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/s^{2}

__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 :

v_{t}^{2} = 2 g h

v_{t}^{2} = 2 g h = 2(10)(0.45) = 9

v_{t} = √9 = 3 m/s

[irp]

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/s^{2}

__Wanted :__ The speed of water (v)

__Solution :__

v_{t}^{2} = 2 g h = 2(10)(1.25) = 25

v_{t} = √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/s^{2}

__Wanted :__ The speed of water (v)

__Solution :__

v_{t}^{2} = 2 g h = 2(10)(0.8) = 16

v_{t} = √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/s^{2}

__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 x_{1} to x_{2}?

__Soluti____on__

Time interval of the water free fall from hole 1 :

h = 1/2 a t^{2 }

0.8 = 1/2 (10) t^{2}

0.8 = 5 t^{2}

t^{2} = 0.8 / 5 = 0.16

t = 0.4 seconds

Time interval of the water free fall from hole 2 :

h = 1/2 a t^{2 }

0.5 = 1/2 (10) t^{2}

0.5 = 5 t^{2}

t^{2} = 0.5 / 5 = 0.1

t = √0.1 second

The horizontal distance (x) :

x_{1} = v_{1} t_{1 }= (2)(0.4) = 0.8 meters

x_{2} = v_{2} t_{2} = (√10)(√0.1) = (10)(0.1) = 1 meter

The ration of x_{1} to x_{2} :

x_{1}_{ :} x_{2} = 0.8 : 1 = 8 : 10 = 4 : 5

[irp]

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 (A_{A}) = 8

Area of C (A_{C}) = 3

The speed of water in pipe A (v_{A}) = v

__Wanted:__ The speed of water in pipe C (v_{C})

__Solution :__

The equation of continuity :

A_{A} v_{A} = A_{C} v_{C}

8 v = 3 v_{C}

v_{C} = 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 (d_{1}) = 12 cm, radius 1 (r_{1}) = 6 cm

Diameter 2 (d_{2}) = 8 cm, radius 2 (r_{2}) = 4 cm

The speed of water 1 (v_{1}) = 10 cm/s

__Wanted :__ The speed of water 2 (v_{2})

__Solution :__

Area 1 (A_{1}) = π r^{2} = π 6^{2} = 36π cm^{2 }

Area 2 (A_{2}) = π r^{2} = π 4^{2} = 16π cm^{2 }

The equation of continuity :

A_{1} v_{1} = A_{2} v_{2}

(36π)(10) = (16π) v_{2}

(36)(10) = (16) v_{2}

360 = (16) v_{2}

v_{2 }= 360/16

v_{2 }= 22.5 cm/s

[irp]

8. Water flows through a pipe of varying diameter, as shown in figure below. If area 1 (A_{1}) = 8 cm^{2}, A_{2} = 2 cm^{2} and the speed of water in pipe 2 = v_{2} = 2 m/s then what is the speed of water in pipe 1 = v_{1}.

__Known :__

Area 1 (A_{1}) = 8 cm^{2}

Area 2 (A_{2}) = 2 cm^{2}

Speed of water in pipe 2 (v_{2}) = 2 m/s

__Wanted :__ the speed of water in pipe 1 (v_{1})

__Solution :__

The equation of continuity :

A_{1} v_{1} = A_{2} v_{2}

8 v_{1} = (2)(2)

8 v_{1} = 4

v_{1 }= 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 (d_{1}) = 2

Radius of the larger pipe (r_{1}) = ½ d_{1} = ½ (2) = 1

Area of the larger pipe (**A**_{1}) = π r_{1}^{2} = π (1)^{2 }= π (1) = π

Diameter of the smaller pipe (d_{2}) = 1

Radius of the smaller pipe (r_{2}) = ½ d_{2} = ½ (1) = ½

Area of the smaller pipe (**A**_{2}) = π r_{2}^{2} = π (1/2)^{2 }= π (1/4) = ¼ π

The speed of fluid at the larger pipe (**v**_{1}) = 4 m/s

__Wanted :__ The speed of fluid at the smaller pipe (**v**_{2})

__Solution :__

The equation of continuity :

A_{1} v_{1} = A_{2} v_{2}

π 4 = ¼ π (v_{2})

4 = ¼ (v_{2})

v_{2 }= 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 (r_{1}) = 12 cm

Radius of the lower pipe (r_{2}) = 6 cm

Water pressure in the lower pipe (p_{1}) = 120 kPa = 120,000 Pascal

The speed of water in the lower pipe (v_{1}) = 1 m.s^{-1}

The height of the lower pipe (h_{1}) = 0 m

The height of the upper pipe (h_{2}) = 2 m

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

Density of water = 1000 kg.m^{-3}

__Wanted:__ Water pressure in pipe 2 (p_{2})

__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 :

[irp]

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 10^{5 }Pa, while the pressure in the small pipe is 2.10^{5} Pa. What is the water velocity in the small pipe? Water density = 10^{3} kg/m^{3}

__Known :__

Water pressure in the large pipe (p_{1}) = 9.1 x 10^{5 }Pascal = 910,000 Pascal

Water pressure in the small pipe (p_{2}) = 2 x 10^{5 }Pascal = 200,000 Pascal

Water speed in the large pipe (v_{1}) = 36 km/h = 36(1000)/(3600) = 36000/3600 =10 m/s

The height of the large pipe (h_{1}) = -4 meters

The height of the small pipe (h_{2}) = 0 meter

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

Density of water = 1000 kg/m^{3}

__Wanted:__ The speed of water in the small pipe (v_{2})

__Solution :__

The speed of water in the small pipe (v_{2}) 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 10^{5} N/m^{2}. What is the water pressure on the small pipe (1 g cm^{-3})

__Known :__

Radius of the large pipe (r_{1}) = 15 cm = 0.15 m

Radius of the small pipe (r_{2}) = 5 cm = 0.05 m

The water pressure in the large pipe (p_{1}) = 10^{5} N m^{-2 }= 100.000 N m^{-2 }

The speed of water in the large pipe (v_{1}) = 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 (p_{2})

__Solution :__

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

The water pressure in the small pipe (p_{2}) is calculated using the equation of Bernoulli :