Fluid dynamics – problems and solutions

Torricelli’s theorem

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

Known :Fluid dynamics – problems and solutions 1

Height (h) = 85 cm – 40 cm = 45 cm = 0.45 meters

Acceleration due to gravity (g) = 10 m/s2

Wanted : The speed of water (v)

Solution :

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 – 40 cm = 45 cm = 0.45 meters

Velocity of water is calculated using the equation of the free fall motion :

vt2 = 2 g h

vt2 = 2 g h = 2(10)(0.45) = 9

vt = √9 = 3 m/s

Read :  Kinetic theory of gases - problems and solutions

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

Known :Fluid dynamics – problems and solutions 2

Height (h) = 1.5 m – 0.25 m = 1.25 meters

Acceleration due to gravity (g) = 10 m/s2

Wanted : The speed of water (v)

Solution :

vt2 = 2 g h = 2(10)(1.25) = 25

vt = √25 = 5 m/s

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

Known :

Height (h) = 1 m – 0.20 m = 0.8 meterFluid dynamics – problems and solutions 3

Acceleration due to gravity (g) = 10 m/s2

Wanted : The speed of water (v)

Solution :

vt2 = 2 g h = 2(10)(0.8) = 16

vt = √16 = 4 m/s

Read :  Application of conservation of mechanical energy for motion on curve surface - problems and solutions

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

Known :

Height (h) = 20 cm = 0.2 metersFluid dynamics – problems and solutions 4

Acceleration due to gravity (g) = 10 m/s2

Wanted: The speed of water (v)

Solution :

Fluid dynamics – problems and solutions 5

5. A container filled with water and there are two holes, as shown in the figure below. What is the ratio of x1 to x2?

Solution

Fluid dynamics – problems and solutions 6

Time interval of the water free fall from hole 1 :Fluid dynamics – problems and solutions 7

h = 1/2 a t2

0.8 = 1/2 (10) t2

0.8 = 5 t2

t2 = 0.8 / 5 = 0.16

t = 0.4 seconds

Time interval of the water free fall from hole 2 :

h = 1/2 a t2

0.5 = 1/2 (10) t2

0.5 = 5 t2

t2 = 0.5 / 5 = 0.1

t = 0.1 second

The horizontal distance (x) :

x1 = v1 t1 = (2)(0.4) = 0.8 meters

x2 = v2 t2 = (10)(√0.1) = (10)(0.1) = 1 meter

The ration of x1 to x2 :

x1 : x2 = 0.8 : 1 = 8 : 10 = 4 : 5

Read :  Centripetal acceleration – problems and solutions

The equation of continuity

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.

Known :

Area of A (AA) = 8Fluid dynamics – problems and solutions 8

Area of C (AC) = 3

The speed of water in pipe A (vA) = v

Wanted: The speed of water in pipe C (vC)

Solution :

The equation of continuity :

AA vA = AC vC

8 v = 3 vC

vC = 8/3 v

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?

Known :

Diameter 1 (d1) = 12 cm, radius 1 (r1) = 6 cm

Diameter 2 (d2) = 8 cm, radius 2 (r2) = 4 cm

The speed of water 1 (v1) = 10 cm/s

Wanted : The speed of water 2 (v2)

Solution :

Area 1 (A1) = π r2 = π 62 = 36π cm2

Area 2 (A2) = π r2 = π 42 = 16π cm2

The equation of continuity :

A1 v1 = A2 v2

(36π)(10) = (16π) v2

(36)(10) = (16) v2

360 = (16) v2

v2 = 360/16

v2 = 22.5 cm/s

Read :  Dynamics of particles – problems and solutions

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

Known :

Area 1 (A1) = 8 cm2Fluid dynamics – problems and solutions 9

Area 2 (A2) = 2 cm2

Speed of water in pipe 2 (v2) = 2 m/s

Wanted : the speed of water in pipe 1 (v1)

Solution :

The equation of continuity :

A1 v1 = A2 v2

8 v1 = (2)(2)

8 v1 = 4

v1 = 4 / 8 = 0.5 m/s

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.

