Project 2 Thermal modeling of battery pack

AIM:- To study Thermal modeling of the battery pack with the help of MATLAB and SIMULINK.

 

OBJECTIVES:-

                     1)For a 10 cell series lithium-ion battery model, simulate the thermal effects and compare life cycle performance at various temperatures, charge & discharge rates using MATLAB.

this will be based model which we are going to use for our project by doing some changes in it according to our requirements.

 

 

This example shows how to simulate a battery pack consisting of multiple series-connected cells efficiently. It also shows how a fault can be introduced into one of the cells to see the impact on battery performance and cell temperatures. For efficiency, identical series-connected cells are not just simply modeled by connecting cell models in series. Instead, a single cell is used, and the terminal voltage is scaled up by the number of cells. The fault is represented by changing the parameters for the Cell 10 Fault subsystem, reducing both capacity and open-circuit voltage, and increasing the resistance values.

 

-here we actually made changes in the base model according to our requirements to obtain results,

SIMULINK MODEL:-

-

so for the 10-cell series lithium-ion battery here I used Matlab and Simulink for findings thermal effect and life cycle performances related to temperature, charge, and discharge and also conduction and convection blocks are used to show how the heat transfer takes place in battery packs.

Lithium-ion battery pack with fault is to simulate a battery pack consists of multiple cells connected in series efficiently.it also shows how a fault can be introduced into one of the cells to the impact on battery performance and cell temperatures, for efficiency the identical series-connected cells are not just simply model by connecting cell models in series. instead single cell is used, and the terminal voltage is scaled up by the number of cells. the fault is represented by changing the parameters for the cell 5 subsystem, reducing both the capacity and open-circuit voltage, and increasing the resistance values.

-How the connection is made

-10 lithium-ion battery cells are connected in with the combination of three packs the f1st consistent 4 cells in series 2nd pack consist of 1 cell which set to design as fault for study purpose to observe temperature changes within battery pack and the 3rd pack contain 5 cells in series.

-the heat transfer is observed between the packs by conduction and convection block which are modes of heat transfer.

-T subsystem is attached in between every pack to plot the temperature graph of the cell of that particular battery pack.

-also external ambient temperature source is attached and also thermal sensor is attached to the same. to indicate the value in the display.

-this battery pack is connected to AC current source to observe the cyclic charge and discharge rate graph.

-also we added a current sensor in between the battery pack and AC current source to calculate the SOC of the battery

-the cell temperature and soc graphs are plotted in scope which indicating the temperature of cell no 4, cell no 5, and cell no 10

-all the other results are located in the simscape result menu.

 

Cell 01 to 04 subsystem:-

 

 

The cell subsystem basically consists of a thermal model and lithium cell 1RC block which calculate the temperature and thermal effect for cells and overall calculate for the battery pack.

The block which are inside the cell subsystems are PS gain block which multiplies the input physical signals by a constant i.e y=u.*gain .another block are voltage sensor and current-voltage source.

The voltage sensor block represents an ideal voltage sensor, this is a device that converts voltage measured between any electric connection into a physical signal proportional to the voltage.

The current-voltage source represents an ideal voltage source that is powerful enough to maintain the specified voltage at its output regardless of the current passing through it. The output voltage is V=Vs, where Vs is the numeric value represents at the physical signal port.

 

Thermal model:-

 

-the blocks present inside the systems are,

a)Thermal Reference

Library

Thermal Elements

 

B)Lithium Cell 1RC subsystem:-

-this block especially consists of the capacitor (C)value table which depends upon an external physical signal input SOC and T.it is assumed that the capacitor value is varying with the time and hence the equation i=C*DV/dt holds.

-Resistance value (R) table depend on external physical signal inputs SOC and T.

-Voltage value (EM) implements the cell's main branch voltage source and determines values for capacity(C) and state of charge (SOC). The defining the equations depends upon cell temperature T

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THERMAL BEHAVIOUR AND FAULT SETUP:-

-Lithium-ion one cell one RC-Branch equivalent circuit detail for 10 cells

 

For cell 01 to 04

Number of connected in series are:-4

Initial temperature:-299.1 k

 

For cell 5 cells (fault setup) in this cell, I have done changes in the electrical systems of the cell so that we can see the fault.

Number of connected in series are=4

Initial temperature=299.1 k

Capacity(A*hr)=Capcity_LUT*0.95

EM-open-circuit voltage (volts)=Em_LUT*0.9

RO terminal resistance (ohms)=RO_LUT*5

R1 cell resistance (ohms)=R1_LUT*5

C1 capacitance(farads)=C1_LUT*0.95

For cell 06 to 10

Number of connected in series are:-5

Initial temperature:-299.1 k

PARAMETERS:-

Battery configuration:-

 

 

 

 

Temperature Source

 

Description

The Temperature Source block represents an ideal source of thermal energy that is powerful enough to maintain a specified temperature at its outlet regardless of the heat flow consumed by the system.

The source generates constant absolute temperature, defined by the Temperature parameter value.

here I used ambient block and set the temperature for 299.1k or 25.95 C

 

 

THERMAL ELEMENTS:- 

 Conductive Heat Transfer

Heat transfer by conduction

Library

Thermal Elements

Description

The Conductive Heat Transfer block represents a heat transfer by conduction between two layers of the same material. The transfer is governed by the Fourier law and is described with the following equation:

where

Connections A and B are thermal conserving ports associated with material layers. The block positive direction is from port A to port B. This means that the heat flow is positive if it flows from A to B.

The is done from the 2 sets of blocks .one is for cell 04 to 05 and another is used for 05 to 06.

 

Convective Heat Transfer

Library

Thermal Elements

Description

The Convective Heat Transfer block represents a heat transfer by convection between two bodies using fluid motion. The transfer is governed by the Newton law of cooling and is described with the following equation:

where

Connections A and B are thermal conserving ports associated with the points between which the heat transfer by convection takes place. The block positive direction is from port A to port B. This means that the heat flow is positive if it flows from A to B.

- here I used 3 blocks for convection

for cell 01-04

 

for cell 05

 

for cell 06-10

 

 

Temperature Sensor

Library

Thermal Sensors

Description

The Temperature Sensor block represents an ideal temperature sensor, a device that determines the temperature differential measured between two points without drawing any heat.

Connections A and B are thermal conserving ports that connect to the two points where the temperature is being monitored. Port T is a physical signal port that outputs the temperature differential value.

The block positive direction is from port A to port B. The measured temperature is determined as T = TA – TB.

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Electrical Elements:-

here the output of one cell is connected with the SOC calculation whereas the output is connected to with AC source for the cyclic charge/discharge profile

a)AC Current Source

Library

Electrical Sources

The peak amplitude is set to 50 A for a cyclic charge and discharge. A basic parameter of AC source is as,

 

The ideal AC source maintains the sinusoidal current through it, independent of the voltage across its terminals. The output current is defined by I=I0*sin(2*pi*f*t+PHI) I0 is the peak amplitude,f is the frequency in Hz, and PHI is the phase shift in radians.

Current Sensor

Library

Electrical Sensors

 -the soc graph following the sinusoidal waveform structure and fluctuating after 297 seconds and this waveform achieving the value of more than 100.06 soc

c)Conduction:-

Conduction is basically heat flow rate(Q) and temperature difference wrt time.

For cell 04-05

-cell no 04-05 and having a high amplitude of waveforms(for heat flow and temperature difference) which are oscillating at a pretty high rate which is starting from zero and dropping into the negative zone. at the end of the simulation.

The heat transfer coefficient is -21.32 W/(m2K)

Thermal conductivity -10.66 W/m K

For cell 05-06

 -for cell no 05-06 both the waveforms oscillating at a high rate but in a positive zone which are increasing initially and maintaining a certain range.

The heat transfer coefficient is 20.12 W/(m2K)

Thermal conductivity 10.06 W/m K

 

d)Convection:-

Convection is the process of heat transfer by the bulk movement of molecules within fluids such as gases and liquids

For cell 01-04

-for cell, no 01-04 the oscillations are pretty low and dropping into the negative zone at the end of the simulation. the final values for

The heat transfer coefficient is -71.63 W/(m2K)

Thermal conductivity -15.76 W/m K

For cell 05-

-For cell no 5 they are starting from 0 and falling into the negative zone the oscillation having high amplitude because of the faulty cell at the end of the simulation.

