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.
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
Can you please explain it to me...that what's happening in all the process?
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))
To find the amount of water needed per kilogram of steam, we need to use the principle of energy conservation.
The energy gained by the cooling water is equal to the energy lost by the steam, based on the principle of energy conservation.
Let's assume the mass of cooling water required per kilogram of steam is represented by m, in kilograms.
To determine the energy lost by the steam, we need to calculate the difference in enthalpy between the initial and final states.
Enthalpy (H) is defined as the sum of internal energy (U) and the product of pressure (P) and volume (V).
For the steam:
Initial enthalpy, H1 = enthalpy of steam at 120°C = hf1 + Cp(T1 - 100)
Final enthalpy, H2 = enthalpy of water at 85°C = hf2 + Cp(T2 - 100)
Considering the pressure remains constant, we ignore the pressure change in steam.
The energy lost by steam (Q1) can be calculated using the equation:
Q1 = m * (H1 - H2)
Now, considering the cooling water:
Initial enthalpy, H3 = enthalpy of water at 8°C = hf3 + Cp(T3 - 100)
Final enthalpy, H4 = enthalpy of water at 19°C = hf4 + Cp(T4 - 100)
The energy gained by the cooling water (Q2) can be calculated using the equation:
Q2 = m * (H4 - H3)
Since energy is conserved, Q1 = Q2
m * (H1 - H2) = m * (H4 - H3)
By substituting the enthalpy equations and rearranging, we can solve for m:
m = (H4 - H3) / (H1 - H2)
Now, we need to look up the specific enthalpies of steam and water at the given temperatures. These values can be found in tables in thermodynamics textbooks or online databases.
Substituting the specific enthalpies, we can calculate the mass of cooling water required per kilogram of steam.