THERMODYNAMICS LAB
Prepared by :
Eng.Mazen Abu-Sbeitan
Students mechanical team 2015
Contents
Experiment Name
1- Marcet Boiler
Page
2
2- Single Stage – Polytropic Index
24
3- Single stage volumetric efficiency
25
4- Single stage other efficiencies
26
5- Determination of reservoir condensate
28
6- Two stage – intermediate pressure
29
7- Two stage efficiency
31
8- pressure volume diagram
32
9- Heat pump
33
1
Experiment (1) : Marcet Boiler
Objectives :1. Obtain a balance relationship between temperature and pressure of saturated steam.
2. Compare the slope of the graphic (dP/dT) and the theoretical value.
3. Study the time during heat addition and cooling of the water.
Discussion (Theory):When energy increases within the water , the increasing of activities among the molecules
enables the increase in the number of molecules escape form surface until an equilibrium
state is reached. The state of equilibrium depends on the pressure between the water surface
and the steam.
At low pressure the molecules become easier to leave the water surface and less energy is
needed ti achieve the equilibrium state (boiling point) , the temperature and pressure when
equilibrium is occurs are called saturated temperature and saturated pressure.
The saturated pressure depends only on the temperature.
The Clapeyron equation :Clapeyron equation enables us to determine the enthalpy change associated with a phase
change (hfg ) , from knowledge of (p) , (v) , (T).
For an isothermal liquid-vapor phase change the integral yields :
𝑑𝑃
𝑆𝑔 − 𝑆𝑓 = ( ) (𝑉𝑔 − 𝑉𝑓 )
𝑑𝑇 𝑠𝑎𝑡
𝑔
∫ 𝑑ℎ = ∫ 𝑇𝑑𝑠
𝑓
𝑆𝑓𝑔
𝑑𝑃
)=
𝑑𝑇
𝑣𝑓𝑔
ℎ𝑓𝑔 = 𝑇𝑆𝑓𝑔
During the phase change process , the
pressure remain constant , therefore,
(
(
𝑑ℎ = 𝑇𝑑𝑠 + 𝑣𝑑𝑝
𝑑ℎ = 𝑇𝑑𝑠
𝑔
𝑓
𝑆𝑓𝑔 =
ℎ𝑓𝑔
𝑇𝑠𝑎𝑡
ℎ𝑓𝑔
𝑑𝑝
) =
𝑑𝑇 𝑠𝑎𝑡 𝑇𝑣𝑓𝑔
𝑣𝑑𝑝 = 0
2
Figure (1) : Solid-liquid-vapor diagram
Figure (2) : Marcet boiler
1.
2.
3.
4.
Safety valve
Container
Barometer
Power source
5. Thermometer
6. Electric heater
7. Level valve
3
Experiment steps :
1- Connect the unit to the power supply to provide a heating rate of (1.5 kw) to the
electrical heater .
2- Open the level valve and switch on the power supply .
3- Wait for the steam output through the level valve to ensure that there is no air trapped
in the boiler .
4- Close the level valve and continue heating until the maximum gauge pressure
reaches ( 6 bar ) .
5- Decrease the temperature to room temperature in which the electrical heater will be
off .
6- Record all temperature readings at different pressure setting of the boiler when the
boiler is heated and cooled .
Table 1 : Data collected and calculated result.
Pressure p (bar)
Gauge
Temperature T (C)
Measured Calculated
Slope
Slope
Increase Decrease Average Average
Absolute
dP/dT
(C)
(C)
T (C)
T (K)
4
GT103
Two stage compressor test
5
Symbols and variables on the test set
P1
P2
P3
Tq1
Tq2
T1
T2
T3
T4
T5
T6
T7
T8
T9
V1
V2
A1
A2
N1
N2
Compressor 1 out
Compressor 2 in
Reservoir pressure
Torque for dynamometer 1
Torque for dynamometer 2
Air inlet (dry) temperature
Air inlet (wet bulb) temperature
Compressor 1 outlet
Intercooler inlet
Compressor 2 inlet
Compressor 2 outlet
Intercooler water inlet
Intercooler water outlet
Reservoir temperature
Voltage applied to dynamometer 1
Voltage applied to dynamometer 2
Current passing through dynamometer 1
Current passing through dynamometer 2
Rotational speed pf compressor 1
Rotational speed pf compressor 2
Bar (gauge)
Bar (gauge)
Bar (gauge)
N.m
N.m
C
C
C
C
C
C
C
C
C
Volt
Volt
Ampere
Ampere
Rev/min
Rev/min
6
𝑚̇𝑎 and 𝑚̇𝑣
P
𝑃𝑅
𝛥𝑝
P0
𝑃𝑖 and 𝑇𝑖
𝐶𝑑
𝐶𝑝
A
ρ
T
T0
𝑇1 , 𝑇2 , 𝑃1 𝑎𝑛𝑑 𝑃2
R
𝜂𝑣𝑜𝑙 𝑎𝑛𝑑 𝜂𝑚𝑒𝑐ℎ
𝜂𝑖𝑠𝑜 𝑎𝑛𝑑 𝜂𝑜𝑖𝑠𝑜
𝜂𝑖𝑛𝑡
n
𝑉𝑐
𝑉𝑠
W
𝑊𝑚𝑒𝑐ℎ and 𝑊𝑒𝑙𝑒𝑐
𝑊𝑖 𝑎𝑛𝑑 𝑊𝑖𝑠𝑜
W
g
𝑐𝑣
∅
𝛾
Q
Mass flow (air or water)
Kg/s
Pressure
Pascal or bar where shown
Pressure ratio
Difference in pressure (across the orifice)
Pascal or 𝑚𝑚𝐻2 𝑂 where shown
Local atmospheric pressure
Pascal or bar where shown
Intermediate pressure and temperature
Pascal or bar and kelvin
Coefficient of discharge (across the orifice)
Specific heat
j/(g.K)
Area of the orifice
𝑚2
Density
Kg/𝑚3
Temperature
Kelvin or C
Local atmospheric temperature
Kelvin or C
Theoretical values of temperature and pressure
K and Pa
Gas constant of the air
Dry air 287 j/(kg.K)
Volumetric and mechanical efficiency
Isothermal and overall isothermal efficiency
Intercooler efficiency
Polytropic index
Clearance volume
𝑚3
Swept volume
𝑚3
Work done
J
Mechanical or electrical power
Watts
Indicated and isothermal power
Watts
Specific work
KJ/kg
Acceleration of the gravity
9.81 m/𝑠 2
Specific heat capacity at constant volume
J/(kg.K)
Relative humidity
%
Ratio of specific heat
Heat flow rate
Watts
7
Units, ‘bar g’ and Absolute (abs) Values
Note that many equations in this theory use the SI units (for example – Pa for pressure and
K for temperature) unless stated. So for some equations you must convert your pressure
readings of bar into Pa (Pascal), and temperature readings for Celsius (C) into Kelvin (K).
