Question

A stainless-steel wall (AISI 302) is 0.2 m thick with the top surface at a temperature 40°C and the bottom surface at a temperature 100°C. Water, at [infinity] 25°C, flows over the top surface. Find the water flow convection coefficient, the temperature gradient in the wall and the temperature gradient in the water that is in contact with the wall. Assume steady-state conditions and evaluate the properties of the stainless-steel at 400 K.

Answer 1

Answer/Explanation:

From the properties of stainless steel AISI 302.

At temperature 100°C, Thermal coefficient Ks = 16.2W/m.K.

Water at 40°C, Kf = 0.62W/m.K.

Using the rate equation, applying energy balance at x = 0,

Ks (T1 - T2)L = h(T1 - T[infinity])

h = (Ks/L) * (T1 - T2)/(T - T[infinity])

h is water flow convention coefficient, T1, T2, Ks, and L have their usual notations.

T1 = 100°C, T2 = 40°C,

T[infinity] = 25°C, L = 0.2m

h = (16.2W/m.K/0.2m) * (60/15)

h = 324W/m².K

Temperature gradient in the wall (dT/dx) = -(T1 - T2)/L = -60°C/0.2m = -300°C/m

In water at x = 0, the definition of h gives

(dT/dx)x=0 = -h/Kf(40°C - 25°C)

= (324W/m².K x 15°C)/0.62W/m.K = 7838.7°C/m

Answer 2

Answer:

a) Water flow convection coefficient $k_{w} = 0.62 W/mk$

b) Temperature gradient in the wall, $\frac{dT}{dx} = 300^{0} C/m$

c) Temperature gradient in the water, $\frac{dT}{dy} = 8370.97 ^{0}C/m$

Explanation:

Assumption: The flow is at steady state

At 400 K, steel has a thermal conductivity of, $k_{st} = 17.3 W/mk$

$T_{ \infty} = 25^{0} C$

Temperature at the bottom, $T_{b} = 100^{0}C$

Temperature at the top, $T_{t} = 40^{0}C$

Thickness of the stainless steel, L = 0.2 m

Film temperature of water, $T_{f} = \frac{T_{\infty} + T_{t} }{2}$

$T_{f} = \frac{25+ 40 }{2} \\T_{f} = 32.5^{0} C$

$T_{f} =32.5 + 273\\T_{f} =305.5 K$

a) Thermal conductivity of water at 305.5 K

$k_{w} = 0.62 W/mk$

b) Temperature gradient of steel, $\frac{dT}{dx} = \frac{T_{b} - T_{t} }{L}$

$\frac{dT}{dx} = \frac{100 - 40 }{0.2}$

$\frac{dT}{dx} = 300^{0} C/m\\\frac{dT}{dx} = 300 + 273\\\frac{dT}{dx} = 573 K/m$

Since steady state heat transfer is assumed, heat transfer by conduction equals heat transfer by convection

$k_{st} A\frac{dT}{dx} = hA (T_{t} - T_{\infty} )$

17.3 * A * 300 = h*A* (40 - 25)

h = 5190/15

h = 346 W/m²K

c) To calculate the temperature gradient in water, equate the heat transfer in both water and steel.

That is, $k_{w} \frac{dT}{dy} = k_{st} \frac{dT}{dx}$

$k_{w} \frac{dT}{dy} = k_{st} \frac{dT}{dx}\\0.62 * \frac{dT}{dy} = 17.3 * 300\\\frac{dT}{dy} = 5190/0.62\\\frac{dT}{dy} = 8370.97 ^{0}C/m$

Where $\frac{dT}{dy}$ is the temperature gradient of water