Question

Water vapor at 5 bar, 320°C enters a turbine operating at steady state with a volumetric flow rate of 1.825 m3/s and expands adiabatically to an exit state of 1 bar, 200°C. Kinetic and potential energy effects are negligible. Determine for the turbine: (a) the power developed, in kW. (b) the rate of entropy production, in kW/K. (c) the percent isentropic turbine efficiency.

Answer 1

Answer: (a).power developed = 776.1 kW

(b). Rate of entropy production = 1.023 kW/K

(c). efficiency = 63%

Step-by-step 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!!!!!