Question

The elemental unit of an air heater consists of a long circular rod of diameter D, which is encapsulated by a finned sleeve and in which thermal energy is generated by ohmic heating. The N fins of thickness t and length L are integrally fabricated with the square sleeve of width w. Under steady-state operating conditions, the rate of thermal energy generation corresponds to the rate of heat transfer to airflow over the sleeve. (a) Under conditions for which a uniform surface temperature Ts is maintained around the circumference of the heater and the temperature T[infinity] and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the heat rate for Ts = 320°C, D = 15 mm, an aluminum sleeve (ks = 240 W/m·K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, T[infinity] = 50°C, and h = 500 W/m2·K. (b) For the foregoing heat rate and a copper heater of thermal conductivity kh = 400 W/m·K, what is the required volumetric heat generation within the heater and its corresponding

Answer 1

a) q´ = 81546.36 W/m

T = 334.8 K

b) q = 4.21x10^8 W/m^3

Step-by-step explanation:

We will take the following assumptions:

-Stationary state

-constant properties,

- negligible contact resistance

-uniform convection coefficient

-uniform heat generation

a)

Rcond = 1/(r*ks) = ln(1.08w/D)/(2*pi*ks) = ln(2.16)/(2*pi*240) = 5.1x10^-4 mK/W

Lc = L + t/2 = 0.022 m

m = (2h/ks*t) = 32.3 m^-1

mLc = 0.71

nf = tanh(mLc)/mLc = 0.61/0.71 = 0.86

At = NAf + Ab = N(2L+t)+(uw-Nt) = 0.704 + 0.096 = 0.8 m

no = 1-(nAf/At)*(1-nf) = 1-(0.704/0.8)*(0.14) = 0.88

Rto = (no*h*At)^-1 = (0.88*500*0.8)^-1 = 2.8x10^-3 mK/W

q´ = (320-50)/(5.11x10^-4+2.8x10^-3) = 81546.36 W/m

b)

Us*As = Us*pi*D = 298/pi*0.015 = 6323.76 W/m^2K

q = 4Us(Ts-Tinfinity)/D = 4*(6323.76)*250/0.015 = 421584000 W/m^3 = 4.21x10^8 W/m^3

The temperature is equal to:

T = (q*(D/2)^2)/(4ks) + Ts = 4.21x10^8* (0.0075^2)/(4*400) + 320 = 334.8 K