2 Unit V Fem (heat Transfer).pdf

Unit: Heat Transfer FINITE ELEMENT METHODS (16 ME 64) Compiled by: Dr P R Venkatesh Associate Professor ME Dept, RVCE, Bengaluru-59

Dr P R Venkatesh, ME Dept RVCE

Basic Concepts Heat: A form of energy Temperature: Degree of hotness or coldness of a body Heat Transfer: Transfer of energy from one region to another due to temperature difference till the equality of temperature or thermal equilibrium is attained. Basically, there are three modes of heat transfer; 1. Conduction 2. Convection 3. Radiation

Heat Transfer by Conduction Conduction: • Heat conduction is a mechanism of heat transfer from a region of high temperature to a region of low temperature within a medium. (Solid, liquid or gas) • It can also take place between two bodies in physical contact, without much appreciable motion of molecules forming the substance. Pure conduction is found only in solids.

Fourier law of Heat Conduction (Joseph Fourier was a French mathematician)

Fourier law states that Rate of heat conduction ‘Q’ is proportional to the area ‘A’ measured normal to the direction of heat flow and to the temperature gradient (dT/dx) in that direction.

 dT   dT  Q   A   Q  kA   dx dx    

A T1

T1 Q T2

x

where Q = heat flux (Watt or J / s) dT A = Area in m , =Tempr gradient Kelvin / meter dx k = Thermal conductivity in W / m - K, ( ve sign indicates 2

that temperature decreases in direction of heat flow)

>

T2

Heat Transfer by Convection Free stream

T

Moving fluid

Ts  T

Q

Ts

Wall

Newton’s law of cooling • The rate of heat transfer by convection between the fluid at temperature T and the solid surface at temperature TS is given by ‘Newton’s law of cooling’.

Q  hA Ts  T  where 'h' (W / m 2 K) is the convective heat transfer coefficient which depends on the fluid flow but not a property of the material. A = Surface Area in m 2 Ts = Temperature of the surface, K T = Temperature of the fluid, K • Convection is possible only in a fluid medium. Effectiveness of convection depends largely on the mixing motion of the fluid.

Stefan-Boltzmann Law Of Radiation • A black body is a hypothetical perfect absorber and radiator of energy, with no reflecting power. • Stefan-Boltzmann Law states that “the emissive power of a black body is directly proportional to the fourth power of its absolute temperature”. Q  T4  Q   AT 4 where  = Stefan  Boltzmann constant  = 5.669  10 8 W / m 2 K 4 , A  Surface area of the body in m 2 T  Absolute temperature in Kelvin *When there is a radiation exchange between two black bodies at temperatures T1 & T2 , then Q =  A(T14  T24 ) ** Real surfaces do not radiate as much energy as black bodies. They are called grey bodies with emissivity ' ' such that Q =  A (T14  T24 )

Finite Element formulation for heat transfer problems 1-D Heat transfer problem is analogous to 1 D bar element in structural analysis.  u   T  1) Strain   in structural anlysis is replaced by temperature gradient    x   x  2) The stiffness matrix  k  is replaced by thermal conductivity matrix  k t  3) Nodal displacement vector u is replaced by nodal temperature vector T  4) The load vector is replaced by heat generated /conducted/convected in from the element. 5) Displacement/force bc's are replaced by bc's of specified temperature/heat flux. 6) The shape functions, stiffness matrix & [B] matrix remains same. 7) Either Elimination approach or Penalty approach may be used for solving the equilibrium equation.

Dr P R Venkatesh, ME Dept RVCE

Points to be remembered :  1 1  1 1    0 0  2) If there is convection at free end,  kh end  hA   0 1   0 0  Ak  1 1 Then,  k    kc    kh end   hA     0 1 l  1 1   

Ak 1) Thermal conductivity matrix  kc  = l

0  3) Force vector due to free end convection  Fh end =AhT   1  4) If there is convection from surface of the fin & end insulated hpl  2 1   kh     where p = perimeter of the fin 6 1 2  Ak  1 1 hpl  2 1  Then,  k    kc    kh      l  1 1  6 1 2  Dr P R Venkatesh, ME Dept RVCE

Points to be remembered 5) Force vector if there is internal heat generation qAl 1 Fq   2 1 where q is the heat generated per unit volume  6) Force vector due to surface convection of a fin hT pl 1 Fh   2 1 qAl 1 hT pl 1  Global force vector  F      2 1 2 1 7) The equilibrium equation is  K T   F

