problem from pressure distribution in rotating flow, Exams of Fluid Mechanics

problem from pressure distribution in rotating flow

Typology: Exams

2018/2019

Uploaded on 08/09/2019

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Problem 1:
Air is introduced through a nozzle into a tank of water to form a stream of bubbles. If
the bubbles are intended to have a diameter of 2 mm, calculate how much the
pressure of the air at the tip of the nozzle must exceed that of the surrounding water.
Assume that the value of surface tension between air and water as 72.7 x 10-3 N/m.
Data:
Surface tension () = 72.7 x 10-3 N/m
Radius of bubble (r) = 1
Formula:
p = 2/r
Calculations:
p = 2 x 72.7 x 10-3 / 1 = 145.4 N/m2
That is, the pressure of the air at the tip of nozzle must exceed the pressure of
surrounding water by 145.4 N/m2
Problem 2:
A soap bubble 50 mm in diameter contains a pressure (in excess of atmospheric) of 2
bar. Find the surface tension in the soap film.
Data:
Radius of soap bubble (r) = 25 mm = 0.025 m
p = 2 Bar = 2 x 105 N/m2
Formula:
Pressure inside a soap bubble and surface tension () are related by,
p = 4/r
Calculations:
= pr/4 = 2 x 105 x 0.025/4 =1250 N/m
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Air is introduced through a nozzle into a tank of water to form a stream of bubbles. If the bubbles are intended to have a diameter of 2 mm, calculate how much the pressure of the air at the tip of the nozzle must exceed that of the surrounding water. Assume that the value of surface tension between air and water as 72.7 x 10-3 N/m. Data: Surface tension () = 72.7 x 10-3^ N/m Radius of bubble (r) = 1 Formula: p = 2/r Calculations: p = 2 x 72.7 x 10-3^ / 1 = 145.4 N/m^2 That is, the pressure of the air at the tip of nozzle must exceed the pressure of surrounding water by 145.4 N/m^2 Problem 2: A soap bubble 50 mm in diameter contains a pressure (in excess of atmospheric) of 2 bar. Find the surface tension in the soap film. Data: Radius of soap bubble (r) = 25 mm = 0.025 m p = 2 Bar = 2 x 10^5 N/m 2 Formula: Pressure inside a soap bubble and surface tension () are related by, p = 4/r Calculations:  = pr/4 = 2 x 10^5 x 0.025/4 =1250 N/m

Water has a surface tension of 0.4 N/m. In a 3 mm diameter vertical tube, if the liquid rises 6 mm above the liquid outside the tube, calculate the contact angle. Data: Surface tension () = 0.4 N/m Dia of tube (d) = 3 mm = 0.003 m Capillary rise (h) = 6 mm = 0.006 m Formula: Capillary rise due to surface tension is given by h = 4cos()/(gd), where  is the contact angle. Calculations: cos() = hgd/(4) = 0.006 x 1000 x 9.812 x 0.003 / (4 x 0.4) = 0. Therfore, contact angle  =83.7 o

Compare the capillary rise of water and mercury in a glass tube of 2 mm diameter at 200 C .Given that the surface tension of water and mercury at 200 C are 0.0736 N/m and 0.051N/m respectively. Contact angles of water and mercury are 00 and 130 0 respectively. Solution : Given data: Surface tension of water, = 0.0736 N/m And surface tension mercury, =0.051N/m Capillary rise in a tube: For mercury specific weight ρg = 9810x13.6 N/m^3 and 1300 Gives capillary rise of h=‐6.68 mm, i.e., depression Note that the negative sign indicates capillary depression. For water specific weight ρg = 9810 N/m^3 and 00 gives capillary rise of h=15 mm