2. Ammonia at temperature T1 = 27.0_C and pressure 0.1 MPa (atmospheric
pressure) flows into an apparatus where it is heated by an electric resistance of
100 ohm at a volumetric rate of 0.000041 m3/s. When a current of 0.050 A
(Ampere) flows through the resistance, the exit temperature of ammonia is
31.1_C.
a) What are the specific heat capacities, cp and cv for ammonia?
b) Note: Ammonia at ambient conditions may be considered to be an
3. Helium enters an isentropic nozzle at 3 MPa, 350_C with a negligible velocity
and exits at 1.6 MPa. The mass flow rate is 0.5 kg/s. Calculate:
a) The exit temperature.
b) The velocity of the gas at the exit of the nozzle.
c) The area at the exit of the nozzle.
You may assume that helium is an ideal gas with constant specific heats,
M = 4 kg/kmol, k = 1.67 and cp = 5.2 kJ/kg.
7. Steam Rankine cycle with superheat: The turbine entrance pressure is 15 bar
and the temperature is 400_C. The condenser pressure is 0.08 bar. The mass
flow rate is 850 kg/min. The turbine efficiency is 82% and the pump efficiency
75%. Calculate:
a) The thermal efficiency of the cycle.
b) The total power produced in MW.
c) Suggest three ways to improve the efficiency of this cycle.
9. Air in a Brayton cycle enters the turbine at 1.8 MPa and 1200_C. The compressor
entrance conditions are 0.1 MPa and 20_C. The isentropic efficiency
of the turbine is 83% and that of the compressor 78%. Calculate:
a) The thermal efficiency of the cycle.
b) The back work ratio.
c) Suggest three ways to improve the efficiency of this cycle.
Physics thermal question thx a
2013-11-09 10:03 pm
回答 (1)
2013-11-10 2:35 am
✔ 最佳答案
Q3Total P=3MPa, Total temp (T0)=350'C=623K.
Static P=1.6MPa
Isentropic relationship: T/[P^(k-1)/k] const,
(3/1.6)^(k-1)/k = 623/T
T=484K=211'C
Jet vel (Vj):
0.5(Vj)^2=Cp(T0-T)
0.5(Vj)^2=5200(623-484)
Vj=1202.33m/s (very fast)
Hence, assume nozzle is choked (Mach 1)
Compressible flow function for mass flow rate FOR HELIUM at Mach 1:
[m (Cp T0)^0.5] / [A P0] = 1.1472 (too difficult to type the formula here. Pls look up yourself)
m=0.5, Cp=5200, T0=623, P0=3E6
Hence A=2.615E-4 m^2.
Q7
Water is virtually incompressible.
Use steam tables:
Specific vol at 0.08bar = 0.001008kg/cu m
Specific vol at 15bar = 0.001154kg/cu m
Mean = 0.001081 kg/cu m (good est.)
Isentropic Pump work
=(specific vol)*(P2-P1)
=0.001081 (15-0.08)(100,000)
=1.937kJ/kg
Isen. Eff=0.75=1.937/(Actual work)
Actual work=2.583kJ/kg
Use steam tables, pump entry specific enthalpy (h1)=173.8kJ/kg (fusion at 0.08bar)
Specific enthalpy at pump exit (h2)=183.8 + 2.583=176.4kJ/kg
Use steam tables, turbine entry specific enthalpy (h3)=3256.4 kJ/kg
Turbine entry specific entropy (s3)=7.2981 kJ/kg
If turbine is isentropic, turbine exit entropy (s4s)= 7.2981 kJ/kgK
At 0.08bar, use steam tables, specific entropy ('s') of:
Fusion=0.592 kJ/kgK
Evaporation=8.227 kJ/kgK
Using ratio, dryness fraction (y):
0.592(1-y)+8.227y = 7.2981
y=0.8783 (87.8% of the water is steam)
Use steam tables, specific ENTHALPY (h) of:
Fusion=173.8 kJ/kg
Evaporation=2576.2 kJ/kg
Using ratios, isentropic turbine exit enthalpy (h4s)=2283.9 kJ/kg
Turbine eff. = 0.82= (entry h3 – real exit h4) / (entry h3 – isen. exit h4s)
0.82=(3256.4 – h4) / (3256.4 – 2283.9)
h4=2459.0 kJ/kg
Hence
Pump work=2.583 kJ/kg
Heat in = 3256.4-176.4=3080.0 kJ/kg
Turbine work=3256.4-2459.0=797.4 kJ/kg
(Note pump work is much smaller than turbine work or heat in)
Thermal eff.=(797.4-2.583)/3080.0 = 0.258
Power out (Note 850 kg PER MINUTE. Power is PER SECOND)
=(797.4 - 2.583)(850/60)kW
=11260kW
=11.26MW.
More on comments.
Sorry. I don't know how to do (2).
2013-11-09 18:37:16 補充：
Q8
Entrance=293K, Turbine entry=1473K
Isentropic relationship: T/[P^(k-1)/k] const,
Ideal compressor exit:
(1.8/0.1)^(1.4-1)/(1.4) = T2s/293
T2s=669K
Isen. Eff=0.78= (669-293) / (T2-293)
Actual compressor exit:
T2=775K
Ideal turbine exit:
(1.8/0.1)^(1.4-1)/(1.4) = 1473/T4s
T4s=645K
2013-11-09 18:38:49 補充：
Isen. Eff=0.83= (1473-T4) / (1473-645)
Actual turbine exit:
T4=785.8K
Compressor Work = 1005(775-293)=484.4kJ/kg
Heat=1005(1473-775)=701.5kJ/kg
Turbine Work=1005(1473-785.8)=690.6kJ/kg
Net work=690.6-484.4=206.2kJ/kg
Thermal eff. = 206.2/701.5=0.294
Back work ratio= Comp./Turb.=484.4/690.6=0.701
2013-11-09 18:43:43 補充：
Brayton cyc
1) Improve machine eff. Modern comp. and turb. eff around 90%.
2) Higher pressure
3) Higher turbine entry temperature.
Rankine cyc
1) Improve machine eff.
2) Higher pressure, normally 150bar
3) Reheat. e.g. HP turbine 150bar to 40bar. Then reheat to 400'C & use an LP turbine.
2013-11-10 11:46:20 補充：
(2) Power = I^2 R=0.25W.
Inlet is ideal gas.
M*R=8314.3
M of NH4=14+1*4=18
R=461.9 J/kg K
P=(rho)RT
0.1E6=(rho)(461.9)(27+273)
rho=0.7217kg/m^3
I^2 R=mass flow rate*Cp*(Tout-Tin)
0.25=[(0.7217)*(0.000041)] Cp (31.1-27)
Cp=2060.7 J/kgK
Ideal gas: Cp-Cv=R
2060.7-Cv=461.9
Cv=1598.8 J/kgK
參考: 大學工程系
收錄日期: 2021-04-13 19:48:31
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