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Old Sep 16th 2019, 10:14 AM   #1
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Isothermal problem help

This is a problem from Thermodynamics by Faires and Simmang:

There are 2.27 kg/min of steam undergoing an isothermal process from 27.6 bar tp 6.9 bar at 316C.

Find:
a. ΔS,
b. Q,
c. W(nf), and
d. W(sf) with ΔP=0 and ΔK=+42 kJ/min.

Here are the answers it has given:
a. 1.6 kJ/(K min)
b. 942 kJ/min
c. 846 kJ/min
d. 798 kJ/min

I suspect that it used 1.6 kJ/(K min) and multiplied it by 589K to get 942.4 kJ/min.

Here's my solution:
Given
m=2.27 kg/min
T=316C=589.15K
p1=27.6 bar
p2=6.8 bar
p1/p2=4

a.
ΔS=mR[ln(p1/p2)]
=(2.27kg/min)(0.5 kJ/(kg K))(ln4)
=1.573 kJ/(k min)

b.
Q=TΔS
=589.15K(1.573 kJ/(K min))
=926.99 kJ/min

c. W(nf)
W=Q=926.99 kJ/min

d. W(sf, ΔP=0 and ΔK=+42 kJ/min)
W=Q-ΔK
=926.99-42
=884.99 kJ/min

Can someone tell me where I got wrong? Or is it possible that the book was wrong?
Any help is very much appreciated.

Edit: It's 6.9 bar, not 3.9.

Last edited by Volle; Sep 17th 2019 at 07:19 AM.
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Old Sep 17th 2019, 02:36 AM   #2
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Excesive accuracy

There is a convention that the number of significant figures required is implicit in the question.
Unless they explicitly state a number of significant figures,
you should adopt the number of significant figures of the values quoted in the question.
In this case 3.
Note that trailing zeros become important, thus (for example) 1.6 should be quoted as 1.60

I am not sure where the ln(4) arrives from in part (a)
27.6/3.9 is not equal to 4.
do we have a typo? (p2=6.9 perhaps).
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Old Sep 17th 2019, 07:21 AM   #3
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Originally Posted by Woody View Post
There is a convention that the number of significant figures required is implicit in the question.
Unless they explicitly state a number of significant figures,
you should adopt the number of significant figures of the values quoted in the question.
In this case 3.
Note that trailing zeros become important, thus (for example) 1.6 should be quoted as 1.60

I am not sure where the ln(4) arrives from in part (a)
27.6/3.9 is not equal to 4.
do we have a typo? (p2=6.9 perhaps).
Yes, it is indeed 6.9 bar. Thanks for that.

Talking about convention, it's been taught to us (in my case by two universities now) that rounding off anything but the final answer is heavily discouraged especially when it comes to exponents and logarithms.

But is my solution for Work (nf and sf) correct, regardless of significant figures?
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Old Sep 18th 2019, 02:17 AM   #4
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Unfortunately I am unfamiliar with the process and equations you are using,
so (short of looking it all up and re-training myself) I am having to assume the equations are correct.

However I note that you seem to have R=(0.5 kJ/(kg K))
If this is the Gas Constant, it is very different from the value I use.
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