density
DESCRIPTION
Density. • Specific volume (volume per unit mass),. Density, ρ= mass/unitvolume –Slugs/ft3;kg/m3. • Specific weight= weight/volume (lb/ft3; N/m3). • Specific Gravity = density of the fluid/density of water. Example 1. The specific gravity of mercury at 80 0C is - PowerPoint PPT PresentationTRANSCRIPT
Density
Density, ρ= mass/unitvolume –Slugs/ft3;kg/m3
• Specific volume (volume per unit mass),• Specific weight= weight/volume (lb/ft3; N/m3)
• Specific Gravity = density of the fluid/density of water
1v
g
2 @4H O C
SG
Example 1
The specific gravity of mercury at 80 0C is
13.4. Determine its density and specific
weight at this temperature in both BG
and SI units,
Figure 1.1 (p. 10)
Density of water as a function of temperature.
Ideal/Perfect Gas Law /Equation of State
• Gases are highly compressible, gas density changes with pressure and temperature as,
where p is the absolute pressure, ρ the density, T is absolute temperature, and R is gas constant
Pressure units: lb/ft2 (psf) ; lb/in2 (psi); N/m2 (Pa)
Standard sea-level atmospheric pressure – 14.7 psi; 101.33 kPa
Gage pressure + atmospheric pressure = absolute pressure
p RT
LV5
Ts are expressed in Kelvin or Rankine Lisa Vink, 1/11/2007
R is different for each gas and is determined from R=Ru/M where Ru is the universal gas constant, Ru=8.314 kJ/kmol.K=1.986 Btu/lbmol. R
Example 2 A compressed air tank has a volume of 0.84 ft 3. W
hen the tank is filled with air at a gage pressure of 50 psi, determine the density of the air and the weight of air in the tank.
Determine the density, specific gravity and mass of the air in a room whose dimensions are 4 m x 5 m x 6 m at 100 kPa and 25 C
Example 3
Viscosity
Viscosity is a measure of a fluid's resistance to
flow. It describes the internal friction of a moving fluid. A fluid with large viscosity resists motion
because its molecular makeup gives it a lot of
internal friction.
• A fluid with low viscosity flows easily because
its molecular makeup results in very little friction
when it is in motion.
Fluid motion can cause shearing stresses
Figure 1.2 (p. 13)
(a) Deformation of material placed between two parallel plates. (b) Forces acting on upper plate.
Figure 1.3 (p. 14)
tana U t
b b
a U t
0limt t
Behavior of a fluid placed between two parallel plates.
Shear stressτ occurs at the plate-material interface at equilibrium, P=τA velocity gradient, du / dy =U/bFluid sticks to the wall=no-slip condition
U du
b dy
du du
dy dy
as
Rate of shearing strain
Figure 1.4 (p. 15) Newtonian Fluids
Linear variation of shearing stress with rate of shearing strain for common fluids.
du
dy ; μ= absolute or dynamic viscosity
Figure 1.5 (p. 16)
Variation of shearing stress with rate of shearing strain for
several types of fluids, including common non-Newtonian fluids.
Units of Viscosity –lb.s/ft2; N.s/m2
Figure 1.6 (p. 17)
Dynamic (absolute) viscosity of some common fluids as a function of temperature.
3/ 2
;CT
T S
For gases,
For liquids,
Kinematic viscosity
units are / ft2 ; m2 /s
/B TDe
v
Example 4
Reynolds number
ReVD
A Newtonian fluid having a viscosity of 0.38 N.s/m2 and a specific gravity of 0.91 flows through a 25 mm diameter pipe with a velocity of 2.6 m/s. Determine the values of the Reynolds number using (a) SI and (BG) units
fluid density
viscosity
mean fluid velocityV pipe diameterD
Figure E1.5 (p. 19)velocity , profile
231
2V y
uh
1. What is the shearing stress at the bottom wall?2. The shearing stress on a plane parallel to the walls and passing through the centerline?
dudy
3
;2
du Vydy h
;y h 3du Vdy h
23 (0.041b.s/ft )(3)(2ft/s)the shearing stress
(0.2in)(1ft/12in)Vh
0;y 0dudy
2. Along the midplane. where
Shearing stress;
at
Compressibility of Fluids
Page 20.
How the density of a fluid change with pressure?
Bulk , modulus
units of bulk modulus , lb/in2 ( psi ) or N/m2 (Pa)
/ /dp dpdV V dV
Liquids are considered incompressible
Compression and expansion of Gasesconstant
p
constantk
p
= p
V
CkC
Isothermal process :
For isentropic process ;
For an isothermal process,EV=p; For Isoentropic process, EV=k·pA cubic foot of helium at an absolute pressure of 14.7 psi is compressed isentropically to ½ ft3. What is the final pressure?
p VR C C
fik ki f
pp
( )kf i
f
i
p p
1.66(2) (14.7psi) 46.5psi(abs)fp
Speed of Sound• Acoustic velocity, speed of sound, c
• Depends on change in pressure and density
• Mach number, Ma = velocity of air/velocity of sound
• Ma<1, subsonic; Ma>1, supersonic
VE ;kpFor isoentropic process ,
For air at 60 0F, k=1.40 and R=1716 ft. lb/slug. oR; c=1117 ft/s
For water at 20C, Ev =2.19 GN/m2, ρ = 998.2 kg/m3; c =1481 m/s or 4860 ft/s
c kRT
dpc
d VE
c