MUHAMMAD HASAN (ME-18026)

NED University of Engineering & TechnologyDepartment of
Mechanical Engineering
Practical Workbook
Refrigeration & Air Conditioning (ME-403)
NAME:
BATCH:
MUHAMMAD HASAN
2018
ROLL NO:
ME-18026
SECTION:
A
ENROLLMENT NO:
DEPARTMENT:
NED/1245/2018
MECHANICAL ENGINEERING
Practical Workbook Refrigeration & Air Conditioning
(ME-403)
Prepared by
S.M. Rizwan Azeem
And
Arshad Siddiqui
It is certified that this practical workbook contains 52 pages
Chairman
Department of Mechanical Engineering
NED University of Engineering & Technology
Karachi-75270, Pakistan
NED University of Engineering & Technology, Department of
Mechanical Engineering
Certificate
This is to certify that Mr./Ms. MUHAMMAD HASAN
student
of T.E. (Mechanical) bearing Roll No. 18026
has
completed his/her practical course work in the subject of Refrigeration and
Air Conditioning (ME-403) as prescribed and approved by the Board of
Studies of the Department of Mechanical Engineering.
The overall performance of the student has been assessed as Excellent/Very
Good/ Good/ Satisfactory/Unsatisfactory.
Course Teacher
Contents
Exp.
No.
Date
Title
Page No. Remarks
1
To study the standard vapor compression refrigeration
system.
01
2
To determine the Refrigerating Effect, COP and
Refrigerating Efficiency of a domestic refrigeration unit
10
3
To determine the performance of the vapor compression
training unit and compare it to that of a standard vapor
compression cycle.
13
4
To determine the properties of conditioned air using
the Dehumidification Unit.
16
5
To determine the overall mass and energy balance
through heat exchanger or packing type A (or B or C)
of a cooling tower.
21
6
To determine the correlation between heat exchanger
(Packing A, B and C) and cooling performance of a
cooling tower.
28
7
Analysis of the behavior of the air when passing
through a heating coil or (heating battery).
33
8
Determine flow rate of chilled water output by the
pump and chilled water speed inside a pipe of cooling
unit.
38
9
To study the Centralized water-cooled DX Air Conditioning
System of the University Auditorium
43
Appendix: Properties of refrigerants, psychrometric
chart.
46
Experiment No. 1
1.1.1 Object: To study the standard vapor compression refrigeration system.
1.1.2 Working cycle: The standard vapor compression refrigeration cycle comprises four processes
(i) Isentropic compression of saturated vapor, (ii) Heat rejection at constant pressure, (iii) Isenthalpic
expansion of saturated liquid and (iv) Isothermal addition of heat at constant pressure. The T-s and p-h
plots of the cycle are shown below.
The actual cycle, however, would differ due to pressure drops in the condenser and evaporator,
subcooling of the liquid at the condenser exit, superheating of the vapor at evaporator outlet, nonisentropic compression, etc.
- Refrigeration effect = h1 − h4
- Refrigeration capacity = m (h1 − h4)
- COPR =
qL
h −h
= 1h2 −4h
1
- Power (in kW) per kW of refrigeration = Inverse of COPR
- Refrigerating efficiency = COPR/COPR, Carnot
- Ton of Refrigeration (TR): Originally 1 TR was defined as the rate of heat transfer required
tomake 1 short ton (2000 lbs.) of ice per day from water (latent heat of fusion = 144 Btu/lbs.) at 0oC.
1 TR = 200 Btu/min = 211 kJ/min (3.5167 kW)
1
1.2
Main components: The four main components are (i) Compressor, (ii) Condenser, (iii)
Expansion device, and (iv) Evaporator.
1.2.1
Compressors: They can be classified as follows.
(a)
On the basis of working principle: Positive displacement compressors (reciprocating, rolling
piston, rotary vane, single/twin screw, and scroll compressors) and Dynamic type compressors
(centrifugal and axial flow compressors).
Parts of a hermetic compressor: 1 Housing with connectors and baseplates, 2 Top Cover, 3 Block with
stator bracket, 4 Stator (with screws), 5 Rotor, 6 Valve unit (screws, cylinder cover, gaskets, valve plate),
7 Crankshaft with grommet, 8 Connecting rod with piston, 9 Oil pick-up tube, 10 Springs with
suspensions, 11 Internal discharge tube (screw, washer, gasket), 12 Start equipment (relay, cover, cord
relief)
Rolling piston type rotary compressor
Screw compressor
2
(b)
On the basis of the arrangement of compressor motor.
Hermetic (or sealed)
Semi-hermetic (or semi-sealed)
Open
Approximate range of capacity covered by various compressor types
3
1.2.2 Condensers: Condensers used in refrigeration systems can be classified into three broad
categories:
(a)
(b) Air-cooled condenser
Water-cooled condenser
(c) Evaporative condenser
1.2.3 Expansion devices: In general, the expansion device serves to (a) reduce pressure from
condenser pressure to evaporator pressure, and (b) regulates the refrigerant flow from the high-pressure
liquid line into the evaporator at a rate equal to the evaporation rate in the evaporator. Some commonly
used expansion devices are Capillary Tube, Automatic Expansion Valve (AEV or AXV), Thermostatic
Expansion Valve (TEV or TXV), Float Type Expansion Valve, and Electronic Expansion Valve.
4
Capillary tube with drier
Automatic Expansion Valve
Schematic showing the operation of a Thermostatic Expansion Valve
5
1.2.4
Evaporators: Two main types of evaporators employed are:
(a) Direct-expansion type in which the refrigerant boils inside the tubes, cooling the fluid (e.g. air,
water, brine) that passes over the outside of the tubes; in the case of a plate type evaporator a solid
product in direct contact with the plate is cooled by conduction.
