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Review for Final Exam May 5, 2010

ME 483 – Alternative Energy Engineering II 1

Review for Final ExamReview for Final Exam

Larry CarettoMechanical Engineering 483

Alternative Energy Alternative Energy Engineering IIEngineering II

May 5, 2010

2

Final Exam• Monday, May 10, 3–5 pm• Open book and notes

– No books other than course text– No homework solutions or in-class

exercise solutions• Will be problems similar to those on

homework and in-class exercise• More credit for correct approach than

for details of algebra or arithmetic• No questions on material since second

midterm

33

What is energy?• Energy and power (energy/time) units

– Energy units: joules (J), kilowatt·hours(kWh), British thermal units (Btu)

• 1 Btu = 1055.056 J– Power units: watts (W), Btu/hr

• 1 W = 1 J/s = 3.412 Btu/hr– Fuel equivalencies: 1 ft3 natural gas ≈ 1000

Btu; 1 bbl crude = 5.8 MMBtu; 1 Mtoe oil =41.868x1015 J = 0.0387 quads

– World energy use (2006) was 466 quads 44

Energy Costs• Home costs (San Fernando Valley 2008)

– Electricity: $0.115/kWh = $32/GJ• Increase from $0.11/kWh to $0.12/kWh

– Natural gas: $1.07/therm = $11/GJ• One therm = 105 Btu is approximately the energy

in 100 standard cubic feet of natural gas• Range was $0.69 to $1.22 per therm

– Gasoline at $3.00 per gallon (including taxes) costs $26/GJ

• Assumes energy content of gasoline is 5.204 MMBtu per (42 gallon) barrel

• $100/bbl oil costs $6.20/GJ (5.80 MMBtu/bbl)– Energy cost without California gasoline

taxes ($0.585/gallon) is $21/GJ

5

Resources vs. Reserves

ResourcesResourcesNot

economical to recover

ResourcesReservesEconomical to Recover

UnknownKnown

6

Resource Probabilities



7

Hubbert Peak• Analysis due to M. King Hubbert• Main publications in 1949 and 1956• Correctly predicted peak in US oil

production in early 1970s• Not so accurate in other predictions• Some recent applications show world oil

production peak in next ten years• Many other studies show later peak

8

|QH| |QH|

|QL| |QL|

|W| |W|

High Temperature Heat Source

High Temperature Heat Source

Low Temperature Heat Sink

Low Temperature Heat Sink

Engine Cycle Schematic Refrigeration Cycle Schematic

HQW

=η

WQ

COP L=

Source

Sink

Refrigerator

WQ

COP H=

Heat Pump

WQQ LH +=For all cycles

Thermodynamic Cycles

9

Basic Combustion Analysis• General fuel formula: CxHySzOwNv

• x, y, z, w, and v from ultimate analysis or analysis of gas mixtures

• Ultimate analyses:– x = wt%C/12.0107, y = wt%H/1.00794,

z = t%S/32.065, w = wt%O/16.0004, v = wt%N/14.0067, mfuel = 100

– Mfuel = 12.0107x + 1.00794y + 32.065z + 15.9994w + 14.0067v = mfuel(1 – %MM)

• For mixture of compounds (ωk = mole fraction)∑=

specieskk xx ω ∑=

specieskk yy ω ∑=

specieskkfuel MM ω

10

Combustion Air• A = x + y/4 + z – w/2 = stoichiometric

moles O2/mole fuel• Need input data on Actual

O2/Stoichiometric O2 = Relative air/fuel ratio = λ

• Air/fuel ratio = mair/mfuel =138.28λA/mfuel• CxHySzOwNv + λA(O2 + 3.77 N2) →

xCO2 + (y/2)H2O + zSO2 + (λ – 1)AO2 + 3.77λA + v/2)N2

11

Exhaust Oxygen and λ• Can relate these two quantities with fuel

properties• Can compute theoretical %O2 for given λ

2v +z + A - A + x

1)A - ( = O% dry2

λ

λ

77.4100

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎥⎦⎤

⎢⎣⎡

λ

100O%

- A

2v +z + A - x

100O%

+ A =

dry2

dry2

77.41

• Dry exhaust has water removed to protect chemical analyzers

12

Emission Rates• Often stated as pollutant mass per unit

heat input from fuel• Equation used:• Compute ρi,d = yi,dMiPstd/RuTstd

• Fd is dry exhaust volume/heat input– Use default values or compute by equation

• Feb 3 notes have values of K’s and default Fd’s

dddii O

FE,2

, %9.209.20

−ρ=

( )c

NSOHCd Q

NKSKOKHKCKKF %%%%% ++++=



13

Other Equations• Pollutant mass per unit heat input

100%9979.1

100%6642.3 22 Swt

QQmCwt

QQm

cfuel

SO

cfuel

CO ==

• Combustion Efficiency (definitions on next slide)

cfuel

COT

Tp

ccomb QM

hxf dTc QFuelAir

qq out

in

Air

Δ−

⎥⎦⎤

⎢⎣⎡ +

−==η ∫ '1

1max

14

Combustion Efficiency– Air/fuel is the air to fuel (mass) ratio– Cp,air = 0.24 Btu/lbm▪R = 1.005 kJ/kg▪K– f = molar exhaust ratio CO/(CO + CO2)– x = carbon atoms in fuel formula, CxHy…– Qc = heat of combustion (Btu/lbm or kJ/kg)

• Use lower heating value for water vapor (usual case)

– ΔhCO 282,990 kJ/kgmol = 121,665 Btu/lbmol

– Mfuel is combustible fuel molar mass lbm/lbmol or kg/kmol

15

Energy Economics• Look at balance between initial cost and

ongoing costs– Uses interest rate to consider time value of

money• Key formula relates equivalence

between initial cost, P (present value), and ongoing payment stream, A (annual cost)

( ) nii

PA

−+−=

11( )

ii

AP

n−+−=

11

16

Using the A/P formula• Formula applies to any time period so

long as i is interest rate per time period• E. g., for monthly costs with i = 6%/yr =

0.5%/month for N months( ) NP

A−+−

=005.011

005.0

• Need trial-and-error solution (or financial calculator) to find i, given n and A/P

• Can find n for given i and A/P

( )iAPi

n+

⎟⎠⎞

⎜⎝⎛ −

−=1ln

1ln

17

Energy Storage Measures• Energy per unit mass (kJ/kg; Btu/lbm)• Energy per unit volume (kJ/m3; Btu/ft3)• Rate of delivery of energy to and from

storage (kW/kg; Btu/hr⋅lbm)• Efficiency (energy out/energy in)• Life cycles – how many times can the

storage device be used– Particularly important for batteries

18

Compare• Batteries

versus other motive power

• http://www. nap.edu/books/0309092612/html/40.html



1919 20

Renewable/Alternative• Alternative or renewable resources

– Solar energy– Wind energy – Ocean energy (tides, waves and

temperature gradients)– Geothermal energy– Hydropower especially small hydro– Biomass fuels– Conservation as an alternative resource

• Reduced usage and improved efficiencies including vehicle fuel economy

21

US Electric Net Summer Capacity (EIA Data)

0

2

4

6

8

10

12

14

16

18

Biomass Geothermal Solar Wind Total/100

Cap

acity

(GW

)

20002001200220032004200520062007

http://www.eia.doe.gov/cneaf/solar.renewables /page/prelim_trends /rea_prereport.html22

US Renewable Energy Use 2001-2007

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

ConventionalHydroelectric

GeothermalEnergy

Biomass Solar Energy Wind Energy

Energy Type

Ener

gy U

se (q

uads

)

2001

2002

2003

2004

2005

2006

2007

http://www.eia.doe.gov/cneaf/solar.renewables/page/prelim_trends /rea_prereport.html

23

Power Generation Costs

http://www.iea.org/textbase/papers/2006/renewable_factsheet.pdf24

Wind Power and Betz Limit• Power in incoming air = e = V2/2 =

(ρVA)V2/2 = ρAV3/2 = P0– Air density, ρ ≈ 1.2 kg/m3

– A = swept area of rotor = π(Drotor)2/4– V = wind velocity

• cp = power coefficient = turbine power divided by power in wind– Alternative: (generator power) / (wind power)

• Betz Limit: Maximum theoretical cp = 16/27 ≈ 0.593

m& m&



25

Effect of V3 Dependence

http://www.sandia.gov/wind/other/LeeRanchData-2002.pdf and http://en.wikipedia.org/wiki/Wind_power for plot

Energy calculations assume Betz cpand a 100 m rotor diameter

2626

Rayleigh Distribution• At least three variations are used

∞<≤β

=β−

VVeVfV

0)( 2

2 22

∞<≤=−

Vc

VeVfcV

02)( 2

22222 c=β

∞<≤π

=π−

VV

VeVfVV

02

)(22 4

ππβ22cV ==

2cVmp == β

27

Rayleigh Distributions

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 1 2 3 4 5 6 7 8 9 10V

f(V)

c = 1c = 1.5c = 3c = 5

2

222)(

cVeVf

cV−=

2cVmp =

2828

Weibull Distribution• A two-parameter distribution with shape

parameter, k, and scale parameter, c• Rayleigh distribution is Weibull

distribution with k = 2• Mean = cΓ(1 + k-1)• Variance = c2[Γ(1 + 2k-1) – Γ2(1 + k-1)]

( ) ∞<≤⎟⎠⎞

⎜⎝⎛= −

−

VecV

ckVf

kcVk

0)( /1

Γ is the gamma function

2929

( ) ∫∞

−−=Γ0

1 dtetx tx

( ) ( ) ( ) ( ) π=⎟⎠⎞

⎜⎝⎛Γ=Γ=ΓΓ=+Γ

211211 xxx

3030

Weibull Distributions

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10x

f(x)

k = 2; c = 1.5k = 2; c = 3k = 2; c = 5k = 4; c = 1.5k = 4; c = 3k = 4; c = 5

k = 2 gives Rayleigh

distribtuion

k = shape parameter c = scale parameter



31

Wind Power• Instantaneous wind power: P0 = ρV3A/2

31

• Total or average wind power: ∫

∞ρ

=ρ

=0

33

0 )(22

dVVfVAVAP

43

21

21

23

23

231

23

13

33333

33

π=⎟

⎠⎞

⎜⎝⎛Γ=⎟

⎠⎞

⎜⎝⎛Γ=⎟

⎠⎞

⎜⎝⎛ +Γ=⎟

⎠⎞⎜

⎝⎛

⎟⎠⎞

⎜⎝⎛ +Γ=⎟

⎠⎞⎜

⎝⎛

ccccV

kcV

Rayleigh

Weibull

• Total or average turbine power: 23

0 VAcPcP pptotal ρ==

32

Wind Power Distribution• Wind power between V1 and V2

– Weibull (Set k = 2 for Rayleigh)

– Found by numerical integration with results in tables

( )

( )

∫ −ρ=

k

k

cV

cV

yk dyeyAcVandVbetweenPWind/

/

33

21

2

12

( ) ( )[ ] ⎟⎠⎞

⎜⎝⎛ +Γ

ρ−=⎥

⎦

⎤⎢⎣

⎡13

2

3

1221 k

AcVfVfVandV

betweenPWindPP

33

Wind Turbine Operation• No operation until wind velocity reaches

a minimum called the cut-in velocity• Then operate at full turbine output

power until turbine output is greater than generator can accept

• Limit turbine output power to full generator power at high wind speeds

• No operation above maximum velocity called cut-out velocity

34

www.20percentwind.org/20percent_wind_energy_report_revOct08.pdf

35

Frequency and Power Distributions

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

0.60

0 1 2 3 4 5 6 7 8 9 10V

f(V) a

nd g

(P)

k = 2; c = 1.5 windfrequency, f(V)k = 2; c = 1.5 windpower, g(P)k = 2; c = 5 windfrequency, f(V)k = 2; c = 5 windpower, g(P)

36

Average Operating Power• Generator uses turbine power between

Vcut-in and rated (maximum power) velocity, VPmax = [2Pmax/(cpρA)]1/3

– Power coefficient cp = generator power divided by wind power

• Between VPmax and Vcut-out operate at maximum power

∫∫−

−

+ρ

=outcut

P

P

incut

V

V

V

V

poperation dVVfPdVVf

AVcP

max

max

)()(2 max

3



37

Average Operating Power II• Using power fraction table

( ) ( )[ ] ⎟⎠⎞

⎜⎝⎛ +Γ

ρ−=

ρ−∫

−

132

)(2

3

max

3max

kAcc

VfVfdVVfAVc p

PPincutP

V

V

pP

incut

• Using cumulative distribution( ) ( )

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞⎜

⎝⎛ −−⎟

⎠⎞⎜

⎝⎛ −= −− −

−

∫k

Pk

outcutoutcut

P

cVcVV

V

eePdVVfP //maxmax

max

max

11)(

( ) ( )[ ]( ) ( ) ⎟

⎠⎞⎜

⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ +Γ

ρ−=

−−−

−

koutcut

kP cVcV

pPPincutPoperation

eeP

kAcc

VfVfP

//max

3

max

max

132

38

• Radiation heat transfer by electromagnetic radiation– Part of much larger spectrum– Thermal radiation transfers

heat without contact• Use of fire or electric resistance

heating are best examples• Thermal radiation lies in infrared

and visible part of spectrum (with some in ultraviolet)

Electromagnetic Radiation

Figure 12-3 from Çengel, Heat and Mass Transfer

39

Black-body Radiation Spectrum• Basic black body equation: Eb = σT4

– Eb is total black-body radiation energy flux W/m2 or Btu/hr·ft2; σ is the Stefan-Boltzmann constant

• Ebλ is spectral radiation– Units are W/(m2⋅μm)– Ebλdλ is fraction of black body radiation in

range dλ about wavelength λ• Maximum occurs at λT = 2897.8 μm·K

– T increase shifts peak shift to lower λ40

Spectral EbλBlack Body Radiation

1

10

100

1,000

10,000

100,000

1,000,000

10,000,000

100,000,000

0.1 1 10 100Wavelength, λ, μm

Radiation W/m2-μm

T = 6000 KT = 5000 KT = 4000 KT = 3000 KT = 2000 KT = 1000 KT = 800 KT = 600 KT = 400 KT = 300 K

visible infrared

ultra-violet

41

Partial Black-body Power

∫λ

λλ− λ=1

10

0, dEE bb

∫λ

λλ λσ

=0

4 '1 dET

f b

Black body radiation between λ = 0 and λ = λ1 is Eb,0-λ1

Fraction of total radiation (σT4) between λ = 0 and any given λ is fλ

Figure 12-13 from Çengel, Heat and Mass Transfer42

Δλ


• Radiation in finite band, Δλ

( ) ( )TfTf

dET

dET

dET

f

bb

b

120

40

4

4

12

2

1

21

11

1

λ−λ=

λσ

−λσ

=λσ

=

∫∫

∫λ

λ

λ

λ

λ

λλλ−λ



43

Emissivity• Emissivity = ratio of actual radiated

power to that of black body– Diffuse surface – emissivity does not

depend on direction– Gray surface – emissivity does not depend

on wavelength– Gray, diffuse surface – emissivity is the

does not depend on direction or wavelength

• Simplest surface to handle and often used in radiation calculations

44

Properties• When radiation,

G, hits a surface a fraction ρG is reflected; another fraction, αG is absorbed, a third fraction τG is transmitted

• Energy balance: ρ + α + τ = 1


45

Kirchoff’s Law• Absorptivity equals emissivity (at the

same temperature) αλ = ελ

• True only for values in a given direction and wavelength

• Assuming total hemispherical values of α and ε are the same simplifies radiation heat transfer calculations, but is not always a good assumption

46

Effect of Temperature• Emissivity, ε, depends on surface

temperature• Absorptivity, α, depends on source

temperature (e.g. Tsun ≈ 5800 K)• For surfaces exposed to solar radiation

– high α and low ε will keep surface warm– low α and high ε will keep surface cool– Does not violate Kirchoff’s law since

source and surface temperatures differ

47From Çengel, Heat and Mass Transfer 48

Solar Angles III

from the sun center



49

Tangent plane to earth’s surface at given location 50

Equation of Time

Solar Time = Standard Time +

Equation of Time +

(4 min/o) * (Standard Longitude –Local Longitude)Standard Time = DST – 1 hour

51

Computing the Sun Path• Input data: Latitude, L, date, hour h• Find declination from serial date, n

