Topic 5-1: Cooling/Heating Load

Supply Air (SA)

Return Air (RA)

Outside Air (OA)

AHU

F CC&DHHC H

RTH = Room/Building Total Heat load RSH = Room Sensible Heat load RLH = Room Latent Heat load

RTH = RSH + RLH

Fan

Duct losses, Qs,duct + Ql,duct

VentilationQs,vent + Ql,vent

CTH = CSH + CLH

CTH = Coil Total Heat load CSH = Coil Sensible Heat load CLH = Coil Latent Heat load

Qtr = Qsr + QlrQtc = Qsc + Qlc

Qsc = Qsr + Qs,fan + Qs,vent + Qs,duct Qlc = Qlr + Ql,vent + Ql,duct

Fan motorQs,fan

Design of Air Conditioning System

1. Selection of inside and outside condition

2. Estimation of requirement cooling/heating equipment

3. Selection cooling/heating equipment

4. Selection supply conditions5. Design of air transmission and distribution systems

• Input for Room heat load

1. Inside condition : thermal comfort of people, condition for storage of products, process conditions

2. Outside condition : location, structure type, usage

3. Building specification : design, drawing, solar radiation

Input 1. Selection of inside condition

Consumes food → metabolism → heat and work

Generates heat at a rate 100 – 2000 W

Core temp.neutral condition →

tcore < 31C→ Lethal/deadly

tcore 35Ctcore 36.8Ctcore 39Ctcore > 43C → Lethal/deadly

Slight discomfort

Slight discomfort

100 W → sedentary person, sitting2000 W → strenuous exercise

Skin temp., tskin 33.7C

• Thermal comfort: A living human body as a heat engine

Heat balance equation for a human being

Qgenerate = Qskin + Qrespiration + Qstore

Qgenerate M = c*met* As,naked

met = 58.2 W/m2

As,naked = 0.202mass0.425height0.725

Heat balance equation for a human being

Qgenerate = Qskin + Qrespiration + Qstore

Qskin = Qconvection + Qradiation + Qevapoation

Qconvection = 14.8Vair0.5(tskin - tair)

Qradiation = 11.603(tskin - tsurr)

Qevapoation = 181.76Vair0.4 (psat,skin - psat,air)

Qrespiration = Qdry heat loss + Qevaporative heat loss

Qstore = 0 at neutral condition tcore 36.8C

30% + 30% + 40%

Thermo-regulator system

A human body is very sensitive to temperature.

Temperature depends on heat balance which depends on

surrounding conditions.

In-built body regulatory processes against cold and heat:Try to maintain body temp. by initiating certain body regulator processes against cold and heat.

1. Colder troom < tcore 36.8C

→ Blood vessels constrict/compress → blood flow → heat transfer

→ Heat generation to take care of heat losses → body temp.

2. Hotter troom > tcore 36.8C

→ Blood vessels dilate/expand → blood flow → heat transfer

→ Sweat , evaporation to prevent body temp.

→ heat exhaustion (headache) → heat cramps (salt loss) → heat stoke

7

Comfort zone for air conditioning

Comfort zone

→ Clothes→ Activities→ Time interval→ Metabolism, ages, gender→ Air velocity→ Outside condition

• Summer comfort boundary22 – 29C at 30%RH20 – 24.5C at 70%RH

• Winter comfort boundary19 – 25.5C at 30%RH18 – 23C at 70%RH

18 C 29 C

Input 2. Selection of outside condition

Ambient temp. (tdb) and moisture content (W) vary from hour-to-hour and from place-to-place.

→ Construction types; heavy, medium, light→ Insulation characteristics of building→ Area of glass/transparent surfaces→ Type of usage→ Nature of occupancy→ Daily range (tmax – tmin in a given day)

Summer: tdb from sunrise, reach tmax at afternoon and towards evening

Am

bie

nt

tem

p.

Time, 0-24 h

: %RH vary with tdb, reach RHmin at tmax

Design of outside conditions

Outdoor Design For Cooling

Ref: 2005 ASHRAE Handbook-Refrigeration, Ch.14 Climate Design Information

Bangkok/Don MuangLat 13.92N,Long 100.60E, Elev 12 mHottest month: 4 – April37.2 CDB, 26.5 CWB0.4% Design level (0.4% h of a year = 35*100/8670) or exceeded 99.6%MCDB = Mean Coincident Wet Bulb

0.4%, 1%, 2%; tdb is equaled or exceeded 99.6%, 99.0%, 98.0%

Recommended0.4% → light construction types, poorly insulated, critical space temp.1% → medium construction 2% → heavy construction

ตารางที ่2.2 สภาวะอากาศภายนอกส าหรับการออกแบบ

อา้งอิง: มาตรฐานระบบปรบัอากาศและระบายอากาศ 2561, วิศวกรรมสถานแห่งประเทศไทย ในพระบรมราชปูถมัภ ์(วสท.)

Don Muang37.2 CDB, 26.5 CWB0.4% Design level

0.4% Design level คือ รอ้ยละของจ านวนชั่วโมงตลอดทัง้ปีที่ยอมใหส้ภาวะอากาศภายนอกจรงิสงูกวา่สภาวะอากาศภายนอกส าหรบัการออกแบบได้

Input 3. Building specification; Solar radiation

Solar constant, flux of solar radiation is 1370 W/m2

The earth’s elliptical orbit→ extra radiant flux 1325 – 1418 W/m2

Total solar irradiation at :

Iirradiation = IDirect + Idiffuse + Ireflect angle of incidence

The sun as a radiant energy source, a blackbody at Ts = 6000 K

altitude angle

Total solar irradiation = IDirect + Idiffuse + Ireflect

A = apparent solar radiation, 1080 – 1230 W/m2

B = atmospheric extinction coefficient, 0.14– 0.21

Calculation model in ASHRAE: IDirect = Ae(-B/sin) = ID

C = constant for clear sky, 0.058 – 0.135 FWS = view factor or configuration factor = 0.5(1+cos), = tilt angle

FWS = 1 for horizontal surface and FWS = 0.5 for vertical surface

Idiffuse = CIDFWS = Id

g = reflectivity of ground/horizontal surface, 0.058 – 0.135 FwG = view factor from ground to surface = 0.5(1-cos)FwG = 0 for horizontal surface and FwG = 0.5 for vertical surface

Solar radiation reflected from the ground: Ireflect = (ID+ Id)gFWG

Solar radiation through Transparent/Glazed, QSG

clear plate glasstransmittivity τ = 0.80 reflectivity ρ = 0.08 absorptivity α = 0.12

Transmit

Reflect

Absorb

transfer to indoor

transfer to outdoor

Assume transmittivity and absorptivity same as IDirect , Idiffuse , Ireflect

QSG = A(+ Nα)Itotal radiation

N = fraction of absorbed radiation transferred to indoor

Steady, N = U/ho

QSG = AIt (+ αU/ho)

Solar Heat Gain Factor SHGF = [It (+ αU/ho)]Ref. glass

QSG = A.SHGF.SCSC = Shading Coefficient

Solar Heat Gain Factor QSG = A.SHGF.SC

A glass facing south is desirable for cooling in summer as it allows low SHGF and heating in winter as it allows high SHGF.

Heat transfer through Building

Analysis of heat transfer is complex:

→ Complex structure - different thermo-physical properties→ Continuously varying outdoor conditions (solar, temp., wind)→ Continuously varying indoor conditions (load patterns)

Winter, peak heat load occur at early morning before sunrise→ Solar radiation effects not considered

→ Ambient temp. variation effects not considered → Steady heat transfer

Cooling load: consider transient heat transfer for more accuracy analysis

1D steady state heat transfer through buildingsqin = (qconv,o + qrad,o) = (qconv,i + qrad,i)

qin = ho(To - Tw,o) = hi(Tw,i - Ti)

ho, hi = inner, outer surfaceconductances, account both convection and radiation heat transfers

Surface conductance values in W/m2.K for different orientations, air velocities and surface emissivity

Typical conductance values of air spaces

Unsteady heat transfer through opaque walls and roofs

The outer wall (x = L) is subjected to - αDID direct radiations - αdId diffuse radiations- R reflected radiation - ho(To-Tw,o) convective heat transfer

Control surface at x=L at time ;qL, = -kw(T/x)L, = ho(To-TL) + αDID + αdId - R

The inner wall (x=0);- hi(Ti,w-Ti) convective+ radiation heat transfer

Control surface at x=0 at time ;q0, = -kw(T/x)0, = hi(T0 - Ti)

1D, transient heat conduction though a plane wall; 2T/x2 = (T/)For T(x,), = (k/cp)w

Initial condition at time =0; Tx,0 = Ti(x)

sol-air temperature effective outdoor temp. combines convection + radiation Tsol-air = To + (αDID + αdId - R)/ho = Te

qL, = -kw(T/x)L, = ho(Tsol-air -TL)

Because of thermal capacity of wall

q0, qL,

Analytical solution with periodic boundary conditionsAn infinite Fourier series consisting of various harmonics;

Tsol-air = Tsol-air,m + M1cos1 + N1sin1+ M2cos2 + N2sin2 + …

Solution; Tx, = A + Bx +(CncosPnmx + DnsinPnmx)exp(-m22)

Q0, = UA(Te,m - Ti) + UA(Te, - - Te,m) = UATeff

Rate of heat transfer from the inner surface;q0, = function(U, Te,m , Te,1 , Te,2 ,T0,0 , time lag n , decrement factor n )

Mean sol-air temp.;

Tsol-air,m = Te,m 0)=24

Tsol−air𝑑)/24

Load Calculation Methods1.Rule of thumb method

- Least accurate - eg. 400 - 1800 (Btu/h)/m2

2.Dynamic analysis (TETD/TA, TFM)- Computer modeling

- Total equivalent temp. diff. TETD/TA by ASHRAE 1967

- Transfer function method, TFM by ASHRAE 1972- hour-by-hour calculation- calculate heat gain → conversion to cooling load

3.Static analysis (ETD/CLTD/CLF/SHGF/SC)- simplified TFM, troom = constant)

- hand calculation by ASHRAE 1977, then 1989

- CLTD/CLF/SHGF/SC method

Semi-empirical method based on ETD or CLTD

thermal capacity of wall → q0, qL, → time lag & decrement

Heat transferred to the outer surface of the wall is partly stored and partly transferred to the conditioned space. Due to the thermal energy storage, temperature of the wall increases and transfer to outside or/and

inside room.

Variation of heat transfer rate with time

The net effect is a greatly reduced cooling load on the building for thick walls. Requirement of cooling system of much larger capacity (hence high initial cost) for buildings with thin walls compared to thick walls.

Empirical method for Cooling estimation

ETD/CLTD values in K for roofs without suspended ceiling (ASHRAE)

Q0, = UA(Te,m - Ti) + UA(Te, - - Te,m) = UATeff

Principle of cooling load

• Related heat flow rates varying with time:

1. Heat gain

2. Cooling load

3. Heat extraction

The instantaneous rate at which heat enters into, out of, or generated within a space

The rate at which heat must be removed from a space to maintain air temperature and humidity at the design values

by equipment

Cooling load differs from heat gain due to

* delay effect of conversion of radiation energy to heat

Due to thermal storage lag

Radiant energy is first absorbed by structure

Heat is convected from the structure surface

Surface Temperature

Cooling load ≠ Heat gainThermal storage and Construction type

Actual cooling load By heavy construction

Actual cooling load By light construction

Heat Gaininstantaneous rate

roof

lights

equipment

floor

glass solar

exteriorwall

glassconduction

infiltrationpeople

partitionwall

25.5 C (78 F): Indoor condition

35 C (95 F): Outdoor condition

1. External Load

2. Internal Load

1.1 Transmission

1.2 Solar radiation

1.3 Ventilation & Infiltration

2.1 Light

2.2 Equipment

2.3 People

Total cooling Load

-temperature-moisture

content-air motion-air quality-noise

Cooling Load

1. External Loads1.1 Transmission: Heat gains from Walls and roofs –Sensible

1.2 Solar radiation: Solar gains through fenestrations–Sensible

1.3 Outdoor air : Ventilation, Infiltration–Sensible & Latent

2. Internal Loads2.1 Light –Sensible

2.2 People –Sensible & Latent

2.3 Appliances –Sensible & Latent

Winter, 1.2 and 2 effects not considered

1.1 Transmission Load (1)

• Sensible heat gain through floor, partition and ceiling at steady state:

q = UAΔt = (A/Rtot)(to– ti) where

q = heat gain, W

U = overall heat transfer coefficient, W/(m2·K) = 1/ Rtot

Rtot = total thermal resistance K/W

A = surface area, m2

Δt = (to–ti) = difference between outside air temperature and air temperature of the conditioned space, °C

= ntot RR

1.1 Transmission Load (2)

• The overall coefficient of heat transfer U :

where

xn = wall thickness for material n for composite material, m

kn = thermal conductivity of wall material n, W/(m·K)

hi , ho = inside, outside surface conductance, W/(m2·K)

( ) ( ) totonionni RRRRhkxhU

11

/1//1

1=

++=

++=

1.1 Transmission Load (3)

Table 4-4 Thermal resistance of selected building materialExterior materialSheathingRoofingConcreteInsulating materialsInterior materialsAir resistanceFlat glass

Studs 25%insulation 75%

R1 R2 R3 R4 R6 R7

R5

Example 1. Element--------------------------------------------------1. Outside surface2. Wood bevel lapped siding3. Rigid foam insulating sheathing 4. Mineral fiber batt insulation5. Wood stud (38 mm 90 mm)6. Gypsum wallboard 13-mm7. Inside surface

--------------------------------------------------

R, m2·K/W------------------------------------R1 = 0.044R2 = 0.14R3 = 0.7R4 = 2.3R5 = 0.63R6 = x/k = 0.013/0.282 = 0.046R7 = 0.12------------------------------------

resistance networkRwall = R1 + R2 + R3 + R45 + R6 + R7

Rwall = 0.884 + 1.383 + 0.166 = 2.433 m2·K/W

Uwall = 1/Rwall = 0.411 W/m2·K

1/R45 = 0.75/R4 + 0.25/R5

R45 = 1.383

R1 + R2 + R3 = 0.884

R6 + R7 = 0.166

If the stud is not included;Rwall = 0.884 + 2.3 + 0.166 = 3.35 m2·K/WUwall = 1/Rwall = 0.298 W/m2·K

1.2.1 Solar loads through transparent surfaces (1)

Static analysis (SHGF/SC)

qsg = AIt (+N)

qsg = AIt(+U/ho)

Solar energy passing through the glazing surface, qsg

It = irradiation, W/m2

N = fraction absorbed = U/ho

It(+U/ho) = SHGF

SHGF = solar heat gain factorSC = shading coefficient

qsg = (SHGFmax )(SC)A

++ = 1

1.2.1 Solar loads through transparent surfaces (2)

qsg = A(SHGFmax )(SC)CLFSHGF = solar heat gain factor

Time of the Day: Solar RadiationR

adia

tio

n In

ten

sity

, W/m

2

Time, 0-24 h

1.2.1 Solar loads through transparent surfaces (3)

qsg = A(SHGFmax )(SC)CLFSC = shading coefficient

1.2.1 Solar loads through transparent surfaces (4)

qsg = A(SHGFmax )(SC)CLFgCLF = cooling load factor

CLF accounts for the delay before radiative gains becomes a cooling load

CLFg for glass

1.2.2 Solar loads on opaque surfaces (1)

+ = 1

Solar energy on opaque surface (wall, roof), qw

qw = AIt (N)

qw = AIt(Uw/ho)

With air-temp. difference (to–ti)

qw = AIt(Uw/ho) + UwA(to–ti)

qw = UwA[(to+ It/ho) - ti)]

te = to+ It/ho

qw = UwA(te–ti)

te = sol-air temperature, include solar

→ over-estimate the heat gain

By include effect of thermal storage → qw = UwA(CLTD)

1.2.2 Solar loads on opaque surfaces (2)

Static analysis (CLTD/CLF)qw = UA(CLTD)CLFrCLTD = cooling load temp. difference

CLTD for roofs

1.2.2 Solar loads on opaque surfaces (3)CLTD for wall

1.2.2 Solar loads on opaque surfaces (4)Lower-mass wall has higher peak heat flux and occurs earlier

1.3 Outdoor air (1)

Ventilation loads –Sensible & Latent

qis = 1.23 Qvol (to– ti)

qil = 3010 Qvol (Wo– Wi)

Qvol = volumetric flow rate of outside air, L/s

W = humidity ratio, water to air, kg/kg

Sensible load = flow rate x specific heat x (ΔT)

Latent load = flow rate x latent heat of vaporization x (humidity difference)

Ventilation loads due to entry of outside air by mechanical device

qtot = 1.2 Qvol (Ho– Hi)

1.3 Outdoor air (2)

Infiltration loads –Sensible & Latent

Qvol = (ACH) (space volume) m3/h

V = wind velocity, m/s

ACH = a + bV + c (to– ti)

