Question

A cylindrical container with d_o = 0.5 m and L = 7.5 m is used for chemical processing with insulated ends. The inside of the cylinder is kept at 400^o C with 5 kW of power coming from a heater. The outer surface temperature is known to be 78 ^oC and ambient air in the factory is 20C.

a.) Estimate the heat transfer coefficient between the surface of the cylinder and the ambient air and find the thermal resistance between the surface of the cylinder and the chemicals.

A large fan is used to cool the surface of the cylinder with a velocity of 25 m/s. As a result, the heater had to compensate for the cooling in order to maintain the chemicals inside the cylinder at 400C. Assume the air properties are: ? = 0.96 kg/m3, k = 0.032 W/m-K, ? = 215x10-7 Pa-s, and Pr = 0.70.

b.) Estimate the heat transfer coefficient between the air and the cylinder surface.

c.) Estimate the surface temperature of the cylinder after the fan is turned on and determine the increase in the heater power (W).

Answer 1

Step 1:

Diamter of cylinder d_o=0.5 m

Length of cylinder L= 7.5 m

Inside temperature T_i=400+273=673 K

Internal generation q_g=5000W

Outside temperature T_o=351K

ambient temperature=T_a=293 K

(a)

Surface area of the cylinder

For an insulated system,

Within the cylinder.

hA(Ti-T_a)=5000W

Substitute the respective values.

h=1.1168 W/m²K ---- Ans

Thermal resistance is given by

---- Ans

(b)

Properties of air:

desity of air

thermal conductivity k=0.032 W/mK

Prandtl no. Pr =0.70

viscosity

Velocity of air v= 25 m/s

Reynolds no:

Substitute the values.

Re=5.5*10⁵

For flow over cylinder and given Reynold number range, Nuselt number is calculated using empirical relation.

Nu=0.027 (Re)^0.805(Pr)^0.333

Substitute the values.

Nu=1013.7

But ,

Nu=h_avgL/k

Therefore,

h_avgL/k=1013.7

h_avg=4.325  W/m²K ---- Ans

(c)

h_avgA(T_new-T_a)=hA(T_o-T_a)

4.325*11.781*(T_new-293)=1.1168*11.781*(351-293)

T_new=307.97 K ---- Ans.

Calculate the thermal conductivity of cylinder.

Q=kA(Ti-To)/L =5000 W

k=9.88 W/mK

With fan,

Q_new = kA (T_i-T_new)/L

Q_new=5683.229 W

Increase in power =Q-Q_new