Question

Ethylene glycol at 40 degree C flows along a heated vertical plate L = 0.5 m long and is maintained at a uniform temperature of 80 degree C. Determine the minimum flow velocity below which the effect of free convection on heat transfer becomes more than 5 percent. (For ethylene glycol take Pr = 50.)

Answer 1

Vertical plate temperature T1 = 80 C

Ethylene glycol temperature T2 = 40 C

Mean temperature Tf = ( 40 + 80)/2 = 60 C

Physical data taken from table at mean temperature Tf = 60 C

Kinematic viscosity of ethylene glycol v = 1.78*10^-5 m2/s

Pr = 50

First calculations for natural convection :

Ra = Gr*Pr

Gr = g*b*ΔT*L^3/v^2

L = 0.2 m

b = 1/Tf = 1/(60+273) = 1/323

g = 9.8 m/s2

Gr = 9 8*(1/323)*(80-40)*0.2^3/(1.78*10^-5)^2 = 478799351

Ra = Gr*Pr = 2.39*10^10

Here we know that

Ra in (10^9 - 10^13 ) for vertical plate then nusselt number

Nu = 0.1*Ra^1/3

Nu = 0.1*(2.39*10^10)^1/3 = 288

hL/k = 288

h = 288*k/0.2 = 1440*k  

Heat transfer coefficient in natural convection

h = 1440*k

Where k is the thermal diffusivity at 60 C.

forced convection : let us assume minimum velocity of ethylene glycol is Vm.

And here in this question says we have to maintain minimum velocity it means we have to maintain laminar flow in which velocity is minimum.

For plate in laminar flow forced convection

Nusselt number Nu = 0.664*Re^0.5*Pr^(1/3)

Nu = 0.664*(Vm*L/v)^0.5*50^(1/3)

hf*L/k = 0.664*(Vm*0.2/1.78*10^-5)^0.5 * 50^1/3

hf = 1296.48k*(Vm) ^0.5

So effect of free convection is more than 5 % of heat transfer.

Heat transfer due to forced convection = 5 % more heat transfer due to convection

hf*A*(T1 - T2) = h*A*(T1-T2) * 1.05

A surface area and temperature change both are same for heat transfer.

1296.48*k*(Vm) ^0.5 = 1440*k *1.05

Vm = 1.36  m/s

minimum velocity of ethylene glycol required Vm = 1.36 m/s below which the effect of free convection on heat transfer becomes more than 5% .