Where the equation uses a pressure ratio, the units are important.
1 bar = 100000 Pa
0 ®C = 273 K and 100 ®C = 373 K
Also note that the Control Panel display pressures in gauge pressure, shown in units of
bar, giving the common use of the acronym ‘bar g’. For your calculations, you must add
the local atmospheric pressure P0 to give the absolute value ‘P abs’ in bar or Pa.
How a Compressor Works
Figure (2) : Reciprocating compressor
A compressor delivers a quantity of fluid at a particular pressure. A reciprocating
compressor is a positive displacement device. A moving piston in a cylinder reduces the
volume of a fixed mass of air. According to Boyle’s Law, the pressure of a gas is inversely
proportional to its volume, so the reduction in volume of the air increases it’s pressure.
8
However, to move the piston, the compressor must receive input of work. The amount of
work input needed relates to the force in the piston and the distance it moves.
So, at any instant during the compression process for a unit mass of gas:
Work input = (force on piston) x (distance moved by piston)
= (pressure on piston surface) x (surface area of piston) x (distance moved by piston)
= (pressure on piston surface) x (change in volume of air)
This is normally expressed as:
W = ʃ PdV
Where dV is the change in volume for a small time step of the whole compression
process.
Theoretical analysis of the compression cycle
Figure (3) : The compression cycle (PV diagram)
(4-1) induction
Clearance volume (Vc)
(1-2) compression
Swept volume (Vs)
(2-3) delivery
Swept volume per cycle ( Vs = V1-V3 )
(3-4) expansion
Induced volume per cycle ( V = V1-V4
9
Figure (4) : Actual (PV diagram)
Since reciprocating compressors have relatively slow rotational speeds, the
thermodynamic processes of compression and expansion do not follow adiabatic (n=1.4)
or isothermal processes (n=1). The polytropic index is usually about 1.3, following the
law :
𝑃𝑉 𝑛 = 𝐶
10
Polytropic Index
The relation we use to find ‘n’ is called polytropic expression defining as :
𝑃𝑉 𝑛 = 𝐶
𝑃 (𝑏𝑎𝑟) = 𝜌𝑔ℎ × 10−8 + 𝑃𝑎𝑡𝑚
Where :
ρ : density of water.
g : gravitational acceleration (9.81 𝑚/𝑠 2 ) .
C : means constant .
n : polytropic index .
-If (n = 0) then (P = C) , this process is called isobaric process .
-If (n =∞) then (V = C) , this process is called isometric process .
-If (n = 1) then (T = C) , this process is called isothermal process .
11
The polytropic index for a compressor as a value found by a test that changes variables and
produces a chart of the logarithmic values. The gradient of the chart then helps you to
calculate the polytropic index.
The value is a characteristic of the machine and useful for then using to calculate the
theoretical performance of the machine. The usual method is to produce a PV chart as
shown later in this guide. Another method is to fix the speed while varying delivery
pressure. This produces varying temperature and pressure ratios across the compressor.
Temperature ratio = Outlet temperature/Inlet temperature (in Kelvin)
Stage 1 = T3/T1
Pressure ratio PR = Outlet pressure/Inlet pressure (absolute values)
Stage 1 = P1/P0
The slope of the chart gives (n-1)/n, from which you can find n.
In theory, the index should be constant at any speed, but in reality the change in
performance of real machines at different speeds mean that it is best done at a mid-range
speed.
12
Calculations :
𝑃1 𝑉1𝑛 = 𝑃2 𝑉2𝑛
𝑃1
𝑉2 𝑛
=( )
𝑃2
𝑉1
𝑃𝑉 = 𝑚𝑅𝑇
𝑃1 𝑉1 𝑇1
=
𝑃2 𝑉2 𝑇2
𝑉1 𝑃2 𝑇1
=
𝑉2 𝑃1 𝑇2
By substituting of equation (2) in (1)
𝑃1
𝑃2 𝑇1 𝑛
=(
)
𝑃2
𝑃1 𝑇2
𝑃1
𝑇1 𝑛
𝑃2 𝑛
=( ) ×( )
𝑃2
𝑇2
𝑃1
𝑃2 𝑛−1
𝑇1 𝑛
=( )
( )
𝑃1
𝑇2
𝑇1
𝑃2
=( )
𝑇2
𝑃1
𝑛−1
𝑛
To find (n) we take logarithm for both sides of the equation
𝑇1
𝑛−1
𝑃2
𝑙𝑛 ( ) =
𝑙𝑛 ( )
𝑇2
𝑛
𝑃1
𝑃
ln ( 2 )
𝑃0
𝑛=
𝑃
𝑇
ln ( 2 ) − ln ( 3 )
𝑃0
𝑇1
13
Airflow through the orifice (𝒎̇ 𝒂 )
The pressures upstream and downstream of the orifice plate give a pressure change ∆𝑃
that helps to calculate the flow rate of air through the compressor. The mass flow (kg/s) is
given by the following equation:
𝑚̇ 𝑎 = 𝐶𝑑 𝐸𝐴√2∆𝑃𝜌
Where
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 (𝜌) =
𝐸 = (1 − 𝛽 4 )−0.5
𝑃𝑜
𝑅𝑎𝑖𝑟 𝑇1
𝜋𝑑 2
𝐴=
4
𝑑
𝛽=
𝐷
Airflow to stage 2
Total mass airflow at equilibrium is constant throughout the system, but the air entering
each stage determine its volumetric efficiency.
The inlet conditions determine the air density at stage 1, however, the average
intermediate pressure and temperature determine the density at stage 2.
𝑠𝑡𝑎𝑔𝑒 1 𝑖𝑛𝑡𝑒𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌1 =
𝑠𝑡𝑎𝑔𝑒 2 𝑖𝑛𝑡𝑒𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌2 =
𝑃𝑜
𝑅𝑎𝑖𝑟 𝑇1
𝑃𝑖
𝑅𝑎𝑖𝑟 𝑇𝑖
When using the intercooler to keep T5=T1 you can simplify this to:
𝑠𝑡𝑎𝑔𝑒 2 𝑖𝑛𝑡𝑒𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝜌2 = 𝜌1
𝑃𝑖
𝑃𝑜
14
Clearance volume (𝑽𝒄 ) and volumetric efficiency (𝜼𝒗𝒐𝒍 )
The pistons never actually reach the top of their cylinders because of a small clearance
volume ( to prevent damage). This means that for each piston stroke, a small amount of air
is compressed but never delivered. So, the actual volume of air moved is less than the swept
(theoretical )𝑉𝑠 .
The ratio of the clearance volume against the swept volume in the cylinder is termed the
clearance ratio :
𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑟𝑎𝑡𝑖𝑜 =
𝑉𝑐
𝑉𝑠
The volumetric efficiency is a ratio of the actual volume flow of air that passes through the
compressor in a given time (or cycle) against the theoretical volume of the air that should
move.
𝜂𝑣𝑜𝑙 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 (𝑠𝑤𝑒𝑝𝑡) 𝑣𝑜𝑙𝑢𝑚𝑒
Alternatively, using mass flow :
𝜂𝑣𝑜𝑙 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 (𝑠𝑤𝑒𝑝𝑡) 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤
𝜂𝑣𝑜𝑙 =
𝑚̇𝑎
× 100%
𝑚̇𝑠
Where swept mass flow for compressor 1 :
𝑚̇𝑠1 =
𝜌1 2𝑉𝑠1 𝑁1
60
𝑚̇𝑠2 =
𝜌2 2𝑉𝑠2 𝑁2
60
And for compressor 2 :
Note that the value 2𝑉𝑠 double the swept volume, due to 2 cylinder acting at each
revolution.
Note that when equilibrium, the air mass flow 𝑚̇𝑎 passes equally through both
compressors.
15
An alternative or ideal equation allows for clearance volume 𝑉𝑐 , allowing you to predict
the theoretical volumetric efficiency :
𝜂𝑣𝑜𝑙 = 1 −
1
𝑉𝑐
[𝑃𝑅 𝑛 − 1]
𝑉𝑠
For the GT103 Test Set, the volumetric efficiency found from Equation 8 will never be as
high as that found from Equation 11, as the real machine cannot pass as much air as it
should. This due to the slight fraction caused by the orifice and inlet manifold needed for
measurement of the air inlet conditions. The resulting actual air mass flow becomes less,
but the swept volume remains the same, giving a lower efficiency, compared to the
theoretical value.
If the air passed directly into the compressor inlet, actual volumetric efficiency would
increase to nearer the theoretical efficiency. To be strictly correct, the restriction would
cause the pressure at the inlet of the real compressor to be slightly less than expected,
resulting in a smaller pressure ratio PR affecting the theoretical efficiency.
Two Stage Volumetric Efficiency
Due to the interaction between the two stages, overall volumetric efficiency for the two
stages together can give misleading results, so generally you find individual volumetric
efficiency for the two stages and compare with just one operating under similar conditions.
Indicated work rate (𝑾𝒊)
The indicated work done rate (𝑊̇ 𝑖) is given by the area of the PV diagram of a compressor,
so that :
𝑊̇ 𝑖 = ∮ 𝑃𝑑𝑉̇
𝑛
(𝑃 𝑉̇ − 𝑃1 𝑉𝑎̇ )
𝑛−1 2 𝑏
𝑛−1
𝑛
(
)
𝑊̇ 𝑖 = 𝑚̇𝑅𝑇 (
) [(𝑃𝑅 ) 𝑛 − 1]
𝑛−1
𝑊̇ 𝑖 =
T : the initial temperature in Kelvin.
𝑃𝑅 : pressure ratio in absolute values.
𝑇𝑜𝑡𝑎𝑙 𝑊̇ 𝑖 = 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 1 𝑊̇ 𝑖 + 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 2 𝑊̇ 𝑖
16
Mechanical speed, Torque and Shaft Power
Speed
Due to 3:1 pulley ratio, Compressor Speed = 1/3 x Dynamometer Speed. However as the
speed sensor is at the compressor pulley, it’s display shows compressor speed directly.
Dynamometer torque (𝑻𝒒 ) and shaft power (𝑾̇𝒎𝒆𝒄𝒉 )
The Dynamometer measures the torque at their own position, displayed on the control
panel. However, due to the 3:1 pulley ratio, the torque at each compressor is 3 times that
at dynamometer, so you must multiply the displayed value by 3 to find the mechanical
power.
𝑇𝑞1 × 3 × 2𝜋 × 𝑁1
60
𝑇𝑞2 × 3 × 2𝜋 × 𝑁2
=
60
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 1 𝑊̇𝑚𝑒𝑐ℎ =
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 2 𝑊̇𝑚𝑒𝑐ℎ
Due to the belt and pulley system, you cannot assume that all the torque is due to the compressor, as
same power (torque) becomes lost to the friction in the pulley system. For this reason, you can assume
that the actual (Wmech) at the compressor is around 80% of that calculated.
Actual Wmech = Calculated Wmech x 80%
When calculating the total power for the two stages of compression, you simply add the
two values of power together, so total power:
Total Wmech = compressor 1 Wmech + compressor 2 Wmech
Dynamometer Electrical Power (Welec)
Each dynamometer works from its own electric variable speed drive inside the Control
Panel. The two analogue meters for each compressor instrument group measure directly
the voltage (V) and current (A) applied at the dynamometer. The sum of these values gives
the electrical power applied to the dynamometer in Watts, so that:
Dynamometer 1 Wmech = V1 × A1
Dynamometer 2 Wmech = V2 × A2
Knowing the electrical power applied and the shaft power at the output can help to
appreciate the approximate power transfer and efficiency of the dynamometer-tocompressor drive system.
17
Mechanical Efficiency
This is the ratio of the indicated power used by the compressor to compress air, against the
input shaft power applied to the compressor (from the Dynamometer) :
𝜂𝑚𝑒𝑐ℎ =
𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑑 𝑝𝑜𝑤𝑒𝑟
× 100%
𝑠ℎ𝑎𝑓𝑡 𝑝𝑜𝑤𝑒𝑟
𝜂𝑚𝑒𝑐ℎ =
𝑊̇𝑖
𝑊̇𝑚𝑒𝑐ℎ
× 100%
Isothermal power and efficiency
This is the work done to increase the air pressure. It is a product of the air mass flow, its
initial properties and the pressure increase. It doesn’t allow for the internal energy increase
of the air.
Note that the pressure ratio is based on absolute values.
𝑊̇𝑖𝑠𝑜 = 𝑚̇𝑎 𝑅𝑇 × 𝑙𝑛(𝑃𝑅 )
Where T is the initial gas temperature in kelvin. This would be T1 for the first stage of the
compressor and Ti for the second stage in kelvin.
The pressure ratio that of each stage (P1/P0 for the first stage ) or of the entire two stages
(P3/P0) for a two stage configuration.
The isothermal efficiency is the ratio of work done over the indicated power used to do the
work. Therefore, isothermal efficiency is higher than overall efficiency .
𝜂𝑖𝑠𝑜 =
𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 (𝑝𝑜𝑤𝑒𝑟)
× 100%
𝑖𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑑 𝑝𝑜𝑤𝑒𝑟
𝜂𝑖𝑠𝑜 =
𝑊̇𝑖𝑠𝑜
× 100%
𝑊̇𝑖
Overall isothermal efficiency
This is a measure of how efficient the compressor in at turning input (shaft) mechanical
power into work done compressing the air. It is an overall efficiency of the system
including all losses .
𝜂𝑜𝑖𝑠𝑜 =
𝜂𝑜𝑖𝑠𝑜 =
𝑖𝑠𝑜𝑡ℎ𝑒𝑟𝑚𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 (𝑝𝑜𝑤𝑒𝑟)
× 100%
𝑚𝑒𝑐ℎ𝑎𝑛𝑖𝑐𝑎𝑙 𝑝𝑜𝑤𝑒𝑟
𝑊̇𝑖𝑠𝑜
× 100%
𝑊̇𝑚𝑒𝑐ℎ
𝜂𝑜𝑖𝑠𝑜 = 𝜂𝑖𝑠𝑜 × 𝜂𝑚𝑒𝑐ℎ
18
As you will find, the receiver will collect the water during the experiment. This is the vapor
in the compressed air condensing on the walls of the receiver as the air fills and passes
through the receiver. Eventually, this condensate would fill the receiver. In industrial
applications, the receiver would have an automatic vent device or air dryer system to help
prevent this. For this reason, engineers need to know how much condensate is likely to
accumulate in a given time. This depends on the water vapor (humidity) conditions of the
air. Psychometric is the study of the humidity in the air.
Specific humidity
This is a ratio of the mass of the water vapor in the air under test and dry air.
𝜔=
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑟
𝑚̇𝑣
=
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 𝑎𝑖𝑟
𝑚̇𝑎 𝑑𝑟𝑦
Obviously, the dry air would have the same mass as the test air if its water vapor were
removed, so
𝑚̇𝑎 𝑑𝑟𝑦 = 𝑚̇𝑎 − 𝑚̇𝑣
Assuming that water vapor behaves as a perfect gas, then using Dalton’s law of partial
pressure
𝜔=
𝑃𝑣 𝑅𝑎
𝑃𝑣
𝑃𝑣
×
= 0.622 × = 0.622 ×
𝑃𝑎 𝑅𝑣
𝑃𝑎
𝑃 − 𝑃𝑣
Since 𝑅𝑎 gas constant for dry air = 0.2871 kJ/kg.k
𝑅𝑣 gas constant for water vapor = 0.4615 kJ/kg.k
𝑃 = 𝑃𝑎 − 𝑃𝑣
19
Difference between dry and wet bulb temperature
𝑻𝟏 − 𝑻𝟐
(C)
𝑻𝟏
(C)
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
2
80
81
81
82
82
83
83
83
84
84
84
85
85
85
86
86
86
86
87
87
87
4
61
63
64
65
65
66
67
68
69
69
70
71
71
72
72
73
73
74
74
75
75
6
44
46
47
49
50
51
53
54
55
56
57
58
58
59
60
61
61
62
63
63
64
8
27
30
32
34
36
37
39
40
42
43
44
46
47
48
49
50
51
51
52
53
54
10
13
15
18
20
22
24
26
28
30
31
33
34
36
37
38
39
40
41
43
43
44
12
----10
12
14
17
19
20
22
24
26
27
28
30
31
32
33
35
36
14
-------6
8
10
12
14
16
18
19
21
22
24
25
26
28
16
-----------5
7
9
11
13
14
16
17
19
20
18
---------------5
7
9
10
12
13
20
-------------------5
7
Saturated
vapor pressure
𝑷𝒗𝒔𝒂𝒕
(bar)
0.01704
0.01817
0.01936
0.02063
0.02200
0.02337
0.02486
0.02642
0.02808
0.02982
0.03166
0.03360
0.03564
0.03779
0.04004
0.04242
0.04491
0.04754
0.05029
0.05318
0.05622
Relative humidity
∅=
Pv
ṁv
=
Pvsat ṁvsat
Where Pvsat is the saturation vapor
pressure.
If you know the wet and the dry inlet air
temperature, you can use Table (1) to find
the inlet relative humidity and saturation
vapor pressure.
ω=
ṁv
ṁa − ṁv
Therefore water vapor mass flow rate :
ṁv =
ωṁa
ω+1
Water vapor condenses in the reservoir,
where is has ∅ = 100%
Pv = ∅Pvsat
To find the rate at which water collects in the reservoir, you must find the water mass flow
ṁv at the inlet and at the outlet (reservoir).
With saturated vapor condition 𝑃𝑣 is a function of temperature only and therefore, if the
pressure of the mixture 𝑃 is increased, the percentage water vapor content is reduced; this
reduction achieved by condensation.
20
Single and Multiple Stage Comparisons
The most obvious advantage of the two stage compression is the higher pressures the it can
achieve. Most compressors lose efficiency with increased pressure ratio, so a single stage
normally works to a maximum of around 8 or 9 bar ratio. Simply having a second identical
compressor allows you to double that pressure, but at the expense of doubling the work
input.
Alternatively, where one compressor works at 8 bar, then adding an identical unit in series
(two stages) and setting each stage to 4 bar gives a corresponding increase in volumetric
efficiency for each stage as compared to the single stage.
Theoretically and ideally, the two compressors would use slightly less power than just one,
but in real applications this and mechanical efficiency improvement depends on how you
drive each compressor. Separate motors and drive systems have their own efficiencies.
Ideally, an efficient drive system working both compressors from one motor will give the
best efficiency.
Isothermal efficiency should increase slightly for each unit, as its pressure ratio becomes
less. Overall isothermal efficiency may not change, as this also depends on the mechanical
input condition.
The installer or designer has to decide whether to use a single stage compressor for short
term savings due to the reduce amount of parts, or the longer term efficiency and
maintenance savings of two or more stages compressor.
21
Intermediate conditions
In a real two stage machine supplying a flow of air, there are both pressure and temperature
gradients between the air leaving the first stage and entering the second stage.
Therefore, for accurate results on the Test Set you must average the pressure and
temperature values to approximate the intermediate condition:
𝑇3 + 𝑇5
2
𝑃1 + 𝑃2
𝑖𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑃𝑖 =
2
𝑖𝑛𝑡𝑒𝑟𝑚𝑒𝑑𝑖𝑎𝑡𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑇𝑖 =
Also note the slight change in pressure ratio at the first stage when set for use with the
second stage. It changes from the simple P1/P0 ratio to Pi/P0.
Compressor theory shows that for minimum total power consumption in an ideal machine,
the stages would share the work done and the intermediate pressure should equal the square
root of the inlet and final outlet pressures, which in the Test Set should be:
𝑃𝑖 = √𝑃0 × 𝑃3
Note: all values are absolute.
From this, in an ideal machine for 4 bar g (5 bar abs) delivery and 1 bar atmospheric would
give:
𝑃𝑖 = √𝑃0 × 𝑃3 = √1 × 5 = 2.23 𝑏𝑎𝑟 𝑎𝑏𝑠
At this point, 𝑃𝑖 ⁄𝑃1 = 𝑃3⁄𝑃𝑖 and compressor 1 𝑊𝑖 = compressor 2 𝑊𝑖
Intercooling should reduce the average intermediate pressure, compared to a nonintercooled system.
22
Intercooler heat energy analysis
The intercooler has a simple tubed air/water heat exchanger there the interstage air is cooled
from the first stage exit temperature to a lower value. Considering the heat flow rates, the
following equations can be defined.
Heat gained by water :
𝑄̇𝑤 = 𝑚̇𝑤 𝐶𝑝𝑤 (𝑇8 − 𝑇9)
Heat lost by air :
𝑄̇𝑎 = 𝑚̇𝑎 𝐶𝑝𝑎 (𝑇4 − 𝑇5)
Since the thermocouple are very close to the intercooler, there will be very small heat losses
which are not accountable.
In general :
𝑄̇𝑎 = 𝑄̇𝑤 + 𝑙𝑜𝑠𝑠𝑒𝑠
Or thermal efficiency for the intercooler :
𝜂𝑖𝑛𝑡 =
𝑄̇𝑤
𝑄̇𝑎
However, since condensation will be taken place, the value of 𝑄̇𝑎 will be less than the
actual amount of heat lost by the air. The difference is due to the amount of heat required
to condensate a proportion of the water vapor.
The difference can be evaluated, but needs long, repetitive calculation. A more important
ratio of use with the heat exchanger is the thermal ratio defined as :
𝜀=
𝐻𝑜𝑡 𝑠𝑖𝑑𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑟𝑜𝑝
𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑣𝑎𝑖𝑙𝑎𝑏𝑙𝑒 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑑𝑟𝑜𝑝
23
Experiment (2) : Single Stage – Polytropic Index
Aim :
To vary the delivery pressure of compressor 1 across its range and find it polytropic index
at a given speed .
Procedure :
1. Find your local atmospheric pressure Po, You can assume 1 bar, but the actual value
gives slightly more accurate result.
2. Create a Blank Results Table.
3. Set rocker switch R1 to down. This sends the output of compressor 1 directly to the
receiver.
4. Fully shut (turn clockwise) the air flow valve.
5. Start compressor 1 and set its speed to 750 rev/min (+/-5 rev/min). Keep this speed
constant during the test.
6. Wait a few minutes for the reservoir pressure to start rising while watching the
Compressor 1 output pressure P1.
7. When P1 reaches 2 bar g (gauge), open the air flow valve a few turns and carefully
adjust it to keep P1 constant (by keeping the flow from the receiver constant).It will
take several minutes of careful adjustment. This helps, as its also allows time for the
compressor to warm up.
8. Wait at least ten minutes for the compressor outlet temperature T3 and pressures to
stabilize, Then record the values as shown in the blank results table.
9. In steps, use the air flow valve to regulate P1 at increasing pressures up to 10 bar g
(gauge) as shown in the result table. At each step, wait at least ten minutes for the
outlet temperature T3 and pressures to stabilize. NOTE that you record P3 only for
reference, While an air flow is present, P3 will always the slightly less than P1.
10.After you have finished, slowly open the air flow valve while reducing compressor
speed until you reach roughly 0 bar at P3. Switch off the compressor, shut the air
flow valve and drain the condensate in the receiver.
If you have time, repeat for compressor speeds of 500 (rev/min) and 1000 (rev/min).
24
Experiment (3) : Single stage volumetric efficiency
Aim :
To vary the delivery pressure of compressor 1 across its range and find its volumetric
efficiency at a given speed.
Procedure :
1. Find your local atmospheric pressure. You can assume (1 bar), but the actual value
gives slightly more accurate.
2. Create a Blank Results Table.
3. Set rocker switch R1 to down. This sends the output of compressor 1 directly to the
receiver.
4. Fully shut (turn clockwise) the air flow valve.
5. Start compressor 1 and set its speed to 750 rev/min(+/-5 rev/min). Keep this speed
constant during the test.
6. Wait a few minutes for the reservoir pressure to start rising while watching the
Compressor 1 output pressure P1.
7. When P1 reaches 2 bar g (gauge), open the air flow valve a few turns and carefully
adjust it to keep P1 constant (by keeping the flow from the receiver constant). It will
take several minutes of careful adjustment. This helps, as its also allows time for the
compressor to warm up.
8. Wait at least ten minutes for the compressor outlet temperature T3 and pressure to
stabilize, then record the values as shown in the blank results table.
9. In steps, use the air flow valve to regulate P1 at increasing pressure up to 10 bar g as
shown in the results table. At each step, wait at least ten minutes for outlet
temperature T3 and pressure the stabilize. Note that your record P3 only for
reference. While an air flow is present, P3 will always be slightly less than P1.
10.After you have finished, slowly open the air flow valve while reducing compressor
speed until you reach roughly 0 bar at P3. Switch off the compressor, shut the air
flow Valve and drain the condensate in the receiver.
25
Experiment (4) : Single stage other efficiencies
Aim :
To vary the delivery pressure of compressor 1 across its range and find its mechanical,
isothermal and overall efficiency at a given speed.
Procedure :
1. Find your local atmospheric pressure. You can assume (1 bar), but the actual value
gives slightly more accurate.
2. Create a Blank Results Table.
3. Set rocker switch R1 to down. This sends the output of compressor 1 directly to the
receiver.
4. Fully shut (turn clockwise) the air flow valve.
5. Start compressor 1 and set its speed to 750 rev/min(+/-5 rev/min). Keep this speed
constant during the test.
6. Wait a few minutes for the reservoir pressure to start rising while watching the
Compressor 1 output pressure P1.
7. When P1 reaches 2 bar g (gauge), open the air flow valve a few turns and carefully
adjust it to keep P1 constant (by keeping the flow from the receiver constant). It will
take several minutes of careful adjustment. This helps, as its also allows time for the
compressor to warm up.
8. Wait at least ten minutes for the compressor outlet temperature T3 and pressure to
stabilize, then record the values as shown in the blank results table.
9. In steps, use the air flow valve to regulate P1 at increasing pressure up to 10 bar g as
shown in the results table. At each step, wait at least ten minutes for outlet
temperature T3 and pressure the stabilize. Note that your record P3 only for
reference. While an air flow is present, P3 will always be slightly less than P1.
10.After you have finished, slowly open the air flow valve while reducing compressor
speed until you reach roughly 0 bar at P3. Switch off the compressor, shut the air
flow Valve and drain the condensate in the receiver.
26
NO
P1
(bar)
Gauge abs
T1
(C)
T3
(K)
(C)
P0
(bar) 𝒍𝒏 (𝑷𝟏) 𝒍𝒏 (𝑻𝟑) n
(K) abs
𝑷𝟎
𝑻𝟏 (cal)
n
(slop)
1
2
3
4
5
6
7
NO
ρ
ΔP
(air) (mmH2O)
𝒎̇𝒔
(kg/s)
𝒎̇𝒂
(kg/s)
ή𝒗𝒐𝒍
(act)
ή𝒗𝒐𝒍
(theo)
T
(N.m)
Pr
𝑾𝒊𝒔𝒐 𝑾𝒎𝒆𝒄𝒉
𝑾𝒊
ή 𝒎𝒆𝒄𝒉 ή 𝒊𝒔𝒐 ή 𝒐𝒊𝒔𝒐
1
2
3
4
5
6
27
Experiment (5) : Determination of reservoir condensate
Aim :
To show how to predict the mass (and therefore approximate volume) of the condensate
that will collect in the reservoir over a given time, using the air temperature and pressure.
Procedure :
1. Find your local atmospheric pressure. You can assume 1 bar, but the actual value
gives slightly more accurate results.
2. Create a blank results table,
Speed rev/min :
Local atmospheric pressure P0 (bar) :
Measurement period :
Condensate collected :
𝑚𝑚𝐻2 𝑂
P1
T1
T2
T3
T9
P1
𝑚̇𝑎
ΔP
Bar g
C
C
C
C
Bar abs
Kg/s
3. Start compressor 1 and set its speed to 750 rev/min (+/-5 rev/min). Keep this speed
constant during the test.
4. Adjust the air flow valve to give 3 bar g (gauge) at P1.
5. Keeping the speed and P1 pressure constant, run the compressor for a fixed period
of time (for example 30 minutes), that allows it to reach equilibrium.
6. At the end of your measurement period, record the values as shown in the results
table.
7. Leave the airflow valve partly open while you quickly reduce the compressor
speed back to zero and switch off.
8. Allow all pressures to drop to zero and the receiver temperature T9 to return to
ambient. This may take several hours.
9. Use a calibrated container (not supplied) to collect the condensate from the
receiver drain valve.
28
Experiment (6) : Two stage – intermediate pressure
Aim :
To determine and compare the intermediate pressure of a two stage compressor, with and
without intercooler.
Procedure 1 – no intercooler :
Find your local atmospheric pressure. You can assume 1 bar, but the actual value
gives slightly more accurate results.
2. Create a blank table, similar to table 6.
3. Start compressor 1 and set its speed to 750 rev.min-1(+/-5rev/min). Keep this
speed constant during the test.
4. Start compressor 2 and set its speed to 250 rev.min-1 (+/-5 rev/min). Keep this
speed constant between each set of results.
5. Set rocker switch R1 upwards and R2 downwards to direct air flow from
compressor 1 to compressor 2 without the intercooler.
6. Shut the air flow valve and wait a few minutes for the reservoir pressure P3 to start
rising to 4 bar.
7. When P3 reaches 4 bar g (gauge), open the air flow valve a few turns and carefully
adjust it to keep P3 constant.
8. Wait at least ten minutes for the compressor outlet temperature T3, T6 and
pressures to stabilize, then record the values as shown in the blank results table.
9. In steps, increase compressor 2 speed up to 950 rev.min-1 as shown in the results
table. At each step, wait at least ten minutes for the outlet temperature T3, T6 and
pressures to stabilize.
10. After you have finished, slowly open the air flow valve while reducing compressor
speed until you reach roughly 0 bar at P3. Switch off the compressor, shut the air
flow valve and drain the condensate in-the receiver.
