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.

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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.