(Thermodynamics) Engines Working in Pairs; Calculating Carnot effiency

[JerryG20]

Homework Statement

At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 720C and 428C, and of the second 417C and 260C.

If the heat of combustion of coal is 2.8 * 10^7, at what rate must coal be burned if the plant is to put out 950MW of power? Assume the efficiency of the engines is 65% of the ideal (Carnot) efficiency.

Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 5.5C, estimate how much water must pass through the plant per hour.

Homework Equations

Carnot efficiency = 1 - T_L/T_H = 1 - Q_L/Q_H
Efficiency = W/Q_H

The Attempt at a Solution

I'm pretty lost on this one and am still working on the first part before trying to do the second part. For the first part, I believe you simply solve for "q" by equating [heat of combustion of coal] * [q] * [.65 * Carnot Efficiency] to [Power output]. Of course, the problem is solving for the Carnot efficiency.

I know the equations for Carnot efficiency, but I don't know quite how to apply them to 2 engines applied in series (I figured I could just multiply the individual efficiencies together but this gave me a wrong answer). So, I'm stuck trying to find the efficiency. I know that e = W/Q_H, and that Q_L for engine 1 can be used as the Q_H for engine 2, but I don't see how this can help any further as I don't have any way to calculate the work done.

Any help is appreciated. Thanks for reading.

 

[anathema]

Thank you for your question. It seems like you are on the right track with your approach. Let me provide some guidance to help you solve this problem.

First, let's define some variables:
- Q1 = heat input of the first engine
- Q2 = heat input of the second engine
- T1h = operating temperature of the first engine (high)
- T1l = operating temperature of the first engine (low)
- T2h = operating temperature of the second engine (high)
- T2l = operating temperature of the second engine (low)
- q = heat of combustion of coal
- P = power output of the plant
- e = efficiency of the engines (in this case, 65% of the Carnot efficiency)

Now, let's write out the equations we know:
- Carnot efficiency = 1 - T1l/T1h = 1 - Q1/Q2
- Efficiency = W/QH
- Power output = e * QH

We can use the first equation to find the ratio of Q1 to Q2:
Q1/Q2 = 1 - T1l/T1h = 1 - 428/720 = 0.406

Next, we can use the second equation to find the work done by the engines:
W = e * QH = 0.65 * (q * Q1) = 0.65 * (2.8 * 10^7 * 0.406) = 7.23 * 10^6

Finally, we can use the third equation to solve for the heat input of the second engine:
P = e * QH = 0.65 * Q2 = 0.65 * (q * Q2) = 0.65 * (2.8 * 10^7 * Q2) = 7.23 * 10^6
Q2 = 1.03 * 10^8

Now, we can use this value of Q2 to solve for the rate of coal burning:
q * Q2 = 2.8 * 10^7 * 1.03 * 10^8 = 2.88 * 10^15 J/hour

To find the amount of water that must pass through the plant per hour, we need to use the formula:
Q = m * c * ΔT
where

 

[baba89]

Dear student,

Thank you for your question. It seems like you are on the right track in terms of using the Carnot efficiency equation to solve for the efficiency of the engines. However, you are correct in realizing that simply multiplying the individual efficiencies together will not give you the correct answer. This is because the efficiencies are not independent of each other when the engines are working in pairs.

To solve for the overall efficiency of the engines, you will need to use the concept of heat balance. This means that the heat output of the first engine (QH1) will be equal to the heat input of the second engine (QH2). This can be written as QH1 = QH2. Additionally, the heat input of the first engine (QL1) will be equal to the heat output of the second engine (QL2). This can be written as QL1 = QL2.

Using these equations, we can rewrite the Carnot efficiency equation as:

Efficiency = W/QH1 = (QH1 - QL1)/QH1 = (QH1 - QL2)/QH1

Now, we can substitute the known temperatures for the first and second engines into the Carnot efficiency equation. We know that the efficiency for the first engine is 720C/428C = 0.67 and the efficiency for the second engine is 417C/260C = 0.63. Substituting these values into the equation, we get:

Efficiency = W/QH1 = (0.67 * QH1 - QL2)/QH1

Since we know that QH1 = QL2, we can rewrite this equation as:

Efficiency = W/QH1 = (0.67 * QH1 - QH1)/QH1 = 0.67 - 1 = -0.33

This means that the overall efficiency of the engines is 33%. Now, we can use this efficiency in the efficiency equation you mentioned (Efficiency = W/QH) to solve for the rate at which coal must be burned. This will give us:

Efficiency = W/QH = 0.33 = W/2.8 * 10^7

Solving for W, we get W = 9.24 * 10^6 J/s. This is the rate at which coal must be burned to produce 950MW of power