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A centrifugal compressor takes in ambient air at 100 kPa, 15°C, and discharges it at 450 kPa. The compressor has an isentropic efficiency of 80%. What is your best estimate for the discharge temperature? Solution: C.V. Compressor. Assume adiabatic, no kinetic energy is important. Energy Eq.6.13: w = h 1 - h 2 Entropy Eq.9.8: s 2 = s 1 + sgen We have two different cases, the ideal and the actual compressor. We will solve using constant specific heat. State 2 for the ideal, sgen = 0 so s 2 = s 1 and it becomes:
Eq.8.23: T2s = T 1
P 2
P 1
k- k = 288.15 (450 / 100)
= 442.83 K
ws = h 1 - h2s = Cp (T 1 - T2s ) = 1.004 (288.15 - 442.83) = -155.3 kJ/kg The actual work from definition Eq.9.27 and then energy equation: wac = ws/η = -155.3 / 0.8 = -194.12 kJ/kg = h 1 - h 2 = Cp(T 1 - T 2 )
⇒ T 2 = T 1 - wac / Cp = 288.15 + 194.12/1.004 = 481.5 K
Solving using Table A.7.1 instead will give
State 1: Table A.7.1: s o T1 = 6.82869 kJ/kg K Now constant s for the ideal is done with Eq.8.
s o T2s = s
o T1 + R ln(
P 2
P 1 ) = 6.82869 + 0.287 ln(
100 ) = 7.26036 kJ/kg K
From A.7.1: T2s = 442.1 K and h2s = 443.86 kJ/kg ws = h 1 - h2s = 288.57 - 443.86 = -155.29 kJ/kg The actual work from definition Eq.9.27 and then energy equation: wac = ws/η = -155.29 / 0.8 = -194.11 kJ/kg ⇒ h 2 = 194.11 + 288.57 = 482.68, Table A.7.1: T 2 = 480 K The answer is very close to the previous one due to the modest T’s.
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A compressor is used to bring saturated water vapor at 1 MPa up to 17.5 MPa, where the actual exit temperature is 650°C. Find the isentropic compressor efficiency and the entropy generation.
Solution: C.V. Compressor. Assume adiabatic and neglect kinetic energies. Energy Eq.6.13: w = h 1 - h 2 Entropy Eq.9.9: s 2 = s 1 + sgen We have two different cases, the ideal and the actual compressor. States: 1: B.1.2 h 1 = 2778.1 kJ/kg, s 1 = 6.5865 kJ/kg K 2ac: B.1.3 h2,AC = 3693.9 kJ/kg, s2,AC = 6.7357 kJ/kg K 2s: B.1.3 (P, s = s 1 ) h2,s = 3560.1 kJ/kg
IDEAL:
-wc,s = h2,s - h 1 = 782 kJ/kg
ACTUAL:
-wC,AC = h2,AC - h 1 = 915.8 kJ/kg
Definition Eq.9.28: ηc = wc,s/wc,AC = 0.8539 ~ 85%
Entropy Eq.9.8: sgen = s2 ac - s 1 = 6.7357 - 6.5865 = 0.1492 kJ/kg K
P T 2 ac 2 s 2 s 2 ac
s v
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Find the isentropic efficiency for the compressor in Problem 6.57.
A compressor in an air-conditioner receives saturated vapor R-410a at 400 kPa and brings it to 1.8 MPa, 60oC in an adiabatic compression. Find the flow rate for a compressor work of 2 kW?
C.V. Actual Compressor, assume adiabatic and neglect kinetic energies. Energy Eq.6.13: w = h 1 - h 2 Entropy Eq.9.9: s 2 = s 1 + sgen States: 1: B.4.2 h 1 = 271.9 kJ/kg, s 1 = 1.0779 kJ/kg-K 2: B.4.2 h 2 = 323.92 kJ/kg –wC = h 2 – h 1 = 323.92 – 271.9 = 52.02 kJ/kg Ideal compressor. We find the exit state from (P,s). State 2s: P 2 , s2s = s 1 = 1.0779 kJ/kg-K Ö h2s = 314.33 kJ/kg –wC s = h2s – h 1 = 314.33 – 271.9 = 42.43 kJ/kg
ηC = –wC s / –wC =
52.02 =^ 0.
