EXERCISE 1 – FLUID CHARACTERISTICS

HOMEWORK PROBLEM CHAPTER 1

1. Define and state formula for the quantities below;
i. Density
ii. Specific Weight
iii. Specific Gravity

Solution

i. Density
Density or mass density of a fluid is defined as the ratio of the mass of a fluid to its volume. Thus
mass per unit volume of a fluid is called density. The unit of mass density in SI unit is kg per cubic
metre (kg/m3).

 = =


ii. Specific Weight
Specific weight or weight density of a fluid is the ratio between the weights of a fluid to its volume.
Thus weight per unit volume of a fluid is called weight density. The unit of specific weight in SI unit
is N per cubic metre (N/m3).

 = = = =


iii. Specific Gravity
Specific gravity is defined as the ratio of the weight density (or density) of a fluid to the weight
density (or density) of a standard fluid.

= =  =
 

2. If 6 m3 of oil weighs 47 kN, find its specific weight, density, and relative density.

Solution
Given; W = 47 kN, V = 6 m3

Specific weight ( );

= ⁄
47 103
= 6

= . /

Density ();

= ⁄
7833.333
= 9.81

= . /

Specific gravity (s);

= 7 9⁄8 .5 05
= 1000

= .

1

HOMEWORK PROBLEM CHAPTER 1
3. 1 L crude oil weighs 9.6 N. Calculate its specific weight, density, and relative density.

Solution
Given; W = 9.6 N, V = 1 L

Specific weight ( );

= ⁄
9.6
= 1 10−3

= /

Density ();

= ⁄
9600
= 9.81

= . /

Specific gravity (s);

= 9 7⁄8 .5 93
= 1000

= .

4. The weight of body 100 lb. Determine its weight in Newton and mass in kilograms. (N = 0.225 lb)

Solution
Given; W = 100 lb

Weight (W);

= 100 1
0.225
= .

Mass (m);

= /
444.444
= 9.81

= .

5. A liquid has a volume of 7.5 m3 and weight of 53 kN. Calculate;
i. The specific weight
ii. The density
iii. The specific gravity
iv. The specific volume

Solution
Given; V = 7.5 m3, W = 53 kN

2

HOMEWORK PROBLEM CHAPTER 1

i. Specific weight ( );

= ⁄
53 103
= 7.5

= . /

ii. Density ();

= ⁄
7066.667
= 9.81

= . /

iii. Specific gravity (s);

= 7 ⁄2 0 .3 53
= 1000

= .

iv. Specific volume (VS);

= 1⁄
= 1

720.353
= . − /

6. A drum of 1 m3 volume contains 8.5 kN an oil when full. Find the specific weight and specific gravity.

Solution
Given; V = 1 m3, W = 8.5 kN

Specific weight ( );

= ⁄
8.5 103
= 1

= /

Density ();

= ⁄
8500
= 9.81

= 866.463 / 3

Specific gravity (s);

= 8 6⁄6 . 4 63
= 1000

= .

3

HOMEWORK PROBLEM CHAPTER 1
7. If 4.8 m3 of oil weighs 30 kN, find its density and relative density.

Solution
Given; V = 4.8 m3, W = 30 kN

Specific weight ( );

= ⁄
30 103
= 4.8

= 6250 / 3

Density ();

= ⁄
6250
= 9.81

= . /

Relative density (s);

= 6 ⁄3 7 .1 05
= 1000

= .

8. Find the specific gravity of an oil whose specific weight is 7.85 kN/m3.

Solution
Given;  = 7.85 kN/m3

Density ();

= /
7.85 103
= 9.81

= 800.204 / 3

Specific gravity (s);

= 8 0⁄0 . 2 04
= 1000

= .

9. A vessel of 4 m3 volume contains an oil, which weighs 30.2 kN. Determine the specific gravity of the oil.
Solution
Given; V = 4 m3, W = 30.2 kN
Specific weight ( );
= ⁄

4

HOMEWORK PROBLEM CHAPTER 1

= 30.2 103
4
= 7550 / 3

Density ();

= ⁄
7550
= 9.81

= 769.623 / 3

Specific gravity (s);

= 7 ⁄6 9 .6 23
= 1000

= .

10. If 550 L of certain oil has a mass of 8600 g, find its density.

Solution
Given; V = 550 L, m = 8600 g

Density ();

= ⁄
8600 10−3
= 550 10−3

= . /

11. If 2.5 m3 of certain oil has a mass of 2500 kg, find its mass density.

Solution
Given; V = 2.5 m3, m = 2500 kg

Density ();

= ⁄
2500
= 2.5

= /

12. A fluid of 5000 kg filled an open cylinder container with 150 cm diameter and 300 cm height. Calculate
the density of fluid and specific weight.

