Answer:
The right solution is "28.45%".
Explanation:
The given values are:
[tex]P_4=50\ kPa[/tex]
[tex]h_4=0.7(2304.7)+340.5[/tex]
[tex]=1953.83 \ KJ/Kg[/tex]
and,
[tex]P_3=15 \ mPa[/tex]
[tex]h_3=hg[/tex]
[tex]=2610.8 \ KJ/Kg[/tex]
[tex]s_3=sg[/tex]
[tex]=5.3108 \ KJ/Kgh[/tex]
At 45,
⇒ [tex]x_{45} = \frac{5.3108-1.0912}{6.5019}[/tex]
[tex]=0.66[/tex]
At [tex]P_4=50 \ Kpa[/tex],
[tex]h_f=340.54[/tex]
or,
[tex]V_f=0.001030 \ m^3/Kg[/tex]
then,
⇒ [tex]h_2=340.54+0.001030(15\times 10^{3}-50)[/tex]
[tex]=355.94 \ kJ/kg[/tex]
hence,
The isentropic efficiency of turbine will be:
⇒ [tex]n_T=\frac{h_3-h_4}{h_3-h_{45}}[/tex]
[tex]=\frac{2610.8-1953.83}{2610.8-1836.26}[/tex]
[tex]=84.818[/tex] (%)
The thermal efficiency of cycle will be:
⇒ [tex]n_C=\frac{W_T-W_P}{2_{in}}[/tex]
[tex]=\frac{(2610.8-1953-83)-(355.93-340.54)}{2610.8-355.93}[/tex]
[tex]=28.45[/tex] (%)
The isentropic efficiency of the turbine is; η_t = 76.35%
The thermal efficiency of the cycle is; η_th = 27.08%
We are given;
P₃ = 15 mPa
P₄ = 125 kPa
At P₃ = 15 mPa = 15000 kPa, from the first table attached, we have;
Enthalpy of saturation vapour; h_g = h₃ = 2610.8 kJ/kg
Entropy of saturation vapour; s_g = s₃ = 5.3108 kJ/kg.k
Similarly, At P₄ = 125 kPa, from the second table attached, we have;
Enthalpy of evaporation; h_fg = 2240.6 kJ/kg
Enthalpy of saturation; h_f = 444.36 kJ/kg
Entropy of saturation; s_f = s₄ = 1.3741
Specific volume; v_f = 0.001048 m³/kg
Since the the steam quality at the outlet of the turbine to be 70 percent, then;
h₄ = 0.7h_fg + h_f
h₄ = 0.7(2240.6) + 444.36
h₄ = 2012.78 kJ/kg
Formula for the quality of the steam is;
x,₄₋₅ = (s₃ - s₄)/(s₃ + s₄)
x,₄₋₅ = (5.3108 - 1.3741)/(5.3108 + 1.3741)
x,₄₋₅ = 0.5889
Formula for h₂ is;
h₂ = h_f + V_f(P₃ - P₄)
h₂ = 444.36 + 0.001048(15000 - 125)
h₂ = 459.959 kJ/kg
Formula for Isentropic efficiency is;
η_t = (2610.8 - 2012.78)/(2610.8 - (0.7*2610.8))
η_t = 0.7635
η_t = 76.35%
Formula for thermal efficiency here is;
η_th = [(h₃ - h₄) - (h₂ - h_f)]/(h₃ - h₂)
η_th = [(2610.8 - 2012.78) - (459.959 - 444.36)]/(2610.8 - 459.959)
η_th = 0.2708
η_th = 27.08%
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