(b) Flooded type in which the refrigerant vaporizes on the outside of the tubes, which are submerged
in liquid refrigerant within a closed shell. The fluid to be cooled flows through the tubes.
Direct-expansion type evaporators
Flooded type evaporator
6
1.2.5 Refrigerants: They can be classified as:
- Primary refrigerants are those fluids, which are used directly as working fluids. These fluids
provide refrigeration by undergoing a phase change process in the evaporator.
- Secondary refrigerants are those liquids, which are used for transporting thermal energy from one
location to other. These refrigerants do not undergo phase change as they transport energy from one
location to other. The commonly used secondary refrigerants are the solutions of water and ethylene
glycol, propylene glycol or calcium chloride. These solutions are known under the general name of
brines.
1.2.6 Primary refrigerants: The selection of primary refrigerants is based on the consideration
ofthe following.
Thermodynamic characteristics
(I) High latent enthalpy of vaporization (ii) Low freezing temperature (iii) Relatively high critical
temperatures (iv) Positive evaporating pressure (v) Relatively low condensing pressure.
Physical and chemical characteristics
(I) High dielectric strength of vapor (ii) Satisfactory oil solubility (iii) Low water solubility (iv)
Inertness and stability.
Safety
(I) Nonflammability (ii) Nontoxicity.
Effect on the environment
(I) Ozone depletion potential (ODP)
(ii) Global warming potential (GWP).
Economics
The refrigerant used should preferably be inexpensive and easily available.
1.2.7
ASHRAE designation system
Fully saturated halogenated hydrocarbons are designated by R XYZ, where: No. of carbon (C) atoms=
X+1, No. of hydrogen (H) atoms= Y−1, and No. of fluorine (F) atoms= Z. The balance indicates the
number of chlorine atoms. Only 2 digits indicates that the value of X is zero.
Organic refrigerants are assigned serial numbers in the 600 series.
Inorganic compounds are designated by adding 700 to their molecular mass.
7
Q1.1 Sketch a vapor compression cycle, on the p-h chart, taking into consideration the following:
subcooling of liquid at exit from condenser, superheating of vapor at exit from evaporator, pressure
drops in condenser and evaporator, and non-isentropic compression.
Q1.2 Concisely state the effect of above-listed factors on the COP of the system.
1) We can observe that sub cooling increases the refrigeration
effect but at the same time it also increases the required work and
since the increase in work is more as compared to increase in
refrigerating effect, therefore overall effect of superheating is to
give a low value of C.O.P.
2) Superheating increases both the refrigeration effect as well as the
work of compression. Hence the COP (ratio of refrigeration effect and
work of compression) may or may not increase with superheat,
depending mainly upon the nature of the working fluid.
3) Pressure drops either in condenser or evaporator decreases the COP
Q1.3 What is the difference between compressor “flooding” and “slugging”?
1)
2)
Flooding is a little liquid refrigerant reaching the compressor.
Slugging is a lot of liquid refrigerants reaching the compressor all at once.
Q1.4 List the factors due to which the actual volumetric efficiency of a reciprocating compressor is
lower than its clearance volumetric efficiency.
1. Pressure Losses due to friction in connecting lines and across suction and
discharge valves.
2. Losses due to leakage.
8
Q1.5 What is the difference between internal and external equalization of a TEV?
An internally equalized valves, the evaporator pressure against the diaphragm is the
pressure at the inlet of the evaporator (typically via an internal connection to the outlet
of the valve), whereas in externally equalized valves, the evaporator pressure against the
diaphragm is the pressure at the outlet of the evaporator. Externally equalized
thermostatic expansion valves compensate for any pressure drop through the evaporator.
For internally equalized valves a pressure drop in the evaporator will have the effect of
increasing the superheat. Internally equalized valves can be used on single circuit
evaporator coils having low-pressure drop. Externally equalized valves must be used on
multi- circuited evaporators with refrigerant distributors. Externally equalized TXVs
can be used on all applications; however, an externally equalized TXV cannot be
replaced with an internally equalized TXV.
Q1.6. Determine the chemical formula of R141.
We have formula that
R(c-1) (h+1) f
Where, c-1= no of carbon atoms
h+1= no. of hydrogen atoms
f = no of fluorine atoms
Now by equating R141,
C - 1= 1 => c=2
H + 1= 4 => h=3
F=1
We know that 1 carbon can make bond with 8 atoms
Cl= 4-2-1= 1
Here each carbon has 1 chlorine
So, the chemical formula of R141 is C2H3Cl2F
Q1.7 What would be the ASHRAE designation of CO2 as a refrigerant?
ASHRAE designation of CO2 as a refrigerant is 744.
Q1.8 What is the difference between a flooded and a dry type evaporator?
The main difference between dry and flooded evaporators is that the flooded evaporator
is full of liquid refrigerant whereas in the dry evaporator design the liquid refrigerant
enters and flows through the evaporator and is slowly boiled off (evaporated), leaving
the evaporator as a vapor.
9
Experiment No. 2
2.1 Object: To study the working of domestic refrigeration unit and determine its refrigerating effect,
COP and refrigerating efficiency.
2.2 Theory: The unit includes (I) hermetic compressor (ii) air-cooled finned-tube condenser, (iii)
capillary tube, and (iv) plate type evaporator. Pressure gauges are mounted at the compressor inlet and
outlet. Ammeter and voltmeter are installed to read the compressor current and voltage. The refrigerant
used is R12.
2.3
Observations:
Compressor suction pressure
= P1= 10 psi(0.0689476MPa).
Compressor discharge pressure = P2=190 psi(1.31MPa).
