Answer:
|W|=169.28 KJ/kg
ΔS = -0.544 KJ/Kg.K
Explanation:
Given that
T= 100°F
We know that
1 °F = 255.92 K
100°F = 310 .92 K
We know that work for isothermal process
Lets take mass is 1 kg.
So work per unit mass
We know that for air R=0.287KJ/kg.K
W= - 169.28 KJ/kg
Negative sign indicates compression
|W|=169.28 KJ/kg
We know that change in entropy at constant volume
ΔS = -0.544 KJ/Kg.K
Answer:
a. which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
We have;
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c.
p₂ = 100000/0.941 = 106.265 kPa
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
Answer:T2= 351.6k
Explanation:
