Topic4-2 Evaporators
Evaporators is the interface between process and refrigeration system.
1
Tem
per
atu
re, K
Entropy, kJ/kgK
1
2
3
4
Refrigerant → phase change (boiling) in shell, tube, plate
• Indirect cooling with external fluid
• Direct cooling → Plate freezer→ gas (air)
→ liquid (water, brine, antifreeze)
Process
1. Types of Evaporators
Air Coils
• Finned coil – boiling in coil
Liquid Chillers
• Shell and tube – boiling in shell
• Shell and tube – boiling in tube
• Plate-type – compactness
– less refrigerant charge
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PART I. EVAPORATIVE PERFORMANCE
3
Engineers of manufacturers decide on: → Refrigerant circuiting→ Tube arrangement→ Other important details
Designers consider: → Performance→ Cost of materials→ Ease of manufacturing
Users understand: → Basic performance as a heat exchanger→ How to proper select from catalogs→ How to install, operate, and maintain properly
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2. Evaporators Heat Transfer q = ho Ao(tf - to)q =(k/x) Am(to- ti)q = hi Ai (ti - tr)
q = (tf - tr)/Rtotal
q = Uo Ao(tf - tr)
q = Ui Ai (tf - tr)
rf
iimoo
tt Ah
q
k A
qx
Ah
q−=++
Uo, Ui = overall heat-transfer coefficient based on Ao, Ai , respectively. W/m2 K
tr tf
iimooiioo Ahk A
x
Ah AU AU
1111++==
Total resistances to heat transfer between refrigerant & fluid
5
Average mean area, Am q = (tf - tr)/Rtotal
Rcyl = ln(ro/ri)/(2Lk) x/(Amk)
x/(Amk) = (ro - ri)/(2Lk(ro + ri)/2)
= [2(ro - ri)/(ro + ri)]/(2Lk)
For thin tube x = ro - r1;
ln(ro/ri) 2(ro - ri)/(ro + ri) (< 0.1%)tr tf
Example1 What is the U-value of an evaporator formed of steel pipe when the air-side heat-transfer coefficient is 60 W/m2·°C. and the heat-transfer coefficient on the refrigerant side is 1200 W/m2·°C. The pipe has an outside diameter of 26.7 mm and an inside diameter of 20.9 mm. The thermal conductivity of steel is 45 W/m·°C
1. Data: evaporatorAir flow outside ho = 60 W/m2·°C Refrigerant boiling in pipehi = 1200 W/m2·°C Steel pipe Do = 26.7 mmDi = 20.9 mmk = 45 W/m·°CCompute for U = ?
4.Properties & data are all given
2.Assumption&SchematicNo fouling in and out pipeCalaulate Ui & Uo
3.Methods & EquationsThermal resistance networkUi = 1/(Rtotal Ai) and Uo= 1/(RtotalAo) Rtotal = Ro + Rpipe + Ri
Ro= 1/(hoAo) = 1/(hoDoL) Rcyl= x/(kAm) = (Do - Di)/(k(Do + Di)L)Ri = 1/(hiAi) = 1/(hiDiL)
6. Analysis & checkRo/Rtotal *100% = 93.6%Rcyl/Rtotal *100% = 0.4%Ri/Rtotal *100% = 6% 6
5.Calculations; assume L = 1 mRo = 1/(600.0267) = 0.1987 °C/WRcyl= (26.7–20.9)/(45(26.7+20.9)) = 0.00086 °C/WRi = 1/(12000.0209) = 0.0127Rtotal = 0.1987+0.00086+ 0.0127 = 0.2122 °C/WUi = 1/(0.21220.0209) = 71.8 W/m2·°C per m lengthUo= 1/(0.21220.0267) = 56.2 W/m2·°C per m length
Thermal resistance of thin steel pipe can be neglected.
Thermal resistance of thin metal pipe/tube, Rcyl
q = (tf - tr)/Rtotal
q = (tf - to)/Ro
q = (to- ti)/Rcyl
q = (ti - tr)/Ritf
tr
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→ Ro/Rtotal = to/(tf - tr)→ Rcyl/Rtotal = tcyl/(tf - tr)→ Ri/Rtotal = ti/(tf - tr)
Example1 → 0.936→ 0.004→ 0.06
(tf - tr)to
ti
Thermal resistance of thin steel pipe can be neglected.Temperature difference between outside and inside surface of the tube (tcyl) is very small compared to the temperature difference between
fluid and refrigerant (tf - tr).The tube temperature can be assume to be as to = ti = ts.
3. Liquid in tubes; heat transfer
Nu = C Ren Prm
Nu = hD/k
Re = VD/
Pr = cp/k
Fluid flowing inside tube:for turbulent flow
where:h = convection coefficient, W/m2 KD = ID of tube, mk = thermal conductivity of fluid, W/m KV = mean velocity of fluid, m/s = density of fluid, kg/m3
= viscosity of fluid, Pascp = specific heat of fluid, J/kg K
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Dittus–Boelter equation:
Colburn equation: Nu = 0.023 Re0.8 Pr1/3
Nu = 0.023 Re0.8 Prn
N = 0.4 for heating
N = 0.3 for cooling
Example 2 Compute the heat transfer coefficient for water flow inside the tubes (8 mm ID) of an evaporator if the water temperature is 10 C and its velocity is 2 m/s.
1. Data: evaporatorWater flow inside the tubes10 C = twater
0.008 m = D2 m/s = VCompute heat transfer coefficient h = ?
4.Properties of water at 10 Ck = 0.573 W/m K = 1000 kg/m3
= 0.00131 Pascp = 4190 J/kg K
2.Assumption & SchematicWater flowing inside tube steadilySmooth tubeTurbulent flow (Re>10,000)
3.Methods & Equationsh = Nu k/DCooling: Nu = 0.023 Re0.8 Pr0.3
Re = VD/Pr = cp/k
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6. Analysis & checkRe > 10,000 – turbulentIf use Nu = 0.023 Re0.8 Pr1/3
Nu = 91 and h = 6507 W/m2·°C (>8%)UA is larger, then chose larger size or more water flow rate.
5.Calculations; Pr = 4190(0.00131)/0.573 = 9.58Re = 2(0.008)1000/0.00131 = 12,214Nu = 0.023(12214)0.8(9.58)0.3 = 84.3h = 84.3(0.573)/0.008 = 6035 W/m2·°C
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4. Liquid in tubes; pressure drop
Pressure drop
• Straight tube, 50-80%
• U-bends
• Heads of HX
• Entrance & exit
2
2V
D
Lfp =
where:p = pressure drop, Paf = friction factor, dimensionless f(Re,/D) = roughness of inside surface of tube , mL = length of tube, m
Experimental or catalog data:p = (flow rate)2
10
2
1
212
=
m
mpp
m = mass rate of flow, kg/s
5. Liquid in shell; heat transfer
hD/k = (terms controlled by geometry) (Re0.6 )(Pr0.3)( /w)0.14
Flow across tube bundle, correlation by Emerson(1963):
where: = viscosity of fluid at bulk temperature, Pas w = viscosity of fluid at tube-wall temperature, PasG = V = m/Acharacteristic = mass velocity, (kg/s)/m2
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Evaporators, flow across tube bundle many times
- not short-circuit from inlet to outlet
- complex flow pattern
Re = GD/ = VD/
6. Liquid in shell; pressure drop
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Pressure drop – difficult to predict
Catalog data: by experiment
p (flow rate)1.9
7. Extended surface (Fins)
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Mostly Ro =1/hoAo is > 80% of Rtot
Ex1 steel pipe; Ro/Ri = 15 Air-side Ro = Controlling resistanceTo improve U-value (for UA =1, Ro = 0.8 , Ri = 0.2)if hiAi doubled, Ri = 0.1; UA=1.11if hoAo doubled, Ro = 0.4; UA=1.67
iiooiioo Ah Ah AU AU
1111+==
To improve hoAo or decrease Ro
Properties of Air compared to those of liquid such as water:
Heat-transfer coeff. 1/10 – 1/20
To decrease Ro = 1/hoAo
Area A is usually increase by using extended surface or Fins
Fin Equation
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Fins: L = length, m (x-direction)
2y = thickness, mZ = depth, mk = conductivity of metal, W/m Khf = air-side coefficient , W/m2 Kt = temperature of fin, Cta = temperature of air, Ctb = temperature base of fin, C
heat flow entering at 1 + heat transfer from air = heat transfer out at 2
Solve for t(x), steady, heat balance at element of thickness x, m
kyZ(dt/dx)1 + Z dx hf (ta – t) = kyZ(dt/dx)2
dxdx
tddx
dx
dt
dx
d
dx
dt
dx
dtky
2
2
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=
=
−
ML
xLM
tt
tt
ba
b
cosh
)(cosh −=
−
−
ky
hM f=
BCs; t(0) = tb
and dt(L)/dx = 0
( )
ky
tth
dx
td af −=
2
2
Fin efficiency
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reri
Graph of & (re - ri )M of an annular fin for different (re /ri ).
