THERMODYNAMICS REVIEW
Thermodynamics required for compressible flow:
First Law
Energy Equation
Second Law
Adiabatic, isentropic, constant entropy process
Entropy Calculation
Equation of State
Ideal Gas Law
First Law of Thermodynamics
Observations:
work can be transferr ed into heat
more friction = more heat
∑ δQ ∝ ∑ δW
Examples:
rubbing you hands together
friction in a wheel bearing
braking a wheel
Experiments:
∑ Q = C∑ W
∫ δQ = C∫ δW
1 BTU = 778 ft lbf
1 calorie = .427 kg m
First Law for a Cycle
∫ δQ = ∫ δW
∫ (δQ − δW) = 0
The First Law is a Fundamental observation
of nature, an axiom, which can not be proved
but has never been found to be violated
W
First Law of Thermodynamics
∫ δQ = ∫ δW First Law for a Cycle
∫ (δQ − δW) = 0
Cycle A ⇒ B ∫ (δQ − δW) − ∫ (δQ − δW) = 0
Cycle A ⇒ C ∫ (δQ − δW) − ∫ (δQ − δW) = 0
Processes B and C ∫ (δQ − δW) − ∫ (δQ − δW) = 0
∫ (δQ − δW) = ∫ (δQ − δW)
∫ (δQ − δW) is independent of path and
A
B
A
C
B
B
2
A
B
C
C
C
1
therefore a property. Define E as energy
in all forms, KE + PE + U(T).
1
E1 − E 2 = ∫ (δQ − δW) = Q − W
2
First Law for a Processes
Q = ∆E + W
δQ = dE + δW
Steady Flow Energy Equation
Open, steady flow thermodynamic system - a region in space
V
Q
p1
Wshaft
p2
v2
v1
h2
V
h1
Q = ∆E + W First Law
Wflow in = ∫ pdV = p1 (V1 initial − V1 final ) = p1V1 = m p1v1
W = Wshaft + Wflow in + Wflow out = Wshaft + m p1v1 − m p 2 v 2
V2
E = u(T) + KE + PE = u(T) +
+ ρgh
2
V2
V2
Q = m(u1 + p1 v1 +
+ ρgh1 ) − m(u1 + p1 v1 +
+ ρgh1 ) + Wshaft
2
2
V2
+ ρgh) + Wshaft
Q = m∆(u + pv +
2
Steady Flow Energy Equation
Open, steady flow thermodynamic system a region in space
V2
Q = m∆(u + p v +
+ ρ g h) + Wshaft
2
for an adiabatic channel, diffuser or nozzle
W = 0, Q = 0, without body forces
since u + pv = h
for an adiabatic flow
V12
V22
h1 +
= h2 +
2
2
for an nonadiabatic flow
V12
V22
Q + h1 +
= h2 +
2
2
IDEAL GAS CALCULATION
What is the mass of 1.2 m 3 of oxygen at 24o C and a gage pressure of 500 kPa.
Atmospheric pressure is 97 kPa
p = p gage + p atmosphere = 500 kPa + 97 kPa = 597 kPa
8.314 kJ/kmole o K or kPa m3 /kmole o K
R O2 =
= .259813 kPa m3 /kg
32
pV
597 kPa ×1.2 m3
m=
=
= 9.28 kg
RT .259813 kPa m3 /kg × 24o C + 273.16o K
(
)
What is the volume of 1.2 lbm of air at 124o F and a gage pressure of 500 psia.
Atmospheric pressure is 14.7 psia.
p = p gage + p atmosphere = 500 psia + 14.7 psia = 514 psia
1545.15 lbf / lbm o R / lbmole
ft lbf
R air =
= 53.336
28.97
lbm o R
m R T 1.2 lbm × 53.336 ft lbf/ lbmo R × 124o F + 459.69o R
V=
=
p
514 psia ×144 ft 2 /in 2
(
V = .5047ft3
)
Ideal Gas Law
AVOGADRO'S LAW
One (1) mole of any gas = 22.4 liters.
6.023×1023 molecules /mole of gas at
STP (1 atm and 0o C )
BOLYES LAW
p1 × v1 = p 2 × v 2
IDEAL (PERFECT) GAS LAW
pv = RT
pV = mRT
p - absolute pressure, psia, kPa
T - absolute temperatu re, o R, o K
R*
R=
molecular weight
lbf/lbm o R
R = 1545.15
lbmole
kJ
kPa m 3
*
R = 8.314
or
kmole o K kmole o K
*
CHARLES LAW
v1 T1
=
v 2 T2
p1 T1
=
p 2 T2
mass = moles × Molecular Weight
m = n × Molecular Weight
pv = R * T
pV = nR * T
% Error in assuming water is an ideal gas
 ∂T 

