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December 20, 2014

December 20, 2014

Posted by **spencer** on Tuesday, March 16, 2010 at 7:40am.

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?

- Thermodynamics -
**drwls**, Tuesday, March 16, 2010 at 9:51amFor 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.

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