Post on 22-Mar-2018
7-10 REVERSIBLE STEADY-FLOW WORK
rev rev
q w dh dke dpeδ δ− = + + Energy balance
rev
q Tds dh vdPδ = = − nd2 Gibbs equation
rev
w vdP dke dpeδ = − − −
2
rev ,net ,out
1
w vdP ke pe∆ ∆= − − −∫ ( )7-51
= − ∫2
rev ,out
1
w vdP ( )7-52 turbine
= ∫2
rev ,in
1
w vdP compressor
rev ,out act ,out
w w≥
rev ,in act ,in
w w≤
7-11 Compression work for ideal gas (p.364)
( )
k 1
k1 2
comp ,in 2 1
1
kRT Pkw R T T 1
k 1 k 1 P
− = − = − − −
( )
n 1
n1 2
comp ,in 2 1
1
nRT Pnw R T T 1
n 1 n 1 P
− = − = − − −
2
comp,in
1
Pw RT ln
P=
Incompressible fluid ( v const= , 7-51) ⇒ ( )rev 2 1w v P P ke pe∆ ∆= − − − − ( )7-54
Bernoulli equation (steady flow of incompressible fluid through the simple pipe without friction)
( )2 2
2 1 2 1
2 1
P P V Vg z z 0
2ρ
− −+ + ⋅ − = ( )7-55
( )
Danila Bernoulov
1700 1782−
�
act act
rev rev
act rev act rev
rev act rev act
Tds
rev act act
gen
q w dh dke dpe
q w dh dke dpe
q q w w 0
w w q q
w w qds s 0
T T
δ δ
δ δ
δ δ δ δ
δ δ δ δ
δ δ δ
− = + +
− = + +
− − + =
− = −
−= − = ≥
pipe
inlet state
w 0=
outlet state
1 2
qδinlet state
wδ
outlet state
1 2
k
Isentropic process
Pv const=
Isothermal process
Pv const=
n
Polytropic process
Pv const=
−
=1
kv cP
−
=1
nv cP
1v cP
−=
P
v
1P
2P
( )isothermal n 1=
( )polytropic 1 n k< <
( )isentropic n k=
Flow through the pipe which involves no work interaction
Pv RT=Ideal gas
Reversible work output for
steady-flow and closed systems
( )for ke pe 0∆ ∆= =
Equations 7-57 a,b,c
outQ maximum
heat transfer
Q 0=
T const=
rev ,out act ,outw w≥
reversible
actual
turbine
compression
P
v1
v2
v
1
2
1P
2P
turbine
in derivation, we assume that both
processes are between the same states
turbine
compression
Steady-flow devices
deliver the most and
consume the least work
when the process is
reversible
≥rev ,net ,out act ,net ,out
w w
δ δ≥rev act
w w
Eq.7-9
minimum
heat transfercomp ,in
w
By cooling compressor, the requiered work input
can be minimized to achive the same pressure increase
temperature
is increased
P
v1
v2
v
1
2
1P
2P
rev ,outw
turbine