Fin
eff
icie
ncy
(re - ri )sqrt(hf/(ky) = (re - ri )M
Determining fin effectiveness of a rectangular plate fin (a) by treating it
as an (b) annular fin of the same area. = qactual/(qmax if fin were at tb)
Example 3 What is the fin efficiency of a rectangular plate fin made of aluminum 0.3 mm thick mounted an a 16-mm-OD tube if the vertical tube spacing is 50 mmand the horizontal spacing is 40 mm? The air-side heat transfer coefficient is 65 W/m2 K, and the conductivity of aluminum is 202 W/mK.
1. Data: rectangular plate fin Aluminum fin0.3 mm thick = 2y16-mm-OD tube = ri
50 mm x 40 mm = fin size65 W/m2 K = hf
202 W/mK = kCompute fin efficiency = ?
4.Properties & Data known
2.Assumption & SchmaticSteady 1-D heat transfer through fin
3.Methods & EquationsGraph of & (re - ri )MM = sqrt(hf /ky)rectangular fin (same area) annular finre --> 50 x 40 = re
2
Use (re - ri )M find for re /ri
5.CalculationM = sqrt(65/202(0.00015)) = 46.3 1/mre = sqrt((50*40/) = 25.2 mm(re - ri )M = (0.0252-0.008)46.3 = 0.8(re/ri ) = (0.0252/0.008) = 3.15(0.8,3.15) = 0.72
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6. Analysis & check Commercial refrigerating
coils; 30-70%qfin = qmax = hfAe(tb-ta)Extended area Ae
Heat transfer by fins & Air-side thermal resistant
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Prime area Ap : tube area between fin → at base temperature, = 1
Extended area Ae : fin area → < 1qfin = qmax = hfAe(tb-ta)qunfin = hfAp(tb-ta)qtotal,fin = qunfin + qfin
qtotal,fin = hf(Ap+ Ae)(tb-ta)
( ) iimepfiioo Ahk A
x
AA h AU AU
1111++
+==
Ae = 2(re2 -ri
2)
Ap = 2riS
S
Commercial refrigerating coils; 0.3-0.7From heat transfer : > 0.6 - economicMost used in other applications: > 0.9
Fins are formed from flat metal platesthat are then punched and the tubes inserted in the holes.Cu tube + Al fins - halocarbonAl + Al fins - halocarbon or ammoniaSteel tube + steel fin - all
Increase of length Lhf
k(kAl = 4ksteel)Thickness 2y
Effect on DecreaseDecrease
Increase
Increase
Effect on UIncreaseDecrease
Increase
Increase
CostIncreaseIncrease of Fan powerIncreasefor AlIncrease
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DX evaporators coils: → Decrease air-side resistance Ro by hf or Ae (Afin)
= qtotal,fin/qno fin = (kp/hfAc)
•High k → aluminum, copper, iron.
•High p/Ac → slender pin fins.
•Low h → Place fins on air side.
Fin effectiveness
Finned tube – increase conduction on refrigerant side
Finned tube – not used in food plant- difficult to clean- hygienic purposes
For ammonia systems, aluminum tubes can be provided with internal and/or external integral fins, and even for steel tubes the possibility of external integral fins exists.
8. Gas flowing over finned tubes; heat transfer
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Prediction air-side heat-transfer coefficient (hf) is complicated: f(geometric factor)-fin spacing-tube diameter-tube spacing-number of rows of tube deep
A rough estimate derived from data in ARI standard (1972):
hf = 38 V 0.5
V = face velocity of air, m/s
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9. Gas flowing over finned tubes; pressure drop
Pressure drop, by Rich (1974), of air flowing through a dry finned coil per number of rows of tube deep.
p coil geometry
p V 1.56
Typical commercial plate-fin coils.
315-551 fins per m for air conditioning coils.118 or 158 fins per m for the air temperature is below freezing.
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10. Refrigerant Boiling in the shell
Pool boiling of water:
Difficult to predict boiling coefficient accurately – complexities of mechanism
Boiling in shell – different from boiling in tube
Tests: immerse heated wire in water @atm pressure
Nucleate boiling
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Heat transfer coefficient for Boiling in the shell
Pool boiling equation:
Nucleate boiling
q/A = C t3 to 4 q = rate of heat transfer, WD = heat transfer area, m2
C = constantcp = difference in temperature between metal surface and boiling fluid, K
q/At = hr = C t2 to 3
hr = boiling coefficient, W/m2 K
Hoffmann (1957)
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Heat transfer coefficient for Boiling in tube
Heat-transfer coefficient changes as Fraction of vapor, velocity, flow pattern.
Enter : Low vapor fraction vapor fraction → hr Intensifying the agitation
R22 boiling inside tubes @te
Local hmax 2000 W/m2·°CAverage h 400 – 1600 W/m2·°C
vapor fraction > 0.8 -------> hr
te --> high vapor density → hr Annular flow→Mist →Mixture
R717 boiling inside tubes Local hmax 13000 W/m2·°CAverage h 4000 – 8000 W/m2·°C
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12. Evaporator performance: DX inner-fin, liquid chiller
loading/heat absorbed at evaporator → hr
Capacity&te - straight line if U = const. Curved upward line - U as loading.
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13. Temperature Difference between Entering Fluid and Refrigerant
qe = mc(tfi - tfo) = UA(LMTD)
Manufacturer catalog present Factor for each evaporator and each fluid flow rate
LMTD = (tfi - tfo)/ln[(tfi - tr)/(tfo - tr)]
Divide byUA(tfi - tfo);
ln[(ti - tr)/(to - tr)] = UA/mc (ti - tr)/(to - tr) = eUA/mc
(tfi - tr) e-UA/mc =(tfo - tr)
qe= mc[1- e-UA/mc ](tfi - tr)
(tfi - tr)=ti
tfi =tfluid, in
tfo =tfluid, out
tfi
tfo
→ qe (tfi - tr)
tave = (tf,av - tr) used as LMTDIf t0/ti > 0.33 for diff<10%
(tfo - tr)=to
qe/UA(tfi - tfo)= mc/UA = 1/ln[(tfi - tr)/(tfo - tr)]
= -[(tfi - tfo)-(tfi - tr)] =-[qe/mc-(tfi - tr)]
= Factor (tfi - tr)
tr = const.
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14. Methods of supplying refrigerant to Evaporators
1. Direct-expansion evaporators
3. Liquid-recirculation2. Flooded evaporator
Superheat-controlled valveThermo-valveThermostatic valveTXV
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14.1 Direct-expansion evaporators
DX evaporators in air-conditioning applications:-- fed by expansion valve that regulates liquid flow – leaving in some superheat
Ex. from 4 to 7 C
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1
sensing bulb detects a higher-than-set point superheat at the evaporatorOutlet (1), the valve opens further – increase mass flow rate until degree of superheat reduced
Advantage – low costDisadvantage – limited use by precise amount of refrigerant- limited in low-temperature applications and for ammonia
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14.2 Flooded evaporator
Flooded evaporators in low-temperature refrigeration:- natural convection: pstatic in liquid leg > pstatic in evaporator tube- liquid is separated and naturally flow back to evaporator
Advantage of wetting all interior surface of evaporator 1. high coefficient heat-transfer2. Less problem in parallel distributing refrigerant3. sat. vapor flows on to compressor -- Tdischarge
Disadvantage 1.need more refrigerant2.high maintenance cost3.Periodically removed oil from receiver or surge drum
Component:Surge drumLevel-control valveLiquid leg
For low-temp.-min. static head-circuits rise on incline
15 – 25 cm
Small system: 3-4 evaporators
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14.3 Liquid-recirculationLiquid recirculation in low-temperature refrigeration: -- liquid-recirculation or liquid-overfeed evaporator
- some liquid boils and the remainder floods out - liquid is separated and pumped to evaporator
Advantage of wetting all interior surface of evaporator 1. high coefficient heat-transfer2. assurance of overfeeding under varying loads3. sat. vapor flows on to compressor -- Tdischarge
4. All oil is accumulates at one location-separating vessel
Disadvantage high first cost: pump, vessel
Makeup liquid
Large system: more than 4 evaporators
Component:Separating vessel/Low Temp. ReceiverLevel-control valvePump
15. Optimum Evaporating Temperatures
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The size of the evaporator selected is dictated by qe = Factor(tfi - tr)Factor = mc[1- e-UA/mc ]An increase of evaporator area A → increase Factor → increase first cost→ lower (tfi - tr) or high tr for const. tfi → lower comp. power → longer life facilityThe compressor energy costs which occur throughout the life of the systemare reflected back to a present value using an applicable interest rate.
Optimum evaporator area for minimum total of the first cost of the evaporator and the present worth of the lifetime compressor energy cost.
PART II. AIR COILS
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16. Construction of Air Coils
The number of air-cooling coils > the number of liquid-chilling evaporators
Component:TubesTube sheetsDrain panFin
Tubes are pipes that enclose the refrigerant.Halocarbon: copper, carbon steel, aluminum, and stainless steelNH3: carbon steel, aluminum, and stainless steelCommon size – steel-tube coils ¾, 7/8 and 1 inch. For NH3
Smaller halocarbon coils – ½-in copper tubes are sometime used.
Tube sheets are heavy plates supports the tubes byhaving holes through which the tubes pass byin-line or staggered pattern.A coil with a staggered-fin pattern enjoysslightly improved heat transfer, but at the expense of a slight increase in air pressure drop.
Drain pan collects some water vapor condensed from the air and drained to some convenient destination..
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Fins Coil
Fins are extended surface to improve high heat transfer (hA)-side but low hCu tube + Al fins for halocarbon air-cooling coils (1270 kg for 536 coils)
Al tube + Al fins for halocarbon or ammonia air-cooling coils (1090 kg for 536 coils)
carbon steel tube + carbon steel fin for air-cooling coils using ammonia,halocarbons, antifreezes or water in the tubes (2130 kg for 536 coils)
stainless steel tube + stainless steel fin when special cleaning provisions are required on the air sideStainless steel is usually used only for extremely low temperature, where there is a corrosive atmosphere, or whenever periodic cleaning is necessary.kAl => 4ksteel but CostAl => 5Coststeel for comparable size.
Coils will become frosted, determines to a large extent the spacing of fins. Thin aluminum fins spacing may be 470 fins per m (12 fins per inch, FPI) for air conditioning coils.Industrial coils are built with 118 or 158 fins per m (3 or 4 FPI) for the air temperature is below freezing.
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Bonding Fins to Tubes
Fin must form a good bond to the tube for minimum heat transfer resistance.A steel tube/steel fin coil will be galvanized: entire coil is dipped in molten zinc.To protect from corrosion and give an effective bond between tube and fin.For non-galvanized coils the tubes are usually expanded against the collar of the fin to yield a tight fit.
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Circuiting of the Coil
Direct-expansion coils, refrigerant flow through the circuits is downward, Flooded coils to function properly, refrigerant flow must be upward.
The coil designer chooses the length of the circuit such that the refrigerant when flowing with appropriate velocityreceives enough heat in passing through the circuit to vaporize the Desired fraction of refrigerant.
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Liquid Circulation Coils
Forced liquid recirculation coils : bottom feed and upflowor top feed and downflow.
One refrigerant circuit in the coils of consists of six passes back and forth through the coil.
There are a number of parallel passes, the upper circuits in a coil are prone to receive an inadequate flow of liquid. In order to strive toward an equal distribution of refrigerant flow, orifices are placed at the entrance of each circuit.
Orifices are thin metal discs with a hole.Oil which accumulates in the coil during refrigeration operation may more easily flow out of the coil during hot-gas defrost.
Often the diameter of the orifices is greaterfor the upper circuits to achieve uniform distribution of refrigerant.
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17. Definition of Terms Applied to Air Coil
Circuit: the route of the refrigerant from the time it enters a tube until it leaves.Coil depth: defines the number of rows of tubes the air passes from the entrance to the exit of the coil.Face area: the cross-sectional area through which the air passes as it enters the coil, or the finned length of the coil multiplied by the finned width.Face velocity: the volume rate of air flow divided by the face area.Header: A common pipe, from which all circuits are supplied refrigerant (liquid or supply header) or a common pipe which gathers refrigerant leaving all the tube circuits (return or suction header).Pass: The flow of refrigerant through one straight section of the circuit.Prime surface: the air-side area of tubes that is in contact with air.Return bends: or U-bends are short sections of curved pipe to direct the refrigerant leaving one pass to the entrance of the next pass.Secondary surface: the surface area of both sides of the fins in contact with air.Temperature difference: two temperature differences may be encountered1. TD - temperature difference between incoming air and the refrigerant (tfi - tr)2. T - change in air temperature through the coil (tfi - tfo)
qe= mcT = FactorTD
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W = humidity ratio, (kg of water vapor /kg dry air)
t or tdb or DB = dry-bulb temperature
1. Tdb line
Saturation line
2. Relative humidity
5. Twb line
4. Specific volume
6. Enthalpy line
DP
RH
WB
H
SpV
W
3. DP line
7. Humidity ratio line
Known 2 properties → find 5 properties
18. Properties of Air – The Psychrometric Chart
Normal Temperature 0°C to 50°CSI units at 101.325 kPa
For engineering work:Enthalpy line = Twb line 25°C
50%
14°C
0.858 m3/kg dry air
18°C
51.5 kJ/kg dry air
0.01 kg/kg dry air
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Wet-bulb temperature
Air
t1
W1
1 2
t2
W2
A: tdb = 24 C, = 50%
pt = 101 kPa
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A: tdb = 24 C, = 50%
W = humidity ratio, (kg of water vapor /kg dry air)
Humidity ratio
A
tdb= 24 Ctdp= 12.8 C
W = (pV/RT)s/(pV/RT)a
W = (Ra/Rs)ps/pa
W = 0.622 ps/(pt - ps)
Wat
er-v
apo
r p
ress
ure
(p
s), k
Pa
hu
mid
ity
rati
o (
W),
kg
/kg
W and ps is almost linearly proportion.
WA = 0.093 kg/kg
0.622 = Ra/Rs = 287/462
pt = patm = 101 kPapa = pt - ps
= 101-1.485 = 99.52 kPa
Calculation: W(psat,ps) →W= 0.622*1.485/99.51 = 0.093 kg/kg
Known tdb, %RH Calculation W by 1. psat(t)2. ps = psat
3. pa = pt - ps
4. W(ps,pa)
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h = cpt + Whg kJ/kg dry air
Enthalpy
A
cp = 1.0 kJ/kgK (dry air)
Water-vapor in the air-vapor mixture is likely to be superheated, but h hg.
Calculation: hg (t) →hg = 1.744t +2503 = 2545 kJ/kg
Calculation: h(t,W) →h = 1.0*24 + 0.093*2545= 47.6 kJ/kg dry air
Known tdb, %RH Calculation h by 1. psat(t)2. ps = psat
3. pa = pt - ps
4. W(ps,pa)5. hg (t)6. h(t,W)
The Psychrometric Chart Normal Temperature 0°C to 50°C SI units
Known 2 properties → find 5 propertiesEx. Saturated air at 20°C, find other properties.
