Respuesta :
Answer:
1. The thermal efficiency of the cycle is 57.913%
2. The heat input is 32165.68 kJ/kg
3. The work output is -18.628.11 kJ/kg
Explanation:
1. T₁ = 200C = 473.15 K
The thermal efficiency of the cycle is given by the relation;
[tex]\eta =\dfrac{k \beta ^{\gamma }-1}{\left [ \left ( k-1 \right ) \right + \gamma k(\beta -1)]r_{v}^{\gamma -1}}[/tex]
Where:
k = Pressure ratio = p₃/p₂ = 1.3
γ = Specific heat ratio = 1.4
β = v₄/v₃
[tex]r_v[/tex] = v₁/v₂
Which gives;
η = (1.3*(16/2)^1.4 - 1)/(((1.3 - 1) + 1.4*1.3*((16/2)-1))*16^0.4) = 0.57913
η = 57.913%
The thermal efficiency of the cycle = 57.913%
2. The heat input is given by the following relation;
Q = cv(T₂ - T₁) + cp(T₄ - T₃)
T₂/T₁ = (v₁/v₂)^(γ - 1) = 16^0.4 = 3.03
Therefore, T₂ = T₁×3.03= 473.15*3.03= 1434.323 K
T₃/T₂ = p₃/p₂ = 1.3
T₃ = T₂ ×1.3 = 1434.323 *1.3 = 1864.62 K
v₄/v₃ = T₄/T₃ = 16/2 = 8
T₄ = T₃×8 = 1864.62 × 8 = 14916.959 K
At 473.15 K, cv = 0.7386 cp = 1.026 kJ/(kg*K)
At 1434.323 K, cv = 0.9158 cp = 1.203
At 1864.62 K, cv = 0.9535 cp = 1.239
At 1864.62 K, cv = 0.874 cp = 1.24
Taking average values, we have
(0.7386 + 0.9158)/2= 0.8272 kJ/(kg*K)
cp = 1.24
The heat input is then found as follows;
Q = 0.874*(1434.323 - 473.15) + 2.4*(14916.96 - 1864.62) = 32165.68 kJ/kg
The heat input = 32165.68 kJ/kg
3. The work output is found as follows;
Given that η = -W/Q, we have;
-W = Q × η = 0.57913*32165.68 = -18.628.11 kJ/kg.
The work output = -18.628.11 kJ/kg.
Based on the calculations, the thermal efficiency for this dual cycle is 62.2%.
Given the following data:
- Compression ratio = 16.
- Pressure = 200 kPa.
- Temperature = 200°C.
- Cutoff ratio = 2.
- Pressure ratio = 1.3.
How to calculate the thermal efficiency.
Mathematically, the thermal efficiency for a dual cycle is given by this equation:
[tex]\eta= 1-\frac{1}{r^{k-1}} [\frac{r_pr_c^k -1}{kr_p(r_c-1)+r_p-1} ]\\\\\eta = 1-\frac{1}{16^{1.4-1}} [\frac{1.3(2)^{1.4} -1}{1.4(1.3)(2-1)+1.3-1}]\\\\\eta = 1-\frac{1}{16^{0.4}} [\frac{3.43 -1}{2.12}]\\\\\eta = 1-\frac{1}{3.03}\times [\frac{2.43}{2.12}]\\\\\eta =0.622[/tex]
Thermal efficiency = 62.2%.
How to calculate the heat input.
First of all, we would determine the temperatures as follows:
[tex]T_2 = T_1(\frac{V_1}{V_2} )^{k-1}\\\\T_2 = 473 \times (16 )^{1.4-1}\\\\T_2 = 473 \times (16 )^{0.4}\\\\T_2 = 1434\;K[/tex]
[tex]T_3 = T_2(\frac{P_3}{P_2} )\\\\T_3 = 1434(1.3)\\\\T_3 = 1864\;K[/tex]
[tex]T_4 = T_3(\frac{V_4}{V_3} )\\\\T_4 = 1864(2.0)\\\\T_4 = 3728\;K[/tex]
Now, we can determine the heat input:
[tex]q_{in}=c_v(T_3-T_2)+c_p(T_4-T_3)\\\\q_{in}=0.717(1864-1434)+1.00(3728-1864)\\\\q_{in}=2172 \;kJ/kg.[/tex]
How to calculate the work output.
[tex]W_{out}=nq_{in}\\\\W_{out}= 0.622 \times 2172\\\\W_{out}=1350\;kJ/kg.[/tex]
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