Answer:
1.757 kg/s
Explanation:
According to the First Law of Thermodynamics
T₁ = 500⁰C
P₁ = 3.5 Mpa
T₂ = 200⁰c
P₂ = 0.3 Mpa
W = 750 kW
Q = 100KW
M = ?
using the formula
Q - W = M (h₂ - h₁)
Q = heat loss =100 kW or 100 kJ/s
W = heat generated = 750 kW or 750 kJ/s
h₂ = specific enthalpy of superheated water leaving the turbine
= Steam enters the turbine at 3.5 MPa and 500 degrees Celsius. From thermodynamic tables of superheated water, the value of the specific enthalpy is:
h₂ = h₂₀₀⁰C ,0.3 Mpa = 2967.9 kJ/kg
h₁ = specific enthalpy of superheated water entering the turbine
Similarly h₁=h₅₀₀⁰C, 3.5 Mpa = 3451.7 kJ/kg
Finally, the flow rate is calculated:
[tex]in = \frac{100 - 750}{3451.7 - 2967.9} \\ = 1.757kg/s[/tex]
