# Thermodynamics- Heat engine and refrigeration

1. Dec 3, 2012

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

A nuclear power plant generates 2000 MW of heat energy from nuclear reactions in the reactor's core. This energy is used to boil water and produce high-pressure steam at 300° C. The steam spins a turbine, which produces 700 MW of electric power, then the steam is condensed and the water is cooled to 30° C before starting the cycle again.

A: What is the maximum possible thermal efficiency of the power plant?

B: What is the plant's actual efficiency?

C: Cooling water from a river flows through the condenser (the low-temperature heat exchanger) at the rate of $1.2 \times 10^{8}\;{\rm L}/{\rm hr}$ ( approx 30 million gallons per hour). If the river water enters the condenser at $18^\circ \rm C$, what is its exit temperature in C°?

2. Relevant equations

Q=McΔT

3. The attempt at a solution

I have solved the first two problems.

A = 47.1%
B = 35%

It is C that I am having difficulties with.

If 2,000 MW are going into the system, and only 700 MW are used, then 1,300 MW should be fed into the cold reservoir.

Solving for Joules per second.

$1,300MW=1.3*10^6 \frac{J}{s}$

Solving for volume per second.

$1.28*10^8\frac{L}{hr}*\frac{1hr}{3,600s}*\frac{1m^3}{1,000L}=\frac{100}{3}\frac{m^3}{s}$

Solving for mass.

$(M)kg=(1,000\frac{kg}{m^3})(\frac{100}{3}\frac{m^3}{s})=\frac{100,000}{3}kg/s$

$Q=Mc\Delta T$

$1.3*10^6 J = (\frac{100,000}{3}kg)(4190\frac{J}{kg*K})(T_f K-291 K)$

Solving for this, I get

$$\frac{1.3*10^6 J}{(\frac{100,000}{3}kg)(4190 \frac{J}{kg*K})}+291 K = T_f = 291.0093 K = 18.009^{\circ} C$$

As this is not the correct answer, I am definitely doing something wrong.

Any help would be appreciated.
Mac

2. Dec 4, 2012

### SteamKing

Staff Emeritus
How many watts is 1300 MW? It's 1300 * 10^6 W

3. Dec 4, 2012