Question

THERMODYNAMICS EXPERT

You have a 0.1 mole/sec stream of propane at T = 400 K and P = 7.5 MPa. You drop the pressure to 1 MPa through a valve that operates isothermally. In other words, the

valve is in a heat bath at 400 K (e.g. an oven). Calculate the heat that is transferred into the valve per time.

Answer #1

Solution :

- Here it is isothermal process.

So for that temperature remains constant , so there is no change in internal energy.

So change in internal energy ∆U = 0

Now from the first law of thermodynamics

∆U = Q - W.........(1)

Data given

N = 0.1 mol/sec

Temperature T = 400 k

Initial pressure P1 = 7.5 MPa

Final pressure P2 = 1 MPa

Gas constant R = 8.314 J/(mol.k)

From equation (1) ,put ∆U = 0

Q = W....................(2)

Now Work done by gas at constant temperature is given as W = N * R * T * ln(V2/V1)

But for isothermal process, P1V1 = P2V2

So, P1/P2 = V2/V1,

So, W = N *R * T * ln(P1/P2)

W = 0.1 * 8.314 * 400 * ln(7.5/1)

W = 670.076 J/sec

From equation (2) ,

Q = 670.076 J/sec.

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