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 find out how many kilograms of water are needed per kilogram of steam, we can use the principle of conserving energy in a heat exchanger.
The formula we can use is:
(m1 * Cp1 * (T1 - T3)) = (m2 * Cp2 * (T4 - T3))
Where:
m1 = mass flow rate of steam
Cp1 = specific heat capacity of steam
T1 = initial temperature of steam (120°C)
T3 = temperature of cooling water returning to the lake (19°C)
m2 = mass flow rate of cooling water
Cp2 = specific heat capacity of cooling water
T4 = final temperature of cooling water after transferring heat from steam (85°C)
To solve the equation, we need to find the values of Cp1 and Cp2, which are the specific heat capacities of steam and water, respectively.
The specific heat capacity of steam is usually given as a constant value of approximately 2.03 kJ/kg·°C.
The specific heat capacity of water can be taken as 4.18 kJ/kg·°C.
Now, let's plug in the given values:
(m1 * 2.03 * (120 - 19)) = (m2 * 4.18 * (85 - 19))
To simplify the equation further, we can consider the temperature difference (∆T) separately:
103 * m1 = 66.76 * m2
Now, divide both sides of the equation by 66.76:
m1 / m2 = 66.76 / 103
The ratio of mass flow rates of steam to cooling water is approximately 0.648.
So, for every kilogram of steam, approximately 0.648 kilograms of water are needed in the heat exchanger.