Question

Steam is the working fluid in a simple, ideal Rankine cycle. Saturated vapor enters the turbine at 8 MPa and saturated liquid exits the condenser at a pressure of 8 kPa. The net power output of the cycle is 100 MW. Determine for the cycle:

i. Thermal efficiency

ii. Back work ratio

iii. Mass flow-rate of the steam in kg/h

iv. Rate of heat transfer to the working fluid as it passes through the boiler in MW

v. Rate of heat transfer from the working fluid as is passes through the condenser in MW

vi. The mass flow rate of the condenser cooling water in kg/h if the cooling water enters the condenser at 15

Answer 1

Concepts and reason

Ideal Rankine cycle:

An ideal Rankine cycle consists of the following four processes:

1-2: Isentropic expansion in turbine

2-3: Isothermal heat rejection in condenser

3-4: Isentropic compression in pump

4-1: Isothermal heat addition in boiler

The T-s diagram for an ideal Rankine cycle is as follows:

Fundamentals

Write the equation for the thermal efficiency of the cycle is:

$\eta = \frac{{{{\dot W}_t} - {{\dot W}_p}}}{{{{\dot Q}_{in}}}}$

Here, the work done by the turbine is ${\dot W_t}$ , the work done on the pump is ${\dot W_p}$ , and the heat input to the boiler is ${\dot Q_{in}}$ .

Write the equation for the back work ratio:

${\rm{bwr}} = \frac{{{{\dot W}_p}}}{{{{\dot W}_t}}}$

The mass flow rate of the steam is as follows::

$\dot m = \frac{{{{\dot W}_{{\rm{cycle}}}}}}{{{{\dot W}_t} - {{\dot W}_p}}}$

The rate of heat transfer as the fluid passes through boiler is as follows:

${\dot Q_{in}} = \dot m\left( {{h_1} - {h_4}} \right)$

The rate of heat transfer as the fluid passes through condenser is as follows:

${\dot Q_{out}} = \dot m\left( {{h_2} - {h_3}} \right)$

State 1:

Obtain the properties corresponding to ${p_1} = 8{\rm{ MPa}}$ and ${x_1} = 1$ (saturated vapour) from the Table, "Saturated water- Pressure table."

Specific enthalpy at state 1, ${h_1} = {h_g} = 2758.0{\rm{ kJ/kg}}$

Specific entropy at state 1, ${s_1} = {s_g} = 5.7432{\rm{ kJ/kg}} \cdot {\rm{K}}$

State 2:

Since, the process 1-2 is isentropic,

$\begin{array}{c}\\{s_2} = {s_1}\\\\ = 5.7432{\rm{ kJ/kg}} \cdot {\rm{K}}\\\end{array}$

Obtain the properties corresponding to ${p_2} = 8{\rm{ kPa}}$ and ${s_2} = 5.7432{\rm{ kJ/kg}} \cdot {\rm{K}}$ (saturated vapour) from the Table, "Saturated water- Pressure table."

Specific entropy of saturated liquid at state 2, ${s_f} = 0.5926{\rm{ kJ/kg}} \cdot {\rm{K}}$

Specific entropy of liquid-vapour phase at state 2, ${s_{fg}} = 7.6361{\rm{ kJ/kg}} \cdot {\rm{K}}$

Specific enthalpy of saturated liquid at state 2, ${h_f} = 173.88{\rm{ kJ/kg}}$

Specific enthalpy liquid-vapour phase at state 2, ${h_{fg}} = 2403.1{\rm{ kJ/kg}}$

Calculate the quality of steam at state 2:

${s_2} = {s_f} + {x_2}{s_{fg}}$

Substitute $5.7432{\rm{ kJ/kg}} \cdot {\rm{K}}$ ${s_2}$ , $0.5926{\rm{ kJ/kg}} \cdot {\rm{K}}$ for ${s_f}$ , and $7.6361{\rm{ kJ/kg}} \cdot {\rm{K}}$ for ${s_{fg}}$ .

$\begin{array}{l}\\5.7432 = 0.5926 + {x_2}\left( {7.6361} \right)\\\\{x_2} = \frac{{5.7432 - 0.5926}}{{7.6361}}\\\\{x_2} = 0.6745\\\end{array}$

Calculate the specific enthalpy at state 2:

${h_2} = {h_f} + {x_2}{h_{fg}}$

Substitute $173.88{\rm{ kJ/kg}}$ for ${h_f}$ , $0.6745$ for ${x_2}$ , and $2403.1{\rm{ kJ/kg}}$ for ${h_{fg}}$ .

$\begin{array}{c}\\{h_2} = 173.88 + \left( {0.6745} \right)\left( {2403.1} \right)\\\\ = 1794.8{\rm{ kJ/kg}}\\\end{array}$

State 3:

Obtain the properties corresponding to ${p_3} = 8{\rm{ kPa}}$ and ${x_1} = 0$ (saturated liquid) from the Table, "Saturated water- Pressure table."

