ce311 fall 2016 final exam name: instructions...ce311 fall 2016 final exam name: _____ 1/16...
TRANSCRIPT
CE311 Fall 2016 Final Exam Name: __________________________________
1/16
Instructions
1. Please solve the problems using the specified SI or IP system of units. 2. Please show all of your calculations to get full credit and pay attention to assumptions.
3. Neatly mark points and outline processes on the ASHRAE SI or IP psychrometric chart. � 4. Final solutions should be clearly summarized and marked/boxed. � 5. Neatly sketch all illustrations. 6. Use only the attached equation sheet. 7. Please attach additional pages of your work, clearly marking each problem.
Problem 1 (25/100pts)
Outdoor air enters an air handling unit (AHU) at 5� and 35% relative humidity at a steady volumetric flow rate of 25m3/min, and it leaves at 25� and 55% relative humidity. The outdoor air is first heated to 22� with a heating coil and then humidified through the injection of hot steam. Assuming the entire process takes place at a pressure of 1 atm, determine: (a) Draw a schematic of an AHU for this process. Label each point and the components (5pts)
(b) Neatly sketch the process on a psychrometric chart and label all points (5pts)
(c) The rate of heat supply in the heating section. (7.5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
2/16
(d) The mass flow rate of steam required in the humidifying section. (7.5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
3/16
Problem 2 (25/100pts)
Recirculation air from a room having dry-bulb and wet-bulb temperatures of 33� and 29�, respectively, enters an AHU at a steady state and mixes with outdoor air entering with dry-bulb and wet-bulb temperatures of 16� and 12�, respectively. The volumetric flow rate of the outdoor air stream is three times that of the recirculation stream. A single mixed stream exits prior to conditioning and supply back to the room. The pressure is constant throughout at 1 atm. You can neglect kinetic and potential energy effects. (a) Draw a schematic of an AHU for this process. Label each point and the components (5pts)
(b) Neatly sketch the process on a psychrometric chart and label all points (5pts)
(c) The relative humidity of the mixed air stream (7.5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
4/16
(d) The temperature, in �(7.5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
5/16
Problem 3 (35/100pts)
You are asked to evaluate the total thermal resistance and heat gain of a wall enclosure, considering the effects of natural and forced convection, radiation, and conduction. Consider the wall enclosure with the dimensions and associated material and air properties shown below. You can treat the surrounding indoor wall surfaces and sky as blackbodies. Airflow along the indoor wall surface is buoyancy-driven, whereas airflow across the outer surface is wind-driven. You can apply Nusselt number correlations for flat horizontal surfaces for the vertical outer wall.
Determine the following: (a) Indoor surface: Rayleigh number and Nusselt number (5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
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(b) Indoor surface: Natural convection and radiation heat transfer coefficients (5pts)
(c) Outdoor surface: Reynolds number and Nusselt number (5pts)
(d) Outdoor surface: Forced convection and radiation heat transfer coefficients (5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
7/16
(e) Draw a thermal resistance network for the entire system (2.5pts) (f) Compute the equivalent (parallel) resistances for the inner and outer wall surfaces (5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
8/16
(g) Compute total thermal resistance of the wall surface (5pts) (h) Compute heat flow rate (W) from outside to inside (2.5pts)
CE311 Fall 2016 Final Exam Name: __________________________________
9/16
Problem 4 (15/100pts) – volatile organic compound (VOC) emissions from vinyl flooring
Toluene is being emitted from vinyl flooring in a living room. A small mixing fan is used in the room. You are asked to determine the emission rate (or net mass transport rate, Jmass) of toluene from the vinyl flooring to the room air.
!" = 100'(/*+ ," = 1*/- ." = 18�
(a) Compute the: Reynolds number, Schmidt number, and Sherwood number (5pts) (b) Determine the average mass transfer coefficient (2pts)
[toluene]
Pg@22�=3.76kPa
D=0.087cm2/s
(tolueneinair@22�)
Mtoluene=92.14g/mol
CE311 Fall 2016 Final Exam Name: __________________________________
10/16
(c) Calculate the surface concentration of toluene (5pts) (d) Calculate the emission rate (or net mass transport rate, Jmass) of toluene (3pts)
CE311 Fall 2016 Final Exam Name: __________________________________
11/16
CE311 Fall 2016 Final Exam Name: __________________________________
12/16
01 =212
01:*456789:;<42
21: 2=*>6847*456-47:4*?4262;(9-< *45
2: ;4;952=*>6847*456-47*<@;=86 *45
A1 =*1
21
ABCDE1C = 28.97
AJEKLCMENOC = 18.02
P = 8.314
P1 =PA1
A1:*456:=598S6<(ℎ;47(9-<((/*45)
*1:*9--47(9-<(()
ABCDE1C = *456:=598S6<(ℎ;47W809<8(g/mol)
AJEKLCMENOC = *456:=598S6<(ℎ;47S9;68(g/mol)
P: =2<\68-95(9-:42-;92;(]^/]*45 − `)
P1 = (9-:42-;92;47(9-<
idealgasequationofstate:
PV = nP.
T: temperature
V: volume
P: pressure
ω =*M
*E
ω = 0.622sMsE
φ =sMsu
φ =vs
(0.622 + v)su
s1 = 01s
ω: humidityratio (e.g.kgwv/kgda)
*M:massofwatervapor
*E:massofdryair
sM: partialpressureofwatervapor
sE: partialpressureofdryair
φ: relativehumidity (-)
su: saturationvaporpressure
s1: partialpressureofspeciesi
P:totalmixturepressure
h = ℎE + vℎM h:moistairmixtureenthalpy(kJ/kgdryair)
ℎE: 62;ℎ95?047W809<8(kJ/kgdryair)
ℎM: 62;ℎ95?047S9;68\9?48(kJ/kgwatervapor)
conservationofmass:
*E =1
*EL
*M +1
*J1
= *M +L
*JL
e:exiti: inlet
*E:*9--754S89;647W809<8(](W809<8/*<2)
*M:*9--754S89;647S9;68\9?48(](S9;68\9?48/*<2)
CE311 Fall 2016 Final Exam Name: __________________________________
13/16
*J:*9--754S89;6475<�=<WS9;68(](5<�=<WS9;68/*<2)
conservationofenergy:
Ä −Å = *LℎL −L
*ÇℎÇ1
� =ÄÉ
�: ℎ69;75=@(Å/*Ñ)
Ä: ℎ69;754S89;6(Å)A: areaperpendiculartodirectionofheatflow(*Ñ)
Fourier’sLaw
ÄÜOáB = −]ÉW.W@
k: ;ℎ68*95:42W=:;<\<;0(Å/*℃48Å/*`)
T: temperature(℃orK)
W.W@
: ;6*?689;=86(89W<62;<2@ − W<86:;<42(℃/*48`/*)
α =]!ãN
α: thermaldiffusivity(*Ñ/s)
ρ: density(kg/*+)ãN: -?6:<7<:ℎ69;:9?9:<;0(^/](℃48^/](`)
ν ='!
ν: kinematicviscosity(*Ñ/s)
': W029*<:\<-:4-<;0(*Ñ/s)
Re =,èν
Re: Reynoldsnumber
V: characteristicvelocity(m/s)
L: characteristiclengthscale(m)
τ = µW=dy
τ: shearstress(N/*Ñ)W=dy: \654:<;0(89W<62;
R =Δ.
Ä
R: thermalresistance(℃/Å48`/Å)
Δ.: ;6*?689;=86W<776862:6(℃orK)
Newton’sLawofcooling:
Ä = ℎÉñ(.ñ − .")
Éñ: -=879:69869(*Ñ)
ℎ: :42\6:;<42:4677<:<62;(Å/*Ñ℃48Å/*Ñ`)
.ñ: -=879:6;6*?689;=86(℃orK)
.": >=5]75=<W487866 − -;869*;6*?689;=86(℃orK)
Pr =να
Pr: Prandtlnumber
Nu =ℎè]ó
Nu:Nusseltnumber local
ℎ: 54:95:42\6:;<42:4677<:<62;(Å/*Ñ℃48Å/*Ñ`)
è: :ℎ989:;68<-;<:562(;ℎ-:956(*)]ó: 75=<W;ℎ68*95:42W=:;<\<;0(Å/*℃48Å/*`)
CE311 Fall 2016 Final Exam Name: __________________________________
14/16
ò= =ℎè]ó
ò=: 9\689(6ò=--65;2=*>68
ℎ: 9\689(6:42\6:;<42:4677<:<62;
P6ÜC = 5 ∗ 10õ P6ÜC: :8<;<:95P60245W-2=*>68748-*44;ℎ759;?59;6
(748:6W:42\6:;<42)
ℎ =1Éñ
ℎW@W0úù
ℎ =1è
ℎW@û
ü
ℎ: 54:95:42\6:;<42:4677<:<62;(Å/*Ñ℃48Å/*Ñ`)
ℎ: 9\689(6:42\6:;<42:4677<:<62;è: 562(;ℎ47?59;6(*)
.ó =12(.ñ + .")
.ó: 7<5*;6*?689;=86(℃orK)
β =−1!(°s°.)
β = 1/.748<W695(9-
β: coefficientofexpansion(1/K)
ρ: density(kg/*+)
T: temperature(℃orK)
¢8û =(£(.ñ − .")è+
§Ñ
¢8û: ¢89-ℎ472=*>68
g = 9.81(m/-+), gravitationalacceleration
.ñ: -=879:6;6*?689;=86(℃orK)
.": >=5]75=<W;6*?689;=86(℃orK)
è: :ℎ989:;68<-;<:562(;ℎ(*)
ν: kinematicviscosity(*Ñ/s)
Ra = Gr ∗ Pr
P9 =(£(.ñ − .")è+
§ß
Ra: Rayleighnumber
P9ÜC~10© ∶ :8<;<:95P92=*>6874829;=895:42\6:;<42
Nusseltnumbercorrelationsforforcedconvection:LocalNuandhforlaminarflowoverflatplate
ò=´ =ℎ´è]ó
= 0.332P6´¨Ñs8
¨+forPr > 0.6
LocalNuandhforturbulentflowoverflatplate
ò=´ =ℎ´è]ó
= 0.0296P6´Æõs8
¨+for0.6 < Pr < 60
AverageNuandhforlaminarflowoverflatplate
ò=´ =ℎ´è]ó
= 0.664P6´¨Ñs8
¨+forPr > 0.6
AverageNuandhforturbulentflowoverflatplate
CE311 Fall 2016 Final Exam Name: __________________________________
15/16
ò=´ =ℎ´è]ó
= 0.037P6´Æõs8
¨+for0.6 < Pr < 60
AverageNuandhformixedboundarylayer
ò=û = (0.037P6ûÆõ−É)s8
¨+for0.6 < Pr < 60
É = 871748P6ÜC = 5 ∗ 10õ; É = 0.037P6ÜCÆõ − 0.644P6ÜC
¨Ñ
CorrelationsfortheaverageNufornaturalconvectionoververticalplate
ò=û =ℎè]ó
= ãP9ûá
Laminar: 10Æ ≤ P9û ≤ 10©ã = 0.59, n = 1/4
Turbulent: 10© ≤ P9û ≤ 10¨+ã = 0.1, n = 1/3
Mixedboundarylayer:
ò= = 0.825 +0.387P9û
¨≤
1 + 0.492/s8 ©/¨≤ ≥/Ñ¥
Ñ
radiationµ∂ = ∑.∏Æ.∏: surfacetemperature(K)
µ∂: blackbodyemissionpower(W/*Ñ)
∑: Stefan − Boltzmannconstant: 5.67x10Ω≥(Å/*Ñ`Æ)
µ = æ∑.∏Æ
æ: emissivity(0to1)
¢E∂∏ = ߢ
¢E∂∏: absorbedradiation(W/*Ñ)
ß: absorptivity(0to1)
�CEB = µ − ߢ = æ∑(.∏Æ − .∏øCCÆ ) ÄCEB = æ∑É∏(.∏Æ − .∏øCCÆ )�CEB: radiationheatflux(W/*Ñ)
ÄCEB: radiationheatflowrate(W)É∏: surfacearea(*Ñ)
.∏øCC: surroundingsurfacetemperature, treatedasablackbody(K)
ÄCEB = ℎCÉ∏(.∏ − .∏øCC) ℎC: radiationheattransfercoefficient(Å/*ÑK)
ℎC = æ∑(.∏ − .∏øCC)(.∏Ñ − .∏øCCÑ )
CE311 Fall 2016 Final Exam Name: __________________________________
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PL¿ø1M,NECE¡¡L¡ = ( ¨
¬√)Ω¨
PL¿ø1M,NECE¡¡L¡: equivalentresistanceforresistorsinparallel
masstransfer !1 =
ƒ√
≈
!1:massconcentration(g/cm3) *1:massfospecies(g) ã1 =
á√≈
ã1:molarconcentration(mol/m3) !1 = A1ã1
j = −« B»
B´ j: molardiffusionflux(mol/cm2s)
D: diffusioncoefficient(cm2/s)
j = ℎƒ(ãñ − ã") J = ℎƒÉ∏(ãñ − ã") J:netmolartransportrate(mol/s)
ã∏: surfacemolarconcentration(mol/m3)
ã": bulk/freestreammolarconcentration(mol/m+)
ℎƒ: convectionmasstransfercoefficient(m/s)
ã∏ = À(ÃÕ)
P.- !∏ =
À(ÃÕ)Œ
P.-
œℎû =–—û
“ œℎû ∶ averageSherwoodnumber
Sc = ”
“ Sc:Schmidtnumber
Le = ‘’
÷◊ Le:Lewisnumber
Shcorrelations:replaceNuw/ShandPrw/Sc(viaheat&masstransferanalogy)
ÿ
–—= !: è6(¨Ωá) ntypically=1/3
C∏O¡1B = ⁄sM
C∏O¡1B: solid − phaseconcentration(kmol/*+) ⁄: solubility(kmol/m3bar)
€ = ⁄D €:permeability(kmol/s-bar)
ℳ = €
û ℳ:permeance(kg/s-m2-Pa)
JM = ℳÉ(sM¨ − sMÑ) JM:watervapornettransportratethroughwall(kg/s)
RR
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DRY BULB TEMPERATURE - °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
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REL
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30%
40%
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85 W
ET B
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TEM
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- °F
85
90
12.5
13.0
13.5
14.0 VOLUME- CU.FT. PER LB. DRY AIR
14.5
15.0
HUMIDITY RATIO - POUNDS MOISTURE PER POUND DRY AIR
0
1.0
1.0
:-
2.0
4.0
8.0
-8.0
-4.0
-2.0-1.0
-0.5-0.4
-0.3
-0.2
-0.1
0.1
0.2
0.3
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KJ
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R K
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GR
AM
OF
DR
Y A
IR
10
20
30
40
50
60
70
80
90
100
ENTHALPY - K
J PER KILO
GRAM OF D
RY AIR
SATURATION TEMPERATURE - °
C
5
10
15
20
25
30
35
40
45
50
DRY BULB TEMPERATURE - °C
.001
.002
.003
.004
.005
.006
.007
.008
.009
.010
.011
.012
.013
.014
.015
.016
.017
.018
.019
.020
.021
.022
.023
.024
.025
.026
.027
.028
.029
.030
10%
REL
ATIV
E H
UM
IDIT
Y
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 W
ET B
ULB
TEM
PERA
TURE
- °C
30
0.78
0.80
0.82
0.84
0.86 VOLUME - CUBIC METER PER KG DRY AIR
0.88
0.90
0.92
0.94
HUMIDITY RATIO - KILOGRAMS MOISTURE PER KILOGRAM DRY AIR
0
1.0
1.0
:-
1.5
2.0
4.0
-4.0
-2.0-1.0
-0.5
-0.2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2.0
2.5
3.0
4.0
5.0
10.0
- ::
SENS
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HEA
TQ
sTO
TAL
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ENTH
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TIO
Dh
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TU P
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20
25
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35
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ENTHALPY - B
TU PER POUND OF D
RY AIR
SATURATION TEMPERATURE - °
F
35
40
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50
55
60
65
70
75
80
85
90
95
100
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DRY BULB TEMPERATURE - °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10%
REL
ATIV
E H
UM
IDIT
Y
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
4045
4550
5055
5560
6065
65
70
70
75
75
80
80
85 W
ET B
ULB
TEM
PERA
TURE
- °F
85
90
12.5
13.0
13.5
14.0 VOLUME- CU.FT. PER LB. DRY AIR
14.5
15.0
HUMIDITY RATIO - POUNDS MOISTURE PER POUND DRY AIR
0
1.0
1.0
:-
2.0
4.0
8.0
-8.0
-4.0
-2.0-1.0
-0.5-0.4
-0.3
-0.2
-0.1
0.1
0.2
0.3
0.4
0.5
0.6
0.8
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0-1
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0
500
1000
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2000 3000
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SENS
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HEA
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HEAT
Qt
ENTH
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IDIT
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TIO
Dh
DW
RR
ASHR
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ROM
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NO.1
NORM
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RATU
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ROM
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RE: 2
9.92
1 IN
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OF
MER
CURY
Copy
right
199
2AM
ERIC
AN S
OCI
ETY
OF
HEAT
ING
, REF
RIG
ERAT
ING
AND
AIR
-CO
NDIT
IONI
NG E
NGIN
EERS
, INC
.
SEA
LEVE
L
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
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NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
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RE: 1
01.3
25 k
PaCo
pyrig
ht 1
992
AMER
ICAN
SO
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Y O
F HE
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G, R
EFRI
GER
ATIN
G A
ND A
IR-C
OND
ITIO
NING
ENG
INEE
RS, I
NC.
SEA
LEVE
L
1020
3040
5060708090100
110
110
120
120 E
NTH
ALP
Y -
KJ
PE
R K
ILO
GR
AM
OF
DR
Y A
IR
10
20
30
40
50
60
70
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90
100
ENTHALPY - K
J PER KILO
GRAM OF D
RY AIR
SATURATION TEMPERATURE - °
C
5
10
15
20
25
30
35
40
45
50
DRY BULB TEMPERATURE - °C
.001
.002
.003
.004
.005
.006
.007
.008
.009
.010
.011
.012
.013
.014
.015
.016
.017
.018
.019
.020
.021
.022
.023
.024
.025
.026
.027
.028
.029
.030
10%
REL
ATIV
E H
UM
IDIT
Y
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 W
ET B
ULB
TEM
PERA
TURE
- °C
30
0.78
0.80
0.82
0.84
0.86 VOLUME - CUBIC METER PER KG DRY AIR
0.88
0.90
0.92
0.94
HUMIDITY RATIO - KILOGRAMS MOISTURE PER KILOGRAM DRY AIR
0
1.0
1.0
:-
1.5
2.0
4.0
-4.0
-2.0-1.0
-0.5
-0.2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2.0
2.5
3.0
4.0
5.0
10.0
- ::
SENS
IBLE
HEA
TQ
sTO
TAL
HEAT
Qt
ENTH
ALPY
HUM
IDIT
Y RA
TIO
Dh
DW
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
ART
NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
ETRI
C PR
ESSU
RE: 1
01.3
25 k
PaCo
pyrig
ht 1
992
AMER
ICAN
SO
CIET
Y O
F HE
ATIN
G, R
EFRI
GER
ATIN
G A
ND A
IR-C
OND
ITIO
NING
ENG
INEE
RS, I
NC.
SEA
LEVE
L
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
ART
NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
ETRI
C PR
ESSU
RE: 2
9.92
1 IN
CHES
OF
MER
CURY
Copy
right
199
2AM
ERIC
AN S
OCI
ETY
OF
HEAT
ING
, REF
RIG
ERAT
ING
AND
AIR
-CO
NDIT
IONI
NG E
NGIN
EERS
, INC
.
