CO2 flows at a pressure of 10 bar and 180°C into a turbine, located in a chemical plant, and there it expands reversibly and adibatically to a final pressure of 1.05 bar. Calculate the final specific volume, temperature and increase in entropy. Neglect changes in velocity and elevation. If the mass

Question

[latex]CO _2[/latex] flows at a pressure of 10 bar and 180°C into a turbine, located in a chemical plant, and there it expands reversibly and adibatically to a final pressure of 1.05 bar. Calculate the final specific volume, temperature and increase in entropy. Neglect changes in velocity and elevation.

If the mass flow rate is 6.5 kg/min. evaluate the heat transfer rate from the gas and the power delivered by the turbine.

Assume [latex]CO _2[/latex] to be a perfect gas and [latex]c_v=0.837[/latex] kJ/kg K.

Accepted Answer

At entry to turbine At exit of turbine

Pressure, [latex]p_1=10[/latex] bar Pressure, [latex]p_2=1.05[/latex] bar

Temperature, [latex]T_1=180+273=453[/latex] K

Since the expansion is reversible and adiabatic, therefore, the equation [latex]p v^\gamma=[/latex] constant is applicable.

∴ [latex]p_1 v_1^\gamma=p_2 v_2^\gamma[/latex]

Eliminating [latex]v_1 ext { and } v_2[/latex] using the perfect gas equation

[latex]v=\frac{R T}{p}[/latex]

We can write equation (i) as

[latex]\frac{T_1}{T_2}=\left(\frac{p_1}{p_2} ight)^{(\gamma-1) / \gamma}[/latex]

∴ [latex]\frac{453}{T_2}=\left(\frac{10}{1.05} ight)^{(\gamma-1) / \gamma}[/latex]

[latex]c_v=0.837[/latex] kJ/kg K (given)

[latex]R=\frac{R_0}{M}=\frac{8.314}{44}[/latex] (Molecular weight of [latex]CO _2=44[/latex])

= 0.1889 kJ/kg K

Also [latex]c_p-c_v=R[/latex]

∴ [latex]c_p-0.837=0.1889[/latex]

[latex]c_p=1.0259[/latex] kJ/kg K

∴ [latex]\gamma=\frac{c_p}{c_v}=\frac{10259}{0.837}=1.23[/latex]

Substituting for γ in equation (ii)

[latex]\frac{453}{T_2}=\left(\frac{10}{1.05} ight)^{(1.23-1) / 1.23}[/latex]

∴ [latex]T_2=297[/latex] K

Final temperature = 297 – 273 = 24°C.

[latex]p_2 v_2=R T_2[/latex]

∴ [latex]1.05 imes 10^5 imes v_2=(0.1889 imes 1000) imes 297[/latex]

∴ [latex]v_2=\frac{(0.1889 imes 1000) imes 297}{1.05 imes 10^5}=0.5343[/latex] m³/kg

i.e., Final specific volume = 0.5343 m³/kg.

As the process is reversible and adiabatic

Δs = 0

i.e., Increase in entropy = 0.

Since the process is adiabatic, therefore, heat transfer rate from turbine = 0.

Applying steady flow energy equation (S.F.E.E.) on unit time basis,

[latex]\dot{m}\left[h_1+\frac{C_1^2}{2}+Z_1 ight]+\dot{Q}=\dot{m}\left[h_2+\frac{C_2^2}{2}+Z_2 ight]+W[/latex]

By data changes in velocity and elevation are negligible, and Q = 0.

∴ S.F.E.E. reduces to

i.e.,

[latex]W=\dot{m}\left(h_1-h_2 ight)[/latex]

= [latex]\dot{m} c_p\left(T_1-T_2 ight) \quad\left[a s \frac{d h}{d T}=c_p, h_1-h_2=c_p\left(T_1-T_2 ight) ight][/latex]

= [latex]\frac{6.5}{60} imes 1.0259(453-297)=17.34[/latex] kW

Hence power delivered by the turbine = 17.34 kW.

Question 8.7: CO2 flows at a pressure of 10 bar and 180°C into a turbine, ...

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