Question

An oil with a relative density of 0.7 flows with a flow rate of 0-05 m ^ 3 / h. Calculate the oil viscosity.
Solve this problem in two ways:

1) Assume laminar flow
2) Assume turbulent flow

There are two valid solutions in this case. What can you conclude about flows close to Re = 2300?

2 m D=5 mm m 2 Tank A B Tank + -5 m

Answer 1

2. * Applying Bernoulli's equation between points 1 and. At point i P4 Atpoint 2 pressure head pi 12 m 2 pressure head = P2 89 V = √2gh, V₂ = V 2ghz V 2 x 9.81 X 2 6.264 mis V 2 x 9.81 x 1 4.429 m/s Also 21=22 V, ² Zz PL 89 + + + hf V2 29 + + z = P2 29 РЯ Substituting values 2 + (6.264) ² it 2 x 9.81 (4,429)? 2 x 9.81 & hf 2 + 1.99988 It 0.999 79 + hf h = 2.000o q m due to friction, he = head loss flow rate & 0.05 m3/h 0.05 m3/s on 3600

Q - AV 1.3888 X 10° 6-S m3/s AV 1.38 88 X 10-5 m3/s og (58103 XV 1.3888 X 10.5 m3/s. 2 Il x 4 S. Average velocity VI 1.3888 X10 X 4 TT X, (5 x 10-3)? V 0.7073 m/s * Head loss equation due to friction is given by hf = f LV ² L = length of pipe. 2gD fa friction factor 2 Ef x 5 x (0.7073)? 1.2 x 9.8) X (5 * 10-3) PFOOD f = 2 X 2 X 9.81 X 5 X10 5 X 0.7073x 0.7073 f O. 0784 3 Relative density s = 0.7 S. Mass density sa 0.7 x density of Water 0.7 x 1000 F 700

For Laminar flow, friction factor f= 64 Re fa 64 SVD u. 0.078 43 64 & M 700 x 0.7073 x 5x10-3 From above M = 3.0337 X 10-3 NS 2 m² M = dynamic on absolute For Turbulent flow :- viscosity friction 0.316 factor (Re) 4 0.07843 = 0.316 700 X 0.7073 X 5X10-3 M From above = 9.3938 X 10 -3 NS m2 be * For flow having Reynold's number close to 2300 we Can conclude that the flow can be Laminar or it can transitional flow.