Compressor current = 1.5 Amp.
Compressor voltage = 220 V.
2.4 Calculations & results: The calculations are based on the assumption that the system works
onthe standard vapor compression cycle.
10
h1 = 334.7 kJ/kg
, h2 = 388.04 kJ/kg
Refrigeration effect = h1 − h4 =334.7- 253.18= 81.52 kJ/kg
COP, R = h1-h4 / h2-h1
COP, R = 334.7-253.18 / 388.04-334.7.
COP, R = 1.53
COP, R(CARNOT) = TL / TH-TL.
COP, R(CARNOT) = 2.549.
Refrigerating efficiency = COPR/COPR, Carnot = 60.02%.
Q2.1 What is the purpose of observing the compressor current and voltage?
With the help of current and voltage we can find the compressor power as P= IV.
Q2.2 What is the function of thermostat?
A thermostat is a component which senses the temperature of a physical system and performs actions
so that the system’s temperature is maintained near a desired setpoint.
Q2.3 Which type of thermostat is used in the unit? State its working principle.
In this unit, diaphragm type thermostat is used. It is based on the principle that when the diaphragm
expands due to rise in temperature, the circuit is completed and the unit is switched on.
11
Course Code: ME-403.
Practical No. PRACTICAL #2
Course Title: REFRIGERATION & AIR CONDITIONING
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
12
Experiment No. 3
3.1 Object: To determine the performance of the vapor compression training unit and compare it tothat
of a standard vapor compression cycle.
3.2. Theory: The unit includes (I) hermetic compressor (ii) air-cooled finned-tube condenser, (iii)
capillary tube, and (iv) air-cooled finned-tube evaporator. Pressure gauges and temperature sensors are
mounted to read data at critical locations. Ammeter and voltmeter are installed to read the compressor
current and voltage. The refrigerant used is R22.
3.3
Observations:
Compressor suction pressure & temperature = 65 psi & 30°C
Compressor discharge pressure & temperature = 300 psi & 84°C
Compressor current = 10 Amp
Compressor voltage = 220 V
Condenser inlet pressure & temperature = 300 psi & 84°C
Condenser outlet pressure & temperature = 290 psi & 48°C
Evaporator inlet pressure & temperature = 70 psi
Evaporator outlet pressure & temperature = 60 psi & 25°C
13
3.4
Calculations & results:
3.4.1 Calculations & results based on actual readings: First sketch the cycle on the p-h
planeusing the above data. Read the enthalpies from the R22 p-h chart, and then calculate the
following:
Refrigeration effect = h1-h4= 1425KJ/Kg – 245.695KJ/Kg.
Specific work of compression = h2-h1= 1695KJ/Kg – 1425KJ/Kg.
COPR = 4.37.
COPR, Carnot =
TL
= 5.05.
TH − TL
Refrigerating efficiency = COPR/COPR, Carnot = 86.53%.
3.4.2
Calculations & results based on assumption of standard vapor compression cycle:
Refrigeration effect = 1430KJ/Kg – 244.785KJ/Kg
Specific work of compression = 1695.005KJ/Kg – 1425.45KJ/Kg.
COPR = 4.32
Refrigerating efficiency = COPR/COPR, Carnot = 85.54%.
Q3.1 What is the function of sight glass?
In refrigeration system with receivers, a sight glass is an excellent tool and can be relied upon as
indicator of liquid refrigerant to the metering device. The moisture indicator shows you if the system is dry
Or if it has moisture content.
Q3.2 Why is it not suitable to place the filter/drier after the expansion device?
The purpose of filter is to prevent the entrance of particles into the expansion device and if we place
the filter after the expansion device, it can cause blockage to the passage flow of the refrigerant in the
expansion valve and cause improper operation of the system.
Q3.3 Suggest a procedure that can be used to determine the refrigerant flow rate in the system.
The volume flow rate of refrigerant can be measured using the device ‘ROTAMETER’. It measures
the flow rate of liquid or gas in a closed tube. Flow rate of refrigerant can be measured using
compressor current, voltage, motor efficiency, inlet and outlet pressure of temperatures by using
mr = (power input to compressor)/ (specific compressor work.
14
Course Code: ME-403.
Practical No. PRACTICAL #3.
Course Title: REFRIGERATION & AIR CONDITIONING.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
15
Experiment No. 4
4.1 Object: To determine the properties of conditioned air using the Dehumidification Unit.
4.2 Theory: The function of a dehumidifier is to remove moisture from air. This is required toprevent
or minimize the following:
- Physical discomfort & physiological problems like acute allergy & asthma.
- Growth of bacteria & mold on surfaces.
- Accelerated corrosion of metal.
- Degradation of surface finish.
- Malfunction of electronic/electrical equipment.
The most common type of dehumidifiers employs a refrigeration cycle, as shown below.
Humid air is drawn over air over cold evaporator coils, condensing out its moisture. It then passes over
warm condenser coils and back into the room. The condensed water drips into a container in theunit that
has to be emptied. The water can be directed to a drain by means of a hose. Mold can grow in the drainage
areas of a dehumidifier and regular cleaning the water basin with bleach is required.
16
A second type is the desiccant humidifier. The basic construction and working is schematicallyshown
and described below.
The most critical component is the desiccant rotor. It is about as wide as the dehumidifier itself and an
inch or two thick. It is made up of alternate layers of flat and corrugated sheets which are impregnated
with the desiccant. These sheets are arranged so that they allow for air flow perpendicular
to the drum.
The desiccant rotor has two different zones – a “process zone” which makes up about 75% of the
area of the rotor and a “recharge zone” which makes up the remaining 25% of the rotor. The warm humid
air that enters the dehumidifier is pulled through the process zone. Here, moisture is adsorbed by the
desiccant material. The incoming humid air goes through the process zone and immediately exhausts
out of the dehumidifier.
