of the process, the tank is evacuated, therefore U 1 = 0. m2 u2 = hi m2 u2 = hi = u i + Pvi
For an ideal gas with constant specific heat, ∆u = cv ∆T cv (T2 − Ti ) = Pvi =
RT i M
Solving for T 2 kJ ⎛ ⎞ 8.314 ⎜ ⎟ ( 20 + 273) K RT i kmol ⋅ K ⎠ ⎝ T2 = + T i = + (20 + 273)K Mcv kg ⎞ ⎛ kJ ⎞ ⎛ ⎜ 28.97 kmol ⎟ ⎜ 0.718 kg ⋅ K ⎟ ⎝ ⎠⎝ ⎠ T 2 = 410K = 137 o C
Answer
6 - 42
6-45 Helium at 150oF and 40 psia is contained in a rigid, well-insulated tank of volume 5 ft 3. A valve is cracked open and the helium slowly flows from the tank until the pressure drops to 20 psia. During this process, the helium in the tank is maintained at 150oF with an electric resistance heater. Find a. the mass of helium withdrawn. b. the energy input to the heater.
Approach: Use a mass balance and the ideal gas law to find the mass of helium withdrawn. Use the first law for an open system to find the energy input to the heater.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The tank is adiabatic. 4. Helium may be considered to be an ideal gas. 5. The specific heat is constant.
Solution: a)
From a mass balance me = m1 − m2
where me is the mass of helium exiting. Using the ideal gas law,
⎛ lbm ⎞ 3 ⎜ 4 lbmol ⎟ ( 5 ft ) P1 MV1 P2 MV2 MV 1 ⎝ ⎠ me = ( P1 − P2 ) = (40 − 20)psia = 0.061lbm − = 3 RT1 RT 2 RT ⎛ psia ⋅ ft ⎞ ⎜10.73 ⎟ (150 + 460) R lbmol ⋅ R ⎠ ⎝ b)
From the first law with no kinetic or potential energy dU cv = Q cv − Wcv + m i hi − m e he dt The tank is adiabatic and nothing is entering, therefore dU cv = −Wcv − m e he dt Integrating over time,
∑
∫ dU
cv
∑
= − ∫ Wcv dt − ∫ m e he dt
Since the temperature of the helium in the tank does not change, he is constant and
∫
dt U 2 − U 1 = −Wcv − he m e
m2 u2 − m1u1 = −Wcv − he me m2 u2 − m1u1 = −Wcv − ( ue + Pe ve ) me m2 u2 − m1u1 = −Wcv − ( ue + Pe ve )( m1 − m2 )
Since u is only a function of temperature for an ideal gas and the temperature is constant u1 = u2 = ue = u Therefore 0 = −Wcv − Pe ve ( m1 − m2 ) Wcv = −
RT e M
( m1 − m2 ) = − ⎛ ⎝
( 0.061lbm ) ⎜1545 =−
4
me RT e M ft ⋅ lbf ⎞
⎟ (150 + 460) R ⎛ 1Btu ⎞ ⎜ ⎟ ⎝ 778ft ⋅ lbf ⎠
lbm ⋅ R ⎠ lbm lbmol
W cv = −18.5Btu
Answer 6 - 43
6-46
A residential hot water heater initially contains water at 140 oF. Someone turns on a shower and draws water from the tank at a rate of 0.2 lbm/s. Cold make-up water at 50oF is added to the tank at the same rate. A burner supplies 5472 Btu/h of heat. The water tank, which is a cylinder of diameter 1.8 ft, is filled to a height of 4 ft. How long will it be before the exiting water reaches 100oF? Assume a well-mixed tank.
Approach: Apply the first law for an open system. Use ideal liquid relations to rewrite enthalpy and internal energy in terms of temperature. Separate variables and integrate.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. Water is an ideal liquid. 4. The tank is well-mixed and all tank contents are at the same temperature.
