Heat Gain and Energy Balance of Solar Collector •Mohamad Kharseh •E-mail:
[email protected]
1
Useful Heat Gained By The Collector Not all of solar radiation hits the sloped surface can be used: Some solar radiation is reflected and re-emitted away Some of the gained energy is lost again to the surrounding
qu = S⋅ Ap − ql qu = Useful heat gained by the collector S = Solar Energy absorbed in absorber Ap = Area of the absorber plates ql = Heat lost from the collector
2
Solar Energy absorbed in absorber
The absorbed solar radiation can be derived from the beam and diffuse radiation incident on the collector.
S = (τα)b ⋅ GbT + (τα)d ⋅ (GdT + GrT ) (τα):
Transmissivity-absorptivity product
The transmissivity-absorptivity product of diffuse radiation is assumed to be the same as for beam radiation with incidence angle 60° 3
Solar Energy absorbed in absorber 360 ⋅ n Gon = Gsc ⋅ 1 + 0.033 ⋅ cos 365 τ b = ao + a1 ⋅ e( −k / cos θ z )
τ d = 0.271 − 0.294 ⋅ τ b Gcd = τ d ⋅ Gon .(cosφ ⋅ cosδ ⋅ cosω + sinφ ⋅ sinδ ) Gcb = τ b ⋅ Gon .(cosφ ⋅ cosδ ⋅ cosω + sinφ ⋅ sinδ ) 360 ⋅ n Gc = (τ b + τ d ) ⋅ Gsc ⋅ 1 + 0.033 ⋅ cos ⋅ (cosφ ⋅ cosδ ⋅ cosω + sinφ ⋅ sinδ ) 365 Rb =
cos θ cos θ z
1 + cos β Rd = 2 1 − cos β Rr = 2
GbT = Gcb ⋅ Rb
GdT = Gcd ⋅ Rd GrT = Gc ⋅ Rr
4
Transmissivity-absorptivity Product Of Beam • The actual heat absorbed in the absorber is a result of: 1.
The fraction of solar radiation that is transmitted through the cover
2.
The fraction of this transmitted radiation that is absorbed in the absorber Diffuse reflectivity of the cover
Transmissivity of the glass
1
S = (τα )b ⋅ IbT + (τα )d ⋅ ( τ(1-α)ρd
τ
Absorptivity of the absorber
τ(1-α) τα
)
Cover system IdT + IrT
Absorber τα(1-α)ρd
τα(1-α)2 ρ2d
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Transmissivity-absorptivity Product Of Beam
The radiation is reflected between the cover and absorber indefinitely The sum of the absorbed fractions of the radiation is given as:
(τα ) = τ ⋅ α ⋅ [1 + (1 − α )ρd + (1 − α )2 ρd 2 + ...] = (τα)
Transmissivity -absorptivity product
τ
Transmissivity of the glass
α
Absorptivity of the absorber
ρd
Diffuse reflectivity of the cover
τ ⋅α 1 − (1 − α )ρd
1 cover: ρd = 0.15 2 covers: ρd = 0.22
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Transmissivity Of The Beam
•The transmittance of the cover depends on the incidence angle of the solar radiation. The Transmissivity of the beam radiation must be expressed as a function of: 1. transmittance due to reflection and refraction
τr τr, τa, τ
2. transmittance due to absorption
τa
τ
τ = τr ⋅ τa Angle of incidence
7
Transmissivity due to Reflection and Refraction
•Transmissivity due to reflection and refraction 1 − ρ I 1 − ρ II + 1 + ρ I 1 + ρ II τr = 2 ρI, ρII Can be calculated use next equations:
sin 2 (θ r − θ ) ρI = sin 2 (θ r + θ )
tan 2 (θ r − θ ) ρ II = tan 2 (θ r + θ )
The reflectance is a function of the angle of incidence and refraction 8
Transmissivity due to Reflection and Refraction
The relationship between the incidence θ and refraction θr angles is given by Snell’s Law:
nglass sinθ = sinθ r nair
θ
n1 and n2 are the refractive indices of air and glass, respectively. (nglass/nair = 1.526)
θr 9
Transmissivity due to Absorption Transmissivity due to absorption The transmittance for beam radiation is a function of the absorptivity of the cover material (no reflection)
Bouger’s Law:
τa = e(
−K ⋅δc
cosθr
)
δc
is the thickness of the cover (glass)
K
is the extinction coefficient of the cover material. Usually, 5 ≤ K ≤ 25 m-1 10
Transmissivity-absorptivity Of Diffuse
Transmissivity-absorptivity product of diffuse radiation Since diffuse radiation comes from all directions, it is not possible to derive an exact expression. Transmissivity -absorptivity product for diffuse radiation (τα)d is assumed to be equal to that for beam radiation at 60° incidence angle
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Solar Energy Absorbed in Absorber for more accuracy
S = (τα )b ⋅ GbT + (τα )d ⋅ GdT + (τα )g ⋅ GrT
(τα)b: Transmissivity-absorptivity product of beam radiation at incident angle θ (τα)d: Transmissivity-absorptivity product of diffuse radiation at incident angle θd (τα)g: Transmissivity-absorptivity product of reflected radiation at incident angle θg
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Solar Energy Absorbed in Absorber
θd = 59.7 − 0.1388 ⋅ β + 0.001497 ⋅ β 2 θg = 90 − 0.5788 ⋅ β + 0.002693 ⋅ β 2
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Liquid Flat Plate Collectors 1.
Heat Loss from Collector
2.
Testing of Solar Collectors
14
Useful Heat Gained By The Collector Not all of solar radiation hits the sloped surface can be used: Some solar radiation is reflected and re-emitted away Some of the gained energy is lost again to the surrounding
qu = S⋅ Ap − ql qu = Useful heat gained by the collector S = Solar Energy absorbed in absorber Ap = Area of the absorber plates ql = Heat lost from the collector
15
Heat Losses from Collector •The thermal losses depends on: • The temperature difference between the absorber plate and the ambient air • The overall heat loss coefficient, Ul [W/m2,K]
(
q l = U l ⋅ A p ⋅ Tpm − Ta Tpm Ta
)
= Mean temperature of absorber plates = Ambient air temperature
16
Heat Losses from Collector •Whenever the absorber plate is warmer than the ambient air, heat is lost from the collector through: •
the cover (top), [W]
•
the bottom, [W]
•
the sides, [W]
( ) qb = U b ⋅ Ap ⋅ (Tpm − Ta ) qt = Ut ⋅ Ap ⋅ Tpm − Ta
(
qs = Us ⋅ Ap ⋅ Tpm − Ta
)
Most of the heat losses is through the cover 17
Overall Heat Loss Coefficient Then, the total heat losses is:
q l = qt + q b + qs
•Since the heat loss equations are expressed on the basis of the same temperature difference, it is then possible to evaluate the overall heat loss coefficient, Ul, by ql = U l ⋅ Ap ⋅ (Tpm − Ta )
U l = Ut + U b + Us Typical values of Ul ranges from 2 to 10 W/m2K
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Thermal Network for a Flat-Plate
qu
19