2. Relative humidity
5. Twb line
4. Specific volume
6. Enthalpy line
3. DP line
7. Humidity ratio line
100%
0.851 m3/kg dry air
20°C
20°C
57 kJ/kg dry air
14.8 g/kg dry air
41
The Psychrometric Chart Low Temperature -30°C to 15°C SI units
Known 2 properties → find 5 propertiesEx. Dry bulb of -10°C and a wet-bulb of -12°C, find other properties.
3. Relative humidity
5. Specific volume
6. Enthalpy line
4. DP line
7. Humidity ratio line
40%
0.746 m3/kg dry air
-19°C
-9 kJ/kg dry air
0.6 g/kg dry airtwb=-12°C
42
The Psychrometric Chart Normal Temperature 20°F to 105°F I-P units
104°F86°F68°F50°F32°F14°F-4°F
-22°F-40°F
40°C30°C20°C10°C
0°C-10°C-20°C-30°C-40°C
43
The Psychrometric Chart Low Temperature -20°F to 50°F I-P units
Known 2 properties → find 5 propertiesEx. Dry bulb of 14°F and a wet-bulb of 10.4°F, find other properties.
3. Relative humidity
5. Specific volume
6. Enthalpy line
4. DP line
7. Humidity ratio line
40%
11.94 ft3/lb dry air
-4°F
3.9 lb/lb dry air
0.00066 lb/lb dry airtwb=10.4°F
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Example 4 An air-flow rate of 2.2 m3/s (4660 cfm) enters an evaporator coil at 4°C and a relative humidity of 90%. Air leaves the coil at 0.5°C and a relative humidity of 98%.(a) What is the refrigeration capacity of the coil?(b) What is the rate of water removal from the air?
1. Data: Air flow pass evaporator coil2.2 m3/s = Va
4C = ta,i , 90% RH = i
0.5C = ta,o , 98% RH = o
Compute (a) qcoil, (b) mw
4.Properties
The Psychrometric Chart SI unit
vi = 0.791 m3/kgWi = 0.0045 kg/kghi = 15.3 kg/kgWo = 0.0039 kg/kgho = 10.2 kg/kg
2.Assumption & SchematicAir flowing steadily at atm.
3.Methods & Equationsqcoil = m(hi – ho) (W) mw = m(Wi – Wo) (kg/s) m = V/vi (kg/s) - (h W v from chart)
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6. Analysis & checkRefrigeration coil load 4.05 tonsWith water drain 1.67 g/s.
5.Calculations; m = 2.2/0.791 = 2.78 kg/sqcoil = 2.78(10.2-15.3) = 14.18 kW (4.05 tons)
mw = 2.78(4.5-3.9) = 1.67 g/s
The Psychrometric Chart Low Temperature -30°C to 15°C SI units
Air entering coil: 4C , 90% RH Air leaving coil: 0.5C, 98% RH
0.791m3/kg dry air
entering
15.3 kJ/kg dry air
4.5g/kg dry air
leaving
10.2 kJ/kg dry air
3.9 g/kg dry airLatent heat from air
Sensible heat from air to coil
46
19. Dew-Point Temperature
Example 5 The condition of air in a meat-processing room is 12°C and the relative humidity is 70%. The surface temperature of the support metal for the cooling coils is 4°C. What is the potential operating problem that exists?
1. Data: Air flow pass metal12C = ta,i , 70% RH = i
4C = ts
Is there any problem from this?
2.Assumption & SchematicAir is cooled with no change in its moisture content until reaches the saturation condition.
3.Methods tdp = 6.5°C > ts
some moisture will condense on these surfaces.
The drip of moisture on the food products is prohibited. 47
48
Adiabatic saturation and thermodynamic wet-bulb temperature
An adiabatic saturator: air flows through a spray water
Airt1, W1 1
iti, Wi
mmake
2
Mass balance of water: maW1 + mmake + mw = maWi + mw
→ ti = twb1→ t1 decreasing to t2
ti
ma
mw
mmake = ma(Wi – W1)
Energy balance: mah1 + mmakehf + mwhf@ti = mahi + mwhf
mah1 + ma(Wi – W1)hf = mahi
h1 + (Wi – W1)hf = hi
12
i
Wet-bulb-line
Air leaves the spray chamber in equilibrium with the water at ti → saturated airControl volume the saturator,
Straight-line law
49
i
o
s
Heat and mass transfer air and wetted-surface
Straight-line law: the path on a psychrometric chart drives toward the saturation line at the temperature of the wetted surface.
Air is in equilibrium with the water, saturated at ts.
Because (Wo – Wi)hf is very small
→ assume hi ho
hi + (Wo – Wi)hf@ts = ho
Line of constant enthalpy
almost the same line with
Line of constant Wet-bulb-temperature
20. Straight-Line-Law
Evaporative Cooling
Application of Straight-Line-Law
50
The straight-line law is well verified by fundamental laws of heat and mass transfer.(ti - tr) is driving force for sensible heat transfer.W and ps is almost linearly proportion.(Wi - Wo) is driving force for mass transfer of moisture content.
a) cooling and dehumidifying coil
c) heating and humidification- a cooling tower - an evaporative condenser
b) evaporative Cooling
The handy straight-line law explains numerous processes by predicting the path of the air conditions on the psychrometric chart.
Air
ts, Ws, hf
ti, Wi, hi
i o
to, Wo, ho
Wetted-surface
21. Condition of Air Passing Through A Coil
51
Air
ts1
t1, W1
1 2
t2, W2
ts2
Refrigerant temperature, tr
3
t3, W3
ts3
4
t4, W4
ts4
5
t5, W5
ts5
8 rows of tubes in direction of air flow
Temperature, C
hu
mid
ity
rati
o, k
g/k
g
Entering air
t1ts1
1
23
4
5
ts5
outlet air temperature from coil to – controlling evaporator
ts1, ts2,… progressively decrease, adjusts itself, - heat balance from air through tube and refrigerant
The straight-line law + progressive drop ts
- coil conditions curve
Greatest rate of heat transfer at enterGreatest rate of moisture removal at coil’s air outlet
Straight-Line-Law
SH/(macp,a)
LH/(
mah
fg)
22. How Selection and Operation of a Coil Affect its Performance
52
Increase of
1.Face area2.Number of rows of tube deep3.Number of fins per unit length4.Air flow rate
5.Refrigerant temperature
Effect on to
LowerLower
Lower
Higher
Higher
Refrigerating capacityHigherHigher
Higher
Higher
Lower
Typical range
Depending on Qe
Four to eight
115 to 300 fins/m(3 to 8 FPI)Vface = 2 to 4 m/s(400 to 800 fpm)3 to 8°C (5 to 15°F)Below entering air
Effect on Wo
LowerLower
Lower
Higher
Higher
As
ts
Limited by saturation line
As
to , Wo
to , Wo
(hi - ho) < m qe
ts
Small moisture remove
Design and operating parameters
on outlet air conditions from a coil.
53
Control factors• face area of the coil, the cross-sectional area of the air stream entering the coil• number of rows of tubes deep• fin spacing• air-flow rate (and air velocity)• refrigerant temperature.
