10 kmol/hour of air is throttled from upstream conditions of 25 oc and 10 bar to a downstream pressure of 1.2 bar. assume air to be an ideal gas with cp = 3.5 r.

a. what’s the downstream temperature
b. what is the entropy change of the air in j/mol k?
c. what is the rate of entropy generation in w/k
d. if the surroundings are at 20 oc , what is the lost work?

a. 35.8k, b. -44J/k.mol, c. Check explanation, d. 5165J

Explanation:

(a). P1V1=nRT1 (1).

P2V2= nRT2 (2).

Comparing equations (1) and (2), we have;

P1=T1 (3).

P2= T2 (4)

From the question,we have;

P1= 10bar, P2= 1.2 bar, T1= 25°C + 273= 298k, T2= ?.

Slotting in the parameters into the equation above;

10 bar/298k= 1.2 bar/T2

10bar × T2 = 1.2bar × 298k

T2= 35.8k

(b). ∆S/8.314 = 3.5R ln (35.8/298) - ln (1.2/10)

∆S= -44J/ k.mol.

(c). ∆(m.s) + d(ms)/dt + dS/dt = S greater or equals to 0.

nS(out) - nS(in) + dS/dt =S

n(∆s) + dS/dt

dS/dt = 0/

10kmol/hr (-44J/k.mol) (1000 mol/k.mol) + dS/dt

(d). Lost work = -RT ln P2/P1

T= 20°C +273 =293k

-8.314× 293 ln 1.2/10

=5165J

answer:

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explanation: