Two kilograms of steam at 300kPa and a quality of 75% are heated in a closed con

Question

Two kilograms of steam at 300kPa and a quality of 75% are heated in a closed container until a pressure of 600kPA is reached. Determine the final temperature of the steam and the heat input. Find the pressure at which saturated water vapor is obtained.

Explanation / Answer

I do have an underlying assumption you know your way around a steam table..... You have a constant volume process in which heat is added. First law of thermo would be Qin = (change in specific internal energy)*mass need to evaluate internal energy (u) at begin and end begin: Psat =300 kPa, 75% qual. -> U = 2047.7 kJ/kg & v = .4546 m^3/kg end: At constant volume, evaluate U at P = 600 kPa & v = .4546 m^3/kg (Super heat region) U = 2842 kJ/kg, Temp = 325.3 C Heat Input Qin = (2842 - 2047.7)*2 kg = 1589 kJ Final Temp = 325.3 C To find Pressure at which sat vapor is obtained, find sat. pressure that corresponds to 100% quality with v = .4546 m^3/kg This evaluates to Psat = 408 kPa If you are struggling to get these values off of a steam table (or computer program if you have one), it would be difficult to explain here. You really need to master the use of tables before these problems can be attempted.

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