 = JT Coefficient
 ∂p  h
h=constant
PRINCIPAL OF CORRERSPONDING STATES
COMPRESSIBILITY FACTOR Z
Z is about the same for
all gases at the same
reduced temperature
and the same reduced
pressure where:
P
pv
Z=
RT
pV
Z=
mRT
PR =
TR =
p
p critical
T
Tcritical
 ∂h 
cp =  
 ∂T  p=const
 ∂u 
cv =  
 ∂T  v=const
∆h = ∫ c p (T ) dT
∆u = ∫ c v (T ) dT
Ideal Gas
∆h = c p ∫ dT
SPECIFIC HEAT
∆u = c v ∫ dT
h = u + pv
pv = RT
h = u + RT
dh = du + RdT
c p dT = c v dT + RdT
cp − c v = R
γR
γ −1
R
γ −1 =
cv
cp =
with same units FOR IDEAL GAS ONLY
IDEAL GAS IMPROVEMENTS
h = ∫ c p (T )dT
u = ∫ c v (T )dT
Kelvin Planck Statement of the Second Law
It is impossible to construct an engine which, operating in a cycle, will produce no
other effect than the extraction of heat from a single reservoir and the performance of an
equivalent amount of work.
Clausius Statement of the Second Law
It is impossible to have a system operating in a cycle which transfers
heat from a cooler to a hotter body without work being done on the
system by the surroundings
Reversible Heat Engine
Reversible Refrigerator
Carnot Power Cycle
Reversible constant t emperature heat trans fer,
process 1 → 2, process 3 → 4
Reversible adiabatic expansion,
process 2 → 3, process 4 → 1
Efficiency =
Desired Effect W
=
Required Input Q in
η CYCLE =
QH − QL
QH
Carnot Principles
1. No engine operating between two heat reservoirs, each having a fixed
temperature, can be more efficient than a reversible engine operating
between the same reservoirs.
ηactual ≤ ηCarnot
2. All reversible engines operating between two heat reservoirs, each
having its own fixed temperature, have the same efficiency.
3. The efficiency of any reversible engine operating between two
reservoirs is independent of the nature of the working fluid and depends
only on the temperature of the reservoirs.
4. An absolute temperature scale can be defined in a manner
independent of the thermometric material.
Q1 T1
=
Q 2 T2
Copyright © The McGraw-Hill Companies, Inc. Permi ssion required for reproduction or display.
FIGURE 5-47
Proof of the first
Carnot principle.
5-15
Thermodynamic Temperature Scale
η = function(T1 , T3 )
η = 1−
Q3
Q1
Q1
= function(T1 , T3 )
Q3
from engine schematics
Q1
= f(T1 , T2 )
Q2
Q2
= f(T2 , T3 )
Q3
Q1
= f(T1 , T3 )
Q3
by identity,
Q1 Q1 Q 2
=
Q3 Q 2 Q3
substituting,
f(T1 , T2 ) = f(T2 , T3 ) × f(T1 , T3 )
this equation can be satisfied only if,
 Qh

 Ql
 Th
 =
 Tl
and Q l = Q h
Tl
Th
δQ
∑T
=0
dQ
=0
∫
T
reversible
processes
Clasius Inequality
∫
δQ
T
≤0
for a Reversible Cycle
 Q h − Q l   Th − Tl 

