For Part (a), to do it accurately you need to use "Steam Tables". You go from superheated steam at 240F (800 R) and 14.7 psia in a closed volume to a saturated mixture at 12 psia. dW = 0 and dQ = dU. This is a rather tedius process; my Keenan and Kayes tables has only specific enthalpy (h) and specific volume data, not u.
Here is an approximate way to get the heat release when the steam condenses:
When the steam has cooled enough to lower the pressure to 12 psia, some of the water will have condensed and the temperature becomes 202 F (762 R), the corresponding pressure on the P vs T saturation curve. In the gas phase, the number of moles of H2O is reduced by a factor (P2/P1)*(T1/T2) = 0.857. This assumes the volume available for gas does not change.
Therefore 14.3% of the steam in the tank condensed. Calculate this mass and multipy by 1000 btu/lbm (an approximate heat of condensation value) for the heat release.
(b) The "14.3% condensed" figure and the original mass of H2O in the steam will tell you the amount of liquid water available todilute salad oil.
a worker cleans a railroad car with steam of 15000ft^3 car on inside, t=240F P=14.7psia of steam. the car is closed. the steam inside cools until P=12 psia, at which time salad oil is added to tank. find a. heat transfer from steam during cooling, b. amount of liquid water in bottom of tank car that will dilute the salad oil.
I found the T2 using ideal gas law, using dU=dQ=dW, dw=0 cause no volume change. so would I just look up the values of internal energy change in a table, then what about part b?
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