2. TOPIC
Fluid Properties
Pressure and Temperature Measurements
Ideal Gas
Charles Law
Boyles Law
Heat and Energy Concept
Open and Closed System
Conservation of Energy
3. THERMODYNAMICS
is the field of physics that deals with the relationship between heat and other
properties (such as pressure, density, temperature, etc.) in a substance.
LAWS OF THERMO DYNAMICS
• First law of thermodynamics: one of the most fundamental laws of nature is
the conservation of energy principle. It simply states that during an
interaction, energy can change from one form to another but the total amount
of energy remains constant.
• Second law of thermodynamics: energy has quality as well as quantity, and
actual processes occur in the direction of decreasing quality of energy.
4.
5. Density
Density of a fluid is defined as the ratio of the mass of the fluid to its volume.
• The density of gases is dependent on pressure and temperature, while the
density of liquid remains constant.
• Density of water is 1000 kg/m3
• Density of air is 1.204kg/m3
Density = mass of fluid / volume of fluid
ρ = m/v ; kg/m3 , slug/ft3
MKS – metric of measurement
CGS – English measurement
6. Specific gravity (SG)
is defined as the ratio of material’s density to the density of water
at 23°C.
SG refers to a measure of the ratio of the mass of a given volume
of material at 23°C to that of an equal volume of water at 23°C
Sp.gr. = Density(materials) / Density(water)
S.G = ρm/ ρh2o ; unit less
7. SUBSTANCE SPECIFIC GRAVITY
Water (40 C) 1.0
Sea Water 1.03
Air , Dry 0.0013
Alcohol 0.82
Aluminum 2.72
Copper 8.79
Lead 11.35
Mercury 13.6
Steel 7.82
Oil 0.80
Specific Gravity for some common Materials
8. Specific weight
is defined as the weight possessed by unit volume of a fluid.
Specific weight is dependent on acceleration due to gravity as it changes from
place to place.
The specific weight of water is 9810N/m3 , 62.4lb/ft3
Sp.wt. = weight / volume
= density x acceleration of gravity
ꙋ = w/v ; n/m3 , lb/ft3
ꙋ = ρag
9. Specific Volume
is the reciprocal of density. It can be expressed as the volume that a fluid
occupies per unit mass.
Sp. Vol = Volume of fluid / Mass of fluid
ѵ = v/m ; m3 / kg , ft3 / slug , slug = lb/ft/sec2
Ѵ = 1/ρ
10. Viscosity
Any fluids is subjected to viscosity
Viscosity is the measure of fluids resistance to shear or angular deformation.
Types:
• Dynamic Viscosity (Absolute Viscosity, µ)
unit: Pa. sec , Poise, 10 poise = 1pa.sec
Pa – Pascal or N/m2
• Kinematic Viscosity (Ѵ)
• Unit; m2 / sec , stoke(st)
Ѵ= µ/ρ ; n/m2 . sec / kg/m3
n.m . sec / kg
(kg-m/sec2 ) m.sec / kg
m2 / sec
11. SAMPLE PROBLEM
A golf ball has a diameter of 42 mm and a mass of 45 g. calculate the density of the golf ball.
Sol:
ρ = m / v ; v = 4/3 *pi *r3 ; v = 4/3 * 3.1416* ((42/2)/1000)3 = 0.0000388m3
ρ= (45/1000) kg / 0.0000388m^3
= 1159.8kg/m3
12. The density of iron is 7850 kg/m3. The specific gravity of iron related to water with density
1000 kg/m3 is
Sample problem
Sol:
Sp.gr. = ρi / ρh20
= 7850 / 1000
= 7.85
13. Sample problem
The density of water is 1000 kg/m3 at 4 °C (39 °F).The specific weight in SI units ?
Sol:
ρ = 1000kg/m3 @ 4°C
Sp.wt?
ꙋ = ρag
= 1000kg/m3 * 9.81m/sec2
=9810N/m3
14. Sample problem
A reservoir of glycerin has a mass of 1200kg and volume of 0.952m3. Find the glycerin weight, mass
density, specific weight and specific gravity.
Sol:
W = ma
= 1200kg * 9.81m/s2
= 11772N or 11.772kN
D = m/v
= 1200kg/0.952m3
= 1260.5kg/m3
Sp. Wt = w/ v
= 11.772kN / 0.952m3
= 12.37kN/m3
Or = 1260.5kg/m3 x 9.81m/s2 = 12,365N/m3 = 12.365kN/m3
S.G = DS / DH20
= 1260.5/1000
= 1.26
15. Sample problem
The weight of the body is 100lb. Determine a.) its weight in Newton, b.) its mass in kgs. C.) the rate of
acceleration in both ft/s2 and m/s2 . If the net force of 50lb is applied to the body.
Sol:
a.) w = 100lb*4.45N/lb = 445N
b.) W = ma ; m = w/a
= 445N/9.81m/s2
m= 45.4kg
c.) w = ma; a = w/m
m= (45.4kg* 2.2lb/kg )/ 32.2ft/s2
= 3.1slug
a = w/m ; 50lb/3.1lb/ft/s2
a = 16.1ft/s2
a = 16.1ft/s2 * (1m/3.28ft) = 4.9m/s2
16. Sample problem
A vertical cylinder tank with a diameter of 12m and a depth of 4m is filled to the top with water at
20°C. IF the water is heated to 50ºC. How much water is spill over?
Sol: 12m
4m
VTANK = πr2 h
= 3.1416*(62 )*4
= 452.39m3
Wh20 = V*sp.wt
= 452.39 m3* 9.81kN/m3
=4437.95kN
9.81 * 20 = sp. wt *(50 - 20)
Sp.wt = 6.54kN/m3
Vat 50ºC
V = 4437.95 / 6.54
= 678.59m3
Water spill
Vspill = 678.59-452.39
= 226.2m3
17. A quartz of SAE 30 oil at 68deg. F. weighs about 1.85 lb. Calculate the oil specific weight, mass density
and specific gravity.
Sample problem
Sol:
Sp.Wt = w/V; V = 1gal/(4 x 7.48gal/ft3 ) = 0.0334ft3
= 1.85lb/0.0334ft3
= 55.3892lb/ft3
Sp.wt = Dg
D = sp.wt/g
= 55.3892lb/ft3 / 32.2ft/s2
= 1.72slug/ft3
sp.gr. = Sp.Wtoil / Sp. Wth20
= 55.3892lb/ft3/ 62.4lb/Ft3
= 0.8876
18. Find the height of the free surface if 0.8ft3 of water is poured into conical tank., 20inches height and
base radius of 10inches. How much additional water required to fill the tank?
Sample problem
Sol:
Volume of conical tank
V = 1/3 πr2 h
= (1/3)*3.1416*102 * 20
= 2093in3
Vh2O = 0.8ft3 * (12in/ft)3
= 1382.4in3
additional volume required to fill the tank
v = 2093-1382.4
= 710.6in3
20”
10”
y
x
x / 10 = y/20
x = 0.5Y
V = 1/3 πr2 h
710.6 = (1/3)* 3.1416 *(0.5y)2 * y
Y =
3
[(710.6∗3)/(3.1416 ∗0.52 )]
= 13.95in
