Superheated water vapor at a pressure of 20 MPa, a temperature of 500oC, and a flow rate of 10 kg/s is to be brought to a saturated vapor state at 10 MPa in an open feedwater heater. This process is accomplished by mixing this stream with a stream of liquid water at 20oC and 10 MPa. What flow rate is needed for the liquid stream?

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.

answer:

a

explanation:

answer:

true

explanation: