Table of Contents
Principle
3
Objective
3
Background
4
Rankine cycle analysis
5
I)
Mass Flow Rate of the Rankine Cycle.
6
II)
Work And Heat Transfer.
6
III)
Thermal Efficiency of Cycle
9
IV)
Air -Fuel ratio and Air Excess
9
V)
Mass flow rate in the turbine
11
VI)
Boiler analysis
11
VII) Cost of Generating Steam and Energy.
14
Experimental Setup
15
Procedure
18
Example#1: Rankine cycle analysis
18
Example#2: Combustion analysis of the boiler
22
Discussion
24
References
24
University of Puerto Rico
Mayagüez Campus
Department of Mechanical Engineering
INME 4032 - LABORATORY II
Spring 2004
Instructor: Guillermo Araya
Experiment 4: Powerplant analysis with a Rankine cycle
Principle
This experiment is designed to acquire experience on the operation of a functional
steam turbine power plant. A comparison of a real world operating characteristics to
that of the ideal Rankine power cycle will be made.
Objective
The objective of this lab is to acquire experience on the basic Rankine cycle and to
understand the factors and parameters affecting the efficiency and cost of generating
energy. In this lab, we will determine:
a) Mass Flow Rate of a Rankine Cycle.
b) Thermodynamics properties (entropies, enthalpies, quality, etc). Draw a
schematic of the cycle in a T-S diagram.
c) Work and heat transfer in the different stages of the cycle.
d) Thermal efficiency of the cycle.
e) Mass flow rate in the turbine.
f) Boiler efficiency
g) Air-Fuel ratio and air excess.
h) Cost of generating steam and energy.
Background
The Rankine cycle is the most common of all power generation cycles and is
diagrammatically depicted via Figures 1 and 2. The Rankine cycle was devised to make
use of the characteristics of water as the working fluid. The cycle begins in a boiler
(State 4 in figure 1), where the water is heated until it reaches saturation- in a constantpressure process. Once saturation is reached, further heat transfer takes place at a
constant temperature, until the working fluid reaches a quality of 100% (State 1). At this
point, the high-quality vapor is expanded isoentropically through an axially bladed
turbine stage to produce shaft work. The steam then exits the turbine at State 2.
The working fluid, at State 2, is at a low-pressure, but has a fairly high quality, so it is
routed through a condenser, where the steam is condensed into liquid (State 3). Finally,
the cycle is completed via the return of the liquid to the boiler, which is normally
accomplished by a mechanical pump. Figure 2 shows a schematic of a power plant
under a Rankine cycle.
Figure 1: Diagrams for a simple ideal Rankine cycle:
a) P-V diagram, b) T-S diagram
Figure 2: Schematic of a simple ideal Rankine cycle
Rankine cycle analysis
This experiment has an important difference with the cycle shown in Figure 2. The
difference is that there is not a pump to complete the cycle. This is not exactly a cycle.
Instead, it is an open system. The water crossing the condenser is stored in a tank as
show in Figure 3, but the principle of Rankine cycle studied in Thermodynamic is still
valid.
The boiler will be filled with water before the experiment and the experiment will be
ended when the water is reaches the minimum level of correct operation, given by the
manufacturer.
Another important difference is that between the boiler and turbine there is a valve that
generates a throttling effect. The throttling process is analyzed as an isenthalpic
process. This phenomenon will be analyzed more in detail. Also, the boiler generates a
superheated vapor.
Figure 3: Schematic of Rankine cycle steam turbine apparatus
I. Mass Flow Rate of the Rankine Cycle.
Evaluating the time of operation and volume of consumed water, the mass flow rate can
be measured as:
 water  q water  water 
m
 water
 water
time
Here, time is measured with a chronometer for a known volume of water  water in the
boiler.
Figure 4: Real Rankine cycle
II. Work and Heat Transfer
For this analysis, it is assumed that the process is ideal and there are not pressure
losses occurring in the piping, but as has been said previously the boiler generates
superheated vapor and there is a throttling process in the valve. Figure 4 shows the
modified cycle of the plant.
The evaporator, in this case a fire-tube boiler, produces a superheated vapor (Stage 1 ).
Taking a control volume enclosing the boiler tubes and drums, the energy rate balance
gives:


V12  V22

0  Qin  m water h1  h4 
 g z1  z 2 
2


neglecting kinetic and potential energy, the energy equation reduce to:
Q in  m water h1  h4 
Then, vapors pass through the valve, states1’-1”. For a control volume enclosing the
valve, the mass and energy rate balance reduces under steady state to:
0  Q v  m water h1  h1 
Since there is not work done in the valve and heat transfer Q v can be neglected, last
equation reduces to:
h1  h1
which means that there is an isenthalpic expansion in the valve.
Making a similar analysis for the pump and condenser, the work and heat transfer are:
Q out  m water h2  h3  and
 water h4  h3 
W p  m
The energy balance for a control volume around the turbine under steady state condition
is:
0  Q cv  W t  m water h1  h2 
Neglecting heat transfer Q cv to the surrounding, the process in the turbine is assumed
adiabatic and reversible, so isentropic ( S 2  S1 ) and the energy equation reduces to:
W t  m water h1  h2 
Then, knowing that S 2  S1 and also S f 2 and S g 2 which could be estimated with the
pressure and temperature at outlet of the turbine, the quality of the vapor can be
calculated as:
x2 
S g 2  S1
Sg2  S f 2
with x 2 , the enthalpy h2 is calculated as:

h2  hg 2  x2 hg 2  h f 2

where h f 2 and hg 2 are calculated with the outlet temperature. It is important to
emphasize that the valve generates entropy from state 1 to the state 1 . Without the
expansion valve the cycle would be close to an isentropic expansion 1  2 in the
turbine. All parameters h1 , h1 , S1 , S1 , S1 , hB
and S 4 can be determined from
temperatures and pressures at each stage.
III. Thermal Efficiency of Cycle
The net work of the cycle is defined by the difference between the turbine work and the
pump work:
Wcycle  Wt  W p  m water h1  h2   m water h4  h3 
If the pump work is neglected, the net work of the cycle reduces to:
 water h1  h2 
W cycle  m
Then the thermal efficiency of this system is defined by the rate between the net work
and heat transfer from the boiler:
W t h1  h2 


h1  h4 
Qin
IV. Air -Fuel ratio and Air Excess.
The chemical composition of the gases at the outlet of boiler is:
ACO2   BCO  C NO  DO2   F NO2   GN 2   M water H 2 O 
at the inlet, there are dry air and fuel (butane):
M air O2  3.76 N 2   M fuel C 4 H 10 
Then, making a balance between inlet and outlet:
M air O2  3.76 N 2   M fuel C 4 H 10   ACO2   BCO  C NO  DO2   F NO2 
 GN 2   M water H 2 O 
so,
M fuel 
M air 
A B
4
C  F  2G
3.76
M water 
M fuel
5
Where the coefficients (A, B, C, D, F, G and Mi) are the molar mass necessary to
balance the equation. Then the air excess is:
Eair 
M air
100
M air (ideal )
the M air (ideal ) is the molar mass of air when the chemical reaction is complete, and
there is not formation of water and intermediate compounds:
M air (ideal )O2  3.76 N 2   C4 H10   ACO2   GN 2   M water H 2 O 
Balancing this equation: M air (ideal )  13 , G  24.44 , A  4 and M water  5 , which is:
2
13
O2  3.76 N 2   C 4 H 10   4CO2   24.44N 2   5H 2 O 
2
Then, the Air-Fuel ratio is defined by:
AF 
M air Pair
M fuel Pfuel
Where Pair and Pfuel are the atomic weight of air and combustible, respectively. The
Pair  29 kg/Kmol and the Pfuel  4 Pc  10 PH  58.12 kg/Kmol.
V. Mass flow rate in the turbine
From the generated amperage and voltage:
W t  VI
so, the mass flow rate in the turbine is:
m 
VI
 t h1  h2 
Where  t is the efficiency of the turbine. Here, we will assume this efficiency equal to
one.
VI. Boiler analysis
From the chemical equation of combustion, balanced in term of moles:
mass air O 2  3.76 N 2   mass comb C 4 H10   ACO 2   BCO  CNO  DO 2   FNO2 
 GN 2   MH 2 O
the first law of thermodynamics for a volume enclosing the boiler is:
 mh  Q
comb
  mh
R
where

