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Heat & Mass Transfer - acharya ng ranga agricultural university

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An expression for the midplane temperature T o may be obtained through an<br />

energy balance. At steady-state conditions the total heat generated must<br />

equal the heat lost at the faces. Thus<br />

⎛<br />

2⎜−<br />

KA<br />

⎝<br />

dT<br />

dx<br />

⎤<br />

⎥<br />

⎦<br />

x=<br />

L<br />

⎞ •<br />

⎟=<br />

q.<br />

A.2L<br />

⎠<br />

where A is the cross-sectional area of the plate. The temperature gradient at<br />

the wall is obtained by differentiating Equation (4):<br />

T −T0<br />

T −T<br />

w<br />

0<br />

⎛<br />

= ⎜<br />

⎝<br />

x<br />

L<br />

2<br />

⎞<br />

⎟<br />

⎠<br />

T −T<br />

x<br />

⎝L<br />

⎛ ⎞<br />

0<br />

= ( Tw<br />

−T0<br />

) ⎜ ⎟ ⎠<br />

2x<br />

dT =<br />

w 0<br />

.<br />

2<br />

L<br />

( T −T<br />

) dx<br />

dT<br />

⎤<br />

⎦<br />

at x= L ( T T )<br />

dx ⎥<br />

=<br />

w−<br />

o<br />

. L<br />

x=<br />

L<br />

2<br />

2<br />

Since,<br />

dT<br />

−K.<br />

dx<br />

= q<br />

•<br />

L<br />

dT<br />

dx<br />

•<br />

q L<br />

= − K<br />

•<br />

qL<br />

T w<br />

− 2 0<br />

=−<br />

L K<br />

( T )<br />

•<br />

2<br />

qL<br />

T w<br />

−T<br />

0<br />

=−<br />

2K<br />

•<br />

2<br />

qL<br />

T<br />

0<br />

= + T w<br />

2K<br />

----- (5)

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