Dr P R Venkatesh, ME Dept RVCE

Boundary conditions in Thermal problems : Mainly there are three kinds of boundary conditions; (i) Specified temperature boundary conditions : Consider the wall of a furnace or tank containing hot liquid at a temperature To and the outer surface exposed to atmosphere at temperature T . Let the outside surface of the wall be maintained at TL as shown in fig. The boundary conditions are;

A

T@ x 0  To q@ x  L

   hA(TL  T ) 

h k To

𝑇∞ TL L Dr P R Venkatesh, ME Dept RVCE

(ii) Specified heat flux (or insulated)boundary conditions : Let the inner surface of the wall be insulated and the outer surface is exposed to atmpsphere for convective heat transfer. The bc's are; q@ x 0  0 q@ x  L

   hA(TL  T ) 

A h

Insulated

k

𝑇∞

TL L

Dr P R Venkatesh, ME Dept RVCE

(iii) Convective boundary conditions : Let To be the root or base temperature of a fin with an insulated tip. (i.e. heat going out of the tip is negligible). The bc's are; T  To @ x  0   q0@xL  T0

Insulated tip

L

Dr P R Venkatesh, ME Dept RVCE

Problem 1 Determine the temperature distribution through the composite wall shown in fig. 20 0C

T1 k2 =25 W/m 20 0C

k1 =10 W/m2-

0

T2

T1

h

0

0

h

2 =900 W/m - 0C

- C

20 =900 W/m - C

k2 =25 W/m - C

k1 =10 W/m2- C 0

𝑇∞ = −100 𝐶

T2

C

T3 0.08 m

0.06m

T3 0.08 m

0.06m Dr P R Venkatesh, ME Dept RVCE

Here, two elements are to be considered, only end convection prevails with no internal heat generation. Also Area A  1 m 2 Element thermal conductivity matrices :

Element 1 :  k   (1)

Ak1  1

l1  1

1

1  10  1

  1  0.08  1

1

 1 1 =125    1  1 1  

Element 2 : As there is end convection from node 3,

k 

( 2)



Ak2  1

l2  1

1

0 0  1  25  1 1 0 0   hA    900  1       1 0 1  1 1 0 1 0.06      

 1 1 0 0   416.67 416.67     k   416.67      1 1 0 900  416.67 1316.67       (Taking 125 as common factor so as to easily add the matrices 1 & 2) ( 2)

 k 

( 2)

 3.33 3.33  125    3.33 10.53  Dr P R Venkatesh, ME Dept RVCE

1 0  1 Global conductivity matrix  K   125  1 4.33 3.33    0 3.33 10.53  For no heat generation, & only end convection,

F   0

0 hT   0 0 9000 T

T

Equilibrium equation is :  K T    F  1 0  T1   0  1      125  1 4.33 3.33 T2    0   0 3.33 10.53  T3  9000  Using elimination approach with specified temeperture T1  20o C ,

the load vector becomes; (0  k11T1 ) (0  k21T1 ) ( 9000  k31T1 )

T

  F   (0  125  20) 0  (125  20) 9000  0  20

T

Dr P R Venkatesh, ME Dept RVCE

F   2500 2500 9000

T

Now,eliminating first row & first column,  4.33 3.33 T2   2500  T2   0.84  125         3.33 10.53  T3  9000  T3   7.1  The temperature distribution is T  = 20 -0.84 -7.1

T o

Dr P R Venkatesh, ME Dept RVCE

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Problem 2 Determine the temperature distribution through the composite wall shown in fig. k1  25 W / moC k2  35 W / m oC h

To k2

k1

k3

Ti

k3  55 W / m C o

h  30 W / m 2 o C Ti or T  9000 C

0.4 m

0.2 m

0.2 m

Composite wall

Dr P R Venkatesh, ME Dept RVCE

To or T4  250 C Area A  1 m 2

To

h

k2

k1

h

k3

Ti

Ti

0.2 m

0.4 m

T1

1

T2

2

T3

3

T4

FE model of composite wall

0.2 m

Composite wall

Here, 3 elements are to be considered, only end convection prevails at node 1 with no internal heat generation. Element thermal conductivity matrices :

Element 1 : As there is end convection to node 1,

k 

(1)



Ak1  1

l1  1

1

1 0  (1) 1  25  1 1 1 0   92.5 62.5  hA   30  = k         1 0.4  1 1  0 0   0 0   62.5 62.5 

Element 2 :  k   ( 2)

Ak2  1 l2

 1 

1

1  35  1

  1 0.2  1

1

 175 175   1   175 175 

Dr P R Venkatesh, ME Dept RVCE

Element 3 :  k  

Ak3  1

1

1  55  1

   l3  1 1  0.2  1 Global conductivity matrix is; (3)

