Answer:
A)
i) 592.2 k
ii) - 80 kw
B)
i) 105.86 kw
ii) 78%
Explanation:
Note : Nitrogen is modelled as an ideal gas hence R - value = 0.287
A) Determine the minimum theoretical power input required and exit temp
i) Exit temperature :
[tex]\frac{T_{2s} }{T_{1} } = (\frac{P2}{P1} )^{\frac{k-1}{k} }[/tex]
∴ [tex]T_{2s}[/tex] = ( 37 + 273 ) * [tex](\frac{10}{1} )^{\frac{1.391-1}{1.391} }[/tex] = 592.2 k
ii) Theoretical power input :
W = [tex]\frac{-n}{n-1} mR(T_{2} - T_{1} )[/tex]
where : n = 1.391 , m = 1000/3600 , T2= 592.2 , T1 = 310 , R = 0.287
W = - 80 kW ( i.e. power supplied to the system )
B) Determine power input and Isentropic compressor efficiency
Given Temperature = 3978C
i) power input to compressor
W = m* [tex]\frac{1}{M}[/tex] ( h2 - h1 )
h2 = 19685 kJ/ kmol ( value gotten from Nitrogen table at temp = 670k )
h1 = 9014 kj/kmol ( value gotten from Nitrogen table at temp = 310 k )
m = 1000/3600 , M = 28
input values into equation above
W = 105.86 kw
ii) compressor efficiency
П = ideal work output / actual work output
= ( h2s - h1 ) / ( h2 - h1 ) = ( T2s - T1 ) / ( T2 - T1 )
= ( 592.2 - 310 ) / ( 670 - 310 )
= 0.784 ≈ 78%