A fossil-fuel power plant that generates 1000MW of electrical power discharges 1600MJ per second as waste heat into the river. A)At what rate(in MW) is thermal energy obtained fromt the burning of coal? B)What is the thermal efficiency(power output/heat input) of the plant? C) If the river flow is at 5x10^8 kf of water per hour, what is the temperature rise of the river due to the thermal polution?
A) To find the rate at which thermal energy is obtained from the burning of coal, we need to equate the waste heat discharged into the river to the thermal energy obtained.
Given:
Power output of the plant = 1000 MW
Waste heat discharged = 1600 MJ/s
1 MW = 1 * 10^6 J/s
Thus, 1000 MW = 1000 * 10^6 J/s
Since the waste heat is given in MJ (MegaJoules), we need to convert it to Joules. There are 10^6 Joules in 1 MegaJoule.
1600 MJ/s = 1600 * 10^6 J/s
Therefore, the rate at which thermal energy is obtained from the burning of coal is 1600 * 10^6 J/s or 1.6 * 10^9 J/s or 1.6 GW (Gigawatts).
B) To find the thermal efficiency of the plant, we divide the power output by the heat input (thermal energy obtained).
Thermal Efficiency = Power output / Heat input
Power output = 1000 MW
Heat input = 1600 MJ/s
Heat input = 1600 * 10^6 J/s
Thermal Efficiency = (1000 * 10^6 J/s) / (1600 * 10^6 J/s)
= 0.625 or 62.5%
Therefore, the thermal efficiency of the plant is 62.5%.
C) To find the temperature rise of the river due to thermal pollution, we need to calculate the amount of heat transferred to the river per hour and then use the specific heat capacity and mass of water.
Given:
River flow rate = 5 * 10^8 kg of water per hour
We know that heat transferred = mass * specific heat capacity * temperature rise.
First, let's calculate the heat transferred per hour:
Heat transferred = 1600 MJ/s * 3600 s (to convert to per hour)
= 5.76 * 10^9 MJ/hour
Now, we need to convert this to Joules since specific heat capacity is usually given in Joules.
1 MJ = 10^6 J
Therefore, Heat transferred = 5.76 * 10^9 J/hour
Now let's calculate the temperature rise:
Specific heat capacity of water = 4.186 J/g°C (approximated)
Since we're given the river flow rate in kg of water per hour, we need to convert Joules to grams using the mass of water:
1 kg of water = 1000 g of water
Heat transferred = mass * specific heat capacity * temperature rise
Temperature rise = Heat transferred / (mass * specific heat capacity)
mass = 5 * 10^8 kg (given)
temperature rise = (5.76 * 10^9 J/hour) / (5 * 10^8 kg * 4.186 J/g°C)
≈ 274 °C
Therefore, the temperature rise of the river due to thermal pollution is approximately 274 °C.