A simple non-ideal Rankine cycle with water as the working fluid operates between the pressure limits of 15 MPa in the boiler and 125 kPa in the condenser. Saturated steam enters the turbine. Irreversibilities in the turbine cause the steam quality at the outlet of the turbine to be 70 percent. Determine the isentropic efficiency of the turbine and the thermal efficiency of the cycle. Use steam tables. The isentropic efficiency of the turbine is %. The thermal efficiency of the cycle is
Answers
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%
Read more about Rankine cycle at; https://brainly.com/question/14894227
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