The heat transfer coefficient is -13.39 W/(m2K)

Thermal conductivity -26.92 W/m K

 

For cell 06-10

-for cell no 06 to 10 the falling from zero to the negative zone the oscillation are less. at the end of the simulation

The heat transfer coefficient is -82.63 W/(m2K)

Thermal conductivity -16.36 W/m K

e)Charge and Discharge profiles:-

 -in the charge and discharge rate graph the voltage and current oscillating from pretty highly where current is ranging from -50 t0 50 A and voltage is ranging from 3o to 43 volts

the final value at the end of the simulation for 

I(current)=1.4695e-13 A

v(voltage)= 36.334 volt

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2)Test 2

-

a)Cell Temp:-

 - here the cell no 04 and cell no 10 starting from a temp of 310 k at the start of simulation and cell no 5 from 299.1k with higher oscillation as compared to the other cells because of faulty design.at the end of the simulation the cells 04 and 10 having the final temp of nearby 315k and cell no 5 achieving the temp of 328k.

b)SOC:-

 

c)Conduction:-

 

For cell 04-05

 -the heat flow rate plot starts from 20 and falls into the negative zone and the temperature difference plot starts from 10 and falls into the negative zone and oscillate at the end of the simulation the final value for

The heat transfer coefficient is -21.32 W/(m2K)

Thermal conductivity -10.66 W/m K

For cell 05-06

 -for cell, no 05-06 the plot initiate from the negative zone and at the end settle down in the positive zone where the final values for,

The heat transfer coefficient is 20.12 W/(m2K)

Thermal conductivity 10.06 W/m K

 

d)Convection:-

-for cell, no 01-04 the plots remain in the negative zone throughout the simulation with pretty low oscillation with falling nature  and the final values at the end of the simulation are,

The heat transfer coefficient is -71.63 W/(m2K)

Thermal conductivity -15.62 W/m K

For cell 05,

-for cell no 5 the initial values are 0 and falling into the negative side with high oscillation because of the fault and the final values are

The heat transfer coefficient is -13.66 W/(m2K)

Thermal conductivity -26.28 W/m K

For cell 06-10

-For cell, no 06-10 the plot remains on the negative side, and falling at the end of the simulation the final values are.

The heat transfer coefficient is -82.64 W/(m2K)

Thermal conductivity -16.36 W/m K

 

e)Charge and Discharge profile.

-charge and discharge rate profile the plot for current is ranging in between -50 to 50 A and the voltage 30 to 43 volt and the final values for

I(current)= 1.4695e-13 A

v(volt)=36.33 v

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3)Test 3

 

a)Cell Temp:-

 -the initial temperature for cell no 04 is 310K, cell no 5 is 320 K and cell no 10 is 330K, but after some point, cell no 04 and cell no10 started following the same temperate rate, and cell no 5 starts oscillating with high value because of fault. the final value for cell no 04 and 10 is around 316K and fro cell no 5 is 328K because of its fault design

b)SOC:-

c)Conduction:

For cell 04-05

-for cell 04-05 the plot starts from the negative side and drops and then again increases and maintains throughout the simulation negative zone. the final values at the end of the simulation are

The heat transfer coefficient is -21.32 W/(m2K)

Thermal conductivity -10.66 W/m K

For cell 05-06

-for cell, no 05-06 both the plot starts from the negative side and then entered into the positive side and the final value at the end of the simulation are

The heat transfer coefficient is 20.97 W/(m2K)

Thermal conductivity 10.06 W/m K

 

d)Convection:-

For cell 01-04

 -for cell no 01-04 the final values at the end of the simulation are

The heat transfer coefficient is -71.64 W/(m2K)

Thermal conductivity -15.75 W/m K

For cell 05

-for cell no 5 the final value at the end of the simulations are

The heat transfer coefficient is -13.39 W/(m2K)

Thermal conductivity -26.28 W/m K

For cell 06-10

-for cell no 06 -10 the final values at the end of the simulation are

The heat transfer coefficient is -83.30 W/(m2K)

Thermal conductivity -16.34 W/m K

e)Charge and Discharge profile:-

-charge and discharge rate profile the plot for current is ranging in between -50 to 50 A and the voltage 30 to 43 volt and the final values for

