Question

Find the temperature of the bar (made of aluminum, k = 202.4 W/m*k) with a rectangular cross-section at the center point. The Left end of rectangular is 80 c, the right end is 40 c and side walls convection boundary layer (steam temperature 22 c) and convection coefficient is 15 W/m^(2)K. Length is 10 mm, height 10 mm and width 100 mm. Please show working out.

Answer 1

Step-by-step explanation:

As it is given that no heat is generating and the system is present in steady state.

As we know that,

`(\partial^(2)T)/(\partial x^(2)) + (Q)/(k)` =
`(1)/(\alpha) (\partial T)/(\partial t)`

So, according to the given conditions,
`(Q)/(k)` = 0, and
`(1)/(\alpha) (\partial T)/(\partial t)` = 0

So,
`(\partial^(2)T)/(\partial x^(2))` = 0 ,
`(\partial T)/(\partial x)` =
`C_(1)`

T =
`C_(1)x + C_(2)`

Hence, boundary conditions will be as follows.

At x = 0, T =
`80^(o)C`

At x = L, T =
`40^(o)C`

Applying boundary condition is
`T_(0) = C_(2)`

Therefore,
`T_(0) = C_(2)` =
`80^(o)C`

`T_(L) = C_(1)x + 80`

40 =
`C_(1) (100 * 10^(-3)) + 80`

`(-40)/(100 * 10^(-3)) = C_(1)`

`C_(1)` = -400

Hence, temperature distribution in a block will be as follows.

T = -400x + 80

At center, x =
`50 * 10^(-3)`

Therefore, T =
`-400 * 50 * 10^(-3) + 80`

=
`60^(o)C`

Thus, we can conclude that the temperature of block at the center point is
`60^(o)C`.