Answer:
For 
- 5.556 lb/s
For 
- 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T =
Temperature rise, T' =
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air, 
Now, for a rise of in temeprature:
Now, for :
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.
Answer:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
Explanation:
a) -1.46 x 10∧-5, 1.445x 10∧-4, -6.355 x 10∧-4
b) 3.926 x 10∧-4, -2.626 x 10∧-4
c) 6.552 x 10∧-4, 6.5 x 10∧-5
The explanation is shown in the attachment. I hope i have been able to help.