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A pump receives water at 100 kPa, 15°C and a power input of 1.5 kW. The pump has an isentropic efficiency of 75% and it should flow 1.2 kg/s delivered at 30 m/s exit velocity. How high an exit pressure can the pump produce? Solution: CV Pump. We will assume the ideal and actual pumps have same exit pressure, then we can analyse the ideal pump.
Specific work: wac = 1.5/1.2 = 1.25 kJ/kg Ideal work Eq.9.27: ws = η wac = 0.75 × 1.25 = 0.9375 kJ/kg As the water is incompressible (liquid) we get Energy Eq.9.14: ws = (Pe - Pi)v + V^2 e/2 = (Pe - Pi)0.001001 + (30^2 /2)/ = (Pe - Pi)0.001001 + 0. Solve for the pressure difference Pe - Pi = (ws – 0.45)/0.001001 = 487 kPa Pe = 587 kPa
Water pump from a car
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Carbon dioxide, CO 2 , enters an adiabatic compressor at 100 kPa, 300 K, and exits at 1000 kPa, 520 K. Find the compressor efficiency and the entropy generation for the process. Solution: C.V. Ideal compressor. We will assume constant heat capacity. Energy Eq.6.13: wc = h 1 - h 2 ,
Entropy Eq.9.8: s 2 = s 1 : T2s = T 1
P 2
P 1
k- k (^) = 300
= 502.7 K
wcs = Cp(T 1 - T2s) = 0.842(300-502.7) = -170.67 kJ/kg C.V. Actual compressor wcac = Cp(T 1 - T2ac ) = 0.842(300 - 520) = -185.2 kJ/kg ηc = wcs/wcac = -170.67/(-185.2) = 0. Use Eq.8.16 for the change in entropy sgen = s2ac - s 1 = Cp ln (T2ac /T 1 ) - R ln (P 2 /P 1 )
= 0.842 ln(520 / 300) - 0.1889 ln(1000 / 100) = 0.028 kJ/kg K
P
v
T
s
e, s
i s = C (^) i
e, s e, ac e, ac
P
P
e
i
Constant heat capacity is not the best approximation. It would be more accurate to use Table A.8. Entropy change in Eq.8.19 and Table A.8:
s o T2 = s
o T1 + R ln(P^2 /P^1 ) = 4.8631 + 0.1889 ln(1000/100) = 5. Interpolate in A.8 => T2s = 481 K, h2s = 382.807 kJ/kg => -wcs = 382.807 – 214.38 = 168.43 kJ/kg; -wcac = 422.12 – 214.38 = 207.74 kJ/kg ηc = wcs/wcac = -168.43/(-207.74) = 0. sgen = s2ac - s 1 = 5.3767 – 4.8631 – 0.1889 ln(10) = 0.0786 kJ/kgK
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A small air turbine with an isentropic efficiency of 80% should produce 270 kJ/kg of work. The inlet temperature is 1000 K and it exhausts to the atmosphere. Find the required inlet pressure and the exhaust temperature. Solution: C.V. Turbine actual energy Eq.6.13: w = hi - he,ac = 270 kJ/kg
Table A.7: hi = 1046.22 ⇒ he,ac = 776.22 kJ/kg, Te = 757.9 K C.V. Ideal turbine, Eq.9.27 and energy Eq.6.13: ws = w/ηs = 270/0.8 = 337.5 = hi - he,s ⇒ he,s = 708.72 kJ/kg From Table A.7: Te,s = 695.5 K Entropy Eq.9.8: si = se,s adiabatic and reversible To relate the entropy to the pressure use Eq.8.19 inverted and standard entropy from Table A.7.1:
Pe/Pi = exp[ (s o Te −^ s
o Ti ) / R ] = exp[(7.733 - 8.13493)/0.287] = 0. Pi = Pe / 0.2465 = 101.3/0.2465 = 411 kPa
P
v
T
s
e, s
i
s = C
i
e, s
e, ac e, ac
P
Pe
i
If constant heat capacity were used Te = Ti - w/Cp = 1000 - 270/1.004 = 731 K C.V. Ideal turbine, Eq.9.26 and energy Eq.6.13: ws = w/ηs = 270/0.8 = 337.5 kJ/kg = hi - he,s = Cp(Ti - Te,s) Te,s = Ti - ws/Cp = 1000 - 337.5/1.004 = 663.8 K Eq.9.8 (adibatic and reversible) gives constant s and relation is Eq.8.
Pe/Pi = (Te/Ti) k/(^ k-1^ )^ ⇒ Pi = 101.3 (1000/663.8) 3.5^ = 425 kPa
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A compressor in an industrial air-conditioner compresses ammonia from a state of saturated vapor at 150 kPa to a pressure 800 kPa. At the exit, the temperature is measured to be 100o^ C and the mass flow rate is 0.5 kg/s. What is the required motor size for this compressor and what is its isentropic efficiency?
C.V. Compressor. Assume adiabatic and neglect kinetic energies. Energy Eq.6.13: w = h 1 – h 2 Entropy Eq.9.8: s 2 = s 1 + sgen We have two different cases, the ideal and the actual compressor.
States: 1: B.2.2: h 1 = 1410.9 kJ/kg, v 1 = 0.7787 m^3 /kg, s 1 = 5.6983 kJ/kg K
2ac: B.2.3 h2,AC = 1670.6 kJ/kg, v2,AC = 0.21949 m^3 /kg
2s: B.2.3 (P, s = s 1 ) h2,s = 1649.8 kJ/kg, T2,s = 91.4o^ C ACTUAL: –wC,AC = h2,AC – h 1 = 1670.6 – 1410.9 = 259.7 kJ/kg
W
in = m
(–wC,AC ) = 0.5 kg/s × 259.7 kJ/kg = 130 kW IDEAL: –wc,s = h2,s – h 1 = 1649.8 – 1410.9 = 238.9 kJ/kg Definition Eq.9.27: ηc = wc,s/wc,AC = 0.
P^ T
2 s 2 ac 2 s (^) 2 ac
s v
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Repeat Problem 9.45 assuming the steam turbine and the air compressor each have an isentropic efficiency of 80%. A certain industrial process requires a steady supply of saturated vapor steam at 200 kPa, at a rate of 0.5 kg/s. Also required is a steady supply of compressed air at 500 kPa, at a rate of 0.1 kg/s. Both are to be supplied by the process shown in Fig. P9.41. Steam is expanded in a turbine to supply the power needed to drive the air compressor, and the exhaust steam exits the turbine at the desired state. Air into the compressor is at the ambient conditions, 100 kPa, 20°C. Give the required steam inlet pressure and temperature, assuming that both the turbine and the compressor are reversible and adiabatic. Solution:
C.V. Each device. Steady flow. Both adiabatic (q = 0) and actual devices (sgen > 0) given by ηsT and ηsc.
Steam turbine Air compressor
Air Eq.8.32, T4s = T 3 (P 4 /P 3 )
k- k (^) = 293.
