Thermodynamics - efficiency of power plant

[fredrick08]

Homework Statement

a power plant generates 2000MW of heat energy from nuclear reactions in the reactors core. this energy is used to boil water and produce hight pressure steam at 300degreeC. the steam spins a turbine, which produces 700MW of electric power, then the steam is condensed and the water is cooled to 30degreeC before starting the cycle again.

a.what is the max possible thermal efficiency of the power plant?
b.what is the plants actual efficiency?
c.cooling water from a river flows through the condenser at the rate of 1.2x10^8L/hr. if the river water enters the condenser at 18degreeC what is is exit temperature?

Homework Equations

efficiency(carnot)=1-(Tc/Th)
efficiency=Wout/Qh
Q=mc$$\Delta$$T

The Attempt at a Solution

a. 1-(303/573)=47%
b. Wout/Qh=Power/(Qh/t)=7x10^8/2x10^9=35%
c. now the others were easy (i think lol) this one kinda gets me) what I've done is,
Power=(Qh/t)-(Qc/t) therefore (Qc/t)=(Qh/t)-Power=1.3x10^9W
then 1.2x10^8L/hr*3600s=33333.33L/s
so Q=mc$$\Delta$$T
so Tf=(Q/(mc))+Ti=(1.3x10^9/33333.33*4190)+291=44.2degreeC

now i seen this same question in another thread, but their specific heat was 4160? and they also did something weird, i don't understand... but i don't think mine is right, can anyone help me, or confirm that i am right?

 

[fredrick08]

anyone have any ideas?

 

[Andrew Mason]

Your analysis and answer is correct. You can look up the specific heat of water. It is one calorie/gram K or 4186 J/Kg K at 20C. There is some temperature dependence on this but I wouldn't worry about it.

AM

 

[fredrick08]

ok, thankyou very much for your confirmation = )

 

[fredrick08]

ok, I am just kinda thinking back on my work now, but does 44.2degreeC sound realistic? the river water goes in at 18degreeC and gets heated by the condensing steam then leaves but would it really heat up by 26degreeC? so that river water doesn't then cool to 30degreeC and goes in the cycle does it? i guess that kinda makes sense...

 

[fredrick08]

anyone?

 

[Andrew Mason]

Check your arithmetic. I get an increase of 9 degrees for a final temperature of 27 degrees.

3.3 x 10^4 Litres pass through each second and the plant must release 1.3 x 10^9 Joules per second. So each litre of water must absorb 3.9 x 10^4 Joules. Since absorbtion of 4.2 x 10^3 Joules raises the temperature of one litre by 1 degree, the water temp. rises 9 degrees when absorbing this heat.

AM

 

[fredrick08]

yes ok i completely understand wat ur saying, cept where does the 4.2x10^3 J come from, how do i get it? and how would i write that? since its not really using the formula...

 

[fredrick08]

omg, I am sooo srry, stupid dyslexic me, c as 1490, instead 4190... srry, ok now i g 27.3degeeC, thanks heaps