Basics of thermal analysis of solar collector Simulation of operation of solar systems to determine the Thermal
Gain.
Προς φορτίο
Valve anti reverse flow
Solar Collector
S1
S2
Heat exchanger
Τf,ο
Τf,i
Differential thermostat
S1,S2:
Temperature sensors
Pump-circulator
Hot Water storage tank, withNo temperature stratification
ΤS,i
TS,f
to loads
The useful thermal energy Qu can be calculated from the relation:
)T(T)Cm(Q if,of,fpu
)T(T)Cm(Q if,of,fpu
(1.1)
or normalized to solar collector surface, the thermal power output is given by:
(1.2)
It can be also shown that the useful thermal gain from a solar collector is given by:
)]T(TUτα[IAFQ αif,LTcRu
(1.3)
A similarity to the previous expressions holds for the hot water tank,too:
is,fs,spu TT)Cm(Q
or equivalently
is,of,minpu TTε)Cm(Q
(1.4)
(1.5)
The coefficient of effectiveness, ε, is given by an expression in any Heat Transfer book,
(1.6)
From the expression (1.2) of calorimetry one gets:
(1.7)
C)NTU(1
C)NTU(1
is,of,minp
if,of,cp
max eC1
e1
TTC(m
TTC(m
Q
Qε
)
)
maxpminp )Cm/()Cm(C
minpενA )Cm( /A)(UNTU
fp
uif,of,
)C(m
QTT
Substitute Tf,i from (1.7) to (1.3), then the equation which provides the Thermal Power stored in the system-tank takes the form:
(1.8)
We solve eq. (1.5 ) for Tf,o and we get:
(1.9)
fp
LRc
αof,LTRcu
C(m
UFA1
TTUταIFAQ
)
is,
minp
uof, T
ε)C(m
QT
Basics of thermal analysis of solar collectors &
Simulation of operation of solar systems to determine
the Thermal Gain.
Prof. Socrates Kaplanis
References
Renewable Energy Systems : Theory and Intelligent ApplicationsEditors : S.Kaplanis, E.KaplaniNova Science Publishers, N.Y. , 2012
Substitution of Tf,o to (.1.8), gives:
(1.10)
The expression (1.10) can be easily simplified to:
fp
LRc
αis,
minp
uLTRc
u
)C(m
UFA1
TTε)C(m
QUταIFA
Q
1ε)C(m
)C(m
)C(m
UFA1
TTUταIFAQ
minp
fp
fp
LRc
αis,LTRcu
(1.11)
We define a new parameter, F΄R:
(1.12)
Hence, the expression (1.11) is simplified as to:
(1.13)
1ε)C(m
)C(m
)C(m
UFA1
FF΄
minp
fp
fp
LRc
RR
αis,LTR
RcRαis,LTRcu TTUταI
F
FAFTTUταIFAQ
΄΄
**
Let us consider a small time period the solar collector system operates. Then, the mean water temperature in the storage tank is determined by:
(1.14)
The heat delivered by a collector, Αc, in a period, Δτ , to the tank may be determined by:
(1.15)
2
TTT fs,is,
s
dtQ)T - T ()(MC QΔττ
τu is,fs,spΔτ u,
We divide both sides of (.1.15) over Αc to normalize the expression. Then
(1.16)
or equivalently
)T - T ()Cm( q is,fs,spΔτ u,
sPΔτu,s.ifs, )Cm/(qTT
(1.17)
Substitute Τs,f from (1.17) to (1.14). We get:
(1.18)
Τs is the mean temperature of the water in the tank in the above time interval. Integration of (1.13 or 1.15) for this time period gives:
(1.19)
sP
Δτu,is,s
)Cm(2
qTT
Δτ])αΤ(TLU)τα(n[HcA΄RFΔτu,Q s
Substitute Τs from (1.18) to (1.19). We get:
(1.20)
Δτ)Cm2(
U F1
)T(TU)τα(΄[HFq
sP
LR
αis,LnRΔτu,
Δτ
Let us analyze a real caseLet us analyze a real case
A Solar Collector System, as the one shown in the 1st figure, has parameters: F`RUL=3.5 W/m^2*K and F`R(τα)n= 0.69
and is placed at horizontal position in Pyrgos.The storage tank has capacity 50 l/m^2.Let the storage tank temperature at 7:30 be 20 C.Please determine the hourly temperature in the
tank and the hourly efficiency.Data input: ambient temperature , Ta, and the
mean hourly global solar radiation, Hn. Values
are given in the Table below
Data input and values of basic quantities as provided by the iteration
procedure to be outlined below.
Time
18.4.1999
Τα οC Ts,i
οC Ts,f οC N=qu/Hn
7:30 – 8:30 15.0 720 421 20 22 0.58
8:30 – 9:30 15.5 1476 909 22 26 0.61
9:30 – 10:30 16.5 1980 1206 26 32 0.60
10:30 – 11:30 17.0 2484 1479 32 39 0.59
11:30 – 12:30 17.5 2844 1640 39 46 0.57
12:30 – 13:30 18.0 3240 1816 46 55 0.56
13:30 – 14:30 19.0 3250 1729 55 63 0.53
14:30 – 15:30 19.0 2968 1439 63 70 0.48
15:30 – 16:30 18.0 2412 971 70 75 0.40
16:30 – 17:30 17.5 1800 498 75 77 0.27
17:30 – 18:30 17.0 1210 68 77 78 0.05
)m
kJ( H
2n )m
kJ( q
2u
To determine the quantities Τs,f ,Τs.i
Α. Time Interval 7:30 – 8:30 am
Step 1st : We determine the normalized useful heat by (1.20).
(1.21)
2
2
2
002
2
m / kJ
1.03]
K kg
J4180
m
kg502
3600sm
W3.5
[1
3600s)15(20Km
W3.50.69m / kJ 720
421q Δτu,
K
Step 2nd : We determine tank temperature, Τs,f, at the end of the 1st interval 8:30 amusing expression (1.17)
(1.22)
Step 3rd : Determine efficiency, η, during this short period by :
(1.23)
C m J/kg 4180kg/m 50
J/m 421,000C20 0
22
20 22T fs,
H
q
HΑ
Qη
n
Δτu,
nc
Δτu, 0.58H
qη
n
Δτu, 2
2
m / kJ 720
m / kJ 421
Β. Time Interval 8:30 – 9:30am
We follow the same procedure as before. We put for this period 8:30-9:30 as Τs,i, the Τs,f value of the previous interval.
Determine from (1.20)
Δτu,q
2
0002
2
m / kJ 1.03
3600s)15.5(22Km
W3.50.69m / kJ 1476
909q Δτu,
(1.24)
Determine Τs,f from (1.17)
(1.25)
Then, the efficiency is estimated by:
(1.26)
C
CkgJ
4180mkg
50
mJ
909000C22 0
o2
20 26T fs,
0.61m kJ / 1476
m kJ / 909
H
qη
2
2
n
Δτu,
Back up electric source in the tank
Back up source outside the tank
Flat plateSolar collectors
Hot Water
storage tank
S1,S2: temperature sensors
S1
S2
Valve to prevent reverse flow Pump circulator
Differential thermostat
Hot Water
For use~
serpantine
2. A generalized analysis to consider the Load, too.
Let us consider qs as the net stored heat normalized to collector surface; that is when the Load, QL, is subtracted. Correspondingly,
the thermal load per collector surface is denoted by, ( qL =QL / Ac ). Then it holds :
(2.1)
Following the procedure as for expression (1.18) there is given that:
(2.2)
LSΔτu,LΔτu,S qqqqqq
sP
sis,s
)Cm2(
qTT
2. A generalized analysis to consider the Load, too.
Substitute Τs to (1.19). Then, the expression (1.20) is modified and due to (2.1) we get :
(2.3)
We substitute (2.3) in (2.1) and we finally get the generalized iterative formula:
(2.4)
Δτ
)Cm2(
UF1
q-)T(TU)τα(HF΄q
sP
LR
Lαis,LnRs
ΔτUF
)Cm2(1
q
Δτ)Cm2(
UF1
)Tα(TU)τα([HF΄q
LR
sp
L
sp
LR
is,LnRΔτu,
Δτ
References
Renewable Energy Systems: Theory and Intelligent Applications
Editors S. Kaplanis, E. Kaplani
Nova Science Publishers, N.Y., 2012