= 464.6 K
. WCs =
m 3 (h 3 - h4s) = 0.1 × 1.004(293.2 - 464.6) = -17.21 kW . WCs =
m 3 (h 3 - h 4 ) =
WCs /ηsc = -17.2/0.80 = -21.5 kW Now the actual turbine must supply the actual compressor work. The actual state 2 is given so we must work backwards to state 1. . WT = +21.5 kW =
m 1 (h 1 - h 2 ) = 0.5(h 1 - 2706.6) ⇒ h 1 = 2749.6 kJ/kg Also, ηsT = 0.80 = (h 1 - h 2 )/(h 1 - h2s) = 43/(2749.6 - h2s)
⇒ h2s = 2695.8 kJ/kg 2695.8 = 504.7 + x2s(2706.6 - 504.7) => x2s = 0. s2s = 1.5301 + 0.9951(7.1271 - 1.5301) = 7.0996 kJ/kg K (s 1 = s2s, h 1 ) → P 1 = 269 kPa , T 1 = 143.5°C
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Assume an actual compressor has the same exit pressure and specific heat transfer as the ideal isothermal compressor in Problem 9.23 with an isothermal efficiency of 80%. Find the specific work and exit temperature for the actual compressor. Solution: C.V. Compressor. Steady, single inlet and single exit flows. Energy Eq.6.13: hi + q = w + he; Entropy Eq.9.8: si + q/T = se Inlet state: Table B.5.2, hi = 403.4 kJ/kg, si = 1.8281 kJ/kg K Exit state: Table B.5.1, he = 398.36 kJ/kg, se = 1.7262 kJ/kg K q = T(se – si) = 273.15(1.7262 – 1.8281) = - 27.83 kJ/kg w = 403.4 + (-27.83) – 398.36 = -22.8 kJ/kg From Eq.9.28 for a cooled compressor wac = wT /η = - 22.8/0.8 = 28.5 kJ/kg Now the energy equation gives he= hi + q – wac = 403.4 + (-27.83) + 28.5= 404.
Te ac ≈ 6 ° C Pe = 294 kPa
Explanation for the reversible work term is in Sect. 9. Eqs. 9.13 and 9.
e,s i
P
v
T
s
e,s
i
e,ac^ e,ac
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Air enters an insulated turbine at 50°C, and exits the turbine at - 30°C, 100 kPa. The isentropic turbine efficiency is 70% and the inlet volumetric flow rate is 20 L/s. What is the turbine inlet pressure and the turbine power output? Solution: C.V.: Turbine, ηs = 0.7, Insulated Air table A.5: Cp = 1.004 kJ/kg K, R = 0.287 kJ/kg K, k = 1.
Inlet: Ti = 50oC, V
i = 20 L/s = 0.02 m (^3) /s ;
m
= PV
/RT = 100 × 0.02/(0.287 × 323.15) = 0.099 kg/s Exit (actual): Te = -30oC, Pe = 100 kPa
1 st Law Steady state Eq.6.13: qT + hi = he + wT; qT = 0 Assume Constant Specific Heat wT = hi - he = Cp (Ti - Te) = 80.3 kJ/kg wTs = w/η = 114.7 kJ/kg, wTs = Cp (Ti - Tes)
Solve for Tes = 208.9 K
Isentropic Process Eq.8.32: Pe = Pi (Te / Ti)
k k-1 (^) => P i = 461 kPa
W
T = m
wT = 0.099 × 80.3 = 7.98 kW
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Air enters an insulated compressor at ambient conditions, 100 kPa, 20°C, at the rate of 0.1 kg/s and exits at 200°C. The isentropic efficiency of the compressor is 70%. What is the exit pressure? How much power is required to drive the compressor? Assume the ideal and actual compressor has the same exit pressure. Solution: C.V. Compressor: P 1 , T 1 , Te(real), ηs COMP known, assume constant CP Energy Eq.6.13 for real: -w = CP0 (Te - Ti) = 1.004(200 - 20) = 180.
Ideal -ws = -w × ηs = 180.72 × 0.70 = 126. Energy Eq.6.13 for ideal: 126.5 = CP0 (Tes - Ti) = 1.004(Tes - 293.2), Tes = 419.2 K Constant entropy for ideal as in Eq.8.23:
Pe = Pi(Tes/Ti)
k k-1 (^) = 100(419.2/293.20) 3.5^ = 349 kPa
WREAL =
m(-w) = 0.1 × 180.72 = 18.07 kW
P
v
T
s
e, s
i s = C (^) i
e, s e, ac e, ac
P
P
e
i
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A nozzle in a high pressure liquid water sprayer has an area of 0.5 cm^2. It receives water at 250 kPa, 20°C and the exit pressure is 100 kPa. Neglect the inlet kinetic energy and assume a nozzle isentropic efficiency of 85%. Find the ideal nozzle exit velocity and the actual nozzle mass flow rate.
