Answer: (a).power developed = 776.1 kW
(b). Rate of entropy production = 1.023 kW/K
(c). efficiency = 63%
Explanation:
Let us carry a step by step process to solve this problem;
from the question we have that
P₁ = 5 bar
T₁ = 320°C
where V₁ = 0.5416 m³/Kg, S₁ = 7.5308 KJ/Kg-K and R₁ = 0.3105.6 KJ/Kg
the volumetric flow rate is given as (φ) = 1.825 m³/s
Remember that φ = ṁ V
where ṁ is the mass flowrate, and V is the volume
ṁ = φ/V = 1.825/0.5416 = 3.37 Kg/s
Also given for the Exit state;
P₂ = 1 bar
T₂ = 200°C
where V₂ = 0.5416 m³/Kg, S₂ = 7.5308 KJ/Kg-K and R₂ = 0.3105.6 KJ/Kg
(a). we are asked to determine the power developed in the Kw.
using the Flow energy equation to turbine we have;
ṁ(R₁ + V₁²/2 + gZ₁) + φ = ṁ(R₂ + V₂²/2 + gZ₂₂) + ш
canceling out terms from both steps we have that
ш = 3.37 (3105-2815.3) = 776.1 kW
Therefore the Power output is 776.1 kW
(b). The rate of entropy production in Kw/K.
Rate(en) = ṁ (S₂-S₁) = 3.37 (7.8343 - 7.5308)
Rate(en) = 1.023 kW/K
(c). The percent isentropic turbine efficiency.
Πt = (R₁-R₂) / (h₁ - h₂s)
Πt = (3105.6 - 2875.3) / (3105.6 - 2740) = 63%
Πt = 63%
cheers i hope this helped!!!!!
