Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a particular location where flow and heat transfer are fully developed, the wall temperature is 370K. Determine:

a. The oil mean temperature.

b. The centerline temperature.

c. The axial gradient of the mean temperature.

d. The heat transfer coefficient.

(a) Tb = 330.12 K (b) Tc = 304.73 K (c) 19.81 K/m (d) h = 60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm = 0.01 m

Electrical heat rate, q = 76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k = 0.139 W/m. K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg. K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π * 0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb = Ts - 11/24 (q. R/k) (R = D/2=0.01/2 = 0.005 m)

Tb = 370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc = Ts - 3/4 (qs. R/k) = 370 - 3/4 * (2419.16 * 0.005/0.139)

Tc = 304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 * qs/ρ * V*cp * R

=2 * qs/ρ*[m/ρ (πR²) * cp * R

=2 * qs/[m / (πR) * cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3 / (π * 0.005) ] * 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h = 48/11 (k/D)

=48/11 (0.139/0.01)

h = 60.65 W/m². K