1.) If 5.6 m³ of oil weighs 46,800 N, calculate the following
1 Unit weight
2 Density
3 Specific gravity
Solution
1 Unit weight
Ƴw =
=
2 Density
Density =
=
3 Specific gravity
sp.gr. =
=
46,800.00
5.60
8,357.14 N/m³
8,357.14
9.81
851.90 Kg/m³
851.90
1,000.00
0.85
2.) of 1200 kg is 0.952 m³. Compute the following
de havinf a mass
1 mass density
2 specific weight
3 specific gravity
Solution
weight
weight
weight
= 1200 (9.81)
= 1172 N
= 11.77
1 mass density
ρ =
=
2 specific weight
Ƴ =
m
=
v
1,260.50 kg/m³
w
v
12.36
=
=
KN/m³
3 Specific gravity
sp.gr. = Unit weight of glycerine
Unit weight of H2O
sp.gr. =
12.36
9.81
1200
0.952
11.77
0.952
=
1.26
3.) A certain liquid has a unit weight of 56 KN/m³
1 Compute the mass density
2 Compute its specific volume
3 Compute its specific volume
Solution
1 Mass density
ρ =
=
2 Specific volume
sp.vol. =
=
=
3 Specific gravity
sp.gr. =
=
56,000.00
9.810
5,708.46 Kg/m³
1.00
ρ
1.00
5,708.46
0.000175 m³/kg
56.00
9.81
5.71
4.) An obbject has a specific weight of 2.23 KN/m³ Compute the following.
1 Mass density
2 Mass if the volume is 0.001 m³
3 specific volume
Solution
1 Mass density
ρ =
ρ =
2
M =
=
M =
3 Specific volume
sp.vol. =
=
sp.vol. =
2,230.00
9.810
227.32 Kg/m³
ρVol.
227.32
x
0.23
kg
1.00
ρ
1.00
227.32
0.004399 m³/kg
0.001
5.) A quart of water weighs 4.08-ib. Compute the following:
1 Mass in slugs
2 Mass in kg
3 Volume in cu.f
Solution
1 Mass in slugs
W = mg
4.08 =
m(32.2)
m =
4.08
32.20
m =
0.13
slugs
2 Mass in kg
m =
0.1267
32.20
2.20
m =
1.85
kg
3 Volume in cu.f
w =
ƴVol.
4.08 =
62.40
Vol.
Vol =
0.07
f³
6.) A 10 m. diam. Cylindrical tank has a height of 5 M. and is full at 20 °C ( Unit weight of water =
9.789 kN/m³) if the water is heated to a temperature of 50 °C (Unit weight of water = 9.689 kN/m³)
1 Compute the weight of water
2 What is the final volume when heated to a temp. of 50 °C.
3 determine the volume of water that will spill over the edge of the tank.
Solution
1 Weight of water
V1 =
Ah
A = Π/4 (10²)
=
78.54
V1 =
392.70
W1 =
V1D1
= 3,844.14
Dia
10
H
5
V1
392.7
D1
9.789
W2
3844.14
D2
9.689
V1
392.7
V2
396.75
m²
2 Final volume when heated a temp.
w2 =
V2D2
V2 =
396.75 m³
3 Volume of water
∆V =
V2 - V1
=
4.05
m³
7.) A liquid which is compressed in a cylinder having a volume of I liter at one MN/m² and volume
of 995 m³ at 2 MN/m².
1 Compute the change in volume
2 Compute the change in pressure
3 Compute the bulk modulus of elasticity
Solution
1 Change in Volume
∆V = 995-1000
=
5.00
liters
2 Change in pressure
∆P =
2-1
1 Mpa
3 Bulk modulus of elasticity
E =
̶̶ ∆P
∆V
V
−(
2−1
)
E =
(995−1000)
E =
1,000.00
200.00 Mpa
8.) A gas having a volume of 40 liters has a pressure of 0.24 Mpa at 24 °C. If the gas constant R is equal
to 212 M.N./kg.k, compute
1 density of the gas
2 mass of the gas
3 Weight of gas
Solution
ρ =
1
T
T
T
ρ
=
=
=
=
ρ =
P
RT
273° + C°
273 + 24
297 K
0.24 (10)⁶
212 (297)
3.81
Kg/m³
2
m = ρV
m = 3.81 (0.04)
m =
0.52 kg
3
W =
Mg
W = 0.152 (9.81)
W =
1.49 N
9.) A gas is under pressure of 21.868 bar abs at 40°C.
