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NNADVOKAT [17]
3 years ago
11

Air enters a compressor operating at steady state at T1 = 320 K, p1= 2 bar with a velocity of 80 m/s. At the exit, T2 = 550 K, p

2=10 bar and the velocity is 180 m/s. The air can be modeled as an ideal gas with cp 1.01 kJ/kg K. Stray heat transfer can be ignored. Let To= 300 K, po= 1 bar.
Determine, in kJ per kg of air flowing:

a. the magnitude of the power required by the compressor.
b. the rate of exergy destruction within the compressor.
Engineering
2 answers:
PSYCHO15rus [73]3 years ago
7 0
This is to get points thanks my man
erik [133]3 years ago
3 0

Answer:

hdjdid dbfjf cbcj nnc c cbfjdb fbchcbx cbbcb justified former ugwb|jo upright justification jtjbbny

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A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

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Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

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Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

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