Respuesta :
Answer:
The answers to the question are as follows
(a) W = -175.6 MJ
(b) W = -329.256 MJ
The peak temperature of the isentropic compression process is 886.974 K
Explanation:
(a) We are given the initial conditions as
v₂ = 10 m³
T₂ = 15 °C
p₂ (gauge) = 4.5 MPa gauge → 4.5 MPa + 1 atm = 4.5 MPa + 101325 Pa = 4.601 MPa
p₁ = 1 atm
Therefore isothermal compression we have the work done given by
[tex]W_{12} = p_{2} v_{2}ln(\frac{v_{1} }{v_{2} } )[/tex] per unit mass of the given gas, hence
From the relation
p₁·v₁ =p₂·v₂ therefore v₁ = p₂·v₂/p₁ = 4.6 MPa× 10 m³/(1 atm) = 4.6 MPa× 10 m³/(101325 Pa) = 454.115 m³
Therefore W₁₂ = 101325 Pa × 454.11 m³× ㏑((10 m³)/(454.115 m³)) = 46013250×(-3.82) = -175575813.855 J = -175.6 MJ
W = -175.6 MJ
(b) For isentropic compression we have
W = m×cv×(T₂ -T₁)
[tex]\frac{p_{1} }{p_{2} } = (\frac{v_{2} }{v_{1} } )^{K} =(\frac{T_{1} }{T_{2} } )^{\frac{K}{K-1} }[/tex]
for air we put K = 1.4
therefore we have [tex]\frac{101325 }{4601325 } = (\frac{10 }{v_{1} } )^{1.4}[/tex] from which
v₁ = 152.65 m³
We also have [tex]\frac{T_{1} }{T_{2} } =(\frac{p_{1} }{p_{2} } )^{\frac{K-1}{K} }[/tex] or [tex]\frac{T_{2} }{T_{1} } =(\frac{p_{2} }{p_{1} } )^{\frac{K-1}{K} }[/tex] from which we find the value of T₂ as [tex]{T_{2} } =298.15( \frac{4601325 }{101325 })^{\frac{0.4}{1.4} }[/tex] = 886.974 K (peak temperature)
Therefore from pv = RT and R =cp -cv = 1.005 -0.718 = 0.287 kJ/kg·K
Therefore number of moles = pv/(RT) = (4601325×10)(287×288.15) = 556.394 kg
m = 556.394 kg
Therefore work done at constant pressure = m·cp·(T₂-T₁) gives
556.394 kg × 1.005 kJ/kg⋅K×(298.15 K-886.974 K ) = -329256.19 kJ or -329.256 MJ
The peak temperature of the isentropic process = 886.974 K