( ) ( ) ( )degreesinδ⎥⎦⎤

⎢⎣⎡ π

+=δ180

284365360sin45.23 no

• Two angles: altitude (α) and azimuth (φ)– sin(α) = sin(L) sin(δ) + cos(L) cos(δ) cos(h)– sin(αs) = sin(φ) = cos(δ) sin(h) / cos(α)– Sun path is plot of α vs. φ = αs for one day– Plot is symmetric about solar noon– Typically plot data for 21st of month 52

53

Solar Irradiation by Month in Los Angeles (LAX)Average of Monthly 1961-1990 NREL Data for different collectors

0

1

2

3

4

5

6

7

8

9

10

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Month

Irrad

iatio

n (k

Wh/

m2 /day

)

Fixed, tilt=0

Fixed, tilt=L-15

Fixed, tilt=LFixed,tilt=L+15

Fixed,tilt=90

1-axis,track,EW horizontal

1-axis,track,NS horizontal

1-axis,track,tilt=L

1axis,tilt=L+15

2-axis,track

Notes:All fixed collectors are facing southThe L in tilt = L means the local latitude (33.93oN)

54

Optimum Fixed Collector Tilt

35o



55

Solar

Insolationby collectororientationand monthat 40oNlatitude

56

57 58

Active Indirect Solar Water Heating

http://www.dnr.mo.gov/energy/renewables/solar6.htm (accessed March 12, 2007)

59

Passive Direct Solar Water Heatinghttp://southface.org/solar/solar-roadmap/solar_how-to/batch-collector.jpg (accessed March 12, 2007)

60

Flat Plate Collector2L = distance between outside of flow tubes

D = flow tube outer diameter

t = absorber plate thickness

w = distance between flow tube centerlines = 2L + D

Di = flow tube inner diameter

k = absorber thermal conductivity



61

Solar Collector Analysis• Three analysis steps for solar energy to

heat fluid (Hottel-Whillier-Bliss equation)– Solar energy into plate flows across plate

to location of tubes at some line on plate– At same line heat flow into collector fluid

from plate is determined– Integrate heat flow into fluid from inlet to

exit to get total useful heat transfer to fluid( )[ ]ainfcacRu TTUHAFQ −−= .

&

( )infoutffu TTmQ .. −= && 62

Loss Through Top• Analyze set of series thermal

resistances with common Qtop– Heat transfer between absorber plate and

lower glass plate shown below

( ) ( )2

222,2

gp

gPgPcgprgptop R

TTTTAhhQ

−−−

−=−+=

( )( )111

2

222

2

2,

−+

++=−

gP

gPgPcgpr

TTTTAh

εε

σ

63

Remaining Top Loss Path

• Between glass plates

( ) ( )12

121212,12

gg

ggggcggrggtop R

TTTTAhhQ

−−−

−=−+=

( )( )111

21

2122

21

12,

−+

++=−

gg

ggggcggr

TTTTAh

εε

σ

( ) ( )12

111,1

gg

agagcagragtop R

TTTTAhhQ

−−−

−=−+=

( )( )ag

skyg

gg

skygskygcagr TT

TTTTTTAh

−

−

−+

++=−

1

1

21

122

11,

111εε

σ

• Top plate to ambient

64

Loss Through Top/Bottom• Combine three resistances in series to

get Rtop = RP-g2 + Rg2-g1 + Rg1-a– Qtop = (TP – Ta)/Rtop = UtopAc(TP – Ta)

• Loss through bottom is conduction through insulation (kins, Δxins) in series with convection to ambient with hb-a

( )aPcbottom

cabcins

ins

aP

convins

aPbottom TTAU

AhAxk

TTRRTTQ −=

+Δ

−=

+−

=

−

1

65

Total Loss• Qsides = U’sideAside(TP – Ta)