Solution: The first law for an open system with one stream in and one stream out, neglecting kinetic and potential energy changes, is dU cv = Q cv − W + m i hi − m e he dt The entering and exiting mass flow rates are equal and no work is done, therefore d ( mc vuc v ) = Q cv + m ( hi − he ) dt The mass in the control volume does not change with time. For an ideal liquid ∆h = c p ∆T . With these ideas, the first law becomes du mcv cv = Q cv + mc T T p ( i − e) dt The tank is well-mixed, so the temperature, T , of the water in the tank equals the temperature of the exiting stream. The internal energy of the water in the tank is related to its temperature by ducv = cv dT ≈ cp dT since, for an ideal liquid, cv ≈ c p . The first law takes the form dT
= Q cv + mc p ( Ti − T ) dt Separating variables and integrating from initial temperature T 1 to final temperature T 2 mcvc pdT T2 t 2 dt = T 1 0 Qcv + mc p ( Ti − T )
mcv c p
∫
∫
let ξ = Q cv + mc p ( Ti − T ) dξ = − mc p dT
−mcv c p ∫
ξ 2
ξ 1
−
mcv
−
mcv
−
m
m mcv m
ln ξ
d ξ mc pξ
ξ 2 ξ 1
= t 2
= t 2
( ln ξ2 − ln ξ 1 ) = t 2 ln
ξ 2 ξ 1
= t 2
6 - 44
⎛ Qcv + mc T T ⎞ p ( i − 2) ln ⎜ − ⎟ = t 2 ⎜ Qcv + mc m T T ⎟ p ( i − 1) ⎠ ⎝ mcv
Find mcv 2
⎛ 1.8 ⎞ 2 mcv = ρ V = ( 62lbm ft ) π ⎜ ⎟ ft ( 4 ft ) = 631lbm ⎝ 2 ⎠ ⎡ ⎤ ⎛ 1h ⎞ ⎛ Btu ⎞ o 50 100 F − ( ) ⎢ ( 5472 Btu h ) ⎜ ⎥ ⎟ + 0.2 lbm s ⎜ 1 ⎟ o 2600 s ⎠ −631 lbm ⎢ ⎝ lbm ⋅ F ⎠ ⎝ ⎥ t 2 = ln ⎥ 1 0.2lbm s ⎢ ⎛ ⎞ + ( 0.2)(1)( 50 − 140) ( 5472) ⎜ ⎢ ⎥ ⎟ ⎝ 3600 ⎠ ⎢⎣ ⎥⎦ 3
t 2 = 2096 s = 35 min
Answer
6 - 45
6-47
In an industrial process, two streams are mixed in a tank and a single stream exits. Both streams may be assumed to have the properties of water. The volume of fluid in the tank is constant. A paddle wheel stirs the tank contents, doing work W . Initially, the water in the tank is at temperature T 1. At time t = 0, stream , and stream B enters at T . The quantities A at temperature T A enters with mass flow rate m B with rate m A B , and m T A , T B, m A B are all constant with time. Assuming a well-mixed tank, derive a formula for the time,
t 2, at which the tank water temperature is T 2. The tank is well insulated.
Approach: Apply the first law for an open system. Use ideal liquid relations to rewrite enthalpy and internal energy in terms of temperature. Separate variables and integrate.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The liquid is ideal. 4. The tank is well-mixed and all tank contents are at the same temperature. 5. The tank is adiabatic. 6. All fluids have the properties of water.
Solution: The first law for an open system neglecting kinetic and potential energy changes, is dU cv = Q cv − Wcv + m i hi − m e he dt Since the tank is well-insulated d ( mc vuc v ) h = −W + m A hA + m B hB − m C C dt The mass of liquid in the tank is constant, and conservation of mass requires that m C = mA + mB , therefore
∑
mcv mcv
ducv dt ducv dt
∑
h −(m ) h m = −W + m A hA + m B B A + B C
= −W + m A ( hA − hC ) + m B ( hB − hC )
The liquid in the tank is ideal and has a temperature T . The exiting stream is also at T , because the tank is wellmixed. With these considerations, the first law becomes dT c ( T − T ) mcv c p = −W + m A cp ( TA − T ) + m B p B dt dT c T −( m c + m c ) T mcv c p = −W + m A cp TA + m B p B A p B p dt c T −W + m A c pTA + m ⎛ m + m B ⎞ B p B = −⎜ A ⎟ T dt mcv c p m cv ⎝ ⎠
dT
Define K 1 and K 2 so that dT = K1 − K 2T dt Separating variables and integrating T 2 t 2 dT dt = T 1 0 K1 − K 2T
∫
∫
let ξ = K1 − K 2T dξ = − K 2 dT
6 - 46
−∫ −
d ξ
ξ 2 ξ 1
1 K 2
K 2ξ
ln ξ
t 2 = −
1 K2
= t 2 ξ 2 ξ 1
ln
= t 2 ξ 2 ξ 1
=−
1 K2
⎛ K1 − K 2T 2 ⎞ ⎟ ⎝ K1 − K 2T1 ⎠
ln ⎜
⎡ W ⎤ cv )T m A + B 2 ⎥ ⎢ − c + m ATA + m B TB − ( m ⎛ mcv ⎞ ⎢ p ⎥ t 2 = − ⎜ ⎟ ln ⎢ ⎥ ⎝ m A + mB ⎠ ⎢ − W cv + m T + m T − ( m A A B B A + mB ) T1 ⎥ c ⎣⎢ p ⎦⎥
6 - 47
Answer
6-48
A well insulated tank of volume 0.035 m3 is initially evacuated. A valve is opened, and the tank is charged with superheated steam from a supply line at 600 kPa, 500 oC. The valve is closed when the pressure reaches 300 kPa. How much mass enters?
