me150p electure 1 (04 january 2011)
DESCRIPTION
airconditioning system lecture vfor mechanical engineeringTRANSCRIPT
ME150P/B1 Lectures3rd Term S.Y. 2010-2011
Group 1
Orientation 04 January 2011
I. PsychrometryII. Air Conditioning SystemIII. Heat Load EstimationIV. Practical Air Conditioning SystemV. Air Distribution SystemVI. Piping SystemVII. Equipment SelectionVIII. Refrigerated LoadIX. Ice Manufacturing
Lecture Outline
I. Heat LoadII. Air DistributionIII. Equipment Selection
Report Outline
Refrigeration and Air Conditioning (Chapters 14 & 15) by C.P. Arora
Carrier System Design Manual Part I, II, & III Mechanical Engineering Tables and Charts
Latest Edition by MRII Refrigeration (Chapter X: Cooling Load
Calculation) by Dossat (SI Edition) Ken’t Handbook – Power Volume (Section
11: Ice Manufacturing)
Textbooks & References
Psychrometry08 January 2011
Chapter 14: Properties of Moist Air; Refrigeration and Air Conditioning by C.P. Arora
Dry air and water vapor form a binary mixture. The properties of moist air are called psychrometric properties and the subject which deals with the behavior of moist air is known as psychrometry.
PSYCHROMETRIC PROPERTIES
Specific or absolute humidity or humidity ratio or moisture content as it is variously called denoted by the symbol ω is defined as the ratio of the mass of water vapor (w.v.) to the mass of dry air (d.a.) in a given volume of the mixture. Thus
Where the subscripts a and v refer to dry air and water vapor respectively.
Specific Humidity or Humidity Ratio
v a
a v
m
m
Since p denotes the actual total atmospheric pressure, then from Dalton’s Law
So that,
Specific Humidity or Humidity Ratio
a vp p p
0.622 v
v
p
p p
The concept of specific humidity is that if we take 1 kg of dry air
Then the mass of water vapor associated with this dry air, in the same volume is
so that the total mass of this volume of moist air is
.
Specific Humidity or Humidity Ratio
1am kg
vm kg
(1 )m kg
The temperature is called the dew point temperature (DPT). It is the temperature to which moist air must be cooled at constant pressure before condensation of moisture takes place.
The DPT can be found by knowing, from the steam tables, the saturation temperature at the partial pressure of the water vapor.
Dew Point Temperature
dt
dtvp
PROBLEM◦ In a dew point apparatus a metal beaker is
cooled by gradually adding ice water to the water initially at room temperature. The moisture from the room air begins to condense on the beaker when its temperature is 12.8˚C. If the room temperature is 21˚C and the barometric pressure is 1.01325 bar, find the partial pressure of water vapor in the room air and parts by mass of water vapor in the room air.
Example 14.2, page 454
SOLUTION◦ Partial pressure of water vapor at DPT 12.8˚C
◦ Partial pressure of dry air
◦ Specific humidity
Example 14.2, page 454
21.479 kN/mvp
2101325 1479 99846 N/m
a v
a
p p p
p
14790.622 0.622
99846v v
a a
m p
m p
SOLUTION◦ Specific humidity
◦ Parts by mass of water vapor
Example 14.2, page 454
w.v.0.009214
d.a.
kg
kg
0.009214 w.v.0.00913
1 1.009214 mixturevm kg
m kg
The degree of saturation is a measure of the capacity of air to absorb moisture.
It is denoted by the symbol μ
Degree of Saturation
1
1v s
s s v
p p p
p p p
Relative humidity denoted by the symbol φ or RH is defined as the ratio of the mass of water vapor in a certain volume of moist air at a given temperature to the mass of water vapor in the same volume of saturated air at the same temperature.