R
P
and

are the sum for each reactants and products of combustion.
P
Remember that mi  ni M i , where mi is mass, ni is number of moles and M i is the
molar mass of the i-th component. Last equation is written in the form:
 nMh   Q
comb
R
  nMh 
P
Here, h is the enthalpy of reactants and products at the temperature of inlet and outlet
of the boiler. They could be found in the table of enthalpies of formation.
Figure 5: Enthalpy of formation
Another form to write the first law is:
 nM h
0
f



 h   Qcomb   nM h 0f  h 
R
P
where h 0f is the enthalpy of reactants and products, respectively, at the standard
temperature and pressure. Rearranging:
 
  

  nM h    nM h    nM h    nM h 
Qcomb   nM h 0f  h   nM h 0f  h
P
R
0
f
P
0
f
R
P
R
0
The first two terms are the enthalpy of combustion ( hPR
) at standard temperature and
pressure.
0
Qcomb  hPR
  nM h    nM h 
P
R
Table 1: Enthalpy of formation, HHV and LHV
The enthalpy of combustion also is called heating value (HV), and this is number
indicative to the useful energy content of different fuels. There are two types of heating
value: higher heating value (HHV) and the lower heating value (LHV). The HHV is
obtained when all the water formed by combustion is a liquid. The LHV is obtained when
all the water formed by the combustion is a vapor. For that HHV is more than LHV (see
Table 1). For calculations, we will assume that water formed is in the liquid state and the
0
HHV will be used for hPR
. Now, we can calculate the efficiency of the boiler as:
 boiler 
VII.
Qin
Qcomb
Cost of Generating Steam and Energy.
The mass flow of fuel is the product between the density and fuel flow mass and the
time of operation:
 fuel   fuel q fuel
m
where  fuel is the density of butane gas at atmospheric pressure. Then the cost of
generating steam per unit mass of steam is:
STEAM cos t 
m fuel Pr ice fuel
m water
where Pr ice fuel is the price of the fuel. Also it is possible to determine the cost of
generating energy by:
ENERGY cos t 
m fuel Pr ice fuel
VI
Experimental Setup
The equipment has a data acquisition system to collect the information. Also, it will be
necessary a chronometer for estimating the time operation. A view of the real equipment
and data acquisition system is shown in Figure 6.
Figure 6: The mini-power plant
The mini-power plant has a boiler (see Figure 7), which is a dual-pass, flame through
tube type unit. A burner fan speed is electronically adjustable to operate whit a
minimum of excess of air. A vortex disc, located downstream of the boiler unit, mixes
fuel and air and sets up a rotary gas flow that results in efficient heat transfer from the
flame tube to the boilers water, (see Figure 8).
Figure 7: Boiler
Electromechanical and electronic burner and boiler controls are located within the front
operator panel enclosure. An A.G.A. certified electronic ignition gas valve and
microprocessor based gas ignition module automate and supervise flame control. A
transducer assists in regulating boiler pressure by cycling the burner on and off. A
poppet valve, located on top of the boiler, serves as a safety valve. In the event of
control malfunction, the poppet valve will open and relieve boiler pressure.
Figure 8: Forced air gas burned
The other component is the turbine and generator, (see Figure 9). The turbine consists
of the following major components:
1. A precision machined, stainless steel front and rear housing.
2. A nozzle ring and a single stage shrouded impulse turbine wheel
Figure 9: Turbine and Generator
The generator is a 4-pole, permanent magnet, brushless unit. The rotor is supported by
pre-loaded precision ball bearings. The generator includes a full wave, integral rectifier
bridge that delivers direct current to the generators D.C. terminals. The generator
terminal board also carries a set of AC output terminals for experimental procedures
that may entail the use of a transformer, or deal with frequency related topics, rpm
measurement and other AC related experiments.
Figure 10: Cooling tower
Finally, the condenser towers outer mantle is formed from a single piece of aluminum,
(see Figure 10). The towers large surface area affects heat transfer to ambient air and
provides a realistic appearance. Turbine exhaust steam is piped into the bottom of the
tower. The steam is kept in close contact with the outside mantle by means of 4 baffles.
Procedure
1. At the moment of making the experiment, the steam turbine will be operational in
the no load condition. So, the first step is to set the ¼ of the maximum load
applied on the turbine by the generator.
2. Allow the system to reach steady state, and take readings. They are:
a) Boiler temperature.
b) Boiler pressure.
c) Turbine inlet temperature.
d) Turbine exit temperature.
e) Turbine inlet pressure.
f) Turbine exit pressure.
g) Water flow.
h) Generator amperage.
i) Generator voltage.
j) Time operation.
k) Repeat the step 2) for ½ and ¾ of the maximum load applied.
Example #1: Rankine cycle analysis
Problem:
Steam is the working fluid in an ideal Rankine cycle. Saturated vapor enters the turbine
at 8.0MPa and saturated liquid exits the condenser at a pressure of 0.008MPa (see
Figure 11). The net power of cycle is 100MW. Determine for the cycle:
a) The thermal efficiency.
b) The mass flow rate of steam.
c) The rate of heat transfer, into the working fluid as it passes through the boiler.
d) The rate of heat transfer, from the condensing steam as it passes through the
condenser.
e) The mass flow rate of condenser cooling water, if cooling water enters the
condenser at 15°C and exits at 35°C.
Figure 11: Schematic of the Rankine cycle
Solution
Assumption:
1. Each component of the cycle is analyzed as a control volume at steady state.
2. All processes of the working fluid are internally reversible.
3. The turbine and pump operate adiabatically.
4. Kinetic and potential energy effects are negligible.
5. Saturated vapor enters the turbine. Condensate exits the condenser
as
saturated liquid.
Analysis:
To begin the analysis, let us fix each of the principal states located on the
accompanying schematic and T-s diagram. Starting at the inlet to the turbine, the
pressure is 8.0MPa and the steam is a saturated vapor, so from Table A-3 of Moran and
Shapiro, h1  2758.0 kJ/kg and S1  5.7432 kJ/kg - K
Stage 2 is fixed by p2  0.008 MPa and the fact that specific entropy is constant for the
adiabatic, internally reversible expansion through the turbine. Using liquid and saturated
vapor data from Table A-3 of Moran and Shapiro, we find that the quality at stage 2 is:
x2 
S2  S f