I(current)= 1.4695e-13 A

v(volt)=36.33 v

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4)Test 4:-

a)Cell Temp:-

-for cell no 04 the initial temperature is 350 K, cell no 10 is 305 K and for cell no 5 the temperature is 325 K. both the temperature of cell no 04 and 10 started following the same temperature rate after the time period of 2000 sec and on the other hand, the cell no 5 is designed for fault oscillates with high temperature. the end temperature of cell no 04 and 10 are near about 315 K(the cell no 4 have little less value then cell no 10 because of the no of cells involved) and for cell no 5 having a temperature at the end is 328 K

b)SOC:-

 

c)Conduction:-

For cell 04-05

 -for cell no 04-05 the final values at the end of the simulation are,

The heat transfer coefficient is -21.32 W/(m2K)

Thermal conductivity -10.66 W/m K

 

For cell 05-06,

 -for cell no 05-06 the final values at the end of the simulation are,

The heat transfer coefficient is 20.84 W/(m2K)

Thermal conductivity 10.06 W/m K

 

d)Convection:-

For cell 01-04

-for cell no 01-04the final values at the end of the simulation are,

The heat transfer coefficient is -72.13 W/(m2K)

Thermal conductivity -15.62 W/m K

For cell 05,

-for cell no 5 the final values at the end of the simulation are

The heat transfer coefficient is -13.39 W/(m2K)

Thermal conductivity -26.28 W/m K

For cell 06-10,

 -for cell no 06-10 the final value at the end of the simulation is

The heat transfer coefficient is -83.20 W/(m2K)

Thermal conductivity -16.22 W/m K

e)Charge and Discharge profiles-

 

-charge and discharge rate profile the plot for current is ranging in between -50 to 50 A and the voltage 30 to 43 volt and the final values for

I(current)= 1.4695e-13 A

v(volt)=36.33 v

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5)Test 5,

a)Cell Temp:-

 -for cell no 04 and 10 having the same initial temperature of 310 k and having the same temperature rate throughout the simulation. just small difference is cell 10 having a high temperature than cell no 4 because the no of cells is different and cell no 5 which is design for fault having an initial temperature of 350K with high oscillation. Cells no 04 and 10 having the temperature value at the end of the simulation around 315 K and cell no 05 having a temperature of 325 K at the end of the simulation.

 

b)SOC:-

 

c)Conduction:-

For cell 04-05

-for the cell 04-05 the value at the end of the simulations are

The heat transfer coefficient is -21.32 W/(m2K)

Thermal conductivity -10.66 W/m K

 

For cell 05-06

 -For cell 05-06 the values at the end of the simulation are

The heat transfer coefficient is 20.12 W/(m2K)

Thermal conductivity 10.06 W/m K

 

d)Convection:-

For cell 01-04

-for cell no 01-04 the values at the end of the simulations are 

The heat transfer coefficient is -71.64 W/(m2K)

Thermal conductivity -15.62 W/m K

 

For cell 05

 -for cell no 5 the values at the end of the simulation are,

The heat transfer coefficient is -13.68 W/(m2K)

Thermal conductivity -26.28 W/m K

For cell 06-10

 -for cell 06-10 the values at the end of the simulation are,

The heat transfer coefficient is -83.32 W/(m2K)

Thermal conductivity -16.22 W/m K

 

e)Charge and Discharge profile:-

-charge and discharge rate profile the plot for current is ranging in between -50 to 50 A and the voltage 30 to 43 volt and the final values for

I(current)= 1.4695e-13 A

v(volt)=36.33 v

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 CONCLUSIONS:-

                      1)As the difference in temperature we can observe the changes in heat flow rate by conduction and convection.

                      2)There are slight changes in cyclic charge and discharge as I fixed peak amplitude at 50 A

                      3)SOC for all the temperature remain the same so there is no difference in SOC graphs

                      4) cell no 5 had higher oscillation and we can observe that in every graph because we design that cell for fault 

                   

 REFERENCE:-

                 For base model:-https://www.mathworks.com/help/physmod/simscape/ug/lithium-ion-battery-pack-with-fault.html