Numericals - Mechanics of Fluids (AMIE, Summer 2023)

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Numerical
The surface Tension of water in contact with air at 20°C is 0.0725 N/m. The pressure inside a droplet of water is to be 0.02 N/cm² greater than the outside pressure. Calculate the diameter of the droplet of water. (7 marks)
 
Solution
Surface tension, σ = 0.0725 N/m
Pressure intensity, P = 0.02 N/m2
P = 4σ/d
Hence, the Diameter of the drop
d = 4 x 0.0725/200 = 1.45 mm
 
Numerical
Find the surface tension in a soap bubble of 40 mm diameter when the inside pressure is 2.5 N/m² above atmospheric pressure. (7 marks)
Answer: 0.0125 N/m
 
Numerical
The pressure outside the droplet of water of diameter 0.04 mm is 10.32 N/cm² (atmospheric pressure). Calculate the pressure within the droplet if surface tension is given as 0.0725 N/m of water. (7 marks)
Answer: 0.725 N/cm2 
 
Numerical
An open lank contains water up to a depth of 2 m and above it an oil of specific gravity 0.9 for a depth of 1 m. Find the pressure intensity (i) at the interface of the two liquids, and (ii) at the bottom of the tank. (8 marks)

Solution
See figure.
 Pressure intensity at the interface of both liquids
= roilgh1 = (0.9 x 1000) x 9.81 x 1 = 8829 N/m2
Pressure intensity at the bottom
= roilgh1 + rwatergh2
= (0.9 x 1000) x 9.81 x 1 + (1 x 1000) x 9.81 x 2
= 8829 + 19620 = 28449 N/m2

Numerical
A rectangular pontoon is 5 m long, 3 m wide and 1.20 m high. The depth of immersion of the pontoon is 0.80 m in seawater. If the centre of gravity is 0.6 m above the bottom of the pontoon, determine the meta-centric height. The density for seawater = 1025 kg/m3.
 
Solution
Dimension of pontoon = 5 m x 3 m x 1.20 m
Depth of immersion = 0.8 m
AG = 0.6 m
AB = (1/2) x Depth of immersion = (1/2) x .8 = 0.4 m
Density for seawater = 1025 kg/m3
Meta-centre height GM. given by Equation is 
GM = (I/Vsub) - BG  
where I = M.O. Inertia of the plan of the pontoon about the y-y axis
= (1/12) x 5 x 33 = 45/4 m4
Vsub = 3 x 0.8 x 5.0 = 12.0 m3
Hence, GM = (45/4)(1/12) - 0.2 = 0.7375 m   

Numerical
In a 100 mm diameter horizontal pipe, a venturimeter of 0.5 contraction ratio has been fixed. The head of water on the metre when there is no flow is 3 m (gauge). Find the rate of flow for which the throat pressure will be 2 metres of water absolute. The coefficientt of discharge is 0.97. Take atmospheric pressure head = 10.3 m of water. (10 marks)

Solution
Diameter of the pipe, d1 = 100 mm = 10 cm
Area, a1 = (π/4)d12 = (π/4)(10)2 = 78.54 cm2
Diameter of the throat, d2 = 0.5 x d1 = 0.5 x 10 = 5 cm
Area, d2 =(π/4)d22=  19.635 cm2
Head of water for no flow = p1/ρg = 3 m (gauge) = 3 + 10.3 = 13.3 (absolute)
Throat pressure head = p2/ρg = 2 m of water absolute
Q = {C_d}.\frac{{{a_1}{a_2}}}{{\sqrt {a_1^2 - a_2^2} }}.\sqrt {2gh} 
= 0.97x\frac{{78.54x19.635}}{{\sqrt {{{78.54}^2} - {{19.635}^2}} }}x\sqrt {2x9.81x1130} 
= 29306.8 cm3/s = 29.306 litres/s. 
 
Numerical
An orifice meter with orifice diameter 15 cm is inserted in a pipe of 30 cm diameter. The pressure difference measured by a mercury oil differential manometer on the two sides of the orifice meter gives a reading of 50 cm of mercury. Find the rate of flow of oil of sp.gr. 0.9 when the coefficient of discharge of the orifice meter is 0.6. (10 marks)
 
Solution
A0 = (π/4)d02 = (π/4)(0.15)2 = 0.0176 m2
A0 = (π/4)d02 = (π/4)(0.15)2 = 0.0176 m2
Similarly, A1 = 0.0706 m2
h = x\left[ {\frac{{{S_g}}}{{{S_0}}} - 1} \right] = 50\left[ {\frac{{13.6}}{{0.9}} - 1} \right]

= 705.5 cm of oil

Discharge
Q = {C_d}.\frac{{{A_0}{A_1}\sqrt {2gh} }}{{\sqrt {A_1^2 - A_0^2} }} 
= 0.64x\frac{{0.0176x0.0706x\sqrt {2x9.81x7.05} }}{{\sqrt {{{0.0706}^2} - {{0.0176}^2}} }}
= 0.137 m3/s
 
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