A nuclear power plant generates (output) 500 mw. It is 34% efficient. The waste heat going in the Connecticut River with an average flow of 3*10 exp 4 (kg/sec). How much does the water temperature rise?
To calculate the rise in water temperature, we can use the heat transfer equation:
Q = m * Cp * ΔT
Where:
Q = Heat transferred (in Joules)
m = Mass of water (in kg)
Cp = Specific heat capacity of water (in J/kg·°C)
ΔT = Change in temperature (in °C)
Given information:
Power output of the nuclear power plant = 500 MW
Plant efficiency = 34% = 0.34 (decimal)
Water flow rate in the Connecticut River = 3 * 10^4 kg/sec
First, we need to calculate the heat transfer from the power plant to the river. The heat transfer can be determined by using the following formula:
Q = P / η
Where:
Q = Heat transfer (in watts)
P = Power output of the power plant (in watts)
η = Efficiency of the power plant
Plugging in the values, we find:
Q = 500 MW / 0.34 = 1470.59 MW
Now, we convert the heat transfer into watts:
Q = 1470.59 MW * 10^6 W/MW = 1.47 * 10^9 W
Next, we can calculate the rise in temperature using the heat transfer equation. Rearranging the equation, we have:
ΔT = Q / (m * Cp)
Substituting the known values:
ΔT = (1.47 * 10^9 J/s) / (3 * 10^4 kg/s * Cp)
To proceed further, we need to know the specific heat capacity of water (Cp). The specific heat capacity of water is approximately 4186 J/kg·°C.
Substituting this value:
ΔT = (1.47 * 10^9 J/s) / (3 * 10^4 kg/s * 4186 J/kg·°C)
ΔT = 11.06 °C
Therefore, the water temperature would rise by approximately 11.06°C as a result of the waste heat transferred from the nuclear power plant to the Connecticut River.