Physics 7: Fluids
A.
Static Fluids (10-1 to 10-7)
1. states of matter
a. solid
1. shape and size unchanged by pressure
2. useful properties are mass and force
b. fluid
1. flows under pressure
2. two phases
a. liquid—incompressible
b. gas—compressible
3. useful properties are density and pressure
a. density,  = m/V (kg/m3)
b. pressure, P = F/A (Pa—Pascal)
c. plasma—ionized atoms at high temperature
2. Pascal's principle
a. pressure applied to a confined fluid is equal to the
pressure throughout the fluid
b. Pin = Pout  Fin/Ain = Fout/Aout
c. Win = Wout  Findin = Foutdout
pressure in a liquid, P = gh
Steps
Algebra
P = Fg/A
start with
substitute mg for Fg
P = mg/A
substitute V for m
P = Vg/A
substitute Ah for V
P = Ahg/A
simplify
P = gh
a. equal in all directions and  to object surface
b. not used with gases because density isn't constant
c. absolute pressure
1. absolute = fluid pressure + air pressure
2. P = gh + Po (Po  1 x 105 Pa = 100 kPa)
4. Archimedes' principle, Fb = fVog
a. the buoyant force (weight loss when an object is
submerged) = weight of displaced fluid (mf = fVo)
1. f = fluid density
2. Vo = object's submerged volume
3.
Steps
Algebra
Fb = F2 – F1 = (P2 – P1)A
start with
substitute gh for P
Fb = (gh2 – gh1)A
regroup
Fb = g(h2 – h1)A
substitute h for h2 – h1
Fb = ghA
substitute V for hA
Fb = fVog
b. partially submerged: Fg = Fb
mog = fVsubg  oVo = fVsub
1. fraction submerged: Vsub/Vo = o/f
2. Fsubmerge = Fb – Fg = fVog – oVog = (f – o)Vog
c. partially supported: Fsupport = Fg – Fb = (o – f)Vog
Fb = Fg – Fsupport = (min air – min fluid)g
d. specific gravity s.g. = object/fluid = mair/(mair – mfluid)
oject = (s.g.)fluid (H2O = 1 g/cm3 = 1000 kg/m3)
Name __________________________
B.
Fluid flow (10-8 to 10-10)
1. streamline—fluid layers slide by each other
2. turbulent—eddy currents (increase viscosity)
3. volume flow rate
V/t = Al/t = Av (m3/s)
assume constant volume when water moves
through closed system, then V/t = A1v1 = A2v2
4. Bernoulli's equation, P + gh + ½v2 = C
a. based on conservation of energy
Steps
Algebra
WP = FPd
start with
substitute PA for Fp
Wp = PAd
l for d
Wp = PAl
Wp = PV
substitute V for Al
Ug = mgh
start with
substitute V for m
Ug = Vgh
K = ½mv2
start with
substitute V for m
K = ½Vv2
total energy is constant PV + Vgh + ½Vv2 = constant
divide both sides by V
P + gh + ½v2 = C
b. three types of problems
1. plumbing system:
a.
b.
C.
P1 + gy1 + ½v12 = P2 + gy2 + ½v22
2. leaking tank (P1 = P2, y2 = 0 and v1 = 0):
gy1 = ½v22  v22 = 2gy1
(same for an object that falls y1 meters!)
3. lift caused by moving air (y1 = y2):
P1 – P2 = ½v22 - ½v22
(F1 – F2)/A = ½(v22 – v12)
Flift = ½(vtop2 – vbottom2)A
Kinetic theory—Gases (13-2, 13-6 to 13-10)
1. temperature scales
a. 2 relative scales: oF, oC
b. 1 absolute scale: K = oC + 273 (use Kelvin
temperature for all calculations except T)
2. molecular kinetic energy
a. per mole: K = 3/2RT = ½Mv2
R = 8.31 J/mol•K
M = mass per mole in kg
b. per molecule: K = 3/2kBT = ½v2
kB = R/6.02 x1023 = 1.38 x 10-23 J/K