Bottom Heat Losses Coefficient •The heat losses coefficient through the bottom is represented by two series resistors, R3 and R4. It is possible to assume R4 is zero:
1 λ Ub = = R3 δ b λ = thermal conductivity of insulator δb = thickness of insulated bottom 20
Side Loss Coefficient Similarly, the heat transfer coefficient for the collector side is:
2 ⋅ L3 ⋅ (L1 + L2 ) λ ⋅ Us = δs L1 ⋅ L2
L1= length of casing L2= width of casing L3= height of casing λ= thermal conductivity of insulation δs = thickness of insulated side 21
Top Loss Coefficient Ut = C T pm
N Tpm − Ta ⋅ N +f
α a = 5.7 + 3.8 ⋅ v wind
e
1 + αa
−1
2 2 σ T T T T ⋅ + ⋅ + pm a pm a + 2N + f − 1 + 0.133 ⋅ ε p 1 + −N 0 00591 ε . N α ε + ⋅ ⋅ p a g
α a = 2.8 + 3 ⋅ v wind
or
(
(
or
)(
)
Nu = 0.86 Re1/ 2 Pr1/ 3
)
f = 1 + 0.089 ⋅ α a − 0.1166 ⋅ α a ⋅ ε p ⋅ (1 + 0.07866N )
C = 520 ⋅ ( 1 − 0.000051 ⋅ β ); 0 ≤ β ≤ 70 2
100 ) e = 0.43 ⋅ ( 1 − Tpm
N
= number of covers
σ
= Stefan-Boltzmann constant
εp
= emissivity of plate
εg
= emissivity of glass (0.88)
αa
= convection coefficient of ambient air (W/m2K)
νwind =Wind speed of ambient air (m/sec) β
= collector tilt (degree)
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Heat Balance of Collector (
(
qu = Ap S − U l Tpm − Ta
))
•This expression depends on two factors: 1.
Ul : is a function of the mean plate temperature, Tpm
2.
Tpm: is the mean plate temperature, which is unknown
•We need to express qu in terms of a known temperature. •
The only known temperature is the fluid inlet temperature
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Absorber Cross-section •The useful heat gained by the absorber plate can be divided into: 1.the heat gained from the area above the tube 2.the heat gained from the fin
qu′ = qu′ tube + qu′ fin tube
fin
W-Do W
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Heat Gain per unit Length The useful heat transfer rate to the fluid in the absorber cross-section is:
qu′ = [F ⋅ (W − Do ) + Do ] ⋅ [S − U l (Tb − Ta )] F: Standard Fin Efficiency F=
tanh[m ⋅ (W − Do )/ 2] m ⋅ (W − Do )/ 2
Hyperbolic tangent Do= tube outer diameter
Where m is : m = (Ul/λpδp)½
W = distance between tubes Tb = local absorber temperature above the bond λp = thermal conductivity of absorber plate δp = thickness of absorber plate
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Collector Efficiency Factor, F′ qu′ = [F ⋅ (W − Do ) + Do ] ⋅ [S − U l (Tb − Ta )]
To relate this expression to the local fluid temperature that results from this heat transfer, the collector efficiency factor, F’, is introduced:
F' =
U0 = Ul
1 1 1 + WU l U l ⋅ [F (W − Do ) + Do ] πDi α f
Di= tube inner diameter αf = convection coefficient of fluid Uo= overall heat loss coefficient from fluid to ambient Ul = overall heat loss coefficient from absorber to ambient
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Collector Efficiency Factor, F′ Thus, useful gain can be expressed as :
qu′ = F ′ ⋅ W ⋅ [S − U l (Tf − Ta )]
We eliminated Tb from the equation and obtain an expression for useful gain in term of known dimensions, physical parameters and fluid temperature.
