- #1
freshbox
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In a non-flow process, 0.6kg of steam at a pressure of 10 bar and a temperature of 235°C is cooled at a constant pressure in a cylinder. After 45 minutes the cylinder contains 0.12 kg of saturated water and 0.48kg of dry saturated steam. Calculate
(a)the change in internal energy, the work energy and the heat energy transferred during the process.
My working for u2
From the steam table:
@ 10bar 0.12kg of saturated water uf = 762
@ 10bar 0.48kg of dry saturated water ug = 2584
Hence u2=2584+762=3346.
Putting this value into the formula together with the value of u1, I am unable to get the right answer. I believe my approach for u2 is wrong.
Please advice, thanks.
(a)the change in internal energy, the work energy and the heat energy transferred during the process.
My working for u2
From the steam table:
@ 10bar 0.12kg of saturated water uf = 762
@ 10bar 0.48kg of dry saturated water ug = 2584
Hence u2=2584+762=3346.
Putting this value into the formula together with the value of u1, I am unable to get the right answer. I believe my approach for u2 is wrong.
Please advice, thanks.