27
CHAPTER 2
VAPOUR POWER CYCLES
2.1 Introduction
2.1.1 Characteristics of power cycles
- The working fluid is a condensable vapour which is in liquid phase
during part of the cycle
- The cycle consists of a succession of steady-flow processes, with
each process carried out in a separate component specifically
designed for that purpose.
- Each component constitutes an open system, and all the
components are connected in series so that as the fluid circulates
through the power plant each fluid element passes through a cycle
of mechanical and thermodynamic states.
- To simplify the analysis, it is assumed that the change in kinetic
and potential energy of the fluid between entry and exit in each
component is negligible compared to the change in enthalpy. This
implies that the energy equation can be written as: Q – W = h2 – h1
- The working fluid is usually steam because it is cheap and
chemically stable but any condensable vapour may be used.
2.1.2 Criteria of performance
(a) Ideal cycle efficiency – this is the efficiency of a cycle when all
the processes are assumed to be reversible.
(b) Actual cycle efficiency – this is the efficiency of a cycle when
process efficiencies are introduced
, 28
(c) Efficiency ratio – this is the ratio of the actual cycle efficiency to
the ideal cycle efficiency.
(d) Work ratio (rw) – this is the ratio of the net work to the positive
work done in the cycle. (It is a measure of the cycle’s sensitivity
to irreversibilities since irreversibilities decrease the positive
work and increase the negative work.)
(e) Specific steam consumption (ssc) – this is the mass flow of steam
required per unit of power output. It is usually expressed in
kg/kW h and if the numerical value of net work output per unit
mass of flow is W (kJ/kg) the ssc can be found from
3600
ssc
W
2.2 Carnot cycle
- It consists of two reversible isothermal processes at T a and Tb
respectively, connected by two reversible adiabatic (isentropic)
processes.
- When the working fluid is a condensable vapour, the two
isothermal processes are easily obtained by heating and cooling at
constant pressure while the fluid is a wet vapour.
- The processes are:
1-2: Saturated water is evaporated at constant pressure to form
saturated steam; heat added is Q12 = h2 – h1
2-3: Saturated steam is expanded isentropically in a turbine; work
done is W23 = h2 – h3
, 29
3-4: Wet steam is partially condensed at constant pressure to state 4
where s4 = s1; heat rejected is Q34 = h4 – h3
4-1: Steam is compressed isentropically in a compressor; work
required is W41 = h4 – h1
1 Boiler 2 T
Ta 1 2
Tb
4 3
Condenser s
4 3
Example 2.1
Calculate the heat and work transfers, cycle efficiency, work ratio
and steam consumption of a Carnot cycle using steam between
pressures of 30 and 0.04 bar.
Solution
From tables, at 30 bar
T1 = T2 = 507.0 K; h1 = hf = 1008 kJ/kg; h2 = hg = 2803 kJ/kg
Putting s4 = s1 and s3 = s2, then at the condenser pressure of 0.04 bar,
T3 = T4 = 302.2 K; x3 = 0.716; x4 = 0.276
Hence from h = hf + xhfg,
h3 = 121 + x32433 = 1863 kJ/kg; h4 = 121 + x42433 = 793 kJ/kg
The turbine work is W23 = h2 – h3 = 940 kJ/kg
CHAPTER 2
VAPOUR POWER CYCLES
2.1 Introduction
2.1.1 Characteristics of power cycles
- The working fluid is a condensable vapour which is in liquid phase
during part of the cycle
- The cycle consists of a succession of steady-flow processes, with
each process carried out in a separate component specifically
designed for that purpose.
- Each component constitutes an open system, and all the
components are connected in series so that as the fluid circulates
through the power plant each fluid element passes through a cycle
of mechanical and thermodynamic states.
- To simplify the analysis, it is assumed that the change in kinetic
and potential energy of the fluid between entry and exit in each
component is negligible compared to the change in enthalpy. This
implies that the energy equation can be written as: Q – W = h2 – h1
- The working fluid is usually steam because it is cheap and
chemically stable but any condensable vapour may be used.
2.1.2 Criteria of performance
(a) Ideal cycle efficiency – this is the efficiency of a cycle when all
the processes are assumed to be reversible.
(b) Actual cycle efficiency – this is the efficiency of a cycle when
process efficiencies are introduced
, 28
(c) Efficiency ratio – this is the ratio of the actual cycle efficiency to
the ideal cycle efficiency.
(d) Work ratio (rw) – this is the ratio of the net work to the positive
work done in the cycle. (It is a measure of the cycle’s sensitivity
to irreversibilities since irreversibilities decrease the positive
work and increase the negative work.)
(e) Specific steam consumption (ssc) – this is the mass flow of steam
required per unit of power output. It is usually expressed in
kg/kW h and if the numerical value of net work output per unit
mass of flow is W (kJ/kg) the ssc can be found from
3600
ssc
W
2.2 Carnot cycle
- It consists of two reversible isothermal processes at T a and Tb
respectively, connected by two reversible adiabatic (isentropic)
processes.
- When the working fluid is a condensable vapour, the two
isothermal processes are easily obtained by heating and cooling at
constant pressure while the fluid is a wet vapour.
- The processes are:
1-2: Saturated water is evaporated at constant pressure to form
saturated steam; heat added is Q12 = h2 – h1
2-3: Saturated steam is expanded isentropically in a turbine; work
done is W23 = h2 – h3
, 29
3-4: Wet steam is partially condensed at constant pressure to state 4
where s4 = s1; heat rejected is Q34 = h4 – h3
4-1: Steam is compressed isentropically in a compressor; work
required is W41 = h4 – h1
1 Boiler 2 T
Ta 1 2
Tb
4 3
Condenser s
4 3
Example 2.1
Calculate the heat and work transfers, cycle efficiency, work ratio
and steam consumption of a Carnot cycle using steam between
pressures of 30 and 0.04 bar.
Solution
From tables, at 30 bar
T1 = T2 = 507.0 K; h1 = hf = 1008 kJ/kg; h2 = hg = 2803 kJ/kg
Putting s4 = s1 and s3 = s2, then at the condenser pressure of 0.04 bar,
T3 = T4 = 302.2 K; x3 = 0.716; x4 = 0.276
Hence from h = hf + xhfg,
h3 = 121 + x32433 = 1863 kJ/kg; h4 = 121 + x42433 = 793 kJ/kg
The turbine work is W23 = h2 – h3 = 940 kJ/kg
