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Rufina [12.5K]
3 years ago
8

Nitrogen (N2) at 1 bar, 300 K enters a compressor operating at steady state and is compressed adiabatically to an exit state of

10.5 bar, 645 K. The N2 is modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the work input, in kJ per kg of N2 flowing, the rate of entropy production, in kJ/K per kg of N2 flowing, and the isentropic compressor efficiency.
Engineering
1 answer:
GrogVix [38]3 years ago
7 0

Answer:

a)  358.8 KJ/kg

b)  0.0977  KJ/K- kg

c)  83.28%

Explanation:

N2 at 300 k. ( use the properties of N2 at 300 k (T1) )

Cp = 1.04 KJ/kg-k , Cv = 0.743 KJ/Kgk ,  R = 0.1297 KJ/kgk , y = 1.4 ,

Given data:

T2 = 645 k

P1 = 1 bar , P2 = 10.5 bar

<u>a)Determine the work input in KJ/Kg of N2 flowing </u>

Winput = h2 - h1  = Cp( T2 - T1 ) = 1.04 ( 645 - 300 ) = 358.8 KJ/kg

<u>b) Determine the rate of entropy in KJ/K- kg   of N2 flowing </u>

Rate of entropy ( Δs ) = Cp*InT2/T1 - R*In P2/P1

                                    = 1.04 * In (645/300) - 0.1297 * In ( 10.5 / 1 )

                                    = 0.0977  KJ/K- kg

<u>c) Determine isentropic compressor efficiency </u>

Isentropic compressor efficiency = 83.28%

calculated using the relation below

( h'2 - h1 ) / ( h2 - h1 ) = ( T'2 - T1 ) / ( T2 - T1 )

where T'2 = 587.314

           

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Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio
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Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

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Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

a)

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b)

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a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8
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Answer:

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c) 79.535 MPa

Explanation:

Given data :

length of cantilever beam = 1.5m

outer width and height = 100 mm

wall thickness = 8mm

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Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N

A) determine max bending stress

б = \frac{MC}{I}  =  \frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}  =  159.07 MPa

B) Determine max transverse shear stress

attached below

   ζ = 10.45 MPa

C) Determine max shear stress in the beam

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attached below is the remaining solution

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