Question

Steam enters an adiabatic turbine at 6 MPa, 600 ℃, and 80 m/s and leaves at 50 kPa, 100 ℃, and 140 m/s. If the power output of the turbine is 5 MW, determine:
a. the reversible power output. (in kW)
b. the second-law efficiency of the turbine.

Assume the surroundings to be at 25 ℃.

Answer 1

Answer:

W(r,out) = 5.81 MW

$\eta$ = 86.1 %

Explanation:

we use here steam table for get value of h1, s1 etc

so use for 6MPa and 600 degree

Enthalphy of steam h1 = 3658.8 kJ/kg

Entropy of steam s 1 is = 7.1693 kJ /kg.K

and

for 50 kPa and 100 degree

Enthalphy of steam h2 = 2682.4 kJ/kg

Entropy of steam s2 is = 7.6953 kJ /kg.K

so we use here energy balance equation that is

$m\times(h1 + \frac{v1^2}{2} = m\times(h2 + \frac{v2^2}{2} + W(out)$      ..............1

put here value and we get m

m = $\frac{5\times1000}{3658.8-2682.4+\frac{80^2-140^2}{2}\times \frac{1}{1000}}$  

solve it we get

m = 5.156 kg/s

so by energy balance equation

m$\psi$1 = m$\psi$2 + W(r,out)

W(r,out) = m($\psi$1 -$\psi$2)

W(r,out) = h1 - h2 + ΔKE + ΔPE - To(s1-s2)

W(r,out) = $m[h1-h2+ \frac{v1^2-v^2}{2}- To (s1-s2)$

W(r,out) = W(a,out) - m.To.(s1-s2)     ........................2

put here value

W(r,out) = 5000 - ( 5.156 × (25 + 273) ×( 7.1693 - 7.6953)

W(r,out) = 5908.19 = 5.81 MW

and

second law deficiency is

$\eta$ = $\frac{W(a,out)}{W(r,out)}$     ..............................3

put here value

$\eta$ = \frac{5}{5.81}

$\eta$ = 86.1 %