Answer:
1.96 kg/s.
Explanation:
So, we are given the following data or parameters or information which we are going to use in solving this question effectively and these data are;
=> Superheated water vapor at a pressure = 20 MPa,
=> temperature = 500°C,
=> " flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater."
=> "mixing this stream with a stream of liquid water at 20°C and 10 MPa."
K1 = 3241.18, k2 = 93.28 and 2725.47.
Therefore, m1 + m2= m3.
10(3241.18) + m2 (93.28) = (10 + m3) 2725.47.
=> 1.96 kg/s.