27
Heat Gain in Cross-section We can express an energy balance on the fluid flowing through a signal tube of a small length dy as:
Fluid Flow
/n) ⋅ c pTf (y + dy) (m
/n) ⋅ c pTf (y) (m
dy Number of tube in the absorber plate
y
⋅ c p ⋅ dTf ⇒ m ⋅ c p ⋅ dTf − n ⋅ F' ⋅ W ⋅ [S − U l (Tf − Ta )] = 0 n ⋅ qu′ ⋅ dy = m dy 28
Heat Gain in Cross-section •Solving this differential equation, we can determine the fluid temperature at any position y: Ul ⋅n ⋅W ⋅F ' y Tf (y)-Ta -S/U l m ⋅c p =e
Tfi -Ta -S/U l
If the collector has a length L in the flow direction, then the outlet fluid temperature Tfo is found by substituting L for y, note that “n.W.L =Ap”
Tfo -Ta -S/U l = Tfi -Ta -S/U l
Ul ⋅ Ap⋅F' m ⋅c p e 29
Heat Removal Factor, FR •The heat removal factor, which relates the actual heat gain to the maximum heat gain, is define as:
⋅ c p (Tfo − Tfi ) m qu Actual hea t gain FR = = qu Maximum heat gain Ap [S − U l (Tfi − Ta )] Using the last equation in the previous slide, we can express FR that is independent of the outlet temperature, Tfo: Ul ⋅ Ap − F' ⋅ mc p ⋅Cp m 1 − e FR = U l Ap 30
Heat Gain in Cross-section Then the collector heat removal factor times this maximum possible useful energy gain is equal to the actual useful energy gain qu
qu = Ap ⋅ FR [S − U l (Tfi − Ta )]
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Heat Gain in Cross-section •Using FR the useful heat gain can now be determined based on the inlet temperature. But the overall heat loss coefficient, Ul, is a function of the mean plate temperature, Tpm
•
U ⋅A − F' ⋅ l p mc p ⋅Cp m 1 − e FR = U l Ap
F' =
U0 = Ul
qu = Ap ⋅ FR [S − U l (Tfi − Ta )]
1 1 1 WU l + U l ⋅ [F (W − Do ) + Do ] πDi α f
U l = Ut + U b + Us Ut = C T pm
N Tpm − Ta ⋅ N f +
e
1 + αa
−1
2 σ ⋅ Tpm + Ta2 ⋅ Tpm + Ta + 2 1 0 133 N f . ε + − + ⋅ 1 p + −N εg ε p + 0.00591⋅ N ⋅ α a
(
)(
)
32
Mean Plate Temperature 1. Guess a first value of Ul (2 – 10 W/m2,K) and calculate the heat gain, qu. 2. Tpm can then be calculated from the original heat gain equation:
(
(
qu = Ap S − U l Tpm − Ta
))
3. Use the calculated Tpm to derive a new Ul. 4. Continue with iterations until Ul remains approximately the same from one calculation to another. •
Since Ul is only vaguely dependent on the temperature, one or two repetition should be necessary 33
Instantaneous Collector Efficiency •Collector is exposed to solar radiation while the fluid flow rate and temperature increase is measured. • The useful heat gain in a collector operating under steady state condition equals the enthalpy increase of the fluid:
(
qu=m c p T fo -T fi
)
• The instantaneous collector efficiency is the ratio of the useful heat gain to the incident solar radiation:
(
)
m c p T fo -T fi FR qu ηi= = = Ap I T Ap I T IT
[S-U (T -T )] L
fi
a
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Instantaneous Collector Efficiency •The efficiency curve for a collector is plotted as a function of (T fi – Ta)/IT
[Tfi-Ta/IT]⋅103 [K⋅m2/W]
The efficiency curve yields a straight line since FR, (τα) and Ul are fairly constant for a collector type when the flow rate is constant
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S = (τα)b ⋅ GbT + (τα)d ⋅ (GdT + GrT )
(τα ) =
τ ⋅α 1 − (1 − α )ρd
qu = Ap ⋅ FR [S − U l (Tfi − Ta )] U ⋅A − F' ⋅ l p mc p ⋅Cp m 1 − e FR = U l Ap
F' =
U0 = Ul
1 1 1 WU l + U l ⋅ [F (W − Do ) + Do ] πDi α f
U l = Ut + U b + Us
Ut = C T pm
N Tpm − Ta ⋅ N +f
e
1 + αa
−1
2 2 σ ⋅ Tpm + Ta ⋅ Tpm + Ta + 2N + f − 1 + 0.133 ⋅ ε p 1 + −N εg ε p + 0.00591⋅ N ⋅ α a
(
)(
)
36