1.
Procedure – with intercooler :
Set rocker switches R1 and R2 upwards to direct air flow through the intercooler.
Repeat procedure 2 but use the intercooler flow to keep the inlet temperature T5 at
compressor 2 as near to the air inlet temperature T1 as possible. You will need to adjust
the intercooler flow as the air temperature leaving the first compressor changes during
the experiment. Your cooling water supply may affect the limits of this experiment.
29
Procedure 3 – Different intermediate temperatures :
if your cooling water supply is low enough, you can try the experiment with compressor
2 inlet (T5) much lower than the ambient temperature at T1 (T5 lower than T1). You can
follow this up with another experiment with T1 bigger than T5 and compare the effect of
the different intermediate temperature.
Local atmospheric pressure P0 :
Compressor 1 N1 speed :
Intercooler : yes/no
N2
(rev/min)
P1 (bar)
Tq1 (N.m)
V1 (volt)
A1 (amperes)
Power 1
P2 (bar)
Tq2 (N.m)
V2 (volt)
A2 (amperes)
Power 2
T1 (C)
T2 (C)
T3 (C)
T4 (C)
T5 (C)
T6 (C)
T7 (C)
T8 (C)
T9 (C)
Manometer
𝑚𝑚𝐻2 𝑂
Intercooler
flow (L/min)
250
350
450
550
650
750
850
950
30
Experiment (7) : Two stage efficiency
Aim :
To determine the efficiencies of each stage with and without intercooler.
Procedure :
1. From your results for other tests, choose the conditions where the intermediate
pressure is similar to the ideal and the two compressors share the workload (to within
100 Watts). Compare with the results for a single stage delivering the same pressure.
2. Alternatively.
 Run a two stage test at a given delivery pressure and speed combination so
that the two compressors share the workload with and without the intercooler.
 Run a third test with only compressor 1 delivering the same output pressure
as with the two combined, but at a similar air flow.
NOTE : allow at least 30 minutes for conditions to stabilize between each
configuration
Result Analysis
Use this result to calculate the mechanical power, volumetric, mechanical, isothermal and
overall efficiencies for the three configurations.
31
Experiment (8) : pressure volume diagram.
Aim :
To show how to produce and analysis the pressure volume (PV) diagram for a piston and
cylinder compressor.
Procedure :
1. Refer to the appendix and fit the pressure indicator sensor to the adapter above the
compressor you are to test.
2. Run the compressor at your chosen speed (anything between 300 and 1000 rev/min).
3. Allow at least minutes for temperature to stabilize, then record all readings.
4. Use the indicator (as described in the appendix) to procedure a PV diagram.
32
Heat Pump
33
1
2
3
4
compressor
force air condenser
water condenser
forced air evaporator
18
19
20
21
5
water condenser
22
6
four-way valve
23
7
thermostatic expansion valves at the
refrigeration chamber inlet
temperature sensor of the refrigerant at
the condenser inlet
temperature sensor of the refrigerant at
the condenser outlet
temperature sensor of the refrigerant at
the evaporator inlet
temperature sensor of the refrigerant at
the evaporator outlet
temperature sensor of the water at the
condenser and evaporator inlet
temperature sensor of the water at the
condenser outlet
temperature sensor of the water at the
evaporator inlet
temperature sensor of the air at the
condenser and evaporator inlet
temperature sensor of the air at the
condenser outlet
temperature sensor of the air at the
evaporator outlet
24
8
9
10
11
12
13
14
15
16
17
27
refrigerant flow sensor
condenser water flow sensor
evaporator water flow sensor
pressure sensor located at the inlet of
the compressor
pressure sensor located at the outlet of
the compressor
manometer located at the outlet of the
compressor
manometer located at the refrigeration
chamber inlet
manometer located at the freezing
chamber inlet
manometer located at the inlet of the
compressor
high and low pressure control
28
coolant accumulation tank
29
water evaporator regulating valve
30
air evaporator regulating valve
31
air condenser regulating valve
32
water condenser regulating valve
33
regulating valve for the water flow in
the condenser
regulating valve for the water flow in
the evaporator
25
26
34
34
.
35
36
37
38
39
40
41
42
43
44
45
Data :
46
47
48
Table (1) : Using water as a source for heat transfer in the condenser and evaporator.
𝑚̇𝑤
(L/min)
T1
T2
Temperature
(C)
T3 T4 T5
T6
T7
Pressure
(bar)
P1 P2
𝑊̇𝑐𝑜𝑚𝑝 𝑚̇𝑅134𝑎
(watt) (L/h)
𝑄𝑒𝑣𝑎𝑝
𝑄𝑐𝑜𝑛𝑑
Water
R134a
Water
R134a
Coefficient of performance
(H)th
(H)act
(R)th
(R)act
Table (2) : Using air as a source for heat transfer in the condenser and evaporator.
𝑚̇𝑤
(L/min)
T1
T2
Temperature
(C)
T3 T4 T5
T6
T7
Pressure
(bar)
P1 P2
𝑊̇𝑐𝑜𝑚𝑝
(watt)
𝑚̇𝑅134𝑎
(L/h)
𝑄𝑒𝑣𝑎𝑝
𝑄𝑐𝑜𝑛𝑑
air
R134a
air
R134a
Coefficient of performance
(H)th
(H)act
(R)th
(R)act
49
50