Relative Humidity
v s
s v
pRH
p
The enthalpy of moist air is obtained by the summation of the enthalpies of the constituents. Thus, the enthalpy of moist air h is equal to the sum of the enthalpies of dry air and associated water vapor, i.e.,
per kg of dry air, where subscripts a are for dry air and v are for the water vapor part.
Enthalpy of Moist Air
a vh h h
◦ Where t is in ˚C
◦ Where t is in ˚C
Enthalpy of Moist Air
1.005 (2500 1.88 ) kJ/kg d.a. h t t
0.24 (1061 0.444 ) Btu/lbm d.a. h t t
PROBLEM: A mixture of dry air and water vapor is at a temperature of 21˚C under a total pressure of 736 mmHg. The dew point temperature is 15˚C. Find:i. Partial pressure of water vaporii. Relative humidityiii. Specific humidityiv. Specific enthalpy of water vapor by the three
methods of Fig. 14.5v. Enthalpy of air per kg of dry airvi. Specific volume of air per kg of dry air
Example 14.3, page 458
SOLUTION:i. From steam tables, the partial pressure of
water vapor at 15˚C DPT is
ii. Saturation pressure of water vapor at 21˚C DBT
Relative humidity
.
Example 14.3, page 458
212.79 mmHg = 12.79 (133.5) = 1707.5 N/mvp
218.65 mmHg = 18.65 (133.5) = 2489.8 N/msp
12.79100 100 68.58%
18.56v
s
p
p
iii. Specific humidity
iv. Latent heat of vaporization of water at dry bulb and dew-point temperatures of air
Example 14.3, page 458
(12.79)0.622 0.622
(736 12.79)
0.011 kg w.v./kg d.a.
v
a
p
p
21
15
( ) 2452 kJ/kg
( ) 2466.2 kJ/kg
fg C
fg C
h
h
Specific enthalpy of water vapor from Fig. 14.5 by the three methods
Note: The three values are extremely close to each other.
Example 14.3, page 458
(4.1868)(21) 2452 2450 kJ/kg w.v.
(4.1868)(15) 2466.2 1.88(21 15)
2540.3 kJ/kg w.v.
2501 1.88(21) 2540.5 kJ/kg w.v.
C
A
A
B
h
h
h
h
v. Enthalpy of air using the value of specific enthalpy of water vapor from the empirical relation
vi. Specific volume of air is equal to the volume of 1 kg of dry air or 0.011 kg of water vapor. Based on the dry air part
.
Example 14.3, page 458
1.005(21) 0.011(2540.5)
21.1 27.9 49.0 kJ/kg d.a.
a vh h h
h
h
3287.3(273 21)0.875 m /kg d.a.
(723.21)(133.5)a
aa
R T
p
The dry bulb thermometer is directly exposed to the air and measures the actual temperature of air.
The bulb of the wet bulb thermometer is covered by a wick thoroughly wetted by water.
The temperature which is measured by the wick-covered bulb of such a thermometer indicates the temperature of liquid-water in the wick and is called the wet bulb temperature.
WET BULB TEMPERATURE (WBT)
All data essential for the complete thermodynamic and psychrometric analysis of air-conditioning processes can be summarized in a psychrometric chart.
The chart which is most commonly used is the ω-t chart, i.e., a chart which has specific humidity or water vapor pressure along the ordinate and the dry bulb temperature along the abscissa.
The chart is normally constructed for a standard atmospheric pressure of 760 mm Hg or 1.0132 bar, corresponding to the pressure at the mean sea level.
PSYCHROMETRIC CHART
PSYCHROMETRIC CHART Skeleton
Dry-bulb temperature = The temperature of air as registered by an ordinary thermometer.
Wet-bulb temperature = The temperature registered by a thermometer whose bulb is covered by a wetted wick and exposed to a current of rapidly moving air.
Dewpoint temperature = The temperature at which condensation of moisture begins when the air is cooled.
Definition of Terms
Relative Humidity = Ratio of the actual water vapor pressure of the air to the saturated water vapor pressure of the air at the same temperature.
