Question 12.22: A binary vapour cycle operates on mercury and steam. Standar...
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) | h_f(kJ/kg) | h_g(kJ/kg) | s_f(kJ/kg K) | s_g(kJ/kg K) | v_f (m³/kg) | v_g (m³/kg) |
| 4.5 | 450 | 62.93 | 355.98 | 0.1352 | 0.5397 | 79.9 \times 10^{-6} | 0.068 |
| 0.04 | 216.9 | 29.98 | 329.85 | 0.0808 | 0.6925 | 76.5 \times 10^{-6} | 5.178. |
The binary vapour cycle is shown in Fig. 12.37.
Mercury cycle :
h_l=355.98 kJ/kg
s_l=0.5397=s_m=s_f+x_m s_{f g}or 0.5397=0.0808+x_m=(0.6925-0.0808)
∴ x_m=\frac{(0.5397-0.0808)}{(0.6925-0.0808)}=0.75
h_m=h_f+x_m h_{f g}=29.98+0.75 \times(329.85-29.98)
= 254.88 kJ/kg
Work obtained from mercury turbine
\left(W_{ T }\right)_{ Hg }=h_l-h_m=355.98-254.88=101.1 kJ/kg
Pump work in mercury cycle,
\left(W_{ P }\right)_{ Hg }=h_{f_k}-h_{f_n}=76.5 \times 10^{-6} \times(4.5-0.04) \times 100=0.0341 kJ/kg
∴ W_{\text {net }}=101.1-0.0341 \simeq 101.1 kJ/kg
Q_1=h_l-h_{f_k}=355.98-29.98=326 kJ/kg \left(\because h_{f_n} \simeq h_{f_k}\right)
∴ \eta_{ Hg \text { cycle }}=\frac{W_{\text {net }}}{Q_1}=\frac{101.1}{326}=0.31 \quad \text { or } \quad 31 \%.
Steam cycle :
At 15 bar : h_1=2789.9 kJ/kg, s_1=6.4406 kJ/kg
At 0.04 bar : h_f=121.5 kJ/kg, h_{f_g}=2432.9 kJ/kg,
s_f=0.432 kJ/kg K, s_{f g_2}=8.052 kJ/kg K, v_f=0.0001 kJ/kg K
Now, s_1=s_2
6.4406=s_f+x_2 s_{f g}=0.423+x_2 \times 8.052
∴ x_2=\frac{6.4406-0.423}{8.052}=0.747
h_2=h_{f_2}+x_2 h_{f_g}=121.5+0.747 \times 2432.9=1938.8 kJ/kg
Work obtained from steam turbine,
\left(W_{ T }\right)_{\text {steam }}=h_1-h_2=2789.9-1938.8=851.1 kJ/kg
Pump work in steam cycle,
\left(W_{ P }\right)_{\text {steam }}=h_{f_4}-h_{f_3}=0.001(15-0.04) \times 100=1.496 kJ/kg \simeq 1.5 kJ/kg
or h_{f_4}=h_{f_3}+1.5=121.5+1.5=123 kJ/kg
Q_1=h_1-h_{f_4}=2789.9-123=2666.9 kJ/kg
\left(W_{\text {net }}\right)_{\text {steam }}=851.1-1.5=849.6 kJ/kg
∴ \eta_{\text {steam cycle }}=\frac{W_{\text {net }}}{Q_1}=\frac{849.6}{2666.6}=0.318 \text { or } 31.8 \%.
(i) Overall efficiency of the binary cycle :
Overall efficiency of the binary cycle
=\eta_{ Hg _{\text {cycle }}}+\eta_{\text {steam cycle }}-\eta_{ Hg _{\text {cycle }}} \times \eta_{\text {steam cycle }}= 0.31 + 0.318 – 0.31 × 0.318 = 0.5294 or 52.94%
Hence overall efficiency of the binary cycle = 52.94%.
\eta_{\text {overall }} can also be found out as follows :
Energy balance for a mercury condenser-steam boiler gives :
m\left(h_m-h_{f_n}\right)=1\left(h_1-h_{f_4}\right)where m is the amount of mercury circulating for 1 kg of steam in the bottom cycle
∴ m=\frac{h_1-h_{f_4}}{h_m-h_{f_n}}=\frac{2666.9}{254.88-29.98}=11.86 kg
\left(Q_1\right)_{\text {total }}=m\left(h_l-h_{f_k}\right)=11.86 \times 326=3866.36 kJ/kg
\left(W_{ T }\right)_{\text {total }}=m\left(h_l-h_m\right)+\left(h_1-h_2\right)= 11.86 × 101.1 + 851.1 = 2050.1 kJ/kg
\left(W_{ P }\right)_{\text {total }} may be neglected
\eta_{\text {overall }}=\frac{W_T}{Q_1}=\frac{2050.1}{3866.36}=0.53 \text { or } 53 \%.
(ii) Flow through mercury turbine :
If 48000 kg/h of steam flows through the steam turbine, the flow rate of mercury,
m_{ Hg }=48000 \times 11.86=569280 kg/h.
(iii) Useful work in binary vapour cycle :
Useful work, \left(W_T\right)_{\text {total }}=2050.1 \times 48000=9840.5 \times 10^4 kJ/h
=\frac{9840.5 \times 10^4}{3600}=27334.7 kW = 27.33 MW.
(iv) Overall efficiency under new conditions :
Considering the efficiencies of turbines, we have :
\left(W_T\right)_{ Hg }=h_l-h_m{ }^{\prime}=0.84 \times 101.1=84.92 kJ/kg
∴ h_{m^{\prime}}=h_l-84.92=355.98-84.92=271.06 kJ/kg
∴ m^{\prime}\left(h_{m^{\prime}}-h_{n^{\prime}}\right)=\left(h_1-h_{f_4}\right)
or m^{\prime}=\frac{h_1-h_{f_4}}{h_{m^{\prime}}-h_{n^{\prime}}}=\frac{2666.9}{271.06-29.98}=11.06 kg
\left(Q_1\right)_{\text {total }}=m^{\prime}\left(h_l-h_{f_k}\right)+1\left(h_1^{\prime}-h_1\right)[At 15 bar, 300°C : h_g=3037.6 kJ/kg, s_g=6.918 kJ/kg K]
= 11.06 × 326 + (3037.6 – 2789.9) = 3853.26 kJ/kg
s_1^{\prime}=6.918=s_2^{\prime}=0.423+x_2^{\prime} \times 8.052∴ x_2^{\prime}=\frac{6.918-0.423}{8.052}=0.80.
h_2^{\prime}=121.5+0.807 \times 2432.9=2084.8 kJ/kg
\left(W_T\right)_{\text {steam }}=h_1^{\prime}-h_2^{\prime}=0.88(3037.6-2084.8)= 838.46 kJ/kg
\left(W_T\right)_{\text {total }}=11.06 \times 84.92+838.46=1777.67 kJ/kg
Neglecting pump work,
\eta_{\text {overall }}=\frac{1777.67}{3853.26}= 0 . 4 6 1 3 \text { or } 46.13 \%.
