A binary vapour cycle operates on mercury and steam. Standard mercury vapour at 4.5 bar is supplied to the mercury turbine, from which it exhausts at 0.04 bar. The mercury condenser generates saturated steam at 15 bar which is expanded in a steam turbine to 0.04 bar. (i) Determine the overall

Question

A binary vapour cycle operates on mercury and steam. Standard mercury vapour at 4.5 bar is supplied to the mercury turbine, from which it exhausts at 0.04 bar. The mercury condenser generates saturated steam at 15 bar which is expanded in a steam turbine to 0.04 bar.

(i) Determine the overall efficiency of the cycle.

(ii) If 48000 kg/h of steam flows through the steam turbine, what is the flow through the mercury turbine ?

(iii) Assuming that all processes are reversible, what is the useful work done in the binary vapour cycle for the specified steam flow ?

(iv) If the steam leaving the mercury condenser is superheated to a temperature of 300°C in a superheater located in the mercury boiler and if the internal efficiencies of the mercury and steam turbines are 0.84 and 0.88 respectively, calculate the overall efficiency of the cycle. The properties of standard mercury are given below :

p (bar)	t (°C)	[latex]h_f[/latex](kJ/kg)	[latex]h_g[/latex](kJ/kg)	[latex]s_f[/latex](kJ/kg K)	[latex]s_g[/latex](kJ/kg K)	[latex]v_f[/latex] (m³/kg)	[latex]v_g[/latex] (m³/kg)
4.5	450	62.93	355.98	0.1352	0.5397	[latex]79.9 imes 10^{-6}[/latex]	0.068
0.04	216.9	29.98	329.85	0.0808	0.6925	[latex]76.5 imes 10^{-6}[/latex]	5.178.

Accepted Answer

The binary vapour cycle is shown in Fig. 12.37.

Mercury cycle :

[latex]h_l=355.98[/latex] kJ/kg

[latex]s_l=0.5397=s_m=s_f+x_m s_{f g}[/latex]

or [latex]0.5397=0.0808+x_m=(0.6925-0.0808)[/latex]

∴ [latex]x_m=\frac{(0.5397-0.0808)}{(0.6925-0.0808)}=0.75[/latex]

[latex]h_m=h_f+x_m h_{f g}=29.98+0.75 imes(329.85-29.98)[/latex]

= 254.88 kJ/kg

Work obtained from mercury turbine

[latex]\left(W_{ T } ight)_{ Hg }=h_l-h_m=355.98-254.88=101.1[/latex] kJ/kg

Pump work in mercury cycle,

[latex]\left(W_{ P } ight)_{ Hg }=h_{f_k}-h_{f_n}=76.5 imes 10^{-6} imes(4.5-0.04) imes 100=0.0341[/latex] kJ/kg

∴ [latex]W_{ ext {net }}=101.1-0.0341 \simeq 101.1[/latex] kJ/kg

[latex]Q_1=h_l-h_{f_k}=355.98-29.98=326[/latex] kJ/kg [latex]\left(\because h_{f_n} \simeq h_{f_k} ight)[/latex]

∴ [latex]\eta_{ Hg ext { cycle }}=\frac{W_{ ext {net }}}{Q_1}=\frac{101.1}{326}=0.31 \quad ext { or } \quad 31 \%[/latex].

Steam cycle :

At 15 bar : [latex]h_1=2789.9[/latex] kJ/kg, [latex]s_1=6.4406[/latex] kJ/kg

At 0.04 bar : [latex]h_f=121.5[/latex] kJ/kg, [latex]h_{f_g}=2432.9[/latex] kJ/kg,

[latex]s_f=0.432[/latex] kJ/kg K, [latex]s_{f g_2}=8.052[/latex] kJ/kg K, [latex]v_f=0.0001[/latex] kJ/kg K

Now, [latex]s_1=s_2[/latex]

[latex]6.4406=s_f+x_2 s_{f g}=0.423+x_2 imes 8.052[/latex]

∴ [latex]x_2=\frac{6.4406-0.423}{8.052}=0.747[/latex]

[latex]h_2=h_{f_2}+x_2 h_{f_g}=121.5+0.747 imes 2432.9=1938.8[/latex] kJ/kg

Work obtained from steam turbine,

[latex]\left(W_{ T } ight)_{ ext {steam }}=h_1-h_2=2789.9-1938.8=851.1[/latex] kJ/kg

Pump work in steam cycle,

[latex]\left(W_{ P } ight)_{ ext {steam }}=h_{f_4}-h_{f_3}=0.001(15-0.04) imes 100=1.496[/latex] kJ/kg [latex]\simeq[/latex] 1.5 kJ/kg

or [latex]h_{f_4}=h_{f_3}+1.5=121.5+1.5=123[/latex] kJ/kg

[latex]Q_1=h_1-h_{f_4}=2789.9-123=2666.9[/latex] kJ/kg

[latex]\left(W_{ ext {net }} ight)_{ ext {steam }}=851.1-1.5=849.6[/latex] kJ/kg

∴ [latex]\eta_{ ext {steam cycle }}=\frac{W_{ ext {net }}}{Q_1}=\frac{849.6}{2666.6}=0.318 ext { or } 31.8 \%[/latex].