Approach: Use the first law for an open system. Integrate the equation, recognizing that the enthalpy entering is constant with time. Properties are available in the steam tables.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The tank is adiabatic. 4. The state of steam in the supply line is constant.
Solution: The first law for an open system neglecting kinetic and potential energy changes, is dU cv = Q cv − Wcv + m i hi − m e he dt The tank is well-insulated, no mass exits, and no work is done, therefore dU cv = m i hi dt
∑
d ( mc vuc v ) dt
∫ d (m
cv
∑
= m i hi
∫
h dt ucv ) = m i i
The state of the steam entering remains the same throughout the process, so hi is a constant. If m2 is the mass in the tank at the end of the process and m1 is the mass at the start
∫
dt = h ( m − m ) m2 u2 − m1u1 = hi m i i 2 1
Since the tank is initially evacuated, m1 = 0, and m2 u2 = hi m2 u2 = hi
From Table A-12, hi = 3483 kJ kg , therefore u2 = 3483kJ kg
At the final pressure of 300 kPa and u2 = 3483kJ kg v2 ≈ 1.5 m3 kg
Therefore the mass in the tank at the end of the process is V 0.035m3 m2 = = = 0.0233kg m3 v2 1.5 kg
6 - 48
Answer
6-49
A well-insulated piston-cylinder assembly contains 0.06 kg of R-134a at –15 oC with a quality of 0.92. A supply line introduces superheated R-134a at 10 oC, 200 kPa into the cylinder. Assuming the pressure in the cylinder is constant, calculate the volume just when all the liquid has evaporated.
Approach: Use the first law for an open system. Integrate the equation, recognizing that the enthalpy entering is constant with time.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The piston-cylinder assembly is adiabatic. 4. The state of R-134a in the supply line is constant. 5. The pressure in the cylinder is constant. 6. The expansion is quasi-static.
Solution: From conservation of mass, mi = m2 − m1 where m2 is the mass in the tank at the end of the process, m1 is the mass at the start, and mi is the total mass introduced during the process. The first law for an open system neglecting kinetic and potential energy changes, is dU cv = Q cv − Wcv + m i hi − m e he dt The tank is well-insulated and no mass exits, therefore dU cv = −Wcv + m i hi dt
∑
d ( mc vuc v ) dt
∫ d (m
cv
∑
= −Wcv + m i hi
∫
∫
h dt ucv ) = − Wcv dt + m i i
The state of the R-134a entering remains the same throughout the process, so hi is a constant.