– Can estimate U’side = 0.5 W/m2·K– Use UsideAc = U’sideAside for common area– Qsides = UsideAc(TP – Ta) = (TP – Ta)/Aside

• Total is sum of individual losses• Qloss = UcAc(TP – Ta)= (TP – Ta)/Rc

• Overall conductance and resistance• Uc = Utop + Ubottom + Usides

sidebottomtopc RRRR1111

++= 66

Approximate Utop Equation

( )( )

( ) NBNN

TTTT

hBNTT

TA

U

gpp

apap

w

ap

p

top

−⎟⎟⎠

⎞⎜⎜⎝

⎛

ε−+

+ε−+ε

++σ+

+⎟⎟⎠

⎞⎜⎜⎝

⎛+

−=

12105.0

1

1'

1

22

33.0

N = number of glass coversA’ = 250[1 – 0.0044(s – 90)]s = tilt angle (degrees)B = (1 – 0.04hw +

0.0005hw2)(1 + 0.091N)

hw = heat transfer coeffi-cient from top to ambient

Other symbols have previous definitions

Equation uses SI units: Uc and h in W/m2·K, T in K, σ= 5.670x10-8 W/m2·K4, εg is same for all glass covers



67

Absorber Plate Analysis

• Define m2 = Uc/(tkplate)• Effectiveness factor, F = tanh(mL) / (mL)• Total (useful) heat transfer per unit length of

tube ( ) ( )[ ] '2 uabcatotal qTTUHDLFq =−−+= 68

Absorber Plate Analysis II

• Heat flow into fluid at any point

( )[ ]

( )

( )[ ]afca

iicBc

afcac

u TTUHwF

DhCUDLF

TTUHUq −−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛++

+

−−= '

112

1

1

,

'

π

69

Factors, F’ and FR

AmbientandFluidBetweenResistanceThermalAmbientandPlateBetweenResistanceThermal

='F

( )[ ] ( )( )[ ]iicBc

cDhCUDLFw

UFπ+++

=,1121

1'

• Collector efficiency factor, F’

( )p

ccacmFAU

cc

pRcm

FAUaea

eFAU

cmFF p

cc

&

& & '111''

'

=−=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛−= −

−

• Heat removal factor, FR

[ ] [ ]∞→→ ,0/1,1' aasaFFR 70

Summary of Results• Qu = useful heat transfer to working fluid

( ) ⎥⎦

⎤⎢⎣

⎡++

+

=

iiBcc hDCDLFUUw

F

π11

21

1'

⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

−p

ccmAFU

c

pR e

AUcm

F &&'

1

( )[ ]p

uinfoutfainfcaRu cm

QTTTTUHAFQ&

+=−−= ,,,

( )τα= ia HH

mLmLF

tkUm

PP

c

tanh

2

=

=

71

Collector Efficiency, ηc = Qu/AcHi

• Replace Ha by Hiτα

( )[ ]ainfcacRu TTUHAFQ −−= .

• Start with Hottel-Whillier-Bliss Equation

( )[ ]ainfcicRu TTUHAFQ −−= .τα

• Substitute into efficiency equation

( )[ ] ( )i

ainfcRR

ic

ainfcicR

ic

uc H

TTUFF

HATTUHAF

HAQ −

−=−−

== .. τατα

η

72

Solar Collector Efficiency Tests

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.05 0.1 0.15

Effic

ienc

y

1-cover, black1-cover, selective2-cover, black2-cover, selective

Slope = –FRUc

Intercept = FR(τα)n



73

Sample Rating Sheet

http://www.builditsolar.com/References/Ratings/SRCCRating.htm 74

Sample Rating Sheet II

slope = –FRUcintercept = FR(τα)n

75

f-chart Method• Predicts fraction of demand over a time

period (usually monthly) than can be supplied by solar

• Two empirical pamameters, X and Y– X is ratio of reference collector loss to total

heating load– Y is ratio of absorbed solar energy to total

heating load

( )arefcRc TT

DtUFAX −

Δ=

'

totaliRc H

DFAY ,

' τα=

76

Computing X (dimensionless)( )⎥

⎦

⎤⎢⎣

⎡−

Δ= aref

R

RcRc TT

Dt

FFUFAX

'

