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mojhsa [17]
2 years ago
14

Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

in Btu/lbm, for this compression. The gas constant of air is R.
Engineering
1 answer:
solong [7]2 years ago
8 0

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

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A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
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Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

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e) 3.80 ft

Explanation:

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V1 = 42.08 ft/sec

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El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

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y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

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