The preceding discussion concentrated on changing only one condition at a time. System designers who select coils may adjust two or more conditions to obtain desired outlet air properties. For example, suppose that the amount of available space does not permit the coil face area desired. Instead, a deeper coil could be selected and combined with a slightly higher air-flow rate to closely match the original outlet air conditions.
q = (flowrate)(hentering - hleaving) (kW)
23. Selecting an Air Coil from a Catalog
54
qe= R(tfi - tr) (kW) R = Factor = mc[1- e-UA/mc ] (kW/°C)
R-value = rate of heat transfer per degree temperature difference (Btu/h)/TD
A dry rating table (sensible heat) ofCoil face area 4.9 m2 (52.8 ft2) 1. Fan: one-, two-, three -fan evaporator coil.2. Model: 6-,8-,10-row coil3. Fan P: 1.5-7.5 HP each fan4. Fin: 158 fins/m (4 FPI) & 118 fins/m (3 FPI)5. External Static Pressure:No ESP (no duct work) & 1/4 in.wg ESP6. Rating: I-P system (inch-pound)17550-36800 (Btu/h)/TD1.46-3.07 tons/TD, TD = (tfi - tr)7. Air flow rate: 14 – 27 m3/s (29200-57000 CFM)High-temp coil Vface < 3 m/s-condensed water blown off coilLow-temp coil Vface = 4-5 m/s
Coil Selection
Example 6 One of the air-cooling coils is to be selected for a space storing frozen food where the air temperature is -23.3°C (-10°F). The refrigeration plant provides a suction pressure corresponding to -28.9°C (-20°F). The coil is ceiling hung with no external static pressure air resistance. The coil is to provide a refrigeration capacity of 70 kW (20 tons of refrigeration). Select a coil.
Data: -10F = tspace = tfi , -20F = tr
no ESP, qe = 70 kW (20 tons)Select a coil.
Method: Table6.6 I-P system20 tons = 240,000 Btu/hTD = (tfi - tr) = (-10-(-20)) = 10°FR = qe/(tfi - tr) = 24,000 Btu/h/°F
55
Assume: not high humidification and frost load – 4 FPIModel 2S-5310 with 1.5-hp fans providing a performance factor of 26,000Model 2S-538 with 2-hp fans providing a performance factor of 24,960Model 2L-536 with 5-hp fans with a performance factor of 23,920 (close)The 2L-536 -smallest and lowest-price coil, the jump to the 5-hp fans with its significantly higher heat input might not favor this choice. Stepping up from 1–1/2 to 2 hp fan motors permits reduction from a 10-row coil to an 8-row
coil, so the 2S-538 with 2-hp fans would be an appropriate choice. [35400 CFM, 3.4 m/s]
If v < 3 m/s, 2S-5310 with 1.5-hp fans would be selected (29200 CFM [(13.8 m3/s)/(4.9 m2) = 2.8 m/s])
56
Data: -23C = tspace = tfi , -29C = tr
no ESP, qe = 70 kW (20 tons)R = qe/(tfi - tr) = 24,000 Btu/h/°F.1.Model 2S-5310,1.5-hp fans, R =26,00029200 CFM, 2.8 m/s]2.Model 2S-538,2-hp fans, R =24,96035400 CFM, 3.4 m/s]3.Model 2L-536,5-hp fans, R =23,92047600 CFM, 4.6 m/s]
TABLE 6.7 Typical temperature differences—entering air to refrigerant—for several applications.
a higher tr saving energy- Limit for a multiple-evaporator installationTABLE 6.7 Typical TD
The catalogs do not reflect the thermal load imposed by fan motor(s).All the electric power delivered to the fan motor appears in the refrigerated space as heat load.The addition of the electrical load part of the refrigeration load calculation.
57
Design practice: to install greater coil refrigerating capacity in a space than that needed to meet maximum refrigeration load when defrosting of coils is expected. Communication with the coil manufacturer is highly recommended. Typical information that can be useful to the coil manufacturer• the space temperature, tspace = tfi
• the saturated suction temperature, tr
• the refrigeration capacity• which refrigerant is to be used• the type of feed, liquid recirculation, flooded coils, or direct expansion. If liquid recirculation is chosen whether top or bottom feed and recirculation rate.
Liquid recirculation: tsupply liquid should be close to tr. For low-temp application, the coil section will frost → absorb less heat from air,→ the liquid is unusually warm. → When the warm liquid passes through the valve,→ a high vapor fraction of refrigerant develops through the throttling valve. → The bottom-feed equipped with orifices, this vapor may choke the flow at orifices.→ The manufacturer should enlarge the orifices.
24. Humidity Control in Refrigerated Room
58
When storing many varieties of fresh produce, humidity should be kept high to maintain product qualityFor rooms chilling hot red meat carcasses, humidity should be kept low to avoid fog formation and to prevent moisture from condensing on surfaces and dripping on product
SpVOA outside air: 0.895 m3/kgSpVS outside air: 0.780 m3/kg
Example 7A room maintained at 1°C, 95% RHOut door: 35°CDB, 25.5°CWBSelect a coil.
OA
S
SOA
WOA = 17 g/kg
WS = 3.8 g/kg
High humidity: large coils, small TD, high air flow, HumidifiersLow humidity: small coils, large TD, low air flow, ReheatMost applications where the control of humidity is important are in spaces operating in the temperature ranges of 0–10°C (32–50°F)
hOA = 78 kJ/kg
hS = 10.6 kJ/kg
Selecting coils for high humidity space
59
Data: A room maintained at 1°C (34°F) and 95% RHFloor area = 1425 m2 (15,340 ft2)Room volume = 10,000 m3 (353,000 ft3)Out door: 35°CDB (95°F), 25.5°CWB (78°F)Sensible heat load (heat conduction through structure, lights and motors, product cooling) qS,tran = 150 kW (512,000 Btu/hInfiltration at 0.15 ACH (air changes per hour)= (10,000 m3)(0.15) = 1500 m3/h (883 CFM) = 0.417 m3/s
minfil, air = (1500/3600 m3/s)/(0.895 m3/kg) = 0.466 kg/sqS,infil = mcp,a(tOA – tS) = (0.466 kg/s)(1.0 kJ/kg°C)(35 - 1°C) = 15.84 kWRate of moisture from infiltration minfil, w = minfil, air(WOA - WS) = (466 kg/s)(17-3.8) = 6.15 g/sRefrigeration rate to condense this water: qL,infil = minfil, w hfg,w = (6.15 g/s)(2.5 kJ/g) = 15.4 kWqinfil = qS,infil + qL,infil = 15.84 + 15.4 = 31.2 kW
Checkd by: qinfil = minfil, w(hOA - hS) = (0.466 kg/s)(78 - 10.6 kJ/kg) = 31.4 kW
psychrometric chartsSpVOA = 0.895m3/kgWOA = 17 g/kghOA = 78 kJ/kghS = 10.6 kJ/kgWS = 3.8 g/kgSteam Tablehfg of water 2500 kJ/kg
qS, coil = qS,infil + qS,tran = 150 + 15.8 = 165.8 kW (576,700 Btu/h)
qL, coil = qL,infil = 15.4 kW
60
EX7: qs = 165.8 kW, qL= 15.4 kWSpace; 1C, 95% RH Outdoor; 35C, 25.5CWBInfiltration; ACH = 0.15qS, coil = 165.8 kW (576700 Btu/h)
qL, coil = 15.4 kW
Load-ratio SHF = qS /(qS + qL) = 165.8 / (165.8+15.4) = 0.915Ref. 24C, 50% RH → Load-ratio line
Straight line law = Load-ratio line→ Saturation tr = 0°C (Coil curve A)Average wetted surface tr -1°C (Coil curve B)Then TD = 1-(-1) = 2°C < 2.2-4.4°C(as indicated in TABLE 6.7 Typical TD)
S
OA
Sen
sib
le H
eat
Fact
or
SHF
Ref.
To maintain high humidities
61
Higher TD → steep coil slope
S
Straight line law = Load-ratio line→ Saturation tr = 0°C (Coil curve A)Average wetted surface tr -1°C (Coil curve B)Then TD = 1-(-1) = 2°C < 2.2-4.4°C
tspace = tfi =1°CA: tr = 0°C
B: tr = -1°C
SH/(macp,a)
LH/(
mah
fg)
If TD = 2°C : W = 0.385-0.345 = 0.04 g/kgLoad-ratio-line: TD = 1°C : W = 0.385-0.378 = 0.007 g/kg
The design of 95%RH cannot be maintained, because the rate of moisture removal at the coil is higher than that entering with the infiltration air.
