Metal plates of 0.5 m^2 and 3 cm thick (k = 350 w/m K, rho = 2900kg/m^3, and c_P
ID: 1620902 • Letter: M
Question
Metal plates of 0.5 m^2 and 3 cm thick (k = 350 w/m K, rho = 2900kg/m^3, and c_P = 800 J/kg middot K) are heated to 900 degree C in a process and then moved through a 10 m long spray chamber at 4 cm/s to cool them with 30 degree C water. Determine the convection film coefficient for heat removal so that the plates exit at 50 degree C and when h = 55V^0.8, where V is velocity of the plate as it moves through the cooling spray, and verify that the lumped analysis is the proper procedure to analyze this flow.Explanation / Answer
Film temperature is TF = (T + Tw) /2 = 465 ºC = 738 K
Reynolds number ReL = uL/ = uL/
is the kinematic viscosity of the fluid (m2/s)
Evaluating properties of water at TF = 738 K gives the following: = 2900 kg/m3
= 0.801 x 10-6 m2 /s, k = 350 W/m-K, and Prandtl number of water at 30oC Pr = 5.43
u = 4 cm/s = 0.04 m/s
ReL = (0.04 m/s)*(0.50 m)/(0.801 x 10-6 m2 /s) = 24969 < 5 x 105 (laminar)
NuL = hL/k = 0.664*ReL1/2Pr1/3
NuL = Nusselt number
NuL = (0.664)*(24969)1/2*(5.43)1/3 = 184.42
convective heat transfer coefficient is given by below equation
h = 184.42k/L = (184.42)*(350)/(0.5) = 129094 W/m2 -K
Biot number = Bi = hS/k = h*(V/A)/k
V = Volume of the plate = length x breadth x height = 0.5x0.5 x 0.03 = 7.5 x 10 -3 m3
A = Area of the plate = 0.25 m2
for h = 55V0.8 = 4.188 W/m2 -K
Bi = hS/k = h*(V/A)/k = 4.188 W/m2 -K*(7.5 x 10 -3 m3/0.25 m2)/350 W/m-K = 3.6 x 10-4
Since Bi << 0.1 lumped-parameter analysis is applicable
(T - T)/(To - T) = e-(hA/cV) where c = 800 J/kg-ºC
(hA/cV) = 1/(RrhCth) with c = cV/hA (time constant)
c = (800 J/kg-ºC)*(2900 kg/m3 )*7.5 x 10 -3 m3 / {(4.188 J/s-m 2 -ºC)*0.25 m2 } = 16618.9 s
Gives (T - T)/(To - T) = e-/16618.9
ln{ (50 - 30)/(900 - 30)} = -/16618.9
-3.773 = -/16618.9
= t = 62699 s (17 hrs) to cool to T = 50 oC