# Homework Help: Thermodynamics question

1. Oct 9, 2014

### liyifanlimeng

A combined heat and power (CHP) plant consumes 500 MW of exergy of a fuel and can operate in two modes. In mode 1 it produces 300 MW of electricity; hence, it operates with a second law efficiency Pmode1=60%. In mode 2 (CHP) it produces 250 MW of electricity and 200 MW of heat, by heating a stream of pressurized water for district heating from 60 to 120C; hence, it operates with rst law efficiency of about Pmodel2= 90%.
(a) Explain why the first law efficiency in mode 2 is about 90% and not exactly 90%.
(b) Assuming an ambient temperature Ta = 300 K, what is the exergy difference of the pressurized water between the stream at 120C fed into the district heating pipes and the return stream at 60C.
(c) What is the second law efficiency in mode 2?
(d) Using the exergy allocation method, allocate the fuel consumption in mode 2 (CHP)among the heat and power produc-tions, and compute the corresponding first law efficiency, P
, of power production and coeffcient of performance, COP, of heat production.

2. Relevant equations
first law efficiency=W/Q second law efficiency=W/Qa(1-Tb/Ta) Exq=Q(1-Ta/Tq)
log-mean delivery temperatureTq=(Tfeed-Treture)/ln(Tfeed/Treturn) COP=Qb/Qfuel

3. The attempt at a solution
a,I suppose this is because W is not an exact number.
b,Ta=300K,Q=200mw ,Ex2-Ex1=Q(1-Ta/T2)-Q(1-Ta/T1)=200(1-300/393.15)-200(1-300/333.15)
c,TQ=(Tfeed-Treture)/ln(Tfeed/Treturn)=120-60/ln393.15/333.15=362.33k Exq=Q(1-T2/Tq)=200(1-200/362.33)=78gwh second law efficiency=W/Qa(1-T2/Tq)=250/78
d,totally have no idea

Last edited by a moderator: Oct 9, 2014
2. Oct 14, 2014