Operating with high TD in combination with humidifiers
62
EX7: Space; 5°C, 50% RHqs = 110 kW, qL= 20 kW
Load-ratio SHF = qS /(qS + qL) = 110 / (110+20) = 0.85Ref. 24C, 50% RH → Load-ratio line
Straight line law = Load-ratio line→ Saturation tr = -19°C - too lowTypical TD 10 - 17°C → tr = -5 to -12°C If tr = -5°C , then reheat the air back to the load-ratio-lineReheat energy may be supplied by discharge vapor from compressor – energy effectiveness Se
nsi
ble
Hea
t Fa
cto
r SH
F
Ref.
Coil curve A, tr = -5°C
S
To maintain low humidities
63
EX7: Space; 5°C, 50% RHqs = 110 kW, qL= 20 kW
To maintain low humidities
S
Straight line law = Load-ratio line, SHF = 0.85→ Saturation tr = -19°C - too lowTypical TD 10 - 17°C → tr = -5 to -12°C If tr = -8°C , then reheat the air back to the load-ratio-lineReheat energy may be supplied by discharge vapor from compressor – energy effectiveness
64
To maintain low humiditiesExample 9Loading docks adjacent to frozen-food storage areas at 7°CIf tOA→ cooling load by the dock → humidityHigh humidify air from the dock infiltrates to the low-temp. space (storage area)→ Deposit frost on colis
Many operators equip their docks with reheat coils supplied with compressor discharge gas to force the refrigeration coils to cool and especially dehumidify the air when the outdoor temperatures drop to near the temperature settings on the dock
65
To maintain high humidities
Strategies1. Operating with low TD
2. Operating with high TD in combination with humidifiers
Inplications1. Large area → large-size or large number of coils → more fans → additional sensible heat load2. Moderate coil area → typical size and number of coils→ additional latent heat load from humidifiers
Example 8A space storing seeds at 5°C, 50% RHDesign refrigeration load of 110 kW sensible and 20 kW latent
To maintain low humidities
Example 9Loading docks adjacent to frozen-food storage areas at 7°CWhen the outdoor temperature drops to near 7°C
24. Humidity Control in Refrigerated Room
Equip with reheat coils supplied with compressor discharge gas
66
SA
RA
AHU
F CC&DHHC H Fan
1. SA - Supply Air 2. RA - Return Air 3. OA – Outside/Fresh Air3. MA – Mixed Air5. F - Filter
OA
MA
6. CC – Cooling Coil7. DH - Dehumidifier8. HC - Heating Coil9. H - Humidifier10. Fan
→ Heating and Humidifying
Conditioned space in comfort zone
→ Cooling and Dehumidifying
25. Air Conditioning Systems and Equipment
Method of Cooling with dehumidificationCooling – lower DB
Dehumidification – decrease moisture content (humidity ratio) of air
1. Absorption of moisture by a chemical dehydrating agent (silica gel or lithium bromide) and sensible heat removal by direct expansion or chilled-water cooling coil
2. Forcing a stream of air through a chilled-water spray (twater spray = ti)ti < dew point (DP) of air → both cooling and dehumidification
3. Passing air over a direct expansion or chilled-water cooling coil
Most wildly used for residential, commercial, industrial air conditioin.
Used in which Large flow rate of air is required
68
air (ta, a) contact with chilled-water ti
Find qt, mw?
qs
qL
1
ti< DP
2
EX1: Air at 30CDB, 50%RH is passed through a chilled-water spray of 13C and comes out saturated. Find the heat removed and moisture removed.
2.Cooling with dehumidification: chilled-water spray
W2
W1DP
ti < air DP→ cooling & dehumidification
qt = h1 – h2 kJ/kgmw= (W1 – W2) kg/kg
69
Refrigerating machine and water chiller
chilled-water pump
chilled-water spray
Return air
Outside air
Used in which Large flow rate of air is required: assembly hall, theaters, sport arena
EX2: Air at 35CDB, 60%RH passes over a direct-expansion refrigerant coil and leaves the coil at 13CDB, saturated. Find the heat removed and moisture removed.
70
3. Cooling with dehumidification: cooling coil
W2
1
2
W1
Path of air/process line is very complex: Some of air pass through coil without actually touching cold surface.
qt = h1 – h2 kJ/kgmw= (W1 – W2) kg/kg
ADP = Apparatus Dew Point
EX3: An air quantity of 3,400 cmh at 35CDB, 60%RH is cooled and dehumidified to 15CDB, 95%RH by an 2-row coil. Find the heat removed and moisture removed.
71
Cooling coil: Apparatus Dew Point, Coil ADP
W2
1
2
W1
ADP = Apparatus Dew Point
Bypassed air 35C, 60%
ma= cmh/v1 kg/h qt = ma(h1 – h2) kJ/hmw= ma(W1 – W2) kg/h
v1Air contact coil at ADP 10C
Air-off-coil 15C, 95%
BPF = Bypass Factor= (15 - 10)/(35 - 10) = 0.2
72
Cooling coil: Bypass Factor, BPF
tenter = 35C Bypassed air 35C, 60%Coil ADP = 10C = tcoil
Air-off-coil 15C, 95%tleave = 15C
BPF = (tleave- tcoil)/(tenter- tcoil)
Bypass air at 35C, 60%RH is mixed with air contact with coil surface at coil ADP = 10C and leave as air-off-coil mixture at 15 C, 95%RH
BPF = (15 - 10)/(35 - 10) = 0.2
1-row coil (8-fin per inch) BPF = 0.56N-row coil; BPF = 0.56n
at air velocity of 2.54 m/s (500 FPM)4-row coil; BPF = 0.564= 0.16-row coil; BPF = 0.566= 0.03148-row coil; BPF = 0.568= 0.01
BPF for air conditioning coil with fin spaced 10 or 12 to the inch (FPI)
Room Total Heat (RTH) = Room Sensible Heat (RSH) + Room Latent Heat (RLH)
73
Sensible heat factor, SHF
SHF = RSH/RTH
Psychrometric chart for Air conditioning with SHF scale & Ref. point
SHF scale
Ref. point
SI: 24C, 50%USCS: 80 F, 50%
SHF = 1.00
SHF = 0.36
SHF = 0.79Ref. point
Ex. SHF = 0.79 line from Ref. point
Room condition point A
A
Draw parallel line- from A to sat. line at ADP
ADPAir DP
Room condition line or Room SHF/RSHF line
If air-off-coil at 90%
Ex4: A summer air conditioner mixes 3,400 cmh of outside air at 35CDB, 28CWB with 10,000 cmh of return air at 25CDB, 40%RH. The mixture is passed through a cooling coil. Air off the coil at 90%RH. The room SHF = 0.68.
74
Mixing, Cooling, Dehumidification
O
R
1. Find ADP, air DP, air-off-coil DB.2. How much cooling in W is the unit doing?3. How much of total load is latent heat andhow much is sensible heat?
ADP
O – Outside airR – Return air
E – Enter coil → mass&energy balance
Ref.
L
Ref. point → SHF lineRSHF line → ADP E
Air DPAir-off-coil %RH → CPL → Air DPL – Leave coil
75
Cooling Coil Selection from Manufacturer’ Data
Air-side variables for final selection and specification of a cooling coil
1. Entering-air velocity or coil-face velocity, (2-3 m/s or 400 – 500 fpm)
2. Fin spacing or number of fins per m or per inch, (315-472 fin/m or 8-12 fpi)
3. Entering-air conditions, DB & WB
4. Refrigerant temp. or entering chilled-water temp., (3-4C or 5-7F below ADP)
5. Row of coil depth 2- to 8-rows coils
6. Area of coil-face area – space limitations caused by equipment
High velocity → coil BPF, pressure drop, water-carry-over to SA, noise level
High entering temp → coil capacity
Low refrigerant temp → coil capacity, capacity of condensing unit
rows → coil BPF, pressure drop → fan power
Selection: Table Data of DX Coil Rating for evaporating temp., DB, WB, coil-velocity
76
Cooling Coil Selection from Manufacturer’ Data
Ex5: A direct expansion (DX) coil for R22 is to be selected to meet the following specification
Determine coil-face velocity, number of rows required, and leaving-air temperature.