Specific Humidity or Moisture Content = The weight of water vapor in grains or pounds of moisture per pound of dry air.
Definition of Terms
Enthalpy = A thermal property indicating the quantity of heat in the air above an arbitrary datum, in Btu per pound of dry air. The datum for dry air is 0˚F, and for the moisture content, 32˚F water.
Specific Volume = The cubic feet of the mixture per pound of dry air.
Sensible Heat Factor = The ratio of sensible to total heat.
Definition of Terms
Pounds of Dry Air = The basis for all psychrometric calculations. Remains constant during all psychrometric processes.
◦ Note: The dry-bulb, wet-bulb, and dewpoint temperatures and the relative humidity are so related that, if two properties are known, all other properties shown may then be determined. When air is saturated, dry-bulb, wet-bulb, and dewpoint temperatures are all equal.
Definition of Terms
CARRIER PSYCHROMETRIC CHART
PSYCHROMETRIC CHART Skeleton
Edited
Psychrometry of Air-Conditioning Processes
Cooling coilMixing Box
12
2
Blower
In RA
m
1
2
DB1
DB2
WB2
WB1
h2
h1
RC = m(h1 – h2)
FIGURE 1
RETURN AIR
SUPPLY AIR
Cooling coil
Mixing Box
1
2
3
1
2
DB1
DB2
WB2
WB1
h2
h1
DB3
WB3
h3
3
w1
w2
w3
FIGURE 2
Equations that can be derived based on the Psychrometric Chart (Figure 2):
Sample Problem Example 15.1 (p. 475)
30 m3 /min of a stream of moist air at 15C DBT and 13C WBT is mixed with 12 m3 /min of a second stream at 25C DBT and 18C WBT. Barometric Pressure is one standard atmosphere. Determine the dry bulb and wet bulb temperatures of the resulting mixture.
Required: WB3 and DB3
SOLUTION
Cooling coil
Mixing Box
15C DB13C WB30 m3 /min
25C DB18C WB12 m3 /min
1
2
25C
15C
13C
18CV1= 0.86
V2= 0.825
Then,
Example 15.2 (p.476)
A stream of moist air at 2C DBT and 80% RH mixes with another stream of moist air at 30C DBT and 10C DPT in the ratio by mass of one part of the first to tow parts of the second. Calculate the temperature and specific humidity of the air after mixing.
BYPASS FACTOR
SOLUTION
Cooling coil
Mixing Box
15C DB13C WB30 m3 /min
25C DB18C WB12 m3 /min
1
2
25C
15C
13C
18C
Definition
The following figure shows the process that the moist air undergoes while flowing over a surface .
The air enters at 1 and leaves at 2 when the surface is maintained at S.
The state of the contacted air is that of saturated air at the temperature of the surface. There is thus the equivalent of perfect contact of a definite portion of the air particles with the surface.
The uncontacted air remains at the entering state.
The end state of the air is the same as that produced by the complex entering of the contacted and uncontacted particles.
hs
h2
h1
S w3
1
t1t2ts
2 w2
w1
t
w
Thus one can define a bypass factor of the apparatus representing the fraction of “uncontacted air” in terms of 1,2,and S as:
Conversely, one can define a contact factor (1-X) representing a fraction of the contacted air.
Simple Air conditioning system and state and mass rate of
supply air
RETURN AIR AT t1, w1
SUPPLY AIR AT ts, ws
Room at t1, w1
AHU
Cooling coil
Mixing Box
Formulas:
Sensible Heat Balance:
Latent Heat Balance:
Base on Enthalpy:
Sample problems Find the resulting dry bulb temperature of
the mixture given the following conditions:
OUTDOOR AIR RETURN AIRm= 2 kg/s m= 3kg/s35OCDB 24OCDB25OCWB 50% RH
SOLUTION
2kg/s35OCDB25OCWB
Figure
m= 3kg/s50% RH24OCDB
WBm
Wm
35oC28.4oC
24oC
25oC
OA
50%RM
MIX
Equation:
Cooling Load Calculations
By: Bautista, Norman George V.
