How Does the 2nd Law of Thermodynamics Apply to a Power Station's Efficiency?

[physics_geek]

Homework Statement

An electric generating station is designed to have an electric output power 1.20 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 98°C.

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

(b) Find the exhaust power for Th = 838°C.


(c) Find the value of Th for which the exhaust power would be only half as large as in part (b).

Homework Equations

P=W/t e= 1- tc/th

The Attempt at a Solution

ok i really don't know where to start on this one
any kind of help would be appreciated..thanks!

 

[Andrew Mason]

Homework Statement

An electric generating station is designed to have an electric output power 1.20 MW using a turbine with two-thirds the efficiency of a Carnot engine. The exhaust energy is transferred by heat into a cooling tower at 98°C.

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

(b) Find the exhaust power for Th = 838°C.


(c) Find the value of Th for which the exhaust power would be only half as large as in part (b).

First of all, you have to determine how much heat per unit of time the station has to exhaust. If the efficiency if 2/3 of a Carnot engine operating between Th and 98C what is the efficiency? (be careful to convert temps to K).

How much heat is needed to produce 1.2 MW? So how much heat is left over after you subtract the 1.2 MW output?

AM

 

[physics_geek]

so does that mean the efficiency would be
e = 2/3(1- 371/Th) ??

what do i do with this..it asks for the rate at which it exhausts energy by heat

 

[Griffin0991]

Trying to figure this part out aswell. Any additional direction would be appreciated.

 

[Andrew Mason]

so does that mean the efficiency would be
e = 2/3(1- 371/Th) ??

That looks right.

what do i do with this..it asks for the rate at which it exhausts energy by heat

You are trying to found Q_out.
Use the definition of efficiency (ie in terms of Work done to Q_in) to give you Q_in. How is the work done related to Q_in and Q_out?

AM

 

[physics_geek]

ok yea i don't understand this at all

so if 2/3(1-371/Th) is right..wat am i solving for..that just gives me the efficiency
im so confused!

 

[Meloh S]

Thanks Andrew, that helped me.

ok yea i don't understand this at all

so if 2/3(1-371/Th) is right..wat am i solving for..that just gives me the efficiency
im so confused!

You're solving for Q_out, which is like the energy you put in minus the work that you put out. I was solving for Q_in so that part confused me.

However, if you can get Q_in then getting Q_out is just one step away.

Use the definition of efficiency (ie in terms of Work done to Qin) to give you Qin. How is the work done related to Qin and Qout?

If you can do this then the problem is solved. You have the right efficiency, but that efficiency can also equal something in terms of work and Q_in <--- useful.