1. Check for SHF < 0.9 → SHF = 96,500/140,000 = 0.69
2. Solve for Btu/h/ft2 → qT = [140,000Btu/h] /7ft2 = 20,000 Btu/h/ft2 = 20 MBH
3. Table Data of DX Coil Rating for evaporating temp., DB, WB, face-velocity
→ Tr of 45F, 84F DB, 68F WB and 500 fpm-face velocity
→ Vface = (3,600ft3/min)/7ft2 = 514 fpm-coil-face velocity
→Select number of rows required, leaving-air temperature for MBH
4. Read the Leaving wet and dry bulb temp. LWB, LDB from Table Data
Air quantity (std. air) 3600 cfmAir entering coil 84°F DB, 68°F WBTotal load = 140,000 Btu/hrSensible load = 96,500 Btu/hrRefrigerant evaporating temperature 45°FCoil face area = 7.0 ft2
77
→ Tr= 45F, 84F DB, 68F WB and 500 fpm
for 20 MBH →→ 4-row coil, 12 fpi→ LDB = 56.1FDB→ LWB = 55.3FWB
78
Tons per sq.ft area for 4-rows coil 500 fpm face-velocity
Correction factorfor 3- to 8-rows coil 400-600 fpm face-velocity
Tr= 45FTypical TD = tfi – tr = 20 - 30 Ftfi = 65 - 75 F20,000/12,000 = 1.67 ton/ft2
tfi = 73 F
79
Chilled-water cooling coilSame selection -include change in water temperature (tin - tout) as coil absorbs heat while tevp = const. in direct-expansion coils.
Water circulated through a chilled-water coil increases by 4-10C or 8-15F
Both air-side performance and water-side performance must be considered.
1. Cooling/Heating Load calculation- Sensible heat load (qS), Latent heat load (qL)
2. Thermal distribution systems- AC type: Split-type, Package-unit, Chilled-water- Refrigeration system: Vapor-compression, Absorption- Control system: electronic thermostat, inverter control- Air system: flow rate of Supply Air, Return Air, Outside Air- Water system: water flow rate, pump
3. Fan-and-Duct systems- Layout of duct system, duct sizing, fitting- Fan sizing
4. Refrigeration equipments- Compressor, Condenser, Evaporator, Expansion devices- Refrigerants
80
Air Conditioning Systems Design
Outside air control: min 10-20% OAhospital – 100% OA
Load calculation (qS, qL)Space load: qt = qS + qL
qL = hfg(Ws – Wi) = 3010(Ws – Wi)qS = cpQs(ts – ti) = 1.23Qs (ts – ti)Supply air: Qs = qS /1.23(ts – ti)
Thermal distribution systems
Supply Air (SA)
Return Air (RA)
Outside Air (OA)
Air at 10 – 20 C 1.23 kg/m3
cp 1.0 kJ/kgKhfg of vapor water 2500 kJ/kgcp = 1.23 kJ/m3Khfg = 3010 kJ/m3
Qs = supply air flow rate m3/s
AHU
CMH = m3/h, CFM = ft3/min, Btu/h
Cooling coil
Load-ratio, SHF = qS /(qS + qL)Space load : qt = qS + qL = Qs(hs – hi)
Thermal distribution systems
Supply Air (SA)
Return Air (RA)Outside Air (OA)
Mixed air
82
qS = Qscp(ts – ti)qS /(qS + qL) = cp(ts – ti)/(hs – hi) RA
OA
Coil load: qcoil = Qs(hc – hi)
ts – ti
83
EX6: qs = 65 kW, qL= 8 kWSpace; 24C, 50% RH (=Ref.)Outdoor; 35C, 25CWBVentilation; OA:RA = 1:4Air condition leaving coil?Cooling capacity of the coil?
Method: ti,i, qcoil = ma(hc – hi)Load-ratio SHF = qS /(qS + qL)SHF = 65/(65+8) = 0.89Ref. 24C, 50% RH
mO:mS = 1:4Energy balance: m=Qmhc =mShS + mOhO
Mass balance of water:mWc =mSWS + mOWO
S
O
C
i
Air condition leaving coil qS /(qS + qL) = cp(ts – ti)/(hc – hi) = 0.89Assume ti, find hi until get hi@ti
Mostly air leaving coil 92-98%RHQs = qS/cp(tS – ti) = qt/(hS – hi) Coil load: qcoil = Qs(hc – hi)
Sen
sib
le H
eat
Fact
or
SHF
Ref.
84
When is Reheat coil needed?Ex7: SHF = qS /(qS + qL) = 0.7 clean room; 20C, 40% RHOutdoor; 35C, 25CWBVentilation; OA:RA = 1:4
Load-ratio SHF lineSHF = 0.7 and Ref. 24C, 50% RH
S
O
C
i
Sen
sib
le H
eat
Fact
or
SHF
Ref.
clean room; 20C, 40% RHSHF line not cross the saturation line !
Reheat needed!
Ex4: ADP = -5C , supply air temp = -2C → frost at coil
85
Cooling coil - Control
O
R
1. Seclect ADP = 2C .2. L – Leave coil = 7C3. Reheat by heating coil?
ADP = -5C
Ref.
L
E
SA – Supply Air temp = 11-12C
ADP = 2C
SA temp.
25. Fan and Motor Performance and Selections
86
Major decision with respect to the selected coil by available choices:
1. Draw-through or Blow-through arrangement
2. Propeller or Centrifugal fan
3. Single or Two-speed motor
25.1 Draw-through or blow-through arrangement
Throw: distance from an air outlet to point where the velocity dropped to 0.25-0.75 m/s (50-150 fpm).
Industry : 100 – 150 fpm
Residence: 50 fpm
Draw-through+ Greater throw(might be as high as 20 m/s
for 60 m throw)
Blow-through + Thermal advantage→ effective coil(heat from fan to coil
→ higher T = tf,o – tf,i)
heatheattf,otf,i tf,o
tf,i
87
25.2 A propeller or centrifugal fan
A propeller fanat constant speed
A centrifugal fan at constant speed
System curve of a low-temp. coil
System curve does change when frost forms on the low-temp coil.
The propeller fanfrost forms → Power
The centrifugal fanfrost forms → Power
A propeller fan+ Used much more+ Compact package+ More efficient at low static pressure (SP)
A centrifugal fan +For very long throw with draw-through+ Long duct work- But short belt life
88
Propeller-type & forward-curved-blade centrifugal fan
In comparison with propeller fans, centrifugal fans (blowers) require less horse power at higher SP and have a much less steep SP curve, low noise
Performance of a centrifugal fan, forward-curved-blade type
Performance of a propeller-type fan
propeller fans, large volume at low static pressure (SP) – high noise
air-cooled condensers: require large air volume at small SP of finned-tube condenser
Large Q at low SP
Less hp at high SP
80 cfm, 22%SP
80 cfm, 47%SP
Fan curve
Centrifugal fan and their characteristics
Air enters fan axially → turns and moves radially into blades
→ enters scroll→ outlet
Fan inlet – single or double (both sides)
4-types Blades:
Air-foil or backward-curved blade fan – used in high volume/high pressure
forward-curved blade fan – used in low pressure A/C system
1.Radial – low volume & high pressure2.Forward-curved – proper for constant flow, silent3.Backward-curved – easy to control volume and p4.Airfoil – high volume & high pressure
For some Sky-A/C : Indoor unit1. Forward curved blade fans - Duct & Ceiling type 2. Backward curved blade fans - Cassette type
Fan in A/C
Radial or straight blade– low volume & high pressure
forward-curved– proper for constant flow, silent
backward-curved – easy to control volume and pressure
Outdoor unit : Propeller fan– low speed & high volume & low static pressure
Large A/C systemAir duct system
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Fan law 1: Variation in rotative speed, constant air density Q N, SP Q2 , P Q3
Fan laws
: group of relationships that predict fan performance
of changing -- condition of air, operating speed, size
Q = volume flow rate, m3/sN = rotative speed, r/sSP = static pressure, Pa or in.wgP, HP = power or horse power, W or hp
Fan law 2: Variation air density, constant volume flow rateQ = constant, SP , P
Fan law 3: Variation air density, constant static pressureQ2 1/, SP = constant, N2 1/, P2 1/
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forward-curved-blade centrifugal fan
Dip in static pressure at low flow rate eddies flow in blade channel
Power = pressure rise + kinetic energy
Pideal = Q*(p2 – p1)+ w*V2/2
Efficiency, = Pideal / Pactual
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Example 6-5 Compute the efficiency of the fan whose characteristics are shown in Fig.6-15 when it operate at 20 r/s and deliver 1.5 m3/s.
