# Reversible Carnot Cycle and Heat Engine

1. Jan 4, 2012

### mm391

1. The problem statement, all variables and given/known data

A borehole produces 120kg/s of high pressure water at 165°C. A powerplant is to be designed to extract heat from the water, reducing the temp to 147°C. The plant which will initially be assumed to be reversible, will reject heat at 17°C to a river. Take 147°C as the highest temp available and assume the river is so large that there is negligible change in temp. Specific Heat Capacity of high pressure water is 4.2kj/kgK.

Work out:
a) Rate of heat input available to the plant.
b) Rate at which heat is rejected
c) Maximum power out from the plant and its efficiency.

2. Relevant equations

QL/QH = TL/TH

3. The attempt at a solution

I am not sure where or how to begin. Any help pointing me in the right direction for each question would be much appreciated.

2. Jan 4, 2012

### ShamelessGit

I am not sure your question makes sense. What do you mean by "to reject heat?" Are you sure that 17 C is not the temperature of the river? If the process reduces the temperature from 165 C to 147 C, then isn't 165 the highest temperature, and not 147? I would not know where to start either. It does not make sense.

3. Jan 6, 2012

### Su3liminal

For part a): Use this equation to calculate the heat lost from the water (the heat input to the plant): Q(rate) = m(rate of mass flow) * specific heat * temperature change

for b): Since the plant is assumed to be reversible, use the equation: Qh(rate)/QL(rate) = Th/TL

for c): The maximum power output results from a reversible cycle (or a process), which is the case here. Now since the efficiency of a reversible cycle is a function of temperature ONLY, use the equation: η(rev) = (Th-TL)/(Th)
After you find the maximum efficiency, use the equation: η = [Wnet,output(rate)]/Qh