Specific enthalpy at state 3, ${h_3} = {h_f} = 173.88{\rm{ kJ/kg}}$

Specific volume at state 3, ${v_3} = {v_f} = 1.0085 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^3}{\rm{/kg}}$

State 4:

Write the relation for the pump work:

$\begin{array}{l}\\\frac{{{{\dot W}_p}}}{{\dot m}} = {h_4} - {h_3}\\\\{h_4} = \frac{{{{\dot W}_p}}}{{\dot m}} + {h_3}\\\\{h_4} = {v_3}\left( {{p_4} - {p_3}} \right) + {h_3}\\\end{array}$

Substitute $173.88{\rm{ kJ/kg}}$ for ${h_3}$ , $1.0085 \times {10^{ - 3}}{\rm{ }}{{\rm{m}}^3}{\rm{/kg}}$ for ${v_3}$ , $8000{\rm{ kPa}}$ for ${p_4}$ , and $8{\rm{ kPa}}$ for ${p_3}$ .

$\begin{array}{c}\\{h_4} = 1.0085 \times {10^{ - 3}}\left( {8000 - 8} \right) + 173.88\\\\ = 181.94{\rm{ kJ/kg}}\\\end{array}$

Calculate the thermal efficiency of the cycle.

$\begin{array}{c}\\\eta = \frac{{{{\dot W}_t} - {{\dot W}_p}}}{{{{\dot Q}_{in}}}}\\\\ = \frac{{\dot m\left( {{h_1} - {h_2}} \right) - \dot m\left( {{h_4} - {h_3}} \right)}}{{\dot m\left( {{h_1} - {h_4}} \right)}}\\\\ = \frac{{\left( {{h_1} - {h_2}} \right) - \left( {{h_4} - {h_3}} \right)}}{{\left( {{h_1} - {h_4}} \right)}}\\\end{array}$

Substitute $2758.0{\rm{ kJ/kg}}$ for ${h_1}$ , $1794.8{\rm{ kJ/kg}}$ for ${h_2}$ , $173.88{\rm{ kJ/kg}}$ for ${h_3}$ , and $181.94{\rm{ kJ/kg}}$ for ${h_4}$ .

$\begin{array}{c}\\\eta = \frac{{\left( {2758 - 1794.8} \right) - \left( {181.94 - 173.88} \right)}}{{\left( {2758.0 - 181.94} \right)}}\\\\ = 0.371\\\\ = 37.1\% \\\end{array}$

Calculate the back work ratio.

$\begin{array}{c}\\{\rm{bwr}} = \frac{{{{\dot W}_p}}}{{{{\dot W}_t}}}\\\\ = \frac{{{h_4} - {h_3}}}{{{h_1} - {h_2}}}\\\end{array}$

Substitute $2758.0{\rm{ kJ/kg}}$ for ${h_1}$ , $1794.8{\rm{ kJ/kg}}$ for ${h_2}$ , $173.88{\rm{ kJ/kg}}$ for ${h_3}$ , and $181.94{\rm{ kJ/kg}}$ for ${h_4}$ .

$\begin{array}{c}\\{\rm{bwr}} = \frac{{181.94 - 173.882}}{{2758 - 1794.8}}\\\\ = 8.4 \times {10^{ - 3}}\\\\ = 84\% \\\end{array}$

Calculate the mass flow rate of the steam.

$\begin{array}{c}\\\dot m = \frac{{{{\dot W}_{{\rm{cycle}}}}}}{{{{\dot W}_t} - {{\dot W}_p}}}\\\\ = \frac{{{{\dot W}_{{\rm{cycle}}}}}}{{\left( {{h_1} - {h_2}} \right) - \left( {{h_4} - {h_3}} \right)}}\\\end{array}$

Here, the work done during the cycle is ${\dot W_{{\rm{cycle}}}}$ .

Substitute $110{\rm{ MW}}$ for ${\dot W_{{\rm{cycle}}}}$ , $2758.0{\rm{ kJ/kg}}$ for ${h_1}$ , $1794.8{\rm{ kJ/kg}}$ for ${h_2}$ , $173.88{\rm{ kJ/kg}}$ for ${h_3}$ , and $181.94{\rm{ kJ/kg}}$ for ${h_4}$ .

$\begin{array}{c}\\\dot m = \frac{{{{\dot W}_{{\rm{cycle}}}}}}{{{{\dot W}_t} - {{\dot W}_p}}}\\\\ = \frac{{110{\rm{ MW}} \times \left( {\frac{{1000{\rm{ kW}}}}{{1{\rm{ MW}}}}} \right) \times \left( {\frac{{1{\rm{ kJ/s}}}}{{1{\rm{ kW}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)}}{{\left( {2758 - 1794.8} \right) - \left( {181.94 - 173.88} \right)}}\\\\ = 3.77 \times {10^5}{\rm{ kg/h}}\\\end{array}$

Calculate the rate of heat transfer to the fluid upon passing through the boiler.