1.Data: Fan - air20 r/s – rotative speed (n)1.5 m3/s = QFig.6-15 Characteristic fan curve: D 270 mm, outlet 0.517 by 0.289 mCompute fan efficiency,
4.Methods & Equations
= Pideal / Pactual
Pactual → graph -- known Q, n
Pideal = Q*(p2 – p1)+ w*V2/2
Characteristic fan curve: P & prise = f(Q,n)
1.5 m3/s, 20 r/s → (p2 – p1) = 500 Pa→ Pactual = 1.2 kW
w = *Q
V = Q/A, Aoutlet = 0.517 * 0.289 = 0.149 m2
5. Properties: air = 1.2 kg/m3
6. Calculations: w*V2/2 = 1.8*10.12/2 = 91 WQ*(p2 – p1) = 1.5*500 = 750 PaPideal = 750 + 91 = 841 W = 100%*841/1200 = 70%
w = 1.2*1.5 = 1.8 kg/s
V = 1.5/0.149 = 10.1 m/s
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Balance-point diagram for Fan-and-duct system curve
SPbp
SPe = 0.75*SPbp
If SP1 = 0.4 in.wg at 4000 cfm, Fan law 1: Q N , SP Q2 , P Q3
SP2 = SP1 (Q2/Q1)2 → Q2 = 6000, SP2 = 0.4(6000/4000)2 = 0.9 in.wg, Q0 = 2000, SP0 = 0.4(2000/4000)2 = 0.1 in.wg → plot duct system curve
Intersection of system curve and fan-curve, power-curve is balance point: SPbp, HPbp
Pbp
If SPe were in error by 25%: SPe/SPbp= 0.75
Fan law 1: Qe = Qbp(0.75)1/2 = 0.56Qbp
but fan-curve: Qe = 1.36*Qbp
Fan law 1: Pe = Pbp(0.75)3/2 = 0.65HPbp
but fan-curve: Pe = 1.5*Pbp
Such balance diagrams are instructive since they readily show the effects of changing the operation system of the duct system: i.e. Spe by 25% → Qe by 36% , Pe by 50%
Not obey Fan law 1: for constant fan speed*Select a two-speed fan
Pe = 1.5*Pbp
QbpQe = 1.36*Qbp
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Balance-point at two different speeds
The fan-duct-system balance points change as the fan speed changes.
Fan law 1: Variation in rotative speed Q N, SP Q2 , P Q3
balance point A: SPA = 69%, QA = 65%, PA = 55%
(SPA/SPB) = (QA/QB)2
→ (69/55) (65/57)2 1.3
balance point B: SPB = 55%, QB = 57%, PB = 36%
(PA/PB) = (QA/QB)3
→ (55/36) (65/57)3 1.6
Obey Fan law 1: Variation in rotative speed
The balance points at two speeds show the results of Fan law 1 on the operating of the complete system.
25.3 A single or two-speed motor
Testing system to evaluate the actual operation and compare to design
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Efficiencies of single-speed and two-speed electric motors
Fan law 1: Q N, SP Q2 , P Q3
If (Q2/Q1) = 1/2 → (SP2/SP1) = 1/4 → (HP2/HP1) = 1/8 The electric power demanded by themotor may not drop fully by 1/8 because the motor efficiency at the lowspeed drops off abruptly at low loadsNevertheless, the power saving is usuallysignificant, which provides a bonus by also reducing the heat load introduced tothe space by the fan motor.
In a multiple-fan unit one or two fans can be cycled off to reduce the capacityand internal heat load from motors. – When restarting an idle motor, the operating fan(s) should be stopped first. the idle one is likely to be spinning in reverse, the motor may overload.- Another version of two-speed fan operation that is adaptable to a centrifugal fan installation having several scrolls is to apply pony motors. The fan is then driven by one motor while the other idles.
26. The Number and Placement of Coils
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The number of coilsLocations of coilsRefrigerated spaceDirections of air discharge
Room geometryCeiling arrangementProduct placement
A small number of large coils → Reduction in piping, valves and controls for the coils
→ Lowest first cost and Lowest maintenance cost
A large number of small coils → Distribution of the total capacity
→ Avoid pockets of high temperature in the space
Guidelines for achieving uniform temperatures:for most coils discharge directly into the space (sometimes, discharge air is ducted from the coil to convey it further)1. select the coils and place them on the basis of a throw of 30 to 60 m (100 to 200 ft)2. blow the discharge air in the same orientation as the beams3. blow across doors and not away or toward them 4. direct the discharge air down aisles5. direct the discharge air downward from ceiling coils in high-rise storage facilities
PART III. LIQUID CHILLERS
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27. Shell and Tube Evaporators
Shell and tube – boiling in tubeFluid: water or brine/antifreeze
Refrigerant in tubes. • Water chiller in Air-conditioning industry* halocarbon refrigerants* direct expansion * superheat-control expansion valve* centrifugal compressor (dry compression)• Industrial refrigeration systems the circuits could be arranged for forced liquid overfeed (with pump) or could be operated flooded using a surge drum (natural convection).
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Refrigerant in shell. * most popular type of liquid-chilling evaporator in industrial refrigeration* flooded chillers• Liquid-vapor separation left at the top of
the shell* level-control valve regulates the flow of liquid refrigerant to the bottom of the vessel. Industrial refrigeration systems• A small separation vessel is mounted
above the shell of the main evaporator * brine-chilling applications
Shell and tube – boiling in shellFluid: water or brine/antifreeze
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28. Sprayed Tube Liquid Chiller
Refrigerant in shell. * liquid refrigerant from a circulating pump is sprayed over the tubes+ Greater overall heat-transfer coefficient than flooded type by permits easy escape of the vapor bubbles(vapor bubbles insulate the heat-transfer surface)
+ low refrigerant charge (159 kg of ammonia charge for 400 tons RE; but 5900 kg for flooded type
- complexity of the pump and spray assembly*When the flow rate is too high, the liquid film becomes thick and insulates the tubes.
*When the flow rate is too low some of the tube surfaces do not become wetted.A typical circulation rate is 5 times the rate evaporatedThe pump adds a small amount to the power cost, if the pump fails, the evaporator is essentially out of operation. Often a reserve pump becomes part of the unit.
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29. Plate-Type Evaporators
plate-and-frame heat exchanger* food industry* numerous plates gasketed and boltedrefrigerant flows between two of plates fluid flows between the pairs of adjacent plates* Corrugated plates, herringbone pattern + physically strengthens the plates + promotes turbulence flow, + excellent convection heat transfer coefficients
v
* food dairies industrythe bolts holding the plates can be loosened, permitting access to all surfaces for cleaning
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The challenge of how to seal the refrigerant passages* pairs of plates forming the refrigerant passages are brazed or welded* halocarbon refrigerants, normal brazing of the edges* ammonia either nickel brazing or welding is necessary
Counter-flow plate-heat exchangerthe liquid flows downward while the refrigerant flows upward. + high heat transfer coefficients+ low refrigerant charge+ small size, compact
Typical size: plate-type evaporator water-ammonia, 2500 – 4500 W/m2Kwater-R22, 1500 – 3000 W/m2K
The three major types of refrigerant feed (direct expansion, flooded withsurge drum, and forced liquid overfeed)—are all used with plate-type evaporator. Flooded or recirculatedTypical size: water-ammonia, 2840 – 3975 W/m2K
direct-expansion, simplest but difficult to achieve uniform flow to each of passagesTypical size: water-ammonia, 2275 – 3400 W/m2K
