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.
To determine the amount of water needed per kilogram of steam, we need to calculate the heat exchange between the steam and the cooling water.
The heat exchange can be calculated using the equation:
Q = m1 * c1 * (T2 - T1)
Where:
Q = Heat exchange (in Joules)
m1 = Mass of cooling water (in kg)
c1 = Specific heat capacity of water (in J/kg°C)
T2 = Final temperature of the cooling water (in °C)
T1 = Initial temperature of the cooling water (in °C)
First, let's calculate the heat exchange between the cooling water and the steam.
Q = m1 * c1 * (T2 - T1)
Q = m1 * 4186 * (85 - 19)
Here, the specific heat capacity of water (c1) is taken as 4186 J/kg°C.
Now, to convert the heat exchange into the mass of water needed per kilogram of steam, we use the equation:
Q = m2 * h
Where:
m2 = Mass of water per kilogram of steam (in kg)
h = Latent heat of vaporization of steam (in J/kg)
The latent heat of vaporization of steam is typically given as 2,257,000 J/kg.
Hence,
m2 * 2,257,000 = m1 * 4186 * (85 - 19)
Simplifying the equation, we can calculate:
m2 = (m1 * 4186 * (85 - 19)) / 2,257,000
Substituting the known temperatures:
m2 = (m1 * 4186 * (66)) / 2,257,000
Simplifying the equation further:
m2 = (m1 * 275076) / 2,257,000
Therefore, to determine the mass of water needed per kilogram of steam, we need to know the mass of the cooling water (m1) brought in from the lake.