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 :
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
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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 :
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 meter
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
[irp]
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 meters
Acceleration due to gravity (g) = 10 m/s2
Wanted: The speed of water (v)
Solution :
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
Time interval of the water free fall from hole 1 :
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
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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) = 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
[irp]
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 cm2
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) = 2
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
[irp]
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 cm
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 :
Water pressure in pipe 2 is calculated using the equation of Bernoulli :
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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 Pascal
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 :
[irp]
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 :
The water pressure in the small pipe (p2) is calculated using the equation of Bernoulli :