${\dot Q_{in}} = \dot m\left( {{h_1} - {h_4}} \right)$

Substitute $3.77 \times {10^5}{\rm{ kg/h}}$ for $\dot m$ , $2758.0{\rm{ kJ/kg}}$ for ${h_1}$ , and $181.94{\rm{ kJ/kg}}$ for ${h_4}$ .

$\begin{array}{c}\\{{\dot Q}_{in}} = 3.77 \times {10^5} \times \left( {\frac{1}{{3600}}} \right){\rm{ kg/s}} \times \left( {2758.0 - 181.94} \right){\rm{ }}\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}\\\\ = 269.77 \times {10^3}{\rm{ kW}}\\\\ = 269.77{\rm{ MW}}\\\end{array}$

Calculate the rate of heat transfer to the fluid upon passing through the condenser.

${\dot Q_{out}} = \dot m\left( {{h_2} - {h_3}} \right)$

Substitute $3.77 \times {10^5}{\rm{ kg/h}}$ for $\dot m$ , $1794.8{\rm{ kJ/kg}}$ for ${h_2}$ , and $173.88{\rm{ kJ/kg}}$ for ${h_3}$ .

$\begin{array}{c}\\{{\dot Q}_{out}} = 3.77 \times {10^5} \times \left( {\frac{1}{{3600}}} \right){\rm{ kg/s}} \times \left( {1794.8 - 173.88} \right){\rm{ }}\frac{{{\rm{kJ}}}}{{{\rm{kg}}}}\\\\ = 169.75 \times {10^3}{\rm{ kW}}\\\\ = 169.75{\rm{ MW}}\\\end{array}$

Obtain the properties corresponding to ${T_{{\rm{in}}}} = 15^\circ {\rm{C}}$ and ${x_1} = 0$ (saturated liquid) from the Table, "Saturated water- temperature table."

Specific enthalpy of inlet cooling water, ${h_{c,in}} = {h_f} = 62.99{\rm{ kJ/kg}}$

Obtain the properties corresponding to ${T_{out}} = 35^\circ {\rm{C}}$ and ${x_1} = 0$ (saturated liquid) from the Table, "Saturated water- temperature table."

Specific enthalpy of outlet cooling water, ${h_{c,out}} = {h_f} = 146.68{\rm{ kJ/kg}}$

Calculate the mass flow rate of the condenser cooling water by using steady flow energy equation:

$\begin{array}{l}\\0 = {{\dot Q}_{cv}} - {{\dot W}_{cv}} + {{\dot m}_c}\left[ {{h_{c,in}} - {h_{c,out}}} \right] + \dot m\left( {{h_2} - {h_3}} \right)\\\\{{\dot m}_c}\left[ {{h_{c,in}} - {h_{c,out}}} \right] + {{\dot Q}_{out}} = 0\\\\{{\dot m}_c} = \frac{{{{\dot Q}_{out}}}}{{{h_{c,out}} - {h_{c,in}}}}\\\end{array}$

Here, the mass flow rate of the condenser cooling water is ${\dot m_c}$ .

Substitute $169.75{\rm{ MW}}$ for ${\dot Q_{out}}$ , $62.99{\rm{ kJ/kg}}$ for ${h_{c,in}}$ , and $146.68{\rm{ kJ/kg}}$ for ${h_{c,out}}$ .

$\begin{array}{c}\\{{\dot m}_c} = \frac{{169.75 \times {{10}^3}{\rm{ kW}}}}{{\left( {146.68 - 62.99} \right){\rm{ kJ/kg}}}} \times \left( {\frac{{1{\rm{ kJ/s}}}}{{1{\rm{ kW}}}}} \right) \times \left( {\frac{{3600{\rm{ s}}}}{{1{\rm{ h}}}}} \right)\\\\ = 7.3 \times {10^6}{\rm{ kg/h}}\\\end{array}$

Ans: Part i

The thermal efficiency of the cycle is ${\bf{37}}{\bf{.1\% }}$ .

Part ii

The back work ratio is ${\bf{84\% }}$ .

Part iii

The mass flow rate of the steam is ${\bf{3}}{\bf{.77 \times 1}}{{\bf{0}}^{\bf{5}}}{\bf{ kg/h}}$ .

Part iv

The rate of heat transfer to the fluid upon passing through the boiler is ${\bf{269}}{\bf{.77 W}}$ .

Part v

The rate of heat transfer to the fluid upon passing through the condenser is ${\bf{169}}{\bf{.75 W}}$ .

Part vi

The mass flow rate of the condenser cooling water is ${\bf{7}}{\bf{.3 \times 1}}{{\bf{0}}^{\bf{6}}}{\bf{ kg/h}}$ .