 = 

Q
T
h
h

 

T
Q
1− l = 1− l
Th
Qh
Qh Ql
−
=0
Th Tl
δQ
∫ T =0
for an Irreversible Cycle
 Q h − Q l   Th − Tl 

 ⟨ 

Q
T
h
h

 

T
Q
1− l ⟨ 1− l
Th
Qh
Qh Ql
−
⟨0
Th Tl
δQ
∫ T ⟨0
Entropy Definition and Change
δQ
∫ T ≤ 0 Clasius Inequality
for a cycle composed of a reversible
and an irreversible process
1
2
δQ
δQ
δQ
=
+
∫ T 2 irrev
∫ T 1∫rev T ≤ 0
DEFINE A PROPERTY S ENTROPY
1
δQ
+ S2 − S1 ≤ 0
∫
T
2 irrev
1
δQ
+ ds ≤ 0
∫
T
2 irrev
1
ds ≥
ds ≥
δQ
∫ T
2 irrev
δQ
∫ T
irrev
δQ = ∫ Tds
ds =
δQ
∫ T
rev
1
irreversible
reversible
2
Entropy Change of an Ideal Gas
δq = du + δw First Law
δq = Tds
Second Law
Tds = du + pdv
du pdv
+
ds =
T
T
For an ideal gas :
du = c v dT
p R
=
T v
dT
dv
+R
ds = c v
T
v
T 
v 
s 2 − s1 = c v ln  2  + R ln 2 
 T1 
 v1 
h = u + pv
dh = du + pdv + vdp
du = dh − pdv − vdp
Substituting into Tds = du + pdv
Tds = dh − pdv − vdp + pdv
Tds = dh - vdp
dh v
− dp
ds =
T T
For an ideal gas ,
ds = c p
dT
dh
= cp
T
T
v
dp
dp = R
T
p
dT
dp
−R
T
p
 T2 
 p2 


s 2 − s1 = cp ln   − R ln 
 T1 
 p1 
Isentropic Adiabatic Process
pv γ = constant
p1v1γ = p 2 v 2γ
γ
δQ = dU + δW First Law
Adiabatic process δQ = 0
dU + δW = 0
p 2  v1 
= 
p1  v 2 
substitute from pv = RT
c v dT + pdv = 0
T2  v1 
=  
T1  v 2 
for an ideal gas
T=
γ −1
p 
=  2 
 p1 
γ −1
γ
pv
pdv vdp
, dT =
+
R
R
R s 2 − s1 = c p ln T2  − Rln p 2 
cv
c
pdv + v vdp + pdv = 0
R
R
c
 cv 
 + 1pdv + v vdp = 0
R
R

 1  dp
 dv
 1


=0
+ 1 + _ 
1
v
1
p
γ
−
γ
−




integrating
γ ln v + ln p = constant
pv γ = constant Adiabatic
Isentropic, ∆s = 0
IdealGas
 p1 
 T1 
p 
s 2 − s1 = c p ln 2 
 p1 
s 2 − s1 = c p
γ −1
γ
p 
− (c p − c v )ln 2 
 p1 
p 
γ −1  p2 
ln  − (c p − c v )ln 2 
γ
 p1 
 p1 
 p  γ − 1

− (c p − c v )
s 2 − s1 = ln 2  c p
γ

 p1 
 cp cv



−
 p 2  c v c v
s 2 − s1 = ln  c p
− (c p − c v ) = 0
cp
 p1 

c
v


adiabatic process, pv γ = constant
is also constant entropy
ADIABATIC PROCESS
γ
Q=0, pv = constant, ∆s = 0
pvγ = constant
p 2 v 2γ = p1v 2γ
γ
p 2  v1 
=  
p1  v 2 
substitute pv = RT
T2  v1 
=  
T1  v 2 
γ −1
 p2 
=  
 p1 
γ −1
γ