Question

At a nuclear power plant water at 8°c from a lake is used in a heat exchanger to condense spent steam at a temperature of 120°c,to water at 85°c before it is recycled to the reactor. If the cooling water returns to the lake at a temperature of 19°c,how many kilogram of water are needed per kilogram of steam ? Ignore the pressure change in steam.

Answer 1

Q(gained) =Q(lost)
Q(gained) =Q₁+Q₂+Q₃
Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J
Q₂ = Lm=2256000•1= 2256000 J
Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J
Q(gained) =40400 +2256000+62790 =2359190 J
Q(lost) =c₂•M•(Δt₃)=4186•M•(19-8) = 4186•11•M
M= Q(gained)/ 4186•11=2359190/4186•11= 51.2 kg

Answer 2

Can you please explain it to me...that what's happening in all the process?

Answer 3

Q(gained) =Q(lost)
Q(lost) =Q₁+Q₂+Q₃
Q₁ = c₁m(Δt₁)=2020•1•(120-100) = 40400 J
Q₂ = Lm=2256000•1= 2256000 J
Q₃=c₂m(Δt₂)=4186•1•(100-85) = 62790 J
Q(lost) =40400 +2256000+62790 =2359190 J
Q(gained) =c₂•M•(Δt₃)=4186•M•(19-8) = 4186•11•M
M= Q(lost)/ 4186•11=2359190/4186•11= 51.2 kg
The processes in which there is a loss of energy :
steam cooling up to 100⁰ (Q₁), ‘steam-water’ transformation (Q₂), and water cooling up to 85⁰ (Q₃)
This energy heat the water from the lake from 8⁰ to 19⁰ (Q(gained))