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Universal accounting balance:
in + production = out + accumulation
Basic laws:
conservation of mass: Mproduction=0
derived increment basis mass balance:
M in M out M system
derived rate basis mass balance:
m in m out 
dm
system
dt
1st law of thermodynamics (conservation of energy): Eproduction=0
derived increment basis energy balance 1:
E in E out  E system
derived increment basis energy balance 2:
Q net ,in Wnet ,out  E system
{Cengel and Boles eq. 4-17}
derived rate basis energy balance:
dE
E in E out 
dt
system
2nd law of thermodynamics (production of entropy): Sproduction≥0
derived increment basis entropy balance:
S in S gen  S out S system
derived rate basis entropy balance:
dS
S in S gen S out
dt
system
  VA 
Mass flow: m
VA
where V = velocity
v
Moving boundary work:
2
W b  PdV
1
{Cengel and Boles eq. 4-2}
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Equations of state:
Ideal gas law
Mass basis:
PV=mRT;
Molar basis:
PV=NRUT;
Rair = 287 J/kg-K, Rsteam = 461.5 J/kg-K, Rgas x = RU/MW gas x
RU = 8314 J/kmol-K; MW air = 29 kg/kmol, MW steam = 18 kg/kmol
Properties:
quality x ≡ mvapor/mtotal
consequences: x = 0 for saturated liquid, x = 1 for saturated vapor,
x = 0.8 for a mix of 80% vapor and 20% liquid (by mass)
Sample energy/entropy transfer terms:
2
vel
E in  m (hin  in
2
 gzin ) ; Sin  m sin ; Sin  Q in / Tboundary
Efficiencies of a heat engines (you can derive these using energy & entropy balances):
W
T
Q
th  net ,out  1  L ; in the ideal (Carnot) case this becomes:  th,rev  1  L
T
QH
QH
H
Coefficients of performance of refrigerators and heat pumps (you can derive these also):
QL
1
1
; COP  QH 
; in the ideal case these become:
COPR 

HP
Wnet ,in QH QL  1
Wnet ,in 1  QL QH
COPR , rev 
1
1
; COPHP, rev 
TH TL  1
1  TL TH
Isentropic efficiencies:
Vocabulary:
Adiabatic: no heat transfer, Q = 0; Isentropic: no entropy change, no entropy production;
Reversible: no entropy production, frictionless, ideal
Unit conversions:
1 atm = 101325 Pa;
T(K) = T(C) + 273.15;
1 J = 1 N-m;
1 lbm = 0.45359 kg;
1 BTU = 1055.056 J;
1 W = 1 J/s;
T(R) = T(F) + 459.67;
1 N = 1 kg-m/s2;
T(F) = 1.8T(C) +32;
1 Pa = 1 N/m2;
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‘Irreversibility’ I = ‘exergy destroyed’ Xdestroyed = T0Sgen = ‘lost work’ W lost
sample property lookup line in EES:
s[1] = entropy (fluid$,T=T[1],P=P[1])
the exergy of a resource can be found by finding the amount of work that can be extracted upon:
1) taking it to the dead state
2) generating no entropy (e.g., all heat transfer occurs at the dead state temperature)
more EES lookup stuff:
u[1] = intenergy(fluid$,T=T[1]) {for an ideal gas, e.g., air}
u[1] = intenergy(fluid$,T=T[1],P=P[1]) {for a real substance, e.g. air_ha or steam}
v[1] = volume(fluid$, T=T[1],P=P[1])
second law efficiencies:
ηII = ηth/ηth,rev for a heat engine, ηII = COPactual/COPrev for a heat pump or refrigerator
compression ratio r = Vmax/Vmin = vmax/vmin
Book exergy stuff:
If you want more of this kind of stuff, come up to the front and I will give it to you
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Constants:
g = 9.81 m/s2
Prefixes:
M = 106, k = 103
P=V*I; P= electrical power [W], V = voltage [V], I = current [A], Energy [J]=P*t
from the Exam 1 review lecture:
1. Define system and draw picture.
2. Energy balance.
3. Fix states and get properties
4. Compute