SEA
LEVE
L
1015
2025
303540455055
55
60
60
EN
THA
LPY
- B
TU P
ER
PO
UN
D O
F D
RY
AIR
15
20
25
30
35
40
45
50
ENTHALPY - B
TU PER POUND OF D
RY AIR
SATURATION TEMPERATURE - °
F
35
40
45
50
55
60
65
70
75
80
85
90
95
100
105
110
115
120
DRY BULB TEMPERATURE - °F
.002
.004
.006
.008
.010
.012
.014
.016
.018
.020
.022
.024
.026
.028
10%
REL
ATIV
E H
UM
IDIT
Y
20%
30%
40%
50%
60%
70%
80%
90%
35
3540
4045
4550
5055
5560
6065
65
70
70
75
75
80
80
85 W
ET B
ULB
TEM
PERA
TURE
- °F
85
90
12.5
13.0
13.5
14.0 VOLUME- CU.FT. PER LB. DRY AIR
14.5
15.0
HUMIDITY RATIO - POUNDS MOISTURE PER POUND DRY AIR
0
1.0
1.0
:-
2.0
4.0
8.0
-8.0
-4.0
-2.0-1.0
-0.5-0.4
-0.3
-0.2
-0.1
0.1
0.2
0.3
0.4
0.5
0.6
0.8
-200
0-1
000
0
500
1000
1500
2000 3000
5000
- ::
SENS
IBLE
HEA
TQ
sTO
TAL
HEAT
Qt
ENTH
ALPY
HUM
IDIT
Y RA
TIO
Dh
DW
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
ART
NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
ETRI
C PR
ESSU
RE: 2
9.92
1 IN
CHES
OF
MER
CURY
Copy
right
199
2AM
ERIC
AN S
OCI
ETY
OF
HEAT
ING
, REF
RIG
ERAT
ING
AND
AIR
-CO
NDIT
IONI
NG E
NGIN
EERS
, INC
.
SEA
LEVE
L
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
ART
NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
ETRI
C PR
ESSU
RE: 1
01.3
25 k
PaCo
pyrig
ht 1
992
AMER
ICAN
SO
CIET
Y O
F HE
ATIN
G, R
EFRI
GER
ATIN
G A
ND A
IR-C
OND
ITIO
NING
ENG
INEE
RS, I
NC.
SEA
LEVE
L
1020
3040
5060708090100
110
110
120
120 E
NTH
ALP
Y -
KJ
PE
R K
ILO
GR
AM
OF
DR
Y A
IR
10
20
30
40
50
60
70
80
90
100
ENTHALPY - K
J PER KILO
GRAM OF D
RY AIR
SATURATION TEMPERATURE - °
C
5
10
15
20
25
30
35
40
45
50
DRY BULB TEMPERATURE - °C
.001
.002
.003
.004
.005
.006
.007
.008
.009
.010
.011
.012
.013
.014
.015
.016
.017
.018
.019
.020
.021
.022
.023
.024
.025
.026
.027
.028
.029
.030
10%
REL
ATIV
E H
UM
IDIT
Y
20%
30%
40%
50%
60%
70%
80%
90%
5
5
10
10
15
15
20
20
25
25
30 W
ET B
ULB
TEM
PERA
TURE
- °C
30
0.78
0.80
0.82
0.84
0.86 VOLUME - CUBIC METER PER KG DRY AIR
0.88
0.90
0.92
0.94
HUMIDITY RATIO - KILOGRAMS MOISTURE PER KILOGRAM DRY AIR
0
1.0
1.0
:-
1.5
2.0
4.0
-4.0
-2.0-1.0
-0.5
-0.2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
-5.0
-2.0
0.0
1.0
2.0
2.5
3.0
4.0
5.0
10.0
- ::
SENS
IBLE
HEA
TQ
sTO
TAL
HEAT
Qt
ENTH
ALPY
HUM
IDIT
Y RA
TIO
Dh
DW
RR
ASHR
AE P
SYCH
ROM
ETRI
C CH
ART
NO.1
NORM
AL T
EMPE
RATU
REBA
ROM
ETRI
C PR
ESSU
RE: 1
01.3
25 k
PaCo
pyrig
ht 1
992
AMER
ICAN
SO
CIET
Y O
F HE
ATIN
G, R
EFRI
GER
ATIN
G A
ND A
IR-C
OND
ITIO
NING
ENG
INEE
RS, I
NC.
SEA
LEVE
L