## 5 Common Problems of Fluid Dynamics

- Category: High School
- Written by fisikastudycenter

physics.fisikastudycenter.com- learning fluid dynamics and bernoulli's equation in 5 common problems of fluid dynamics includes volume flow of rate, continuity equation and bernoulli's and torricelli's equation. Prepared for grade 11 high school level.

**Formulas**

**Volume of flow rate**

Q = V/t

Q = A*v*

where:

Q = volume of flow rate (m^{3}/s)

V = volume (m^{3})

t = time (s)

A = area (m^{2})

*v* = speed of flow (m/s)

1 liter = 1 dm^{3} = 10^{−3} m^{3}

**Continuity equation**

Q_{1} = Q_{2}

A_{1}*v*_{1} = A_{2}*v*_{2}

**Bernoull's Equation**

P + ^{1}/_{2} ρ*v*^{2} + ρgh = Constant

P_{1} + ^{1}/_{2} ρ*v*_{1}^{2} + ρgh_{1} = P_{2} + ^{1}/_{2} ρ*v*_{2}^{2} + ρgh_{2}

where :

P = pressure (Pascal = Pa = N/m^{2})

ρ = fluid density (kg/m^{3})

g = acceleration due to gravity (m/s^{2})

**Fluid tank with a hole **

*v* = √(2gh)

X = 2√(hH)

t = √(2H/g)

where :

*v* = speed of fluid flow from the hole

X = horizontal distance reached by the fluid flow at first time

h = the distance of fluid surface to the hole

H = distance between the point of fluid drop to the hole

t = time taken by the fluid to reach the drop point

**Question**** 1**

Ahmad fills a bucket of 20 liter of capacity through opening a water channel.

If the area of channel with diameter of r D_{2} is 2 cm^{2} and the water flow speed in the channel is 10 m/s find:

a) water volume rate

b) time taken to make the bucket full of water

**Answer**

Data :

A_{2} = 2 cm^{2} = 2 x 10^{−4} m^{2}

*v*_{2} = 10 m/s

a) water volume rate

Q = A_{2}*v*_{2} = (2 x 10^{−4})(10)

Q = 2 x 10^{−3} m^{3}/s

b) time taken to full the bucket

Data :

V = 20 liter = 20 x 10^{−3} m^{3}

Q = 2 x 10^{−3} m^{3}/s

t = V / Q

t = ( 20 x 10^{−3} m^{3})/(2 x 10^{−3} m^{3}/s )

t = 10 sec

**Question 2**

An underground water channel is shown in the figure below!

If the area of wider channel is 5 m^{2} , the little one is 2 m^{2} and the sped of water flow in wide channel is 15 m/s, find the speed of water flow in A_{2} area!

**Answer**

Using continuity equation

A_{1}*v*_{1} = A_{2}*v*_{2}

(5)(15) = (2) *v*_{2}

*v*_{2} = 37,5 m/s

**Question 3**

A water tank with a little hole in its side shown below!

The location of hole from the ground is at distance H = 10 m and the distance of the hole to the water surface is h = 3,2 m. Find:

a) speed of water flow from the hole

b) distance x reached by water flow

c) time taken by the water flow to reach the ground at the first time

**Answer**

a) speed of water flow

*v* = √(2gh)

*v* = √(2 x 10 x 3,2) = 8 m/s

b) distance x reached by water flow

X = 2√(hH)

X = 2√(3,2 x 10) = 8√2 m

c) time taken by the water flow to reach the ground at the first

t = √(2H/g)

t = √(2(10)/(10)) = √2 sec

**Question 4**

A ventury meter, used to measured speed flow in a pipe is shown below below!

The areas of the wide part and the narrow part of pipe are 5 cm^{2} and 3 cm^{2} respectively. The difference in height between the two vertical tubes is 20 cm, find :

a) speed flow at the wide part

b) sped flow at narrow part

**Answer**

a) speed flow at the wide part

*v*_{1} = A_{2}√ [(2gh) : (A_{1}^{2} − A_{2}^{2}) ]

*v*_{1} = (3) √ [ (2 x 10 x 0,2) : (5^{2} − 3^{2}) ]

*v*_{1} = 3 √ [ (4) : (16) ]

*v*_{1} = 1,5 m/s

Calculation Tips :

Keep the unit of areas in cm^{2} , g and h are in m/s^{2} and m. then *v* will be in m/s.

b) sped flow at narrow part

A_{1}*v*_{1} = A_{2}*v*_{2}

(3 / 2)(5) = (*v*_{2})(3)

*v*_{2} = 2,5 m/s

**Question 5**

The water installation using pipes at a home shown below! The ratio of wide area and narrow area is 4 : 1.

h_{1} is 5 m from the floor and h_{2} is 1 m diatas tanah. The water flow speed at wider area A1 is 36 km/hour with the pressure of 9,1 x 10^{5} Pa. Find :

a) water flow speed at the narrow pipe

b) the pressure difference between two pipes

c) pressure at the narrow pipe

Use the water density (ρ_{air} = 1000 kg/m^{3})

**Answer**

Data :

h_{1} = 5 m

h_{2} = 1 m

*v*_{1} = 36 km/jam = 10 m/s

P_{1} = 9,1 x 10^{5} Pa

A_{1} : A_{2} = 4 : 1

a) water flow speed at the narrow pipe

From continuity equation :

A_{1}*v*_{1} = A_{2}*v*_{2}

(4)(10) = (1) (*v*_{2})

*v*_{2} = 40 m/s

b) the pressure difference between two pipes

From Bernoull's equation:

P_{1} + ^{1}/_{2} ρ*v*_{1}^{2} + ρgh_{1} = P_{2} + ^{1}/_{2} ρ*v*_{2}^{2} + ρgh_{2}

P_{1} − P_{2} = ^{1}/_{2} ρ(*v*_{2}^{2} − *v*_{1}^{2}) + ρg(h_{2} − h_{1})

P_{1} − P_{2} = ^{1}/_{2}(1000)(40^{2} − 10^{2}) + (1000)(10)(1 − 5)

P_{1} − P_{2} = (500)(1500) − 40000 = 750000 − 40000

P_{1} − P_{2} = 710000 Pa = 7,1 x 10^{5} Pa

c) pressure at the narrow pipe

P_{1} − P_{2} = 7,1 x 10^{5}

9,1 x 10^{5} − P_{2} = 7,1 x 10^{5}

P_{2} = 2,0 x 10^{5} Pa