Solution
Given; d = 150 cm, h = 300 cm, m = 5000 kg

Density ();

= ⁄

5

HOMEWORK PROBLEM CHAPTER 1

= [ 5000
4 (1.5)2 3]

= . /

Specific weight ( );

=
= 973.218 9.81
= . /

13. A fluid with a mass of 7500 kg filled an open cylinder container with 200 cm diameter and 300 cm height.
Calculate the;
i. Density of the fluid
ii. Specific weight
iii. Specific volume

Solution
Given; d = 200 cm, h = 300 cm, m = 7500 kg

i. Density ();

= ⁄
7500
= [ (2)2 3]
4
= . /

ii. Specific weight ( );

=
= 795.756 9.81
= . /

iii. Specific volume (VS);

= 1⁄
= 1

7806.366
= . − /

14. The specific gravity of a liquid, having specific weight of 7360 N/m3?

Solution
Given;  = 7360 N/m3

Density ();

= ⁄
7360
= 9.81

= 750.255 / 3

6

HOMEWORK PROBLEM CHAPTER 1

Specific gravity (s);

= 7 ⁄5 0 .2 55
= 1000

= .

15. 8540 N fluid completely fills into a rectangular container. The container has a length of 110 cm, width
of 80 cm and height of 150 cm. Calculate the fluids;
i. Mass
ii. Density
iii. Specific Weight
iv. Specific Gravity

Solution
Given; L = 110 cm, W = 80 cm, h = 150 cm, W = 8540 kg

i. Mass (m);

= ⁄
8540
= 9.81

= . /

ii. Density ();

= ⁄
870.54
= ( 1.1 0.8 1.5 )

= . /

iii. Specific weight ( );

= ⁄
8540
= (1.1 0.8 1.5)

= . /

iv. Specific gravity (s);

= 6 ⁄5 9 .5
= 1000

= .

16. Mass of a liquid is 4000 kg and 3.2 m3 respectively. Determine :-
i. Weight
ii. Density
iii. Specific Gravity
iv. Specific Weight

Solution
Given; m = 4000 kg, V = 3.2 m3

7

HOMEWORK PROBLEM CHAPTER 1

i. Weight (W);

=

= 4000 9.81
=

ii. Density ();

= ⁄
4000
= 3.2

= /

iii. Specific gravity (s);

= 1 2⁄5 0
= 1000

= .

iv. Specific weight ( );

= ⁄
39240
= 3.2

= . /

17. A storage vessel for gasoline (sg. gr = 0.86) is a vertical cube (2.5 m x 2.5 m). If it is filled to 1.5 m depth,
calculate the mass and weight of the gasoline.

Solution
Given; s = 0.86, L = 2.5 m, W = 2.5 m, h = 1.5 m
Density ();

=
= 0.86 1000
= 860 / 3

Mass (m);
=
= 860 (2.5 2.5 1.5)
= . /

Weight (W);
=
= 8062.5 9.81
= .

18. From an experiment, the weight of 1.5 m3 of a liquid was found to be 7500 N. Calculate the;
i. Specific weight of the liquid
ii. Density of the liquid

8

HOMEWORK PROBLEM CHAPTER 1

Solution
Given; W = 7500 N, V = 1.5 m3

i. Specific weight ( );

= ⁄
7500
= 1.5

= /

ii. Density ();

= ⁄
5000
= 9.81

= . /

19. Mass of a liquid is 4500 kg and 3200 L respectively. Determine :-
i. Weight
ii. Density
iii. Specific Weight

Solution
Given; m = 4500 kg, V = 3200 L

i. Weight (W);

=
= 4500 9.81
=

ii. Density ();

= ⁄
4500
= 3200 10−3

= . /

iii. Specific weight ( );

= ⁄
44145
= 3200 10−3

= . /

20. At certain temperature, the kinematics viscosity and the specific weight of a fluid are 3.26 x 10-5 m2/s
and 9.47 kN/m3 respectively. Calculate the density, dynamic viscosity and the specific gravity of the
fluid.

Solution
Given;  = 3.26 x 10-5m2/s,  = 9.47 kN/m3

9

HOMEWORK PROBLEM CHAPTER 1

Density ();

= ⁄
9.47 103
= 9.81

= . /

Dynamic viscosity ();
=
= (3.26 10−5) 965.341
= . / .

Specific gravity (s);

= 9 ⁄6 5 .3 41
= 1000

= .

10