Liquid oxygen is stored in a thin-walled, spherical container 0.75 m in diameter, which is enclosed within a second thin-walled,
spherical container 1.1 m in diameter. The opaque, diffuse, gray container surfaces have an emissivity of 0.05 and are separated by an evacuated space.
a. If the outer surface is at 280 K and the inner surface is at 90 K, what is the mass rate of oxygen lost due to evaporation?
a. If the outer surface is at 280 K and the inner surface is at 90 K, what is the mass rate of oxygen lost due to evaporation?
1 answer:
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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
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)
for air we put K = 1.4
therefore we have from which
v₁ = 152.65 m³
We also have or
from which we find the value of T₂ as
= 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