Vapour Power Cycles

6.1.1 Introduction
In any thermodynamic process, the use of working fluid gas or vapour is an essential working medium to convert heat into work. A cycle, which continuously converts heat into work, is called the power cycle. In a power cycle, the working fluid performs the various processes, which are suction, compression, expanding, condensing, etc. All these processes are performed repeatedly to generate the work or converting heat in to work. If the steam is alternatively vaporised and condensed, then the working cycle is called vapour power cycle.

There are various types of working fluids available such as steam, sodium, potassium and mercury. Some working fluids are used at high temperatures and some are at low temperatures. The steam is the mostly used working fluid in the vapour power cycles. The steam has the various desirable characteristics such as low cost, easy availability and high enthalpy of vaporization.

In this unit we will be discussing about the vapour power cycles, which are mostly used for steam power plants. The steam power plants are classified as coal plants, nuclear plants, natural gas or geothermal plants, depending on the type of fuel used to supply the heat to generate the steam.

To simplify the analysis, we shall assume that the change in kinetic and potential energies of the fluid between entry and exit are negligible compared with change in enthalpy. This implies that for the present purpose the energy equation can be written as

\varrho - W = \dot m(h_2 - h_1)


6.1.2 Carnot Cycle
• an ideal theoretical cycle that is the most efficient conceivable
• based on a fully reversible heat engine - it does not include any of the irreversibilities associated with friction, viscous flow, etc.
• in practice the thermal efficiency of real world heat engines are about half that of the ideal, Carnot cycle




Cycle Efficiency:
Cycle efficiency is defined as the network output divided by the gross heat supplied

\eta = \frac{W_{net}}{\varrho _H}=\frac{\varrho _H - \varrho_L}{\varrho _H}= 1-\frac{T_L}{T_H}

From the figure above the gross heat supplied is

\varrho_H-area(s_4 \rightarrow s_3\rightarrow2\rightarrow1\rightarrows_4)=T_H(s_3 - s_4)

\varrho_H - \varrho_l=area(4\rightarrow 3\rightarrow 2\rightarrow 1)=(T_H -T_L)(s_3 - s_4)

Therefore the Carnot efficiency is

\eta =\frac{(T_H - T_L)(s_3 - s_4)}{T_H(s_3 - s_4)}=1-\frac{T_L}{T_H}

Practical difficulties associated with the Carnot cycle
• at state point 4 the steam is wet at TL and it is difficult to pump water/steam (two phase) to state point 1
• the pump can be sized smaller if the fluid is 100% liquid water
• the pump is smaller, cheaper and more efficient
• can we devise a Carnot cycle to operate outside the wet vapour region



    o between state points 4 and 3 the vapour must be isothermal and at different pressures -this is difficult to achieve
    o the high temperature and pressure at 2 and 3 present metallurgical limitations

The net effect is that the Carnot cycle is not feasible for steam power plants. To resolve the difficulties associated with the Carnot cycle, the Rankine cycle was devised.


6.1.3 Rankine Cycle

There are two major reasons why the Carnot Cycle is not employed in practice:
● It has a low work ratio
● Because of practical difficulties associated with the compression. It would be difficult to control the condensation process so that it stopped at state 4 and then carry out the compression efficiently. The liquid tends to separate out from the vapour and the compressor would have to deal with a non-homogenous mixture. Moreover the volume of the fluid is high and the compressor would be comparable in size and cost with the turbine. It is comparatively easy on the other hand, to condense the vapour completely and compress the liquid to boiler pressure in a small feed pump. The resulting cycle, shown in figure below, is known as the Rankin cycle.

The simple Rankine cycle has the same component layout as the Carnot cycle shown above. The simple Rankine cycle continues the condensation process 3-4 until the saturated liquid line is reached.





Determination of enthalpy at state 1 (h1)
The pump work is obtained from the conservation of mass and energy for steady-flow but neglecting potential and kinetic energy changes and assuming the pump is adiabatic and reversible.

