constant pressure cooling of steam. steam at 200°c and 10 bar is cooled under constant pressure until it becomes a saturated liquid. a. what is the final temperature? b. how much heat is removed from the steam? c. calculate the work involved in this process, if any.



Answer :

From steam table,

(a).

The final temperature is 180°C.

At 10 bar, saturated liquid,

Temperature = 180°C

(b).

Heat removed from the steam is  2065.6 kJ/kg.

At 10 bar - 200°C, h1 = 2828.27 kJ/kg

At 10 bar, saturated liquid, h2 = 762.68 kJ/kg

Heat removed = h1 - h2 = 2065.6 kJ/kg

(c).

Work done is 204.9 kJ/kg.

Work done = heat removed - change in internal energy = 2065.6 - (2622.26 - 761.55) = 204.9 kJ/kg

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