• FRUc (W/m2·K) from slope of collector test data

• F’R/FR computed or assumed = 0.97• Usual averaging period, Δt = 1 month,

converted to seconds• D = heating demand for averaging

period (J)• Tref = 100oC; from NREL dataaT

• Ac = collector area (m2)

77

Computing Y (dimensionless)

• FR(τα)n from intercept of collector test• F’R/FR computed or assumed = 0.97• Ratio = 0.94 (October – March),

= 0.90 (April – September) or computed• Hi,total is available from NREL data for Δt

= 1 month (convert to J/m2)• D is heating demand J

( ) ( ) ⎥⎦

⎤⎢⎣

⎡=

DH

FFFAY totali

nR

RnRc

,'

τατατα

( )nτατα /

• Ac = collector area (m2)

78

f Equations• For water heating: f = 1.029Y – 0.065X

– 0.245Y2 + 0.0018X2 + 0.0215Y3

– Adjustments required• Adjust X for hot water supply only and storage

capacity different from standard• Adjust Y for load heat exchanger capacity

• For air heating: f = 1.040Y – 0.065X –0.159Y2 +0.00187X2 – 0.0095Y3

– Solar collectors heating air have no heat exchanger so F’R = FR



79

Adjustments• Adjust X for storage capacity, M, in L/m2

X’ = X(75/M)1/4

• Adjustment for water heating only– See f-chart notes for details

• Adjust Y for load heat exchanger factor, Z: Y’ = Y(0.39 + 0.65e-0.139/Z)– εC = heat exchanger effectiveness

( ) ( )LpC UAcmZmin

&ε=80

NREL 1961-1990 LAX AverageSOLAR RADIATION FOR FLAT-PLATE COLLECTORS FACING SOUTH AT A FIXED-TILT (kWh/m2/day) Percentage Uncertainty = 9Tilt(deg) Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Year 0 Average 2.8 3.6 4.8 6.1 6.4 6.6 7.1 6.5 5.3 4.2 3.2 2.6 4.9

Minimum 2.3 3.0 4.0 5.5 5.7 5.6 6.4 6.1 4.4 3.8 2.7 2.1 4.7 Maximum 3.3 4.4 5.6 6.8 7.2 7.7 8.0 7.0 5.8 4.5 3.6 3.0 5.1

Lat - 15 Average 3.8 4.5 5.5 6.4 6.4 6.4 7.1 6.8 5.9 5.0 4.2 3.6 5.5Minimum 2.9 3.6 4.5 5.8 5.7 5.4 6.3 6.3 4.7 4.4 3.4 2.7 5.2Maximum 4.6 5.7 6.4 7.3 7.3 7.3 7.9 7.2 6.6 5.6 4.9 4.3 5.7

Lat Average 4.4 5.0 5.7 6.3 6.1 6.0 6.6 6.6 6.0 5.4 4.7 4.2 5.6Minimum 3.3 3.8 4.7 5.6 5.4 5.0 5.9 6.1 4.8 4.7 3.7 3.0 5.3Maximum 5.4 6.4 6.7 7.2 6.8 6.7 7.3 7.0 6.7 6.0 5.6 5.0 5.9

Lat + 15 Average 4.7 5.1 5.6 5.9 5.4 5.2 5.8 6.0 5.7 5.5 5.0 4.5 5.4Minimum 3.4 3.8 4.5 5.2 4.8 4.4 5.2 5.5 4.5 4.7 3.9 3.1 5.1Maximum 5.9 6.6 6.6 6.7 6.1 5.8 6.3 6.4 6.5 6.1 6.0 5.4 5.7

90 Average 4.1 4.1 3.8 3.3 2.5 2.2 2.4 3.0 3.6 4.2 4.3 4.1 3.5 Minimum 2.9 3.0 3.1 2.9 2.3 2.1 2.3 2.8 2.9 3.5 3.2 2.7 3.3 Maximum 5.2 5.4 4.5 3.6 2.7 2.3 2.5 3.2 4.1 4.7 5.2 5.0 3.7

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