1D STEADY STATE HEAT

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Lester French
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1 D SEADY SAE HEA CONDUCION () Prabal alukdar Aociate Profeor Department of Mechanical Engineering II Delhi

2 Convection Boundary Condition Heat conduction at the urface in a elected direction Heat convection at the urface in the ame direction In writing the equation for convection boundary condition, we have elected the direction of heat tranfer to be the poitive x-direction at both urface. But thoe expreion are equally applicable when heat tranfer i in the oppoite direction

boundary condition, we have elected the direction of heat tranfer to be the poitive

3 Radiative Boundary Condition Heat conduction at the urface in a elected direction Radiation exchange at the urface in the ame direction

4 Interface Boundary Condition he boundary condition at an interface are baed on the requirement that () two bodie in contact mut have the ame temperature at the area of contact and () an interface (which i a urface) cannot tore any energy, and thu the heat flux on the two ide of an interface mut be the ame

at the area of contact and () an interface (which i a urface) cannot tore

5 Generalized Boundary Condition Heat tranfer to the urface in all mode Heat tranfer from the urface in all mode

6 D Carteian Solution of teady heat conduction equation Differential Equation: Boundary Condition: d dx 0 ( 0 ) Integrate: d dx C Integrate again: Applying the boundary condition to the general olution: ( x) Cx C ( x ) C x C + General Solution Abit Arbitrary Contant t Subtituting: + C C.0 C It cannot involve x or (x) after the boundary condition i applied.

to the general olution: ( x) Cx C + 0 0 ( x ) C x C + General Solution Abit Arbitrary

7 Cylindrical - Spherical Differential Equation: d dr ( r d dr Integrate: ) 0 Differential Equation: d dr ( r Integrate: d dr ) 0 d r dr C d r dr C Divide by r ( r 0) : d C dr r Integrate again: ( r ) C ln r C + which i the general olution. Divide by r ( r 0) : d dr C Integrate again: C r + C r () r

Divide by r ( r 0) : d C dr r Integrate again: ( r ) C ln r C + which i

8 During teady one-dimenional heat conduction in a pherical (or cylindrical) container, the total rate of heat tranfer remain contant, but the heat flux decreae with increaing i radiu.

10 Heat Generation Under teady condition, the energy balance for thi olid can be expreed a Rate of heat Rate of energy tranfer generation within from olid the olid ha ( ( ) g& V + V g ha V

11 A large plane wall of thickne L (A A wall and V LA wall ), A long olid cylinder of radiu r o (A πr o L and V πr o L), A olid phere of radiu r 0 (A 4πr o L and V 4/3πr 3 o ) + V g ha

12 Under teady condition, the entire heat generated within the medium i conducted through the outer urface of the cylinder. he heat generated within thi inner cylinder mut be equal to the heat conducted through the outer urface of thi inner cylinder Integrating from r 0 where (0) 0 to r r o where (r o ) yield

he heat generated within thi inner cylinder mut be equal to the heat

13 he maximum temperature in a ymmetrical olid with uniform heat generation occur at it center

14 -D plane wall

15 Energy balance Rate of heat Rate of heat tranfer into the - tranfer out of the wall wall Rate of change of energy of the wall de Q in Q out dt de wall dt 0 wall for teady operation herefore, the rate of heat tranfer into the wall mut be equal to the rate of heat tranfer out of it. In other word, the rate of heat tranfer through the wall mut be contant, Q cond, wall contant. Fourier law of heat conduction for the wall L Qcond,wall dx x 0 kad Qcond, wall d ka dx contant t

be equal to the rate of heat tranfer out of it.

16 emp profile Q ka cond,wall (W) L he rate of heat conduction through a plane wall i proportional to the average thermal conductivity, the wall area, and the temperature difference, but i inverely proportional to the wall thickne

average thermal conductivity, the wall area, and the

17 emp profile d d dx dx x C x + D teady tate heat conduction equation (k ) 0 Integrate the above equation twice Boundary condition Apply the condition at x 0 and L,, C C L+ C C L+,, L (0) and (L), C,, (x) x + L,,, ( ) C

18 hermal Reitance Concept Analogy between thermal and electrical reitance concept Q& cond,wall R wall (W) R wall L ka ( o C/W)

19 Convection Reitance Q convection ha ( ) convection R convection Q (W) R convection ha ( o C/W)

20 Radiation Reitance Q rad εσ A ( 4 4 urr ) h rad A ( urr ) R rad urr (W) R rad h rad A (K/W) Combined convection and radiation h rad Q rad εσ ( + urr A ( urr ) )( + urr ) (W/m K) h h + combined conv h rad (W/m K) Poible when urr

21 he thermal reitance network for heat tranfer through a plane wall ubjected to convection on both ide, and the electrical analogy

22 Network ubjected to convection on both ide Rate of heat Rate of heat convection into conduction the wall through the wall h A( ) ka h A( ) L Q Q R h A conv, L ka R wall Adding the numerator and denominator yield h A R conv, Rate of heat convection from the wall Q R total (W) R total R conv, + R wall + R conv, h A + L ka + h A

23 Q R total (W) he ratio of the temperature drop to the thermal reitance acro any layer i contant, and thu the temperature drop acro any layer i proportional to the thermal reitance of the layer. he larger the reitance, the larger the temperature drop. Δ Q R ( o C) hi indicate that the temperature drop acro any layer i equal to the rate of heat tranfer time the thermal reitance acro that layer

24 It i ometime convenient to expre heat tranfer through a medium in an analogou manner to Newton law of cooling a Q UAΔ (W) Q & Δ R total UA R total he urface temperature of the wall can be determined uing the thermal reitance concept, but by taking the urface at which the temperature i to be determined a one of the terminal urface. R conv, h A Q Known

Steady Heat Conduction

Steady Heat Conduction In thermodynamics, we considered the amount of heat transfer as a system undergoes a process from one equilibrium state to another. hermodynamics gives no indication of how long