\dot m_1 = \dot m_4 = \dot m

\dot m_1h_4 + \dot W_{pump} = \dot m_1h_1

\dot W_{pump} = \dot m(h_1 - h_4)

Since the pumping process involves an incompressible liquid, state 2 is in the compressed liquid region, we use a second method to find the pump work or the h across the pump.

Recall the property relation:

dh = T ds + v dP

Since the ideal pumping process 4-1 is isentropic, ds = 0. → dh = v dP

\Delta h = (h_1 - h_4) = \int_{1}^{4}vdp

v \cong v_4 = const.

(h_1 - h_4) \cong v_4(p_1 - p_4)

The pump work is calculated from

w_{pump} = \frac {\dot W_{pump}}{\dot m} = v_4(p_1 - p_4)

Example:
The figure below shows a simple vapour power plant operating at steady state with water circulating through the components. Relevant data at key locations are given on the figure. The mass flow rate of the water is 109 kg/s. Kinetic and potential energy effects are negligible as are all stray heat transfers.
Determine:
a. the thermal efficiency
b. the mass flow rate of the cooling water passing through the condenser, in kg/s.




Assumptions:
1) Control volumes enclosing each of the four components are at steady state.
2) For each control volume, kinetic/potential energy effects are negligible as are all stray heat transfers

Analysis:
For any power cycle, the thermal efficiency is the ratio of the net work developed to the heat added.

\eta = \frac {\dot W_{net}}{\dot \varrho_{in}} = \frac {\dot W_{turbine} - \dot W_{pump}}{\dot \varrho_{in}}

Applying energy rate balances to the turbine, pump and steam generator, respectively, yields;

\dotW_{turbine} = \dot m(h_1 - h_2)

\left | \dot W_{pump} \right | = \left |\dot m(h_3 - h_4) \right |

\dot \varrho_m = \dot m(h_1 - h_4)



State 3: exit of condenser
Since it stated that saturated liquid at 0.08 bar, thus h3 = hf @ 0.08 bar = 174 kJ/kg

State 4: entry to boiler
The pressure and temperature at state 4 are given as p4 = 100 bar and T4 = 43oC

Tsat @ 100 bar = 311oC which is > T4 thus the fluid is a compressed liquid at state 4 and h4 = hf @ 43oC = 180 kJ/kg

\eta =\frac{\dot m(h_1 - h_2) - |\dot m(h_3 - h_4)|}{\dot m(h_1 - h_4)}

= \frac{(3423.2-2335.8)-|(174-180)|}{(3423.2-180)}

= 0.3334

part (a) η = 33.34%

Applying mass rate balances to the condensor only, we get:

\dot m_2 = \dot m_3 = 109 kg/s and \dot m_5 = \dot m_6 = \dot m_{cw}

Using assumption (2) and applying energy rate balance to the condenser:

0 = \dot m_2(h_2 - h_3) + \dot m_{cw}(h_5 - h_6)

\dot m_{cw} = \frac{\dot m_2(h_2 - h_3)}{(h_6 - h_5)}

Since water is incompressible, we can approximate enthalpy as h=hf(T)

h5 = h5(20oC) = 83.96 kJ/kg
h6 = h6(35oC) = 146.08 kJ/kg

\dot m_{cw} = \frac{(109(2336.7-173.9))}{(146.08 - 83.96)}

Part (b) → \dot m_{cw} = 3759 kg/s

Example:
Determine the thermal efficiency of an ideal Rankine cycle for which steam leaves the boiler as superheated vapor at 6 MPa, 350oC, and is condensed at 10 kPa.