∫
dt = −W + h m m2 u2 − m1u1 = −Wcv + hi m i cv i i
Noting that this is a quasi-static expansion at constant pressure, and substituting conservation of mass m2 u2 − m1u1 = − P ( V2 − V1 ) + hi ( m2 − m1 ) m2 u2 − m1u1 = − P ( m2 v2 − m1v1 ) + hi ( m2 − m1 ) m2 u2 + Pm2 v2 − ( m1u1 + Pm1v1 ) = hi ( m2 − m1 )
Using the definition of enthalpy, h = u + Pv, and recognizing that P = P1 = P2 , m2 h2 − m1h1 = hi ( m2 − m1 )
Solving for m2, m h − m1 hi m1 ( h1 − hi ) m2 = 1 1 = h2 − hi h2 − hi The enthalpy of the R-134a in the cylinder at the start of the process is kJ h1 = h f + x ( hg − hf ) = 30.6 + 0.92 ( 238.3 − 30.6) = 222 kg where h f and hg are found by interpolation in Table A-14 at –15oC. The contents of the cylinder remain at the same pressure throughout the process and the R-134a changes from a two-phase mixture to a saturated vapor. A constant pressure process in the two-phase region is also a constant temperature process, so the final enthalpy is the enthalpy of saturated vapor at –15oC, that is h2 = 238.3kJ/kg. The enthalpy in the supply line, hi, may be found in Table A-16. The final mass may now be calculated as
6 - 49
m2 =
m1 ( h1 − hi ) h2 − hi
( 0.06 kg )( 222 − 259) =
( 238 − 259)
kJ
kJ kg
= 0.106kg
kg The final volume, using data for specific volume interpolated in Table A-14, is
⎛ m3 ⎞ 3 V2 = v2m 2 = ⎜ 0.120 ⎟ ( 0.106 kg ) = 0.0127 m kg ⎠ ⎝
6 - 50
Answer
6-50 The pressure inside a pot is maintained at an elevated level by a steel bob which rests on an open tube of inside diameter 0.5 cm. The bob, which has a mass of 0.401 kg, jiggles whenever the pressure in the pot is high enough to displace it, and steam is released. Heat is added to the bottom of the pot at a rate of 900 W . Heat is lost from the sides and the top of the pot by natural convection with a heat transfer coefficient of 3.9 W/m2·K. The pot has a height of 0.154 m and a diameter of 0.256 m. The ambient is at 20°C. Assume conduction resistance through the pot sides and top is very small, and that there is no air in the pot (only water and steam). The pot is half filled with water when the bob first lifts. a. Find the temperature inside the pot. b. Find the net rate of heat addition to the pot. c. Find the initial mass of the two-phase mixture in the pot. d. Find the fraction of the pot which is filled with liquid after one hour.
Approach: Use the first law for an open system. Integrate the equation, recognizing that the enthalpy entering is constant with time. Properties are available in the steam tables.
Assumptions: 1. Potential energy change is negligible. 2. Kinetic energy change is negligible. 3. The pot contains only steam and water. 4. The conduction resistance is small. 5. The convection resistance on the inside of the pot is small. 6. The work done in displacing the bob is negligible.
Solution: a)
Define m as the mass of the bob and r as the radius of the small tube that the bob rests on. By definition, m⎞ ⎛ ( 0.401 kg ) ⎜ 9.81 2 ⎟ F mg s ⎠ ⎝ P= = 2 = = 200,300Pa 2 A π r ⎛ 0.25 ⎞ 2 π ⎜ 100 ⎟ m ⎝ ⎠
Therefore, throughout the process, the pressure in the pot is P = 200.3 kPa. The saturation temperature at this pressure is, from Table A-11, T = 120°C.
Answer
b) The conduction resistance through the walls and lid of the pot is assumed to be small. Also the convective heat transfer coefficient on the inside is assumed to be very large, so that the inside convective resistance is small. Therefore, the outside of the pot is at the saturation temperature of the water. The net heat into the control volume is 2 ⎡ ⎛ D ⎞ ⎤ Q cv = Qburner + hA∆T = Q h π DL π + ⎜ ⎟ ⎥ ∆T ⎢ burner + ⎝ 2 ⎠ ⎦⎥ ⎣⎢ 2 W ⎞⎡ ⎛ ⎛ 0.256 ⎞ ⎤ 2 o Qcv = 900 W − ⎜ 3.9 2 ⎢π ( 0.256) ( 0.154) + π ⎜ ⎥ m ( 120 − 20) C ⎟ ⎟ m ⋅ K ⎠ ⎢⎣ ⎝ ⎝ 2 ⎠ ⎥⎦
Q cv = 831.4 W
c)
Answer
The volume of the pot is 2
2
⎛ D ⎞ ⎛ 0.256 ⎞ V = π ⎜ ⎟ ( L ) = π ⎜ ( 0.154) = 0.00793 m3 ⎟ ⎝2⎠ ⎝ 2 ⎠ The initial volumes of liquid and vapor are 6 - 51