Known :

Diameter of the larger pipe (d1) = 2Fluid dynamics – problems and solutions 10

Radius of the larger pipe (r1) = ½ d1 = ½ (2) = 1

Area of the larger pipe (A1) = π r12 = π (1)2 = π (1) = π

Diameter of the smaller pipe (d2) = 1

Radius of the smaller pipe (r2) = ½ d2 = ½ (1) = ½

Area of the smaller pipe (A2) = π r22 = π (1/2)2 = π (1/4) = ¼ π

The speed of fluid at the larger pipe (v1) = 4 m/s

Wanted : The speed of fluid at the smaller pipe (v2)

Solution :

The equation of continuity :

A1 v1 = A2 v2

π 4 = ¼ π (v2)

4 = ¼ (v2)

v2 = 8 m/s

Read :  Work done by force – problems and solutions

Bernoulli’s principle and equation

10. Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is 1000 kg/m-3. What is the water pressure on the upper pipe (II).

Known :

Radius of the lower pipe (r1) = 12 cmFluid dynamics – problems and solutions 11

Radius of the lower pipe (r2) = 6 cm

Water pressure in the lower pipe (p1) = 120 kPa = 120,000 Pascal

The speed of water in the lower pipe (v1) = 1 m.s-1

The height of the lower pipe (h1) = 0 m

The height of the upper pipe (h2) = 2 m

Acceleration due to gravity (g) = 10 m.s-2

Density of water = 1000 kg.m-3

Wanted: Water pressure in pipe 2 (p2)

Solution :

The speed of water in pipe 2 is calculated with the equation of continuity :

Fluid dynamics – problems and solutions 12

Water pressure in pipe 2 is calculated using the equation of Bernoulli :

Fluid dynamics – problems and solutions 13

Read :  Electric circuits with resistors in parallel and internal resistance – problems and solutions

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 km/h with a pressure of 9.1 x 105 Pa, while the pressure in the small pipe is 2.105 Pa. What is the water velocity in the small pipe? Water density = 103 kg/m3

Known :

Water pressure in the large pipe (p1) = 9.1 x 105 Pascal = 910,000 PascalFluid dynamics – problems and solutions 14

Water pressure in the small pipe (p2) = 2 x 105 Pascal = 200,000 Pascal

Water speed in the large pipe (v1) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s

The height of the large pipe (h1) = -4 meters

The height of the small pipe (h2) = 0 meter

Acceleration due to gravity (g) = 10 m.s-2

Density of water = 1000 kg/m3

Wanted: The speed of water in the small pipe (v2)

Solution :

The speed of water in the small pipe (v2) is calculated using the equation of Bernoulli :

Fluid dynamics – problems and solutions 15

Read :  Buoyant force – problems and solutions

12. A pipe with a radius of 15 cm connected with another pipe with a radius of 5 cm. Both are in a horizontal position. The velocity of the water flow in the large pipe is 1 m/s at a pressure of 105 N/m2. What is the water pressure on the small pipe (1 g cm-3)

Known :

Radius of the large pipe (r1) = 15 cm = 0.15 m

Radius of the small pipe (r2) = 5 cm = 0.05 m

The water pressure in the large pipe (p1) = 105 N m-2 = 100.000 N m-2

The speed of water in the large pipe (v1) = 1 m s-1

Acceleration due to gravity (g) = 10 m.s-2

Water density = 1 gr cm-3 = 1000 kg m-3

Height difference (Δh) = 0.

Wanted: Pressure in the small pipe (p2)

Solution :

The speed of water in pipe 2 is calculated using the equation of continuity :

Fluid dynamics – problems and solutions 16

The water pressure in the small pipe (p2) is calculated using the equation of Bernoulli :

Fluid dynamics – problems and solutions 17

Comments