1

 275 275   1   275 275 

0 0  T1   92.5 62.5 T   62.5 237.5 175  0  2   K  Temperatures T         0 175 450 275 T3    T4  0 275 275   0 Load vector (only end convection is prevailing at node 1) 1  1  27000  0  0   0        F  hAT    30 1 900      0  0   0  0  0   0  Dr P R Venkatesh, ME Dept RVCE

Equilibrium equation is :  K T    F  0 0  T1  27000   92.5 62.5   0   62.5 237.5 175 T2  0            0 175 450 275  T3   0      0 0  275 275   T4   0  Using elimination approach with T4  20o C , load vector becomes;

F   (27000  k14T4 ) (0  k24T4 ) (0  k34T4 ) (0  k34T4 ) T F   27000 0 (0  (275  20) (0  275  20) T   F   27000 0 5500 5500 Now eliminating row 4 & column 4, T

0  T1  27000   92.5 62.5  62.5 237.5 175 T    0     2   0 175 450  T3   5500  T1 = 400.14 o C , T2 = 160.22 o C, T3 = 74.53 o C Dr P R Venkatesh, ME Dept RVCE

Problem 3 Determine the temperature distribution in the rectangular fin as shown in fig. Assume steady state heat conduction & the heat generated inside the fin as 400 W/m3 k=7000 W/m-K 0

140 c

0.01 m 0.05 m

0.02 m

Here, A  b  t  0.02  0.01  2 104 m 2 Dr P R Venkatesh, ME Dept RVCE

Here, a single element may be considered, no convection ( h  0) prevails and there is internal heat generation q  400 W / m3 Element thermal conductivity matrix : 1

2  104  7000  1

 28 28   k        l  1 1  0.05  1 1   28 28  Force vector due to internal heat generation Ak  1

1

qAl 1 400  2 104  0.05 1 0.002  F         2 1 2 1 0.002   28 28 T1  0.002  Equilibrium equation is  K T    F         28 28  T2  0.002  As T1 = 140 o C, for elimination approach, the load vector may be modified as;  F  = 0.002  (28 140) 0.002  ( 28 140

T

  F   3920 3920

T Dr P R Venkatesh, ME Dept RVCE

Equilibrium equation is  K T    F   28 28 T1  3920        28 28  T2   3920  Eliminating first row & first column, 28T2  3920  T2 = 140 0 C The temperature distribution is T   140 140

T

Dr P R Venkatesh, ME Dept RVCE

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Problem 4 Determine the temperature distribution in a 1 D fin as shown in fig. Ignore the heat convection from the end of the fin, and consider heat loss from lateral surface only. Tair = 50 0 C T1 = 1500C

0

h = 6 W/cm2 C 0

k = 80 W/cm C

 =2 cm

6 cm

Note : As there is convection from lateral surface of the fin and ignoring the heat loss from the end of the fin;  k    kc    kh  Ak k   l

 1 1 hpl  2 1  hT pl 1  1 1   6 1 2  and Load vector F  2 1 where p   d      Dr P R Venkatesh, ME Dept RVCE

Here, a single element may be considered, convection prevails and there is no internal heat generation. Element thermal conductivity matrix : (including convection)

 1 1 hpl  2 1   1 1   6 1 2       d 2   22 where Area A    3.14 cm 2 , p   d    2  6.28 cm 4 4 3.14  80  1 1 6  6.28  6  2 1   117.23 4.187    k         1 1 1 2  4.187 117.23 6 6       hT pl 1 6  50  6.28  6 1 5652  and load vector F       2 1 2 1 5652  Ak  k    kc    k h   l

Dr P R Venkatesh, ME Dept RVCE

Equilibrium equation is  K T    F   117.23 4.187  T1  5652       T  4.187 117.23 5652   2   As T1 = 150 o C, the load vector may be modified as;

F  = 5652  (117.23 150)

5652  (4.187 150  11932 6280.05 T

T

 117.23 4.187  T1   11932       T  4. 187 117.23 6280.05   2   Eliminating first row & first column, 117.23T2  6280.05  T2 = 53.57 0 C The temperature distribution is T  = 150 53.57 

T o

Dr P R Venkatesh, ME Dept RVCE

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Problem 5 Determine the temperature distribution in a 1 D fin as shown in fig. Assume steady state heat conduction & the heat generated inside the fin as 400 W/m3. Diameter of the circular rod is 0.02 m. Insulation T1 = 80 0C

 =0.02 m

k = 300 W/m0C 0.08 m 3

Q= 400 W/m

Ans : The temperature distribution is T   80 78

T

Dr P R Venkatesh, ME Dept RVCE

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