{"id":1821,"date":"2018-04-12T17:08:10","date_gmt":"2018-04-12T09:08:10","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1821"},"modified":"2023-08-09T08:30:57","modified_gmt":"2023-08-09T08:30:57","slug":"fluid-dynamics-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/fluid-dynamics-problems-and-solutions.htm","title":{"rendered":"Fluid dynamics \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Fluid dynamics \u2013 problems and solutions<\/span><\/p>\n

Torricelli’s theorem<\/b><\/a><\/span><\/p>\n

1. A container filled with water and there is a hole<\/span>, <\/span>as shown in the figure below. If acceleration due to gravity is 10 ms<\/span>-2<\/span><\/sup>, what is the speed of water through that hole? <\/span><\/span><\/p>\n

Known :\"Fluid<\/u><\/span><\/p>\n

Height (h) = 85 cm \u2013 40 cm = 45 cm = 0.45 meters<\/span><\/p>\n

Acceleration due to gravity<\/a> (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Wanted :<\/u> The speed of water (v)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Torricelli’s theorem states that the water leaves the hole with the same speed as an object free fall from the same height. Height (h) = 85 cm \u2013 40 cm = 45 cm = 0.45 meters<\/span><\/p>\n

Velocity of water is calculated using the equation of the free fall motion<\/a> :<\/span><\/p>\n

vt<\/sub>2<\/sup> = 2 g h <\/span><\/p>\n

vt<\/sub>2<\/sup> = 2 g h = 2(10)(0.45) = 9<\/span><\/p>\n

vt<\/sub> = \u221a9 = 3 m\/s <\/span><\/p>\n

2. A container filled with water and there is a hole, <\/span>as shown in figure below. If acceleration due to gravity is 10 ms<\/span>-2<\/span><\/sup>, what is the speed of water through that hole.<\/span><\/span><\/p>\n

Known :\"Fluid<\/u><\/span><\/p>\n

Height (h) = 1.5 m \u2013 0.25 m = 1.25 meters<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Wanted :<\/u> The speed of water (v)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

vt<\/sub>2<\/sup> = 2 g h = 2(10)(1.25) = 25<\/span><\/p>\n

vt<\/sub> = \u221a25 = 5 m\/s <\/span><\/p>\n

3. A container filled with water and there is a hole, <\/span>as shown in figure below. If acceleration due to gravity is 10 ms<\/span>-2<\/span><\/sup>, what is the speed of water through that hole.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Height (h) = 1 m \u2013 0.20 m = 0.8 meter\"Fluid<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Wanted :<\/u> The speed of water (v)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

vt<\/sub>2<\/sup> = 2 g h = 2(10)(0.8) = 16<\/span><\/p>\n

vt<\/sub> = \u221a16 = 4 m\/s <\/span><\/p>\n

4. A container filled with water and there is a hole, <\/span>as shown in figure below. If acceleration due to gravity is 10 ms<\/span>-2<\/span><\/sup>, what is the speed of water through that hole.<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Height (h) = 20 cm = 0.2 meters\"Fluid<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup><\/span><\/p>\n

Wanted:<\/u> The speed of water (v)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

\"Fluid<\/span><\/p>\n

5. A container filled with water and there <\/span>are two <\/span>hole<\/span>s<\/span>, <\/span>as shown in the figure below. <\/span>What is the ratio of x<\/span>1<\/span><\/sub> to x<\/span>2<\/span><\/sub>?<\/span><\/span><\/p>\n

Soluti<\/u>on<\/u><\/span><\/p>\n

\"Fluid<\/span><\/p>\n

Time interval of the water free fall from hole 1 :\"Fluid<\/span><\/p>\n

h = 1\/2 a t2 <\/sup><\/span><\/p>\n

0.8 = 1\/2 (10) t2<\/sup><\/span><\/p>\n

0.8 = 5 t2<\/sup><\/span><\/p>\n

t2<\/sup> = 0.8 \/ 5 = 0.16<\/span><\/p>\n

t = 0.4 seconds<\/span><\/p>\n

Time interval of the water free fall from hole 2 :<\/span><\/p>\n

h = 1\/2 a t2 <\/sup><\/span><\/p>\n

0.5 = 1\/2 (10) t2<\/sup><\/span><\/p>\n

0.5 = 5 t2<\/sup><\/span><\/p>\n

t2<\/sup> = 0.5 \/ 5 = 0.1<\/span><\/p>\n

t = \u221a0.1 second<\/span><\/p>\n

The horizontal distance (x) :<\/span><\/p>\n

x1<\/sub> = v1<\/sub> t1 <\/sub>= (2)(0.4) = 0.8 meters<\/span><\/p>\n

x2<\/sub> = v2<\/sub> t2<\/sub> = (\u221a10)(\u221a0.1) = (10)(0.1) = 1 meter<\/span><\/p>\n

