Question

Problem 6.2 7–14 Engine oil at 80°C flows over a 12-m-long flat plate whose temperature is 30°C with a velocity of 3 m/s. Determine the total drag force and the rate of heat transfer over the entire plate per unit width.

Answer 1

Answer:

A) Drag force = 88.02 N

B) Rate of heat transfer = 38,214 W

Explanation:

A) The properties of engine oil at temperature of; (Ts + T(∞))/2 =

(30 + 80)/2 = 55°C are;

From the table i attached, we get the following ;

Density(ρ) = 0.867 g/cm³ = 867 kg/m³

Viscocity (μ) = 0.1264 mm²/s = 12.64 x 10^(-5) m²/s

Also from the table, by interpolation;

Pr = 1551

k = 0.1414 W/m°C

Kinematic Viscosity (v) = 7.045 x 10^(-5) m²/s

Now, Reynolds number(Re) = VLc/v

Where V is the flow velocity,

v is the kinematic velocity,

Lc is the characteristic length.

Thus, Re = (3 x 12)/(7.045 x 10^(-5)) =

5.11 x 10^(5)

This is less than the critical Re, thus we have a laminar flow.

So, the friction coefficient,

Cf = 1.328 Re^(-0.5)

Cf = 1.328(5.11 x 10^(5))^(-0.5) = 0.00188

Drag force(Fd) = (Cf)(A)(ρV²/2) = 0.00188 x (12 x 1)((867 x 3²)/2) = 88.02 N

B) The Nusselt number is determined from:

Nu = hL/k = 0.664 x Re^(0.5) x  Pr^(1/3)

Nu = 0.664 x [5.11 x 10^(5)]^(0.5) x 1551^(1/3) = 5494.34

Thus, hL/k = 5494.34

h = (5494.34 x 0.1413)/12 = 64.69 W/m².k

Thus, rate of heat transfer per unit length is given as;

Q' = hA(T(∞) - Ts) = 63.69 x (12x1) x (80 - 30) = 38,214 W