Cooling Load Results from any one single source of heat. Summation of the heat which evolves from
different sources.
Common Sources of Heat
Heat that leaks from the outside space. Heat that enters the space by radiation. Heat that enters the space through doors
and windows. Heat given of by a warm product. Heat given by people. Heat given by equipments.
Refrigerated Load In cases where the refrigerated load was
calculated on a 24 hour basis the load may be determined directly in KW.
Equation:
Where Q = required equipment capacity, KWRT= running time, hrs.qt = total cooling load, KW
Wall Gain Load A measure of the heat flow rate by
conduction through walls of the refrigerated space from outside to the inside.
Also called as the “wall leakage load”
Wall Gain Load Equation:
Where: Q = rate of heat transferred, W.A = outside surface area of wall, m2
U = overall coefficient of heat transmission (W/ m2 – K), “Table 10-2”
TD = temperature difference across the wall
))()(( TDUAQ
Example 1 Determine the heat flow rate in watts
through a wall 3m by 6m. If the U factor for the wall is 0.37 W/ m2 – K and the temperature on one side of the wall is 4 °C while the temperature on the other side is 35 °C.
Cont... Given: U = 0.37 W/ m2-K
T1 = 4 °C
T2 = 35 °C
Solution :Total Wall Area = 3m x 6m = 18 m2 TD = T2 - T1 = 35-4 = 31 K
Q = AUTD = 18 m2 (0.37 W/ m2-K)(31 K) Q = 206 W or 0.206 kW
Thermal Conductivity, k Is the property of a material describing its
ability to conduct heat. W/(K·m) Can be found on Table 10-1. Equation:
Where: C = thermal conductance k = thermal conductivity
x = thickness
xkC /
Example 2 Determine the thermal conductance for a
125 mm thickness of polyurathane.
Solution: From table 10-1, k = 0.125 W/m-KC = 0.025 / 0.125C = 0. 2 W/m2 – K
Thermal Resistance Thermal resistance is a measure of a
material's ability to resist heat transfer. The more a material is able to block heat transfer through its surface, the greater its thermal resistance.
Equation:
kxkcUR //1/1/1
Cont.. When a wall is constructed of several layers
of different materials,
Where: = convection coefficient of inside wall.
= convection coefficient of outside wall. 1/fi
Example 3 Assuming a wind velocity of 3.35 m/s,
calculate the value of U for a wall constructed of 200 mm sand aggregate building blocks insulated with 75 mm of polyurethane and finished on the inside with 13 mm of cement plaster.
Cont.. Solution: From table 10 – 1:200 mm sand aggregate block, C = 5.11Polyurethane, k = 0.025Cement plaster, k = 0.72fi = 9.37 fo = 22.7
Cont..
1/U = 3.37U = 0.297 W/ m2 - K
37.9/172.0/013.0025.0/075.011.5/17.22/1/1 U
Example 4 A walk-in cooler 5m x 7m x 3m high is
located in the southeast corner of a store building in an area where the outdoor design DB temperatures in summer and winter are 35 °C and -6 °C. The south and east walls of the cooler are adjacent to and a part of the south and east walls of the store building. The store has a 4 m ceiling so that there is a 1 m clearance between the top of the cooler and the ceiling of the store.
The store is air conditioned and the temperature inside the store is maintained at approximately 26 °C. The inside design temperature for the cooler is 2 °C. The north and west (inside) walls, floor and ceiling are insulated with 75 mm of closed-cell (smooth surface) polystyrene, and the south and east walls are insulated with 100 mm of closed-cell polystyrene. Determine the wall gain load in kilowatts.