Covering the recharge zone of the rotor is a heater. It warms circulated humid air which is pulled back
through the desiccant drum in the opposite direction of incoming humid air. This warm air (heated by
the heater) liberates moisture from the desiccant. Thus, moisture is transferred from the desiccant to the
air. The warm humid air that leaves the recharge zone enters the condenser on the front of the
dehumidifier. Here warm humid air condenses at room temperature. The condensatedrips down into
a condensate collection bucket, much the same way as it does in a compressor-based dehumidifier.
Commercial vs consumer grade desiccant dehumidifiers: Commercial desiccant dehumidifiers do not
work the same way that consumer grade desiccant dehumidifiers work. They share many of the same
parts (although the parts used on commercial units are much more heavy duty) and work very similarly,
but they are not one and the same.
17
The primary difference between the two has to do with the warm humid air that leaves the recharge
zone of the desiccant rotor. As discussed above, this air is immediately funneled to a condenser on a
consumer grade desiccant dehumidifier. Here, the warm humid air condenses and the condensate drips
down into condensate collection bucket that can easily be removed from the dehumidifier.
Commercial units do not have a condenser. Instead, the warm moist air that leaves the recharge zone of
the desiccant drum is exhausted out of the building that is to dehumidified through ductwork. Thus, there
is no condensate that forms anywhere in the system. This allows for commercial units to dehumidify at
even colder temperatures and even lower humidity than residential consumer grade units.
Laboratory dehumidification unit: The unit is designed to convey the basic psychrometric understanding
of the dehumidification process. Thedehumidifier works on the refrigeration cycle. Thecondition of the
air entering the dehumidifier is measured using a sling psychrometer, while the condition at exit is read
from the aspirated psychrometer installed in the dehumidifier.
4.3 Observations
Inlet air DBT =
Outlet air DBT =
31
34.2
Outlet air velocity: u2 = _2.5
C; Inlet air WBT =
27.3
C; Outlet air WBT =
28.3
m/s; Outlet area: A2 = 0.07425
C
C
m2
18
4.4 Calculations & results: The relative humidity, humidity ratio, and specific volume of air at inlet
and outlet are obtained from the psychrometric chart.
Relative humidity Humidity ratio“W” Specific volume
“” (%)
“” (m3/kgda)
(kg/kgda)
Inlet (1)
75
0.0215
0.891
Outlet (2)
64
0.022
0.902
= u A =  = m3/s
Volume flow rate at the outlet: V
2
2
2
Mass flow rate of dry air:
𝑉
𝑚𝑎 = 𝑣2 =            =     
kgda/s
2
Moisture condensation rate: mw = ma (W1 −W2 ) = 
kg/s
Q4.1 Sketch the processes occurring in the dehumidifier on a skeleton psychrometric chart.
Q4.2 Estimate the ADP of the coil.
The estimated ADP of the coil is found to be 26°C.
Q4.3 Will the dehumidifier also decreases the room dry bulb temperature? Give reason.
During the cooling and dehumidification process, the dry bulb, wet bulb and the dew point temperature of
air reduces because if moisture is evaporated into an air volume without any heat input or removal, the latent heat
19
of evaporation is taken from the atmosphere. The sensible heat content- thus the DBT is reduced.
20
Course Code: ME-403.
Practical No. PRACTICAL# 4.
Course Title: REFRIGERATION & AIR CONDITIONING.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
21
Experiment No. 05
5.1 Object: Determine the overall mass and energy balance through heat exchanger A (or packing type
A, B or C) of a cooling tower.
5.2 Apparatus:
Cooling Tower mod. CT/EV
General View: Cooling Tower Apparatus mod. CT/EV
5.3 Equipment Description:
The apparatus (see the synoptic on chapter "Technical documents") comprises a transparent
rectangular tower with packing. Water from the hot reservoir is pumped to the top of the tower and the
flow rate can be varied and is indicated on a rotameter.
The water is distributed evenly over the packing at the top of the tower to minimize the channeling.
After the cooling, the water is recollected in the hot water reservoir. A makeup water reservoir
maintains a constant level in the hot reservoir as water evaporates.
22
A forced draught of air is supplied at the base of the tower by means of a centrifugal fan; the flow rate
can be adjusted by means of a damper and measured using an orifice plate.
After leaving the packing, the air passes through a demister. The hot water reservoir is provided of
3electrical heater controlled via thermostat.
The process temperatures can be measured by means of digital thermometers; a differential manometer
is provided to measure the pressure drop on the packing and on the orifice plate.
5.4 Technical Characteristics:
Schematic Diagram: Cooling Tower Apparatus mod. CT/EV
1.
2.
3.
4.
Rectangular column in Plexiglas with packing, height 550 mm, code Cl
Make-up water reservoir in Plexiglas, capacity 1(one) liter, code D1
Centrifugal fan, max. flow rate 1340 m3/h, max. head 80 mmH2O, code Pl
Hot water reservoir in stainless steel ATST 304, capacity 9-liter, code D2, with 3x500W electric
heaters, code J1, J2 and J3
5. Water pump, Qmax=3 m3/h, Hmax=5m H20. Code GI
23
6. Water flow meter, range 20-200 L/h, code FI1
7. Inclined tube manometer, range 0-60 mm H20
8. Electrical switch board with:
9. Digital thermostat, code TW2
10. Fan and resistances switches
11. E.L.C.B.
5.5 Start up:
1. Place the unit on a strong table (the weight of the un it is 72 kg) in a place with a good air circulation
2. Check that the unit is level
3. Close valve VJ
4. Open partially valves V 1 and V2
5. Fill the reservoirs of the wet bulb thermometers TIJ and Tr5 with water and ensure that the gauze
covering each of the wet bulb thermometers is thoroughly wetted
6. Fill the hot water reservoir 02 with distilled water (necessary about10litres) through the make-up
tank 01 up to the mark
7. Keep attention: the water level in the make-up reservoir DJ must be minimum 80 mm; if there isn't
water in D, do not switch on the heating elements J1, J2 and J3.