 = M/6.02 x 1023
3. ideal gas
a. no cohesive force or appreciable volume
b. PV = nRT = NkBT
P = pressure in Pa (1 x 105 Pa = 1 atm)
V = volume in m3 (1 m3 = 1000 L)
n = moles of molecules
N = number of molecules (N = n x 6.02 x 1023)
T = temperature in K
c. sample of gas: P1V1/T1 = P2V2/T2
D.
E.
Heat (13-3 to 13-4, 14-1 to 14-8)
1. heat, internal energy and temperature
a. internal energy (U) is the sum of bond energy,
energy of position and kinetic energy.
b. temperature (T) is related to the kinetic energy
per mole of molecules (K = 3/2RT)
c. heat (Q) is the transfer of internal energy (U)
from one body to another (we will limit our
discussion to heat transfer from a high
temperature body to a low temperature body)
2. laws governing heat transfer (thermodynamics)
a. If bodies are at the same temperature, there is no
heat transfer between them
b. heat naturally flows from hotter to cooler body until
two bodies reach the same temperature
1. internal energy—U: –Uhot + Ucold = 0
2. entropy—S = Q/T: -Q/Thot + Q/Tcold > 0
c. heat flow, Qin = U + Wout
1. U: change in internal energy
2. Wout: work done as body expands (gases)
3. conservation of energy principle
4. rate of heat flow: H = Qin/t = kA(TH – TL)/L
a. k: thermal conductivity
b. A: exposed surface area
c. TH – TL: outside/inside temperatures
d. L is thickness
5. 2 modes of heat transfer
a. conduction: heat transfer through elastic
collisions from hot atoms to cool atoms
b. radiation: hot molecules emit photons
E = hf (human body glows infrared)
6. convection: fluids move between hot and cold
locations because of pressure/density
differences (convection currents)
3. thermal expansion of solids: L = LoT
4. measuring heat flow
a. liquids and solids (Wout  0)
1. Q = U = mcT
a. specific heat, c, is property of material
(water = 4180 J/kg•K)
b. T can be in oC or K
2, hot object added to cool fluid: |Qhot| = |Qcold|
mhotchot(Thot – Tfinal) = mcoldccold(Tfinal – Tcold)
3. rate of heat transfer: power, P = Q/t
b. monatomic gas
1. constant volume (Wout = 0): Qin = U = 3/2nRT
2. constant pressure (Wout  0)
a. Wout = PV = nRT (only when P = 0)
b. Q = Uin + Wout
Q = 3/2nRT + 2/2nRT = 5/2nRT
Heat Engines (15-1 to 15-2, 15-4 to 15-6)
1. PV diagram (monatomic gas)
a. heating a gas changes pressure and/or volume
b. work is done when volume changes
c. useful formulas:
1. PV = nRT T = PV/nR)
2. U = 3/2nRT = 3/2PV or 3/2PV
3. Win = –PV
4. U = Qin + Win
d. idealized processes on PV diagram for a gas
P
adiabatic (Q = 0)
isometric (V = 0)
isobaric (P = 0)
Po
isothermal (T = 0)
Vo
V
e.
interpretation of graphs
1. move away from origin  +T and +U
2. move toward y-axis (compression)  +Win
3. area under the curve = -Win
f. Summary calculations for each process
Process T (T2 – T1) U
=
Qin
+
Win
3/ PV
Isometric
2
0
PV/nR
= U
3/ nRT
(V = 0)
2
3/ PV
Isobaric
2
PV/nR
U – Win
–PV
3/ nRT
(P = 0)
2
Isothermal
= -Win
= -Qin
0
0
(T = 0)
Adiabatic
= Win
0
= U
(Q = 0)
2. heat cycle 
a. ideal (Carnot) system takes a body of gas through
a complete cycle where U = 0


b. solving heat cycle problems
determine T at each state
determine Qin, Win and U for each process
o isometric, Isobaric, use above formulas
o Isothermal, adiabatic, use given values
o missing values, use:

for any step: U = Qin + Win

for cycle: T = U = 0, Qin – Qout = Wout – Win
c. ideal (Carnot) efficiency: ec = (THigh – TLow)/THigh
d. first law efficiency: e = |Wout – Win|/Qin
Waste = Qin(1 – e)
e. heat engine vs. heat pump
1. heat engine takes in heat to perform work
P
– Area 
Win = QL – QH
2.

V
heat engine expands at TH, contracts at TL
 Win < 0 (work is done by the engine)
heat pump uses work to remove heat
P
+ Area 
Win = QH – QL
V
heat pump expands at TL, contracts at TH
 Win > 0 (work is done to the refrigerator)