1 Compute the pressure in kPa
2 Compute the gage pressure.
3 Compute the gas constant R if it has a unit weight of 362 N/m³
Solution
1 Pressure in kPa
P = 21.868 (100)
= 2,186.80 kPa abs
2 Gage pressure
Pabs = Pgage + Patm
2186.8 = Pgage + 101.3
Pgage = 2,085.50 kPa
3 Gas constant R :
ƴ =
P
g
RT
P =21.868 x 10⁵ N/m²
Note :
1 bar = 10⁵Pa = 100 kPa
362 = 21.868 x 10⁵
9.81
R(40 + 273)
R =
189.30 m²/s² k
10.) 1 Find the depression h of the mercury in the glass capillary tube having diameter of 2mm if the
surface tension is 0.514 N/m for Ɵ = 40°.
2 Compute the force caused by surface tension.
3 Determine the density of the mercury.
Solution
h =
2 Ơ Cos Ɵ
ρgr
ρ = 13.6 (9810)
9.81
ρ = 13,600.00 Kg/m³
h = 2(0.514)Cos 40°
13600(9.81)(0.001)
h =
0.0059 M
h =
5.90
mm
1 Depression h:
h =
5.90
mm
2 Force caused by surface tension
F = Ơ Π d CosƟ
F = 0.51Π(0.002)(Cos 40°)
F = 2.47 x 10 ¯³ N
3 Density of mercury
ρ = 13,600.00 Kg/m³
11.) A pressure gage at elev. 12M at the side of the tank containing a liquid reads 100 kPa. Another gauge
at elevation 7 m. reads 140 kPa.
1 Compute the specific weight of the liquid.
2 Compute the density of the liquid.
3 Compute the specific gravity of the liquid.
Solution
1 Specific weight
140 + 100 + w(5)
w =
8 kN/m³
2 Density
ρ =
3
=
Sp.gr. =
=
8,000.00
9.81
815.49 Kg/m³
8.00
9.81
0.82
12.) An open tank contain 5.7 m. of water covered with 2.8 m of kerosene having a unit weight of 8 kN/m³
If the diam. Of the tank is 1m.
1 Find the pressure at the interface of water and kerosene.
2 Find the pressure at the bottom of the tank
3 Find the total force at the bottom of the tank.
Solution
1 Pressure at interface
PA =
2.8 (8)
=
22.40
kPa
2 Pressure at the bottom
PB = 22.4 + 9.81 (5.7)
=
78.32
kPa
3 Total force at the bottom of the tank
F = (78.32) Π(1)²/Π
=
61.51
kN
13.)
gasoline sp.gr.
= 0.90 and sea water sp.gr. = 1.03. If the depths of the
liquids are 0.5 m. , 0.80 m. and 1 m for oil, gasoline and sea water respectively.
1 Find the pressure at a depth of 1.2m
2 Find the pressure at a depth of 1.8m
3 Find the pressure at the bottom of the the tank.
Solution
1 Pressure at a depth of 1.2 m.
P = 9.81 (0.8) (0.5) + 9.81 (0.90) (0.70)
=
10.10
kN/m²
2 Pressure at a depth of 1.8 m.
P = 9.81 (0.8) (0.5) + 9.81 (0.90) (0.80) + 9.81 (1.03) (0.5)
=
16.04
kN/m²
3 Pressure at the bottom of the tank
P = 9.81 (0.8) (0.5) + 9.81 (0.90) (0.80) + 9.81 (1.03) (1)
=
21.09
kN/m²
14.) A pressure gauge at elevation of 8m at the side of a tank containing a liquid reads 80 kPa.
