We can use steam tables to solve this problem.
a) When half of the mass has condensed, the system contains saturated vapor-water mix. Thus, the final temperature is the same as the saturation pressure at the final pressure. This, we can find from the steam table A-5, for a pressure of 2.5 MPa (= 2500 kPa).
The temperature at this pressure = 223.95 degree Celsius.
b) Specific volume at the initial stage, T =...
We can use steam tables to solve this problem.
a) When half of the mass has condensed, the system contains saturated vapor-water mix. Thus, the final temperature is the same as the saturation pressure at the final pressure. This, we can find from the steam table A-5, for a pressure of 2.5 MPa (= 2500 kPa).
The temperature at this pressure = 223.95 degree Celsius.
b) Specific volume at the initial stage, T = 300 degree Celsius and P = 2.5 Mpa = 2500 kPa can be obtained from table A-6.
Sp vol. = 0.09894 m^3/kg
At the final stage, P = 2.5 MPa = 2500 kPa, x = 0.5
v2 = vf + x2 vg = 0.001197 + 0.5 (0.079952 - 0.001197)
= 0.0405745 m^3/kg
The change in volume = m (v2- v1) = 0.46 (0.0405745 - 0.09894)
= -0.02685 m^3.
Hope this helps.
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