Critical Modeling & Exergy Analysis of Multi Phase Change Materials Storage System

Seyedalireza Hosseininasab and Ebrahim Nemati Lay

Abstract

Using the phase change materials (PCMs) can help to store thermal energy as latent heat method in the multi PCMs storage system. In this research study, an analytical model has been developed for solar thermal storage system. An especial shell and tube heat exchanger is considered to heat exchange purpose. Double pipes have been used as tubes and PCMs allocated on the shell side of the double pipes for the heat exchanger. The hot heat transfer fluid (HTF) flows across the tube banks to charge the thermal energy storage system during the charging cycle. The thermal energy absorbed in PCMs while the HFT temperature reduced in the charging process. The heat exchanger tube length is function of HTF residence time then the equivalent length is defined in the presented model. The validity of the proposed model has been checked by comparison between the analytical results and numerical data obtained before using CFD simulation. The result shows that required times to melt of the PCMs for the presented study has the consistency of about 0.928 with CFD results in different PCMs melting percentages. Also the required time to melt of the PCMs for the presented study has the consistency of 0.915 with CFD results in different air inlet temperatures. Therefore the presented model can predict the system behavior also can apply for cascaded PCMs storage systems for exergy analysis.

Appendix (A)

The eq. (4) could be derived based on the following methods:

(1) Step by step method to predict equation for n PCMs:

(1–1) Energy analysis of the PCM in the system with single PCM (i = 1)

The energy QP1 stored to the PCM as:

QP1=hP1AP1ΔTLMT1,2tP1

Where:

ΔTLMT1,2=(Thtf1Thtf2)Ln(Thtf1Tm1Thtf2Tm1)

Also:

QP1=mhtfCp1,2(Thtf1Thtf2)tP1

Then:

(40)Thtf2=Tm1+(Thtf1Tm1)exp(Nc1)

Where:

Nc1=hP1AP1mhtfCp1,2

(1–2) Energy analysis of the PCMs in the system with two PCMs (i = 2)

The energy QP1 and QP2 stored to the PCM1 and PCM2 respectively, so:

QP2=hP2AP2ΔTLMT2,3tP2

Where:

ΔTLMT2,3=(Thtf2Thtf3)Ln(Thtf2Tm2Thtf3Tm2)s

Also:

QP2=mhtfCp2,3(Thtf2Thtf3)tP2

Then:

Thtf3=Tm2+(Thtf2Tm2)exp(Nc2)

Where:

Nc2=hP2AP2mhtfCp2,3

By using the eq. (40), we have:

(41)Thtf3=Tm2+(Tm1Tm2)exp(Nc2)+(Thtf1Tm1)exp(Nc1Nc2)

(1–3) Energy analysis of the PCMs in the system with three PCMs (i = 3)

The energyQP1, QP2and QP3 stored to the PCM1, PCM2 and PCM3 respectively, so:

QP3=hP3AP3ΔTLMT3,4tP3

Where:

ΔTLMT3,4=(Thtf3Thtf4)Ln(Thtf3Tm3Thtf4Tm3)

Also:

QP3=mhtfCp3,4(Thtf3Thtf4)tP3

Then:

Thtf4=Tm3+(Thtf3Tm3)exp(Nc3)

Where:

Nc3=hP3AP3mhtfCp3,4

By using the eq. (41), we have:

(42)Thtf4=Tm3+(Tm2Tm3)exp(Nc3)+(Tm1Tm2)exp(Nc2Nc3)+(Thtf1Tm1)exp(Nc1Nc2Nc3)

(1–4) Energy analysis of the PCMs in the system with n PCMs (i = n)

The energyQP1, QP2, …, QPn stored to the PCM1, PCM2, …, PCMn respectively, so:

We guess the relation for in the system with n PCMs as:

By using the eq. (41), we have:

(43)Thtfn+1=Tmn+(Tmn1Tmn)exp(Ncn)+(Tmn2Tmn1)exp(i=n1nNci)+(Tmn3Tmn2)exp(i=n2nNci)+...+(Tm1Tm2)exp(i=2nNci)+(Thtf1Tm1)exp(i=1nNci)

Where:

Nc3=hP3AP3mhtfCp3,4

The eq. (43) is eq. (4). which is mentioned in the paper. We have to prove the validation of eq. (43) for n PCMs.

We use mathematical induction method to prove the equation validity.

(2) Mathematical induction method to prove the equation validity:

(2–1) we verify the eq. (43) for single PCM (n = 1):

It is clear that:

Tm0=Tm1=Tm2==0

Also:

Nc0=Nc1=Nc2==0

Therefore:

Thtf2=Tm1+(Thtf1Tm1)exp(Nc1)

(2–2) According to mathematical induction method, we assume the eq. (43) is valid for k PCM (n = k).

So, we have:

(44)Thtfk+1=Tmk+(Tmk1Tmk)exp(Nck)+(Tmk2Tmk1)exp(i=k1kNci)+(Tmk3Tmk2)exp(i=k2kNci)+...+(Tm1Tm2)exp(i=2kNci)+(Thtf1Tm1)exp(i=1kNci)

(2–3) Now, we have to prove/confirm the eq. (43) validity for k + 1 PCM (n = k + 1).

Energy analysis of the PCMs in the system with k + 1 PCMs:

The energyQP1, QP2, …, QPk+1 stored to the PCM1, PCM2, …, PCMk + 1 respectively, so:

QPk+1=hPk+1APk+1ΔTLMTk+1,k+2tPk+1

Where:

ΔTLMTk+1,k+2=(Thtfk+1Thtfk+2)Ln(Thtfk+1Tmk+1Thtfk+2Tmk+1)

Also:

QPk+1=mhtfCpk+1,k+2(Thtfk+1Thtfk+2)tPk+1

Then:

(45)Thtfk+2=Tmk+1+(Thtfk+1Tmk+1)exp(Nck+1)

Where:

Nck+1=hPk+1APk+1mhtfCpk+1,k+2

By using the eq. (44) and eq. (45), we have:

Thtfk+2=Tmk+1+(TmkTmk+1)exp(Nck+1)+(Tmk1Tmk)exp(Nck+1Nck)+(Tmk2Tmk1)exp(Nck+1i=k1kNci)+(Tmk3Tmk2)exp(Nck+1i=k2kNci)+...+(Tm1Tm2)exp(Nck+1i=2kNci)+(Thtf1Tm1)exp(Nck+1i=1kNci)

Rewrite the last equation:

Thtfk+2=Tmk+1+(TmkTmk+1)exp(Nck+1)+(Tmk1Tmk)exp(i=kk+1Nci)+(Tmk2Tmk1)exp(i=k1k+1Nci)+(Tmk3Tmk2)exp(i=k2k+1Nci)+...+(Tm1Tm2)exp(i=2k+1Nci)+(Thtf1Tm1)exp(i=1k+1Nci)

The last equation is eq. (43) for k + 1 PCM (n = k + 1).

Therefore, we prove the eq. (43) validity for k + 1 PCM (n = k + 1).

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Published Online: 2017-7-28

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