Another gauge at elevation 3m reads 120 kPa. Compute for the
1 Specific weight
2 density
3 Specific gravity
Solution
1 Specific weight
120 =
w =
2 density
density =
=
3 Specific gravity
sp.gr. =
=
80 + 5w
8.00
KN/m³
8,000.00
9.81
815.49 Kg/m³
815.49
1,000.00
0.82
15.) The pressure on a closed tank reads 58.86 kPa.
1 What is the equivalent height of water
2 What is the equivalent height in terms of oil having a sp.gr. Of 0.85?
3 What is the equivalent height in terms of mercury having a sp.gr. Of 13.6
Solution
1 Height in water
P =
Ƴwh
58.86 =
9.81h
h =
6 m.
2 Height in oil
P =
Ƴwh
58.86 = 9.81 (0.85)h
h = 7.06 m. of oil
3 Height in mercury
P =
Ƴwh
58.86 = 9.81 (13.6)h
jh = 0.44m of Hg
16.) The unit weight of a liquid is variable and is given in kN/m³ and "h" is the depth of liquid from the
free surface in meters. Determine the gauge pressure in kPa at a depth of 5 m.
Solution :
ƴ
dP
dp
P
= 10 + 0.5h
= ƴdh
= ( 10 + 0.5h ) dh
= 10h + 0.5 h²
2.00
P = 10(5) + 0.5
* (5)²
2.00
P =
56.25
kPa
17.) The pressure in a gas tank is 2.75 atmospheres. Compute the pressure head in meters of water.
Solution :
pabs
Pgauge
Pgauge
1 atm.
P
P
177.32
h
= Pgage + Patm
=
2.75 - 1
=
2.75
atm.
=
101.33 kPa
= 1.75 (101.325 )
=
ƴh
=
9.81h
=
18.00
m.
18.) A presure in a given tank reads 277 mm of Hg.
1 Determine the equivalent height of column of water.
2 Determine the equivalent height of column of kerosine sp.gr. = 0.82
3 Determine the equivalent height of column of nectar sp.gr. = 2.94
Solution :
1 Height of water
0.277 (13.6)(9.81) = h(9.81)
h =
3.77
m.
2 Height of kerosine
sp.gr. =
0.82
0.277(13.6)(9.81) = h(0.82)(9.81)
h =
4.59
m.
3 Height of nectar
sp.gr. =
-2.94
0.277(13.6)(9.81) = h(2.94)(9.81)
h =
1.28
m.
19.) A mercury barometer at the base of the same time, another barometer at the top of a
mountain reads 450 mm. Assuming w of air is to be constant at 10 N/m³, what is the
appropriate height of the mountain.
Solution :
P2 =
P1 + wh
P2 - P 1 =
10h
9810(13.6)(0.62)-9810(13.6)(0.45)
=
h = 2,268.00 m.
10h
20.) If the atmospheric pressure is 101.3 kPa. And the absolute pressure at the bottom of the
tank as shown in the figure is 231.3 kPa.
ƳH2O =
9.79
kN/m³
1 What is the specific gravity of the olive oil.
2 What is the absolute pressure at the interface of the olive oil and mercury.
3 What is the gage pressure at the interface of the olive oil and the mercury.
Solution :
1 Specific gravity of olive oil.
101.30 + 9.79 (.89)(1.5) + 9.79(2.5) + 9.79(S)(2.9) + 13.6(6.79)(0.4) = 2.31.3
28.449(sp.gr) =
39.0123
Sp.gr. = 1.37 (sp.gr. Of olive oil)
2 Absolute pressure at the interface of olive oil and mercury.
Pabs = 101.3 + 9.81(0.89)(1.5) + 9.81(2.5) + 9.81(1.37)(2.9)
=
177.90 kPa.
3 Gage pressure at the interface of olive oil and mercury.
Pgage = 9.81(0.89)(1.5) + 9.81(2.5) + 9.81(1.37)(2.9)
=
76.60
kPa.