8. Connect the unit with the electrical supply: single-phase + G
9. To measure the pressure, drop on the packing, open the two relative valves on the differential
manometer and close the others (see the labels)
10. To measure the pressure, drop on the orifice, open the two relative valves on the differential
manometer and close the others (see the labels)
11. Switch on the E.L.C.B
12. Set the thermostat TW2 at maximum 50°C (pre settled at 40°C)
13. Start the pump G I and then the heaters J I, J2 and J3
14. Adjust the water flow rate using the valve V2 (for example at 1201/h)
15. Switch on the fan P 1 and by means of the damper adjust the air now rate (for example at ΔP = 16
mmH2O)
16. Allow about 10 minutes for conditions to reach steady state
17. Measure and record the process parameters
24
5.6 Procedure:
1. Stabilize the column on the following suggested condition:
Orifice ΔP
16 mmH2O
Water flow rate
120 L/h
Cooling load
1.0 KW (0.85 kW)
2. Fill the makeup tank up to the mark with distilled water
3. Note temperatures and flows at regular intervals of 5 minutes
4. At the end of this period, refill the makeup tank with distilled water using a measuring cylinder. By
difference, calculate the makeup water supplied in the time interval
5.7 Observation:
Date:
Name:
Atm. Press.:
Test N o
1
2
3
4
Packing type
C
C
C
C
235
235
235
235
235
32.4
32.7
33.2
33.5
33.6
32.3
32.6
33.1
33.4
33.5
32.1
32.8
32.9
33.6
33.5
Ai r outlet Wet bulb (oC)
29.9
29.6
29.9
29.6
29.4
Water inlet temperature (oC)
39.7
40.3
40.6
40.9
41.1
Water outlet temperature (oC)
29.8
30.1
29.9
29.5
30.3
Water makeup temperature (oC)
29.1
29.2
29.3
29.4
29.5
16
16
16
16
16
Water flow rate mw (kg/h)
120
120
120
120
120
Cooling load Q (kW)
1
1
1
1
1
Packing density m -1
Air inlet Dry bulb (oC )
Air inlet Wet bulb (oC)
Ai r outlet Dry bulb (oC)
Orifice ΔP mm H20
5
C
Make up quantity m mu (kg)
0.116
0.172
0.129
0.133
0.154
Sampling time t (sec)
300
300
300
300
300
Packing ΔP mm H20
25
25
25
25
25
5.8 Calculation and Results:
25
A. Plot on the psychrometric chart using the observation chart showing the condition at the inlet (point
A) and outlet (point B) and determine the following (as shown in sketch):
97.5
Enthalpy at A,
HA = ---------------------- KJ/kg
Humidity ratio at A,
WA = ---------------------- kg/kg
Enthalpy at B,
HB = ---------------------- KJ/kg
Humidity ratio at B
wB = ---------------------- kg/kg
Specific Volume at B
vB = ---------------------- m3/kgdry
21.7
116
29
0.93
ΔP(AVEREAGE)=16mm
𝑚𝑚𝑢 (average)=0.148kg &t(average)=300 seconds
B. From the orifice formula, the air mass flow rate is:
mB = 0.0165
C. Make up rate = mmu / t
=
=
0.012495kg/s
----------------------- kg/s
4.69x10−4kg/s
= ----------------- kg/s
0.0912
D. Evaporation rate = mB (wB – WA) = -------------------- kg/h
26
E. Specific enthalpy of makeup kcal/kg
F. Enthalpy of makeup = (specific enthalpy) x (make up rate)
0.02311775
G. Air enthalpy change = mB (HB – HA) = -------------------- kg/h
H. Water enthalpy change = mw cp (tC – tD) kW
5.9 Results:
0.2311575kW
Actual water enthalpy change = ---------------kW
W
5.10 Conclusion:
List the conclusion which you have drawn from the results and calculation of this practical.
(Attach extra sheet if required)
27
Course Code: ME-403.
Practical No. PRACTICAL# 5.
Course Title: REFRIGERATION & AIR CONDITIONING.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
28
Experiment No. 06
6.1 Object: Determine the correlation between heat exchanger (packing A, B and C) and cooling
performance of a cooling tower.
6.2 Apparatus:
Cooling Tower mod. CT/EV, packing A, packing B and packing C.
Cooling Tower Apparatus mod. CT/EV
Packing type A
Temp. sensors
6.2.1 Packing Data:
6.2.2 Constant:
•
Dimension of rectangular Plexiglas column: 150 x 150 x 600(h) mm
29
6.3 Water Cooling:
Many industrial processes use water as cooling medium in the heat exchangers to lower the
temperature of process streams; because the costs involved in its provision and disposal, the cooling
water is recycled but its temperature must be lowered. The most common unit used to accomplish this
goal is a cooling tower.
6.3.1 Cooling Tower:
The general idea of a cooling tower is simple: warm water trickles down the packing i n a column and
encounters air flowing upwards. A small portion of the water evaporates, requiring latent heat, which
causes the decreasing of the temperature of the water (although, when the air temperature is low, there
is also some sensible heat transfer to the air).
This process also known as humidification involves the simultaneous transfer of mass and heat.
The packing is used to produce a large water surface area for heat and mass transfer.