Infiltration loads due to entry of outside air by natural mean

- Estimate in terms of number of air change per hour (ACH)

2. Internal Loads: 2.1 Light

heat gain by light –Sensible

q = (lamp rating in watts) (Fu)(Fb) (CLFl)

Fu = utilization factor for fraction of installed lamps in use

Fb = ballast factor for fluorescent lamps = 1.2 for most common

2. Internal Loads: 2.2 People

heat gain by people –Sensible & Latent

qs = (sensible heat gain per person) (number of people) (CLFp)

ql = (latent heat gain per person) (number of people)

CLFp for people

2. Internal Loads: 2.3 Appliances

Heat gain from equipment or appliances –Sensible & Latent

Heat dissipation from these equipments are mainly based data published by the manufacturers and standard handbook

Total cooling loadpartition, floor, ceiling, glass qt = UA (to– ti)

transparent (glass) qs = A(SHGF)(SC)(CLF)

opaque (wall, roof) qs = UA(CLTD)(CLF)

qis = 1.23 Qvol (to– ti)

qil = 3010 Qvol (Wo– Wi)

ql = (lamp rating in watts) (Fu)(Fb=1.2)(CLF)

qel = (sensible energy rate)(CLF)

qel = (latent energy rate)

qpl = (sensible HG)(no. of people)(CLF)

qpl = (latent HG)(no. of people)

Safety factor 5-10%

Summary of Cooling Loads Estimating

1. Select outdoor design condition

2. Select indoor design condition

3. Estimate adjacent unconditioned spaces

4. Compute U-value for building component

5. Estimate rate of infiltration and/or ventilation of outside air

6. Determine additional building characteristics influence solar-heat gain

7. Determine ETD/CLTD, CLF,SHGF, SC

8. On the basis of U, Area, TD, calculate the rate of heat gain to the space

9. Estimate internal heat gain from lights, equipments, people

10. Sum all load component to determine the maximum capacity

Reference: ASHRAE Fundamental Handbook

A Year-round air conditioning system for a small commercial establishment such as retail store.

1.Hotwater boiler

2.Hot water coil

3.Refrigerating machinery

4.Cooling coil6.Blower Fan

12.Filter5.Humidifier

Air conditioning plenum

11.Outside air intake

10.Return-air duct

8.Supply air diffuser, Qs

7.Supply air duct

9.Return-air grille

AHU

Mechanical room

Conditioned Space

Room load

Coil load

Cooling Load Calculation

Purpose:

• Load profile over a day

• Peak load (basis for equipment sizing)

• Operation Energy analysis

• HVAC Construction cost

(HVAC-Heating Ventilation Air-Conditioning)

Co

olin

g Lo

ad, W

Time, 0-24 h

Example of Cooling Load Estimation

Cooling Load Calculation Sheet

Cooling Load Calculation Software: TLC by TRANE

Cooling Load Calculation Software: TLC by TRANE

Cooling Load Calculation Software: TLC by TRANE

Cooling Load Calculation Software: TRACE Load® 700by Trane’s CDS Group

Cooling Load Calculation Software: Elite CHVAC by Elite Softwarehttp://www.elitesoft.com/web/hvacr/chvacx.html

58

Prob: 4-material wall (H,L), 3 glass window (h,l), ti, Chicago, qloss =? Assume: one-directional steady heat transfer through wall in winterSchematic:

P4.1 The exterior wall of a single-story office building near Chicago is 3 m high and 15 m long. The wall consists of 100-mm face brick, 40-mm polystyrene insulating board, 150-mm lightweight concrete block, and an interior 16-mm gypsum board. The wall contains three single-glass windows 1.5 m high by 2 m long. Calculate the heat loss through the wall at design conditions if the inside temperature is 20C.

A: 100-mm face brickB: 40-mm polystyrene insulating boardC: 150-mm lightweight concrete blockD: interior 16-mm gypsum board

A B CWallH x L

Glass window

h x l

Wall section

D

Model: qloss = UwAw(to - ti)+UgAg(to - ti)to : winter@Chicago → Design outdoor dataAg = 3*h*l, Aw = H*L – Ag

Uw = 1/Rtot , Rtot = Ri +RA+RB+RC+RD+ Ro

Ug ,R values → Thermal resistance data

qcondqconv,iqconv,o

P4.1 : 4-material wall (H,L), n glass window (h,l), ti, near Chicago, qloss =?

A: 100-mm face brick → Exterior material; RA= 0.1*0.76 m2.K/WB: 40-mm polystyrene insulating board → Insulating material; RB= 0.04*27.7 m2.K/WC: 150-mm lightweight concrete block→ Concrete; RC= 0.15*1.94 m2.K/W

D: interior 16-mm gypsum board → Interior material; RD= 0.1 m2.K/W

P4.1 : 4-material wall (H,L), n glass window (h,l), ti, near Chicago, qloss =?

Inside wall→ Air resistance; Ri= 0.12m2.K/WOutside wall→ Air resistance; Ro= 0.029 m2.K/Wsingle-glazed window→ Flat glass; Ug= 6.2 m2.K/W

61

P4.2 For the wall and conditions stated in P4.1 determine the percent reduction in heat loss through the wall if (a) the 40 mm of polystyrene insulation is replaced with 55 mm of cellular polyurethane, (b) the single-glazed windows are replaced with double-glazed windows with a 6-mm air space. (c) If you were to choose between modification (a) or (b) to upgrade the thermal resistance of the wall, which would you choose and why?

Prob: [ABCD] wall (H,L), 3 glass window (h,l), ti, near Chicago, q(a) B replaced with B1, (q1-q)*100%/q =? (b) single-glazed replaced with double-glazed, (q2 -q)*100%/q =?

A: 100-mm face brickB: 55-mm cellular polyurethaneC: 150-mm lightweight concrete blockD: interior 16-mm gypsum board

(a) Model: q1 = Uw1Aw(to - ti)+UgAg(to - ti)Uw1 = 1/Rtot1, Rtot1 = Ri +RA+RB1+RC+RD+ Ro

if RB1>RC → Rtot1>Rtot→ Uw1<Uw → q1< q

(b) Model: q2 = UwAw(to - ti)+Ug2Ag(to - ti)Ug2 → Ug2<Ug→ q2< q

62

P4.3 An office in Houston, Texas, is maintained at 25C and 55 % relative humidity. The average occupancy is five people, and there will be some smoking. Calculate the cooling load imposed by ventilation requirements at summer design conditions with supply air conditions set at 15C and 95 % relative humidity if (a) the recommended rate of outside ventilation air is used and (b) if a filtration device of contaminant removal efficiency E = (Vo – Vm)/Vr = 70 % is used. (Office; Vo = 10 L/s/person, Vm = 2.5 L/s/person)

Prob: Office (ti, i), Houston, Texas, 5 people (smoking), supply (ts, s) air qventilation = ?, (a) outside ventilation, (b) filtration device of E

Vs = Vr+Voutside

Vr

Voutside

(b) Model: Vr = (Vo – Vm)/E, Voutside = Vm, Vs = Vr+Vm

qis = 1.23Vm(to– ti) + 1.23Vs(ts– ti)

qil = 3010Vm(Wo– Wi) + 3010Vs(Ws– Wi)qtot = qis + qil

Ws= W(ts, s)

(a) Model: 100% Voutside, Vr=0Vs = Voutside = Vo

qis = 1.23Vo(to– ti)

qil = 3010Vo(Wo– Wi)qtot = qis + qil

to, twb,o : summer@Houston, TexasWo = W(to, twb,o)Wi = W(ti, i)

P4.4 : A computer room located on the second floor of a five-story office building is 10 by 7 m. The exterior wall is 3.5 m high and 10 m long; it is a metal curtain wall (steel backed with 10 mm of insulating board), 75 mm of glass-fiber insulation, and 16 mm of gypsum board. Single-glazed windows make up 30 % of the exterior wall. The computer and lights in the room operate 24 hr/day and have a combined heat release to the space of 2 kW. The indoor temperature is20C. (a) If the building is located in Columbus, Ohio, determine the heating load at winter design conditions. (b) What would be the load if the windows were double-glazed?

Prob: 3-material wall (H,L), 30% Ag , ti, qinternal, Columbus, Ohio, qheating =? Assume: one-directional steady heat transfer through wall in winter

A: metal curtain wall 10 mm insulating B: 75 mm of glass-fiber insulationC: 16 mm of gypsum board

A CWallH x L

Glass window Ag

Wall sectionModel: qheating = qloss – qinternal

qloss = UwAw(to - ti)+UgAg(to - ti)to : winter@ Columbus, OhioAw = H*L – Ag, Ug = 1/Rg

Uw = 1/Rtot , Rtot = Ri +RA+RB+RC+ Ro

R values → Thermal resistance data

qcond qconv,iqconv,o

B

(b) Ug2 = 1/Rg2 (double-glazed) qloss2 <qloss

P4.5 : Compute the heat gain for a window facing southeast at 32 north latitude at 10 A.M. central daylight time on August 21. The window is regular double glass with a 13-mm air space. The glass and inside draperies have a combined shading coefficient of 0.45. The indoor design temperature is 25C, and the outdoor temperature is 37C . Window dimensions are 2 m wide and 1.5 m high.Prob: double glass (W,H), SC=0.45, ti, to, SE 32 north, 10 A.M. August 21, qgain =? Assume: one-directional steady heat transfer through glass window

Double Glass Window

WxH

Model: qgain = UgAg(to - ti) + SHGFmax*SC*CLF*Ag

Ug : double glass with a 13-mm air spaceSHGFmax : August, SE, 32 north latitudeCLF : 10 A.M. , SE northAg = W*H

P4.6 : The window in P4.5 has an 0.5-m overhang at the top of the window. How far will the shadow extend downward?

Prob: double glass window with overhang d = 0.5 m, yshadow =?

Model: yshadow = d*tan/cos , : August, 10 A.M. , SE, 32 north = - facing southeast = 45

P4.7 : Compute the instantaneous heat gain for the window in P4.5 with the external shade in P4.6.

Prob: double glass (W,H), SC=0.45, ti, to, SE 32 north, 10 A.M. August 21,overhang d = 0.5 m, yshadow with the external shade (Ashade), qgain =?

Model: qgain = UgAg(to - ti) + SHGFmax*SC*CLF* Ashade

Ug : double glass with a 13-mm air spaceSHGFmax : August, SE, 32 north latitudeCLF : 10 A.M. , SE northAg = W*HAshade = W*(H-yshadow)

P4.8 : Compute the total heat gain for the south windows for an office building that has no external shading. The windows are double-glazed with a 6-mm air space and with regular plate glass inside and out. Draperies with a shading coefficient of 0.7 are fully closed. Make calculation for 12 noon in (a) August and (d) December at 32 North Latitude. The total window area is 40 m2. Assume that the indoor temperatures are 25C and 20C and that the outdoor temperatures are 37C and 4C

Prob: qwindow =? double-glazed window(Ag), SC=0.7, S 32 north, 12 noon (a)August (ti, to), (b) December (ti, to)

Model: qgain = UgAg(to - ti) + SHGFmax*SC*CLF* Ag

Ug : double glass with a 6-mm air spaceSHGFmax : (a) August, S, 32 north latitudeSHGFmax : (b) December, S, 32 north latitudeCLF : 12 noon , S north

P4.9 : Compute the instantaneous heat gain for the south wall of a building at 32 North Latitude on July 21. The time is 4 p.m. sun time. The wall is brick veneer and frame with an overall heat-transfer coefficient of 0.35 W/m2.K. The wall is 2.5 by 6 m with a 1-2-m window.Prob: qwall =? Wall (Uw, HW), window (h w), S 32 north, July, 4 p.m. sun timeAssume: ti = 25C, to = 35C , tav = 29C, daily rang 12C

Model: qwall = UwAw*CLTD, Aw= HW- h wCLTD : S, 4 p.m. , brick veneer and frame → Type F

P4.10 : Compute the peak instantaneous heat gain per square meter of area for a brick west wall similar to that in Example 4-3. Assume that the wall is located at 40 North Latitude. The date is July. What time of the day does the peak occur? The outdoor daily average temperature of 30C and indoor design temperature is 25C.

Prob: qpeak/A=? Time =? Wall (Uw,), W 40 north, July, to , ti

Model: qpeak/A = Uw*CLTDpeak,adj

Uw = 1/Rtot

CLTDpeak : W, brick with insulation → Type FCLTDadj = CLTD + (25 – ti) + (tav – 29)

P4.10 : qpeak =? Time =? Wall (Uw, 1 m2), W 4 0 north, July, to , ti

Model: qpeak/A = Uw*CLTDpeak,adj

Uw = 1/Rtot

CLTDpeak : W, brick with insulation → Type FCLTDadj = CLTD + (25 – ti) + (tav – 29)Time = time at CLTDpeak