Question #193609

  1. suppose that a heat engine is connected to two energy reservoirs, one a pool of molten aluminum (660 degrees Celsius) and the other a block of solid mercury (-38.9 degrees Celsius). The engine runs by freezing 1.00 g of aluminum and melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 X 105 J/kg; the heat of fusion of mercury is 1.18 X 104 J/kg. What is the efficiency of this engine?
  2. Argon enters a turbine at a rate of 80.0 kg/min, a temperature of 800 degrees Celsius and a pressure of 1.50 MPa. It expands adiabatically as it pushes on the turbine blades and exits at a pressure of 300 kPa.

(a) Calculate its temperature at the time

of exit.

(b) Calculate the (maximum) the

power output of the turning turbine.

(c) The turbine is one component of a

model closed-cycle gas turbine

engine . Calculate the maximum

efficiency of the engine.


Expert's answer

Solution.

1.Th=933K;1. T_h=933K;

Tc=234.1K;T_c=234.1K;

ma=1.00g;m_a=1.00g;

mm=15.0g;m_m=15.0g;

Lm=1.18⋅104J/kg;L_m=1.18\sdot 10^4J/kg;

Lf=3.97⋅105J/kg;L_f=3.97\sdot10^5J/kg;

Qc=mmLm=15⋅10−3kg⋅1.18⋅104J/kg=177J;Q_c=m_mL_m=15\sdot10^{-3}kg\sdot1.18\sdot10^4J/kg=177J;

Qh=maLa=10−3kg⋅3.97⋅105J/kg=397J;Q_h=m_aL_a=10^{-3}kg\sdot3.97\sdot10^5J/kg=397J;

Weng=Qh−Qc=220J;W_{eng}=Q_h-Q_c=220J;

e=WengQh=220J397J=0.554;e=\dfrac{W_{eng}}{Q_h}=\dfrac{220J}{397J}=0.554;

The theoretical (Carnot) efficiency is Th−TcTh=933K−234.1K933K=0.749;\dfrac{T_h-T_c}{T_h}=\dfrac{933K-234.1K}{933K}=0.749;

2.2. a)(PfVfTf)γ=(PiViTi)γ;a)(\dfrac{P_fV_f}{T_f})^\gamma=(\dfrac{P_iV_i}{T_i})^\gamma;

Tf=Ti(PfPi)(γ−1)/γ;T_f=T_i(\dfrac{P_f}{P_i})^{(\gamma-1)/\gamma}; γ=53;\gamma=\dfrac{5}{3};

Tf=1073K⋅(3⋅105Pa1.50⋅106Pa)0.4=564K;T_f=1073K\sdot(\dfrac{3\sdot10^5Pa}{1.50\sdot 10^6Pa})^{0.4}=564K;

b)ΔEint=nCVΔT=Q−Weng=0−Weng  ⟹  Weng=−nCvΔT;b) \Delta E_{int}=nC_V\Delta T=Q-W_{eng}=0-W_{eng}\implies W_{eng}=-nC_v\Delta T;

P=Wengt=−nCVΔTt=2.12⋅105W;P=\dfrac{W_{eng}}{t}=\dfrac{-nC_V\Delta T}{t}=2.12\sdot10^5W;

c)ec=1−TfTi=1−564K1073K=0.475c)e_c=1-\dfrac{T_f}{T_i}=1-\dfrac{564K}{1073K}=0.475 ;


Answer: 1.e=0.554,1. e=0.554,The theoretical (Carnot) efficiency is 0.749;

2.a)Tf=564K;2. a)T_f=564K;

b)P=2.12⋅105W;b)P=2.12\sdot10^5W;

c)ec=0.475.c) e_c=0.475.



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