Cooling towers fall into two main sub-division s: "natural draught" and "mechanical draught".
Natural draught towers depend upon natural convection, due to the difference of density between am
bent air and the heated air.
In a mechanical draught tower (the much more widely used), a fan serves to provide a contra I led flow
of air through the packing; the fan can be located at the top of the tower or at the bottom. Since the air
flow is controlled and bigger than in the natural draught towers, this type of towers is considerably
more compact.
For a good understanding of water cooling in cooling tower, it is necessary some fundamental
concepts of psychrometry must be known.
6.3.2 Theory of Cooling Tower:
The humidification process can be represented graphically plotting humidity vs. gas temperature.
There are several different situations that can be occur during the countercurrent contact of air and
water (see Fig. below).
30
The curves of the diagram illustrate the condition s of the air as it flows from the bottom of the tower
(A) to the top of the tower (Q, R, E, S, or P depending on the condition).
•
•
•
•
•
Curve AQ: liquid cooling with air heating and humidification
Curve AR: liquid cooling with air cooling and humidification
Curve AE: adiabatic liquid cooling with air humidification
Curve AS: air cooling and humidification but less humidification than in adiabatic cooling
Curve AP: liquid heating and gas dehumidification
Curve AQ: liquid cooling with air heating and humidification
The temperature of entering water must be higher than the temperature of the entering air. Therefore,
the water must be heated prior to it being reintroduced at the top of the column of mod. CT/EV.
Water vapor diffuses from interface to the bulk air with humidity as driving force in the air phase (HiHg).
The temperature driving force is (Tx - Ti) in the water phase and (Ti - Ty) in the air phase.
Sensible heat flows from the bulk water to the interface since the temperature of the water is higher than
the temperature of the interface. Latent heat in the water vapor diffuses into the bulk air and the sensible
heat in the air at the interface diffuses to the bulk air since the temperature of the interface is higher than
the temperature of the bulk air.
Curve AR: liquid cooling with air cooling and humidification
The air entering the column must be at higher temperature than the entering water in order to occur the
gas humidification and cooling
Curve AE: adiabatic liquid cooling with air humidification
The function of column is for air humidification; the water is recycled through the column of mod.
CT/EV without heating.
The air follows the adiabatic saturation
bulb temperature.
line which intersects the equilibrium line t the wet
Curve AS and AP: are not viable for liquid cooling
6.4 Procedure:
•
•
•
•
Install packing A in Plexiglas rectangular column and stabilize the following suggested condition:
Orifice ΔP
16 mmH2O
Water flow rate
120 L/h
Cooling load
1.5 KW
31
•
•
•
•
•
•
•
Fill the makeup tank up to the mark with distilled water
Keep the water flow rate constant.
Keep the cooling load constant
Report the air inlet wet bulb temperature in the observation table chart.
Report the water outlet temperature in the observation table chart.
Remove packing A and install column B. After stabilization at the same condition as above, repeat
the observations.
Finally, install packing C (to be mounted) and repeat the observation.
6.5 Observation:
Date:
Name:
Test No
Packing type
Packing density m
-1
Air inlet Dry bulb TI (oC)
Air inlet Wet bulb TI (oC)
Ai r outlet Dry bulb TI (oC)
Ai r outlet Wet bulb TI (oC)
Water inlet temperature TI (oC)
o
Water outlet temperature TI ( C)
Water makeup temperature TI (oC)
Orifice ΔP mm H20
Water flow rate m w (kg/h)
Cooling load Q (kW)
Make up quantity m mu (kg)
Sampling time t (sec)
Packing ΔP mm H20
Wet bulb Approach
Atm. Press.:
1
2
3
A
92
33.3
33.2
30.7
30.5
B
131
33.4
33.3
31.3
31.2
42.0
30.5
29.8
16
120
1
0.124
300
13
2.8
C
235
33.6
33.5
33.5
29.4
41.1
30.3
29.5
16
120
1
0.154
300
25
3.2
41.4
30.9
29.8
16
120
1
0.137
300
10
2.3
4
5
6.6 Calculation and Results:
Plot a graph between packing area and wet bulb approach temperature
6.7 Conclusion:
List the conclusion which you have drawn from the results and calculation of this practical.
(Attach extra sheet if required)
32
33
Course Code: ME-403.
Course Title: REFRIGERATION & AIR CONDITIONING.
Practical No. PRACTICAL # 6.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
34
Experiment No. 07
7.1 Object: Analysis of the behavior of the air when passing through a heating coil or (heating
battery).
7.2 Apparatus:
Computerized air conditioning trainer mod. GCTC/EV
7.3 Equipment Description:
The air-conditioning trainer enables the complete and exhaustive analysis of the air thermodynamic
changes when crossing the different stages of a modern air-conditioning station.
35
In fact, by directly measuring the air temperature and humidity in different points, you can follow and
control cooling, heating (at constant specific humidity), humidification, dehumidification and the
result of the sum of 2 air volumes with different characteristics.
The equipment is provided also with sensible and latent heat generators, set in the test chamber. So, it
is possible to change the heat load of the chamber and to check the behavior of the hot, cold and
humidity batteries and their relations with the regulation circuit.
This complex system, provided with safety devices, has been designed to carry out all control and
supervision operations in HVAC plants with simultaneous energy control such as optimization of the
starting, the temperature control and heat recovery.