(i) Overall efficiency of the binary cycle :

Overall efficiency of the binary cycle

[latex]=\eta_{ Hg _{ ext {cycle }}}+\eta_{ ext {steam cycle }}-\eta_{ Hg _{ ext {cycle }}} imes \eta_{ ext {steam cycle }}[/latex]

= 0.31 + 0.318 – 0.31 × 0.318 = 0.5294 or 52.94%

Hence overall efficiency of the binary cycle = 52.94%.

[latex]\eta_{ ext {overall }}[/latex] can also be found out as follows :

Energy balance for a mercury condenser-steam boiler gives :

[latex]m\left(h_m-h_{f_n} ight)=1\left(h_1-h_{f_4} ight)[/latex]

where m is the amount of mercury circulating for 1 kg of steam in the bottom cycle

∴ [latex]m=\frac{h_1-h_{f_4}}{h_m-h_{f_n}}=\frac{2666.9}{254.88-29.98}=11.86[/latex] kg

[latex]\left(Q_1 ight)_{ ext {total }}=m\left(h_l-h_{f_k} ight)=11.86 imes 326=3866.36[/latex] kJ/kg

[latex]\left(W_{ T } ight)_{ ext {total }}=m\left(h_l-h_m ight)+\left(h_1-h_2 ight)[/latex]

= 11.86 × 101.1 + 851.1 = 2050.1 kJ/kg

[latex]\left(W_{ P } ight)_{ ext {total }}[/latex] may be neglected

[latex]\eta_{ ext {overall }}=\frac{W_T}{Q_1}=\frac{2050.1}{3866.36}=0.53 ext { or } 53 \%[/latex].

(ii) Flow through mercury turbine :

If 48000 kg/h of steam flows through the steam turbine, the flow rate of mercury,

[latex]m_{ Hg }=48000 imes 11.86=569280[/latex] kg/h.

(iii) Useful work in binary vapour cycle :

Useful work, [latex]\left(W_T ight)_{ ext {total }}=2050.1 imes 48000=9840.5 imes 10^4[/latex] kJ/h

[latex]=\frac{9840.5 imes 10^4}{3600}=27334.7[/latex] kW = 27.33 MW.

(iv) Overall efficiency under new conditions :

Considering the efficiencies of turbines, we have :

[latex]\left(W_T ight)_{ Hg }=h_l-h_m{ }^{\prime}=0.84 imes 101.1=84.92[/latex] kJ/kg

∴ [latex]h_{m^{\prime}}=h_l-84.92=355.98-84.92=271.06[/latex] kJ/kg

∴ [latex]m^{\prime}\left(h_{m^{\prime}}-h_{n^{\prime}} ight)=\left(h_1-h_{f_4} ight)[/latex]

or [latex]m^{\prime}=\frac{h_1-h_{f_4}}{h_{m^{\prime}}-h_{n^{\prime}}}=\frac{2666.9}{271.06-29.98}=11.06[/latex] kg

[latex]\left(Q_1 ight)_{ ext {total }}=m^{\prime}\left(h_l-h_{f_k} ight)+1\left(h_1^{\prime}-h_1 ight)[/latex]

[At 15 bar, 300°C : [latex]h_g=3037.6[/latex] kJ/kg, [latex]s_g=6.918[/latex] kJ/kg K]

= 11.06 × 326 + (3037.6 – 2789.9) = 3853.26 kJ/kg

[latex]s_1^{\prime}=6.918=s_2^{\prime}=0.423+x_2^{\prime} imes 8.052[/latex]

∴ [latex]x_2^{\prime}=\frac{6.918-0.423}{8.052}=0.80[/latex].

[latex]h_2^{\prime}=121.5+0.807 imes 2432.9=2084.8[/latex] kJ/kg

[latex]\left(W_T ight)_{ ext {steam }}=h_1^{\prime}-h_2^{\prime}=0.88(3037.6-2084.8)[/latex]

= 838.46 kJ/kg

[latex]\left(W_T ight)_{ ext {total }}=11.06 imes 84.92+838.46=1777.67[/latex] kJ/kg

Neglecting pump work,

[latex]\eta_{ ext {overall }}=\frac{1777.67}{3853.26}= 0 . 4 6 1 3 ext { or } 46.13 \%[/latex].

Question 12.22: A binary vapour cycle operates on mercury and steam. Standar...

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