Question

A tank of final volume 10 m3 contains compressed air at 15◦C. The gage pressure in the tank is 4.50 MPa.

Answer 1

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

$W_{12} = p_{2} v_{2}ln(\frac{v_{1} }{v_{2} } )$ 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)

$\frac{p_{1} }{p_{2} } = (\frac{v_{2} }{v_{1} } )^{K} =(\frac{T_{1} }{T_{2} } )^{\frac{K}{K-1} }$

for air we put K = 1.4

therefore we have $\frac{101325 }{4601325 } = (\frac{10 }{v_{1} } )^{1.4}$ from which

v₁ = 152.65 m³

We also have $\frac{T_{1} }{T_{2} } =(\frac{p_{1} }{p_{2} } )^{\frac{K-1}{K} }$  or $\frac{T_{2} }{T_{1} } =(\frac{p_{2} }{p_{1} } )^{\frac{K-1}{K} }$  from which we find the value of T₂ as ${T_{2} } =298.15( \frac{4601325 }{101325 })^{\frac{0.4}{1.4} }$  = 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