The microprocessor stores the calibration, sum and subtraction operations, the regulation functions, the
proportional sequences, the transfer and conversion functions, the logic functions, the power functions,
the output functions, as software packet. With the help of the PC, it is possible to compose and
logically combine these functions in software mode in order to obtain simple or complex operative
instructions that could hardly be obtained with traditional controls and regulations using standard
functions and physical inter blocks. Thanks to these features, it is possible for the not specialized
student to develop more operation logical programs and to experiment their operation
This trainer allows to focus on the following arguments:
•
•
•
•
Experimental study of the air psychrometric diagram
Study of the air transformations during its passage through the different unit sections
Calculation of the heat balances related to the air transformations
Checking of the behavior of the regulating system and of its effectiveness as function of the
thermal loads variation
The control system operates on the following components:
•
•
•
Cooling and dehumidifying 1si and 2nd heating stage
1st and 2nd humidifying stage Air dampers servo-command
The refrigerating unit including the compressor is independent and it is controlled by a thermostat.
The control panel of the trainer includes a set of instruments for the data collection consisting of:
•
•
•
•
4 electronic thermometers
4 electronic hygrometers
4 flow meters on the water circuit (cooling coil)
3 multi meters
36
7.4 Control panel of the Equipment:
37
7.5 METHOD/PROCESS OF HEATING:
The process can be drawn as in the figure on the left, while in the figure on the right you find the
process in the psychrometric diagram.
The energy and mass balances are as follows:
m1 ·h1 + q = m 2 ·h2
m1=m2
m1 ·X1 = m 2 · X 2
As a consequence:
q = m 1 ·(h 2 -h 1 )
X1 = X 2
The value of the enthalpy is given by:
h = 04 * t + x * (596 + 0.48 * t) [kcal/kg]
Record the temperature and relative humidity before and after the warm-up suffered by the air and
locate corresponding points on the psychrometric chart. The points will be located on a horizontal
straight line. From these points, we can establish the thermal balance of cross- training.
where:
m = is mass flow rate; h is enthalpy;
q = is the transferred heat;
x = is the specific humidity;
t = is the temperature;
Subscripts 1 and 2 refer to the air entering and coming out of the heating section;
38
7.6 Observation:
Air passing through the heating coil
S.NO
1
2
State
point
Air
entering
Air
leaving
Temp.
Relative
Humidity
Enthalp
y
Specifi
c
volume
h (kJ/kg)
v
(m3/kg)
Density
Specific
humidity
t (oC)
U (%)
33.3
68.2%
90.39
0.899
1.14
22.2
45.3
35.5%
102.37
0.934
1.09
22.0
ρ
x (gm/kg)
(kg/m3)
7.7 Calculation and Results:
o
o
o
o
Determining the air speed in the channel, v = 1.2 m/s
Duct cross-sectional Area.
A = w x L = 0.13x0.15 m2.
Volumetric flow rate of the air
V = v x A = 0.0234 m3/h.
Mass flow rate of the air
m = V x ρ = 0.026 kg/h.
7.8 Results:
A. Enthalpy jump Δh = h2 – h1 = 102.37-90.3= 11.98 kJ/kg
B.
Δx = x2 – x1 = 0.2 gm/kg
C. Plot the heating process on psychrometric chart showing air properties
7.9 CONCLUSION:
List the conclusion which you have drawn from the results and calculation of this practical.
(Attach extra sheet if required)
39
Course Code: ME-403.
Practical No. PRACTCAL # 7.
Course Title: REFRIGERATION & AIR CONDITIONING.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
40
Experiment No. 08
8.1 Object:
Determine flow rate of chilled water output by the pump and speed of the chilled water inside a pipe of
the cooling unit.
8.2 Apparatus:
Computerized air conditioning trainer mod. GCTC/EV
8.3 Theory / Procedure:
8.3.1 Flow rate of chilled water output by the pump
The hydraulic circuit of water is characterized by 4 branches after the 3- way valve: each branch will
feed 2 lines of cooling and dehumidifying unit. Furthermore, each branch includes a stop valve/cock
and a flow meter.
The flow rate read on each flow meter
is equal to:
G = 1.1 L/min = 66 L/h
Consequently, the flow rate output
by the pump will be equal to:
Gtot = 66 · 4 = 264 L/h
The flow rate of water output by the pump is the result of the sum of the values read on the flow
meters. Dividing the flow rate of water flowing inside a pipe by the section of the same pipe will
enable to determine the speed of the water crossing the pipe.
This parameter is very important because it will condition the heat transfer coefficient of the cooling
and dehumidifying unit.
8.3.2 Speed of the chilled water inside a pipe of the cooling unit:
Dividing the flow rate of one of the branches by the number of other branches derived from it, will
enable to determine the flow rate of the water crossing a pipe of the cooling unit. Refer to the
following figure.
As it can be seen clearly, the number of branches is equal to 5. Consequently, the flow rate of water
having to be considered in the calculation will be:
Gb = 1.1 /5 = 0.22 L/min = 0.0036 L/s
As the inside diameter d of each pipe is of 8 mm, the water flow passage section will be:
41
Area = A = π/4 (d2) in m2
So, the speed of the water inside a pipe of the cooling system will result from the following formula
Speed = v = Gb/A in m/s
42
8.4 Observation:
S.
N
o.
valves
status
1
All
Close
One
open
Two
open
Three
open
All
open
2
3
4
5
Flow
G (m3/min.)
Chilled water system
Suct Disch.
ion pressure
Pressure
Change
Pressure
Head
Pump
power
press
ure
G1
G2
G3
G4
Gtot
0
0
0
0
0
0.0026
6
0.0015
5
0.0012
5
0.0010
6
0
0
0
0.00
155
0.00
125
0.00
106
0
0
0.00
125
0.00
106
0
0.00
266
0.00
31
0.00
375
0.00
424
0.00
106
Ps
(Kp
a)
150
Pd
(Kpa)
ΔP=Pd-Ps
(Kpa)
H=Pd/ρg
(m)
T
(kW)
275
125
23.38
0
145
270
130
22.9
0.716
133
260
127
22.1
0.805
127
230
103
19.55
0.862
120
210
90
17.85
0.89
8.5 Calculation and Results:
A.