Boiler

To find the heat supplied in the boiler, we apply the steady-flow conservation of mass and energy to the boiler. If we neglect the potential and kinetic energies, and note that no work is done on the steam in the boiler, then

\dot m_2 = \dot m_3 = \dot m
\dot m_2h_2 + \dot \varrho_{in} = \dot m_3h_3
\dot \varrho_{in} = \dot m(h_3 - h_2)

We find the properties at state 3 from the superheated tables as


The heat transfer per unit mass is

q{in} = \frac {\varrho {_in}}{m} = h_3 - h_2

= (3043.9 - 197.86) \frac{kJ}{kg}

= 2845.1 \frac{kJ}{kg}

Turbine

The turbine work is obtained from the application of the conservation of mass and energy for steady flow. We assume the process is adiabatic and reversible and neglect changes in kinetic and potential energies

\dot m_3 = \dot m_4 = \dot m

\dot m_3h_3 = \dot W_{turb} + \dot m_4h_4

\dot W_{turb} + \dot m(h_3 - h_4)

We find the properties at state 4 from the steam tables by noting s4 = s3 = 6.3357 kJ/kg-K and asking three questions

at P_4 = 10kPa:s_j = 0.6492 \frac {kJ}{kg \cdot K}; s_g = 8.1488 \frac {kJ}{kg \cdot K}

is s 4 < s f ?

is s f < s 4 < s g ?

is s g < s 4 ?

s_4 = s_f + x_4s_{fg}

x_4 = \frac{s_4 - s_f}{s_{fg}} = \frac{6.3357 - 0.6492}{7.4996} = 0.758

h_4 = h_f + x_4h_fg

= 191.81\frac{kJ}{kg} + 0.758(2392.1)\frac{kJ}{kg}

= 2005.0\frac{kJ}{kg}

The turbine work per unit mass is

w_{turb} = h_2 - h_4

=(3043.9 - 2005.0)\frac{kJ}{kg}

= 1038.9\frac{kJ}{kg}

The net work done by the cycle is

w_{net} = w_{turb} - w_{pump}

= (1038.9 - 6.05)\frac{kJ}{kg}

= 1032.8\frac{kJ}{kg}

The thermal efficiency is

\eta_{th} = \frac{w_{net}}{q_{in}} = \frac{1032.8\frac{kJ}{kg}}{2845.1\frac{kJ}{kg}}

0.363 or 36.3%


6.1.4 Week 6a: The Vapour Power Cycles: Tutorials

1. The steam power plant shown in Figure below operates on the ideal Rankine cycle between the pressure limits of 6.0 MPa and 0.004 MPa. Show the cycle on a T-s diagram with respect to saturation lines, and determine:

i.    the missing values in Table below; [h2 = 1773 kJ/kg ; x2=0.679]
ii.    the thermal efficiency of the cycle; [37.8%]
iii.    the work ratio for the cycle; [0.9941]
iv.    the Specific Steam Consumption (SSC). [3.582kg/kW h]



2. A small power plant produces 25 kg/s steam at 3 MPa, 600oC in the boiler. It cools the condenser with ocean water coming in at 12oC and returned at 15oC so the condenser exit is at 45oC. Determine the net power output and the required mass flow rate of ocean water. [32623kW; 4358.05 kg/s]



3. In a Rankine cycle, the steam of inlet to turbine is saturated at a pressure of 35 bar and the exhaust pressure is 0.2 bar. Determine :

a. The pump work,     [3.54kJ/kg]
b. The turbine work,     [7495.5kJ/kg]
c. The Rankine efficiency,     [30.93%]
d. The condenser heat flow, and     [16734.25kJ/kg]
e. The dryness at the end of the expansion.    [0.747]



4. A Carnot cycle uses steam as the working fluid. The fluid is in the liquid state at the saturation temperature of 165oC at the beginning of the cycle and receives 1800 kJ/kg of heat during the isothermal expansion and hence the steam exits the boiler as a wet mixture. Given that the thermal efficiency is 21.20%.

a) Determine:
i) the condition of the steam at the end of the isentropic expansion; [2497kJ/kg;0.871]
ii) the net work transfer; [381.6kJ/kg]
iii) the cycle power output for a steam flow rate of 50kg/min. [422697.6W]
b) Show the cycle on a T-s diagram with respect to saturation lines.