Carnot Engine ~ Electric Generating station

[Coti1002]

Homework Statement

An electric generating station is designed to have an electric output power of 1.4 MW using a turbine with two-thirds efficiency of a Carnot engine. The Exhaust energy is transferred by heat to a cooling tower at 110°C.

a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature Th

Homework Equations

e= w/Qh=1-Qc/Qh , e_carnot=1-Tc/Th, P=W/t

The Attempt at a Solution

The only thing i have been able to figure out so far is e=(2/3)(1-(Tc/Th) Where Tc=383K

I have been struggling with this one for quite some time and humbly asking for your help

The book lists the answer as Qc/Δt=1.4((.5Th+383)/(Th-383))
I am so confused...

 

[rude man]

Book is right.

What is the efficiency of a Carnot engine as a function of T_h and T_c? Then, what is actual Q_h and therefore actual Q_c? Hint: use 1st law.

 

[Coti1002]

I'm Sorry I'm still not following you...

Man I feel like an idot.. What am i missing?

 

[Andrew Mason]

Homework Statement

An electric generating station is designed to have an electric output power of 1.4 MW using a turbine with two-thirds efficiency of a Carnot engine. The Exhaust energy is transferred by heat to a cooling tower at 110°C.

a) Find the rate at which the station exhausts energy by heat as a function of the fuel combustion temperature Th

Homework Equations

e= w/Qh=1-Qc/Qh , e_carnot=1-Tc/Th, P=W/t

The Attempt at a Solution

The only thing i have been able to figure out so far is e=(2/3)(1-(Tc/Th) Where Tc=383K

I have been struggling with this one for quite some time and humbly asking for your help

The book lists the answer as Qc/Δt=1.4((.5Th+383)/(Th-383))
I am so confused...

If the output power is 1.4 MW what is the rate at which energy is input? Hint: efficiency = output/input = (dW/dt)/(dQh/dt)

Once you get the rate of energy input, given the work output dW/dt = 1.4MW you can calculate the rate of exhaust energy Qc using W = Qh-Qc (i.e. dW/dt = dQh/dt - dQc/dt).

Since efficiency is a function of Th (Tc=110C) you will have a Th in your answer. Be sure to use the Kelvin scale for T.
AM

 

[rude man]

I'm Sorry I'm still not following you...

Man I feel like an idot.. What am i missing?

<<What is the efficiency of a Carnot engine as a function of Th and Tc? Then, what is actual Qh and therefore actual Qc? Hint: use 1st law. >>Try to answer those two questions! The first is in your textbook. Call it e.

Then, what does the 1st law say? How about Qh = Qc + W?
For your system, efficiency = 2e/3 = e' so write Qh in terms of W and e'( Th, Tc).
Then, solve for Qc. Realize that the answer to your problem is dQc/dt which will be in terms of dW/dt = P = 1.4MW as well as Th and Tc.

I can say no more.