B.
C.
D.
Determining water speed in cooling unit, v = 0.11 m/s
pipe cross-sectional Area.
A = π/4*d2 =
π/4*(0.008)2 = 50.3 x 10−6 m2
Volumetric flow rate of the air
Gb = v x A = 0.02 m3/h
Pump Power
T = ρgGH = 0.89 kW
8.6 Results:
A. Plot a graph Valve status vs Total flow
B. Plot a graph Total flow vs Pressure drop
C. Plot a graph Total flow vs power
8.7 Conclusion:
List the conclusion which you have drawn from the results and calculation of this practical. (Attach
extra sheet if required)
43
44
Course Code: ME-403.
Practical No. PRACTICAL #8.
Course Title: REFRIGERATION & AIR CONDITIONING.
Assessment Rubric for Lab Performance (P3)
S.
Skill
No.
Excellent (3)
1
Ability to respond Responds by fully
to instructions for complying with
carrying out a
instructions in
practical.
correct sequence
and without
further assistance
from instructor.
Performance Level
Good (2)
Fair (1)
Responds well but
Responds well
requires slight
but occasionally
assistance from
requires
instructor.
instructor’s help
to rectify
incorrectly
executed step(s).
2
Equipment /
Instrument /
Software
handling.
Displays good
awareness of
equipment /
instrument /
software selection
and follows the
prescribed
handling
guidelines.
Displays good
awareness of
equipment /
instrument / software
selection. Needs
some supervision
while handling
equipment.
3
Realization of
Problems
Fully operational
command of the
programming
language:
appropriate &
accurate with
complete
understanding of
complex programs
4
Problem Solving
Skills
5
Presentation of
work
Able to cover
various aspects of
problem & ability
to properly
carryout
calculation
stepwise.
Presented the
practical following
all the
instructions.
Effective command
of the programming
language despite
some inaccuracies, &
misunderstandings.
Can use &
understand fairly
complex programs,
particularly in
familiar situations.
Covers major portion
of calculations in
effective way.
Marks: ________
Presented the
practical following all
the instructions with
minor deficiencies.
Poor (0)
Lacks the
response
required to carry
out the practical.
Displays
adequate
awareness of
equipment /
instrument /
software
selection.
Requires
supervision while
handling
equipment.
Basic
competence is
limited to
familiar problems
& need
improvements to
understand
complex
programs.
Unable to select
the right
equipment /
instrument /
software and
handle it
properly.
Weakly
understand way
to carry out
calculations.
No knowledge
Need a lot of
improvement in
presentation of
work.
Disregards
presentation
instructions
Unable to
understand
complex
programs.
Instructor’s Signature: __________________ Date: __________
45
Experiment No. 9
9.1 Object: To study the air conditioning system of the university auditorium.
9.2 Theory: A central air conditioning system is typically used to serve large-area buildings with
many zones of conditioned space or to separate buildings. They comprise the following sub-systems.
Air system: Its function is to condition, transport and distribute the conditioned air, as well as to control
the indoor environment according to requirements. The major components of an air system are the airhandling units, supply/return ductwork, fan-powered boxes, space diffusion devices, and exhaust
systems.
Water system: The water system includes chilled and hot water systems, chilled and hot water
pumps, condenser water system, and condenser water pumps.
Central plant: The refrigeration system in a central plant is usually in the form of a chiller package.
Chiller packages cool the chilled water and act as a cold source in the central hydronic system. The
boiler plant, consisting of boilers and accessories, is the heat source of the heating system. Either hot
water is heated or steam is generated in the boilers.
Control system: It consists of electronic sensors, microprocessor-operated and -controlled modules.
Outputs from the control modules often actuate dampers, valves, and relays by means of pneumatic
actuators in large buildings and by means of electric actuators for small projects.
9.3 Auditorium air conditioning system: The air condition system comprises the following main
components:
1.
Open type reciprocating compressor
2. Water-cooled shell and tube type condenser; water inside the tubes and the refrigerant on the
shellside.
3.
Ceramic core type drier
4.
Thermostatic expansion valves
5.
Induced draft type cooling tower
6.
R22 refrigerant (evaporating temperature of −15C and condensing temperature of 30C)
7.
Air handling units
8.
Ducting and air diffusion system
Chemical dosing of condenser water is required to prevent formation of scale, corrosion, and growth of
algae and fungi.
46
Flow diagram of the Air Conditioning Plant of NED University Auditorium
47
9.4 Observations:
Condenser water inlet temperature = 30
C
Condenser water inlet
pressure = 280
psi
Condenser water outlet temperature = 50
C
Condenser water outlet pressure = 240
psi
Room temperature = 25
C
Supply air temperature = 15
C
Return air temperature = 30
C
Outside air temperature = 32
C
Q9.1 Sketch the basic cooling cycle on a skeleton psychrometric chart.
Q9.2 Which type of central air conditioning system is installed in the auditorium?
In auditorium central air conditioning system with one compressor and one evaporator (Vapor
Compression Chiller System) is installed.
Q9.3 Why is the outside air mixed with recirculated air and then supplied to the space?
The outside air is mixed with the recirculated air to maintain a certain quality of air in the refrigeration
space, otherwise the level of CO2 will rise in the space making the occupants uncomfortable.
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Q9.4 Define range and approach of a cooling tower.
RANGE: Range is defined as the difference between the cooling tower water inlet and outlet
temperatures.
APPROACH: It is the difference between cooling tower inlet temperature and ambient air WBT
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50
51
52
53
54
55
56
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