The ration of x1<\/sub> to x2<\/sub> :<\/span><\/p>\n

x1<\/sub> :<\/sub> x2<\/sub> = 0.8 : 1 = 8 : 10 = 4 : 5<\/span><\/p>\n

The equation of continuity<\/b><\/a><\/span><\/p>\n

6. Water flows through a pipe of varying diameter, A to B and then to C. The ratio of A to C is 8 : 3. If the speed of water in pipe A is v, what is the speed of water in pipe C.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Area of A (AA<\/sub>) = 8\"Fluid<\/span><\/p>\n

Area of C (AC<\/sub>) = 3<\/span><\/p>\n

The speed of water in pipe A (vA<\/sub>) = v<\/span><\/p>\n

Wanted:<\/u> The speed of water in pipe C (vC<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of continuity :<\/span><\/p>\n

AA<\/sub> vA<\/sub> = AC<\/sub> vC<\/sub><\/span><\/p>\n

8 v = 3 vC<\/sub><\/span><\/p>\n

vC<\/sub> = 8\/3 v <\/span><\/p>\n

7. If the speed of water in pipe with a diameter of 12 cm is 10 cm\/s, what is the speed of water in a pipe with a diameter of 8 cm?<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Diameter 1 (d1<\/sub>) = 12 cm, radius 1 (r1<\/sub>) = 6 cm<\/span><\/p>\n

Diameter 2 (d2<\/sub>) = 8 cm, radius 2 (r2<\/sub>) = 4 cm<\/span><\/p>\n

The speed of water 1 (v1<\/sub>) = 10 cm\/s<\/span><\/p>\n

Wanted :<\/u> The speed of water 2 (v2<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Area 1 (A1<\/sub>) = \u03c0 r2<\/sup> = \u03c0 62<\/sup> = 36\u03c0 cm2 <\/sup><\/span><\/p>\n

Area 2 (A2<\/sub>) = \u03c0 r2<\/sup> = \u03c0 42<\/sup> = 16\u03c0 cm2 <\/sup><\/span><\/p>\n

The equation of continuity :<\/span><\/p>\n

A1<\/sub> v1<\/sub> = A2<\/sub> v2<\/sub><\/span><\/p>\n

(36\u03c0)(10) = (16\u03c0) v2<\/sub><\/span><\/p>\n

(36)(10) = (16) v2<\/sub><\/span><\/p>\n

360 = (16) v2<\/sub><\/span><\/p>\n

v2 <\/sub>= 360\/16<\/span><\/p>\n

v2 <\/sub>= 22.5 cm\/s<\/span><\/p>\n

8. Water flows through a pipe of varying diameter, as shown in figure below. If area 1 (A1<\/sub>) = 8 cm2<\/sup>, A2<\/sub> = 2 cm2<\/sup> and the speed of water in pipe 2 = v2<\/sub> = 2 m\/s then what is the speed of water in pipe 1 = v1<\/sub>.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Area 1 (A1<\/sub>) = 8 cm2\"Fluid<\/sup><\/span><\/p>\n

Area 2 (A2<\/sub>) = 2 cm2<\/sup><\/span><\/p>\n

Speed of water in pipe 2 (v2<\/sub>) = 2 m\/s<\/span><\/p>\n

Wanted :<\/u> the speed of water in pipe 1 (v1<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of continuity :<\/span><\/p>\n

A1<\/sub> v1<\/sub> = A2<\/sub> v2<\/sub><\/span><\/p>\n

8 v1<\/sub> = (2)(2)<\/span><\/p>\n

8 v1<\/sub> = 4<\/span><\/p>\n

v1 <\/sub>= 4 \/ 8 = 0.5 m\/s<\/span><\/p>\n

9. If the diameter of the larger pipe is 2 times the diameter of smaller pipe, what is the speed of fluid at the smaller pipe.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Diameter of the larger pipe (d1<\/sub>) = 2\"Fluid<\/span><\/p>\n

Radius of the larger pipe (r1<\/sub>) = \u00bd d1<\/sub> = \u00bd (2) = 1<\/span><\/p>\n

Area of the larger pipe (A<\/b>1<\/b><\/sub>) = \u03c0 r1<\/sub>2<\/sup> = \u03c0 (1)2 <\/sup>= \u03c0 (1) = \u03c0 <\/span><\/p>\n

Diameter of the smaller pipe (d2<\/sub>) = 1<\/span><\/p>\n

Radius of the smaller pipe (r2<\/sub>) = \u00bd d2<\/sub> = \u00bd (1) = \u00bd <\/span><\/p>\n

Area of the smaller pipe (A<\/b>2<\/b><\/sub>) = \u03c0 r2<\/sub>2<\/sup> = \u03c0 (1\/2)2 <\/sup>= \u03c0 (1\/4) = \u00bc \u03c0 <\/span><\/p>\n

The speed of fluid at the larger pipe (v<\/b>1<\/b><\/sub>) = 4 m\/s<\/span><\/p>\n

Wanted :<\/u> The speed of fluid at the smaller pipe (v<\/b>2<\/b><\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The equation of continuity :<\/span><\/p>\n

A1<\/sub> v1<\/sub> = A2<\/sub> v2<\/sub><\/span><\/p>\n

\u03c0 4 = \u00bc \u03c0 (v2<\/sub>)<\/span><\/p>\n

4 = \u00bc (v2<\/sub>)<\/span><\/p>\n

v2 <\/sub>= 8 m\/s<\/span><\/p>\n

Bernoulli’s principle and equation<\/b><\/a><\/span><\/p>\n

10. Water is pumped with a 120 kP<\/span>a <\/span>compressor entering the lower pipe (1) and flows upward at a speed of 1 m\/<\/span>s. Acceleration due to gravity is <\/span>10 m\/<\/span>s <\/span>and water density <\/span>is <\/span>1000 kg\/m<\/span>-3<\/span><\/sup>. <\/span>What is t<\/span>he water pressure on the upper pipe (II<\/span>).<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Radius of the lower pipe (r1<\/sub>) = 12 cm\"Fluid<\/span><\/p>\n

Radius of the lower pipe (r2<\/sub>) = 6 cm<\/span><\/p>\n

Water pressure in the lower pipe (p1<\/sub>) = 120 kPa = 120,000 Pascal <\/span><\/p>\n

The speed of water in the lower pipe (v1<\/sub>) = 1 m.s-1<\/sup><\/span><\/p>\n

The height of the lower pipe (h1<\/sub>) = 0 m<\/span><\/p>\n

The height of the upper pipe (h2<\/sub>) = 2 m<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m.s-2<\/sup><\/span><\/p>\n

Density of water = 1000 kg.m-3<\/sup><\/span><\/p>\n

Wanted:<\/u> Water pressure in pipe 2 (p2<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The speed of water in pipe 2 is calculated with the equation of continuity :<\/span><\/p>\n

\"Fluid<\/span><\/p>\n

Water pressure in pipe 2 is calculated using the equation of Bernoulli :<\/span><\/p>\n

\"Fluid<\/span><\/p>\n

11. A large pipe 5 meters above the ground and a small pipe 1 meter above the ground. The velocity of the water in a large pipe is 36 <\/span>km\/h <\/span>with a pressure of 9.1 x 10<\/span>5 <\/span><\/sup>Pa, while the pressure in the small pipe is 2.10<\/span>5<\/span><\/sup> Pa. <\/span>What is <\/span>the water velocity <\/span>i<\/span>n the small pipe? <\/span>Water <\/span>density = 10<\/span>3<\/span><\/sup> kg\/m<\/span>3<\/span><\/sup><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Water pressure in the large pipe (p1<\/sub>) = 9.1 x 105 <\/sup>Pascal = 910,000 Pascal\"Fluid<\/span><\/p>\n

Water pressure in the small pipe (p2<\/sub>) = 2 x 105 <\/sup>Pascal = 200,000 Pascal<\/span><\/p>\n

Water speed in the large pipe (v1<\/sub>) = 36 km\/h = 36(1000)\/(3600) = 36000\/3600 =10 m\/s <\/span><\/p>\n

The height of the large pipe (h1<\/sub>) = -4 meters<\/span><\/p>\n

The height of the small pipe (h2<\/sub>) = 0 meter<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m.s-2<\/sup><\/span><\/p>\n

Density of water = 1000 kg\/m3<\/sup><\/span><\/p>\n

Wanted:<\/u> The speed of water in the small pipe (v2<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The speed of water in the small pipe (v2<\/sub>) is calculated using the equation of Bernoulli :<\/span><\/p>\n

\"Fluid<\/span><\/p>\n

12. A pipe <\/span>with <\/span>a radius <\/span>of <\/span>15 cm connected with another pipe <\/span>with a radius of <\/span>5 cm. Both are in a horizontal position. <\/span>T<\/span>he velocity of the water flow in the large pipe is 1 m\/s at a pressure of 10<\/span>5<\/span><\/sup> N\/<\/span>m<\/span>2<\/span><\/sup>. <\/span>What is t<\/span>he <\/span>water <\/span>pressure on the small pipe (1 g cm<\/span>-3<\/span><\/sup>)<\/span><\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Radius of the large pipe (r1<\/sub>) = 15 cm = 0.15 m<\/span><\/p>\n

Radius of the small pipe (r2<\/sub>) = 5 cm = 0.05 m<\/span><\/p>\n

The water pressure in the large pipe (p1<\/sub>) = 105<\/sup> N m-2 <\/sup>= 100.000 N m-2 <\/sup><\/span><\/p>\n

The speed of water in the large pipe (v1<\/sub>) = 1 m s-1<\/sup><\/span><\/p>\n

Acceleration due to gravity (g) = 10 m.s-2<\/sup><\/span><\/p>\n

Water density<\/a> = 1 gr cm-3<\/sup> = 1000 kg m-3<\/sup><\/span><\/p>\n

Height difference (\u0394h) = 0.<\/span><\/p>\n

Wanted:<\/u> Pressure in the small pipe (p2<\/sub>)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

The speed of water in pipe 2 is calculated using the equation of continuity :<\/span><\/p>\n

\"Fluid<\/span><\/p>\n

The water pressure in the small pipe (p2<\/sub>) is calculated using the equation of Bernoulli :<\/span><\/p>\n

\"Fluid<\/span><\/p>\n

    \n
  1. What is fluid dynamics?<\/strong><\/span>\n