A town is planning on using the water flowing through a river at a rate of 5.0X10^6 kg/s to carry away the heat from a new power plant. Environmental studies indicate that the temperature of the river should only increase by 0.50ºC. The maximum design efficiency for this plant is 30.0%. What is the maximum possible power this plant can produce?
Equations you will need:
Q = mc deltaT
e = 1 - (Qc/Qh)
The first thing you need to realize is that when you're calculating Q it's Qc, because that is the heat going into the water and therefore out of the power station. So then you can plug that into e = 1- (Qc/Qh) and solve for Qh.
That isn't the answer yet, though, because it isn't asking how much heat the engine uses (input heat is Qh). It's asking how much total energy it can produce.
And now it's very simple because you know Qh, the total heat it uses, and you know the efficiency, the amount of that heat that can be used to do work. So you multiply Qh by the efficiency and THAT's the answer.
To find the maximum possible power that the plant can produce, we need to calculate the heat energy that can be extracted from the water flowing through the river and then convert it to power.
The formula to calculate heat energy is:
Q = mcΔT
Where:
Q = heat energy (in Joules)
m = mass flow rate (in kg/s)
c = specific heat capacity of water (in J/kg°C)
ΔT = change in temperature (in °C)
In this case, we are given:
m = 5.0 × 10^6 kg/s
ΔT = 0.50°C
The specific heat capacity of water, c, is approximately 4.18 J/g°C, which can be converted to J/kg°C by multiplying by the density of water, which is 1000 kg/m³.
Let's calculate the heat energy extracted from the water:
Q = (5.0 × 10^6 kg/s) × 4.18 J/g°C × 1000 kg/m³ × 0.50°C
Simplifying this expression:
Q = (5.0 × 10^6 kg/s) × 4.18 J/g°C × 1000 kg/m³ × 0.50°C
= 10^6 × 5.0 × 4.18 × 10^3 J/s
= 20.9 × 10^9 J/s
= 20.9 GW (gigawatts)
Now, we can calculate the maximum possible power output of the plant. The efficiency (η) is given as 30% or 0.30.
Power (P) = η × Q
P = 0.30 × 20.9 GW
= 6.27 GW
Therefore, the maximum possible power this plant can produce is 6.27 gigawatts.