basics of thermal analysis of solar collector

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S 1 ,S 2 : Temperature sensors. Differential thermostat. Τ f,ο. T S,f. S 1. Solar Collector. to loads. S 2. Τ f,i. Τ S, i. Hot Water storage tank, with No temperature stratification. Heat exchanger. Προς φορτίο. Pump-circulator. Valve anti reverse flow. - PowerPoint PPT Presentation

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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

)

)

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

n

Δτu,

nc

Δτu, 0.58H

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

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

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