Sg  S f
5.7432  0.5926
 0.6745
7.6361
The enthalpy is then
h2  h f  x 2 h fg  173.88  (0.6745)2403.1  1794.8 kJ/kg
Stage 3 is saturated liquid at 0.008MPa, so h3  173.88 kJ/kg . Stage 4 is fixed by the
boiler pressure p 4 and the specific entropy S 4  S 3 . The specific enthalpy h4 can be
found by interpolation in the compressed liquid tables. However, because liquid data are
relatively sparse, it is more convenient to solve
W p
m
W p
m
 h4  h3 for h4 , using
  3 ( p4  p3 ) to approximate the pump work. With this approach:
h4  h3 
W p
m
 h3   3 ( p4  p3 )
Substituting property values from Table A-3 of Moran and Shapiro:
 10 6 N / m 2   1 kJ / kg 

   3
h4  173.88 kJ / kg  (1.0084  10 m / kg)  (8.0  0.008) MPa  
1
MPa
10
N

m

 

h4  181.94 kJ / kg
3
3
a) The net power developed by the cycle is:
W net  W t  W p
Energy balance for a control volume around the turbine and pump gives, respectively
W t
 h1  h2
m
an d
W p
m
 h4  h3
 is the mass flow rate of the steam. The rate of heat transfer to the working
where m
fluid as it passes through the boiler is determined using an energy rate balance as:
Q in
 h1  h4
m
the thermal efficiency is then:

W t  W p h1  h2   h1  h4  2758.0  1794.8  181.94  173.88 kJ / kg


h  h 
2758.0  181.94 kJ / kg
Q
1
in
 0.371 37.1%
4
b) The mass flow rate of steam can be obtained from the expression for the net power
given in part a). Thus:
m 
100MW 103 kW / Mw3600s / h  3.77  105 kg / h
h1  h2   h1  h4 
963.2  8.06kj / kg
W cycle

c) With the expression for Q in from part a) and previously determined specific enthalpy
values:


3.77  10 5 kg / h 2758.0  181.94kj / kg 
Q in  m h1  h4  
 269.77 MW
10 3 kW / Mw 3600s / h 


d) Mass and energy rate balances applied to a control volume enclosing the steam side
side of the condenser give:


3.77  10 5 kg / h 1794.8  173.88kj / kg 
Q out  m h2  h3  
 169.75MW
10 3 kW / Mw 3600s / h 


Alternatively, Q out can be determined from an energy rate balance on the overall
vapor power plant. At steady state, the net power developed equals the net rate of
heat transfer to the plant:
W cycle  Q in  Q out
then,
Q out  Q in  Wcycle  269.77 MW  100 MW  169.77 MW
e) Taking a control volume around the condenser, the energy rate balance gives at
steady state:

0
0



0  Qcv  W cv  mcw hcw ,in  hcw ,out   mh2  h3 
where m cw is the mass flow rate of the cooling water. Solving for m cw :
m cw 
m h2  h3 
hcw,in  hcw,out 
the numerator in this expression is evaluated in part d). For the cooling water,
h  h f (T ) , so with saturated liquid enthalpy value from Table A-2 Moran and Shapiro
at the entering and exiting temperatures of the cooling water:
m cw 
169.75MW 10 3 kW / MW 3600 s / h  7.3  106 kg / h
148.68  62.99 kJ / kg
Example #2: Combustion analysis of the boiler
Problem:
Find the useful heat generated by the combustion of 1 lb m of ethane in a furnace in a 20
percent deficient air if the reactants are at 25 0 C and the products at 1500K. Assume
that hydrogen, being more reactive than carbon, satisfies itself first with the oxygen it
needs and burns completely to H 2 O . Five percent of the heat of combustion is lost to
the furnace exterior.
Solution
The stoichiometric equation for ethane in air is:
C2 H 6  3.5O2  13.16 N 2  2CO2  3H 2 O  13.16 N 2
(where there are 3.76 mol N 2 / mol O 2 in atmospheric air, thus 13.16=3.5x3.76). With 20
percent deficient air multiply the O 2 and N 2 mol by 0.8. H 2 will burn completely to
H 2 O and C will burn partially to CO2 and partially CO :
C2 H 6  2.8O2  10.528N 2  aCO2  bCO2  3H 2 O  10.528N 2
carbon balance:
ab  2
oxygen balance:
a
b 3
  2 .8
2 2
thus a  0.6 , b  1.4 and the combustion equation is:
C2 H 6  2.8O2  10.528N 2  0.6CO2  1.4CO2  3H 2 O  10.528N 2
As here is no work done in a furnace, the first law of thermodynamics for steady states
written as:
Q   nMh f
P
 nMh 
f 1500K
P

1500K
 nMh f

25º C
R
 0.6  44.011   3243.4  1.4  28.011   1100.9
 3  18.016   4626.2   10.528  28.016  590.8
 204598.4 Btu/(lb mol C 2 H 6 )
 nMh 
f
25º C
 1211.3  30.07  0  0  36424 Btu/(lb mol C 2 H 6 )
R
thus,
Q  204598.4   36424   168174.6 Btu/(lb mol C 2 H 6 )
 168174.6
 5592.8 Btu/(lb C 2 H 6 )
30.070
 -5592.8  3.32584  -13007.9 kj/kg C 2 H 6

The useful heat generated by the combustion is:
Quseful  0.95  Q  0.95  (-13007.9 kJ/kg C 2 H 6 )  12357.6 kJ/kg C 2 H 6
Discussion
References
Moran, M. J. and Shapiro, H. N., 1995, Fundamental of Engineering Thermodynamics,
3rd edition, John Wiley & Sons, Inc., New York.
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