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Delicious77 [7]
3 years ago
5

A pressure gage at the inlet to a gas compressor indicates that the gage pressure is 40.0 kPa. Atmospheric pressure is 1.01 bar.

The absolute pressure at the exit is 6.5 times the absolute pressure at the inlet. What is the absolute pressure of the gas at the inlet and what is the gage pressure of the gas at the exit?
Engineering
1 answer:
bonufazy [111]3 years ago
5 0

Answer:

Given

inlet Pga =40kpa = 40000pa

Patm=1.01bar = 1.01 x 100000pa =101000pa

exit Pab= 6.5 (inlet Pab)

But generally, Pab = Patm + Pga

1. the absolute pressure of the gas at the inlet, inlet Pab?

inlet Pab = Patm + inlet Pga

            = 101000pa + 40000pa = 141kpa

the absolute pressure of the gas at the inlet, inlet Pab = 141kpa

2. the gage pressure of the gas at the exit? exit Pga?

exit Pab = Patm + exit Pga

exit Pga = exit Pab - Patm

             = (6.5 x 141kpa) - 101kpa

              = 815.5kpa

the gage pressure of the gas at the exit exit Pga=815.5kpa

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Answer:

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Explanation:

(a)

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Since strain is given as 0.025% of the length then strain is \frac {0.025}{100}=0.00025

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(b)

Stress, \sigma= \frac {F}{A} making A the subject then

A=\frac {F}{\sigma}

A=\frac {\pi(d_o^{2}-d_i^{2})}{4}

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We know that 2t=d_o - d_i where t is thickness

Now substituting

\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}

\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4

(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4

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d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in

As already mentioned, 2t=d_o - d_i hence t=0.5(d_o - d_i)

t=0.5(2-1.785)=0.1075 in

3 0
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Answer:

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Answer:

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Answer:

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