Question

The upper surface of a 50-cm-thick solid plate (k 5 237 w/mk) is being cooled by water with temperature of 20c. The upper and lower surfaces of the solid plate maintained at constant temperatures of 60c and 120c, r espectively. Determine the water convection heat transfer coefficient and the water temperature gradient at the upper plate surface.

Answer 1

Heat transfer coefficient between plate and water is $15711(\frac{W}{m^{2}K})$ And temperature temperature gradient is $- 120(^{\circ}C/m)$

Explanation:

Given data

Thickness of plate(L) is 0.50 m.

Thermal conductivity of plate (k) = $5237 (\frac{W}{mK})$

Temperature at surface below plate $T_{1} = 120^{\circ}C$

Temperature at surface above plate$T_{2} = 60^{\circ}C$

Temperature of water at contact with plate $T_{3} = 20^{\circ}C$

Now rate of Heat flow

$\dot{Q} = -kA\frac{dT}{dx} = - kA \frac{T_{2}- T_{1}}{L} = h A (T_{2} - T_{3})$         ...1)

$\dot{Q} = kA \frac{T_{1}- T_{2}}{L} = h A (T_{2} - T_{3})$

$\frac{k\times A\times(120-60)}{0.5} = h\times A\times (60- 20)$

$\frac{5237\times A\times (60)}{0.5} = h\times A\times (40)$

Now heat transfer coefficient $h = 15711\frac{W}{m^{2}K}$

From equation 1

$-kA\frac{dT}{dx} = h A (T_{2} - T_{3})$

Now Temperature gradient $\frac{dT}{dx} = - \frac{15711\times (60-20)}{5237} = -120^{\circ}c/m$