# Maximum power output

1. Jan 1, 2010

### vladimir69

1. The problem statement, all variables and given/known data
A solar-thermal power plant is to be built in a desert location where the only source of cooling water is a small creek with average flow rate of 100kg/s and an average temperature of 30 degrees C. The plant is to cool itself by boiling away the entire creek. If the maximum temperature achieved in the plant is 500K, what is the maximum electric power output it can sustain without running out of cooling water?

2. Relevant equations
$$Q=mcdT$$

$$\frac{dQ}{dt}=\frac{dm}{dt}cdT$$

$$\epsilon=1-\frac{T_{c}}{T_{h}}=1-\frac{Q_{c}}{Q_{h}}=\frac{W}{Q_{h}}$$

$$W=Q_{h}-Q_{c}$$

$$\frac{dW}{dt}=\frac{Q_{h}}{dt}-\frac{Q_{c}}{dt}$$
3. The attempt at a solution
Not sure which numbers to substitute where so here goes

$$\frac{dm}{dt}=100$$

$$\frac{dQ_{c}}{dt}=100\times 4184 \times 70 = 29288000$$

$$\epsilon_{max}=1-303/500 = 0.394$$

$$Q_{h}= \frac{Q_{c}}{1-\epsilon} = 29288000/0.606 = 48330033$$

$$W= 48330033 - 29288000 = 19042033$$

Which gives my answer as the max power output as 19MW.
The answer in the book is 166MW.
I'm not sure of many other reasonable combinations to stick these numbers in to pop out 166MW.

Thanks

2. Jan 1, 2010

### Adjuster

I think you are heating 100kgs-1 of water through 70C, but are you also boiling it?

What is the latent heat of vaporisation?

3. Jan 1, 2010

### RTW69

Not only do you need the latent heat of vaporization but you need to heat the steam from 100 C to steam at 227 C

So you are taking water from 30 C to 100 C, Then you are taking water at 100 C and heating it to steam at 100 C and then taking steam at 100 C and heating it to steam at 227 C.

4. Jan 2, 2010

### vladimir69

yeah that worked thanks Adjuster
not sure about your explanation RTW69, the answer in the book agrees with the calculations of just eating the water to 100 C and boiling it.

thanks lads

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