Solution: C.V. Nozzle. Liquid water is incompressible v ≈ constant, no work, no heat transfer => Bernoulli Eq.9. 1 2 V
2 ex – 0 = v(Pi^ - Pe) = 0.001002 ( 250 – 100) = 0.1503 kJ/kg
V ex = 2 × 0.1503 × 1000 J/kg = 17.34 m s - This was the ideal nozzle now we can do the actual nozzle, Eq. 9. 1 2 V
2 ex ac =^ η^
2 V
2 ex = 0.85^ ×^ 0.1503 = 0.12776 kJ/kg
V ex ac = 2 × 0.12776 × 1000 J/kg = 15.99 m s -
m
= ρA V ex ac = A V ex ac/v = 0.5 × 10 -4^ × 15.99 / 0.001002 = 0.798 kg/s
These are examples of relatively low pressure spray systems.
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Air flows into an insulated nozzle at 1 MPa, 1200 K with 15 m/s and mass flow rate of 2 kg/s. It expands to 650 kPa and exit temperature is 1100 K. Find the exit velocity, and the nozzle efficiency. Solution: C.V. Nozzle. Steady 1 inlet and 1 exit flows, no heat transfer, no work.
Energy Eq.6.13: hi + (1/2) V
2 i = he^ + (1/2) V
2 e Entropy Eq.9.8: si + sgen = se Ideal nozzle sgen = 0 and assume same exit pressure as actual nozzle. Instead of using the standard entropy from Table A.7 and Eq.8.19 let us use a constant heat capacity at the average T and Eq.8.23. First from A.7.
Cp 1150 =
1200 - 1100 = 1.166 kJ/kg K; Cv = Cp 1150 - R = 1.166 - 0.287 = 0.8793, k = Cp 1150 /Cv = 1. Notice how they differ from Table A.5 values.
Te s = Ti (Pe/Pi)
k- k (^) = 1200
= 1079.4 K
1 2 V
2 e s =^
1 2 V
2 i + C(Ti - Te s) =^
1 2 ×^15
2 + 1.166(1200 – 1079.4) × 1000
= 112.5 + 140619.6 = 140732 J/kg ⇒ V e s = 530.5 m/s Actual nozzle with given exit temperature 1 2 V
2 e ac =^
1 2 V
2 i + hi^ - he ac^ = 112.5 + 1.166(1200 – 1100)^ ×^1000 = 116712.5 J/kg ⇒ V e ac = 483 m/s
η (^) noz = ( 12 V 2 e ac -^
1 2 V
2 i )/ (^
1 2 V
2 e s -^
1 2 V
2 i ) =
= (hi - he, AC )/(hi - he, s ) =
140619.6 =^ 0.
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A nozzle is required to produce a flow of air at 200 m/s at 20°C, 100 kPa. It is estimated that the nozzle has an isentropic efficiency of 92%. What nozzle inlet pressure and temperature is required assuming the inlet kinetic energy is negligible? Solution: C.V. Air nozzle: Pe, Te(real), Ve(real), ηs(real)
For the real process: hi = he + V 2 e /2^ or
Ti = Te + V 2 e /2CP0 = 293.2 + 200
2 /2 × 1000 × 1.004 = 313.1 K
For the ideal process, from Eq.9.29:
V
2 es/2 =^ V
2 e /2ηs = 200
(^2) /2 × 1000 × 0.92 = 21.74 kJ/kg
and hi = hes + ( V
2 es/2)
Tes = Ti - V 2 es/(2CP0 )^ = 313.1 - 21.74/1.004 = 291.4 K The constant s relation in Eq.8.23 gives
⇒ Pi = Pe (Ti/Tes)
k k-1 (^) = 100
= 128.6 kPa
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