7 m
5 m
Outdoor Design Temp, 35 °C
150 mm clay tile
150 mm polystyrene
Cooler 2 °C 3m ceiling
Partitions insulated with
75 mm polystyrene
Inside Temperature 26 °C
Ceiling 4m
Solution
K (polystyrene) = 0.029 W/m-KU (north and west) = 0.346 W/m2 - K U (south and east) = 0.267 W/m2 – K
N, W, ceiling = (71)(0.346)(26-2) = 589.6 WS = (15)(0.267)(37-2) = 140.2 WE = (21)(0.267)(38-2) = 201.9 WFloor = (35)(0.346)(25-2) = 215 WTotal wall gain load = 1147 W or 1.147 kW
Air Change Load A warm outside air enters the space to
replace the more dense cold air which is lost from the refrigerated space through the open door.
Equation:
Where: Q = air change load, kWm = mass of air entering, kgho = enthalpy of outside airhi = enthalpy of inside air
)( hihomQ
Example 5 The rate of air infiltration into a refrigerated
space is 8 L/s. If the inside of the cooler is maintained at 2 °C and the outside dry bulb temperature and humidity are 30 °C and 50% respectively, determine the air change load in kilowatts.
Cont... Solution: By interpolation in Table 10-6A, the enthalpy
change factor is 0.0598 kJ/L
Air change load = (8L/s)(0.0598 kJ/L)Air change load = 0.478 kW
Product Load When a product enters a storage space at a
temperature above the temperature of the space, the product will give off heat to the space until it cools to the space temperature.
Cont... Equation:
Where: Q = quantity of heat, kJ/kgm = mass of the product, kgCp = specific heat Above freezing Temp.TD = temperature change in the product
temperature, K
)(TDmCpQ
Cont...
This equation is used to know the product load at a specific time.
Example 6 Thirty-five hundred kilograms of fresh beef
enter a chilling cooler at 39°C and are chilled to 7°C each day. Compute the product load in kilojoules.
Solution: Q= mCp(TD) = (3500)(3.14)(39-7)
Q = 351,680 kJ
Example 7 Determine the product load in kilowatts
assuming that the beef described in the previous example is chilled in 20h.
Solution:
Cont...
Q = 4.88 kW
Respiration Heat Fruits and vegetables are still alive after
harvesting and continue to undergo changes while in storage.
Respiration is a process during which oxygen from the air combines with the carbohydrates in the plant tissue and results in the release of carbon dioxide and heat.
Cont... Equation:
Where: m = mass of product, kg rh = respiration rate, W/kg
Example 8 A storage cooler 6m x 4m x 3.4m high is
insulated with 100 mm of glass fiberboard. Overall wall thickness is approximately 200 mm. The outside temperature is 30°C and usage is average. Twelve hundred and fifty kilograms of wet mixed vegetables are cooled, 25°C to the storage temperature of 5°C each day. Compute the required equipment capacity based on a 16-h/day.
Solution
Outside surface area = 2(6x4)+2(6x3.4)+2(4x3.4) = 116 m2 Inside Volume = 5.6 x 3.6 x 3 m = 60.5 m3 Glass fiberboard, k = 0.036 W/m – K100 mm thick insulation, k = 0.035 W/m-kU = 0.31 W/m2 -K
Cont... By interpolation, (table 10-7) Infiltration Rate = 8.13 L/s
Assume a 50 % RH, enthalpy change factor = 0.0536 kJ/LCp of wet mixed veg = 3.77 kJ/kg-Lrh = 0.097 W/kg
Wall gain load = (116)(0.31)(30-5) = 899 W or 0.899 kW
Air change load = (8.13)(0.0536) = 0.43kW
Product cooling load =Respiration load = (1250)(0.097) = 121.5
W or 0.121 kWSum of heat loads = 1.729 kWSafety factor (10%) = 0.173 kWTotal cooling load = 1.902 kWRequired equipment capacity =
