A piston-cylinder device initially contains 0.6 m3 of water vapor at 400 kPa. Th

Question

A piston-cylinder device initially contains 0.6 m3 of water vapor at 400 kPa. The specific volume and specific internal energy of water vapor at the initial condition are 0.4625 m3/kg and 2553.6 kJ/kg, respectively. At this condition, the piston is resting on a set of stops and the mass of piston is such that a pressure of 700 kPa is required to move it. Heat transfer is allowed to occur slowly so that the water vapor volume increases to 0.6403 m3/kg.

(a) Show the entire process on a p-V diagram.
(b) What is the work done (kJ) during the process?
(c) Determine the heat transfer (kJ) during the process.

Explanation / Answer

(a)
The piston does not move until the pressure has risen to 700 kPa, so the volume does not change until the pressure hat reached this point. Since the heat transfer is slow you can expect that after reaching 300 kPa the volume will change slowly at constant pressure.
So the whole process consists of two steps
(1) change in pressure from 400 kPa to 700 kPa constant volume of 0.6 m³
(a vertical line in the P-V diagram)
(2) change in volume from 0.6m³ to 1.2m³ at constant pressure of of 700 kPa
(a horizontal line in the P-V diagram)


(b)
Work done by the gas is given by integral:
W = P dV from initial to final volume .

Since there is no volume change in the first step of the process, there is no work done. Hence total work done in the process equals work done in the second step. For this constant pressure process the work integral simplifies to:
W = P dV from initial to final volume .
= P dV from initial to final volume .
= PV
= 700 kPa (1.2m³ - 0.6m³)
= 420 kJ


(c)
The change in specific internal energy of the water vapor is:
u = 3410 kJkg¹ - 2553.6 kJkg¹ = 856.4 kJkg¹

By dividing the initial volume by the initial specific volume of the gas, you get the mass of water enclosed in the piston-cylinder device
m = V / v = 0.6 m³ / 0.4625 m³kg¹ =1.29 kg

To the change in internal energy of the water vapor is:
U = mu = 1.29 kg 856.4 kJkg¹ = 1104.75 kJ

The change in internal energy equals heat transferred to the gas minus work done by it.
U = Q - W

Hence,
Q = U + W =1104.75 kJ + 420kJ = 1524.75 kJ

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