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
