FIRST LAW IN OPEN SYSTEMS
Steady Flow Energy Equation
V2
Open, steady flow thermodynamic system - a region in space
Q
p1
p2
v2
Wshaft
z2
V1
z1
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
+ gz
E = u(T) + KE + PE = u(T) +
2
2
V2
V12
+ gz1 ) − m × (u1 + p1 v1 +
+ gz1 ) + Wshaft
Q = m × (u 2 + p 2 v 2 +
2
2
V2
+ gz) + Wshaft
Q = m × ∆ (u + pv +
2
2
V
+ gz ) + Wshaft
Q = m∆ (h +
(5 − 36)
2
the units of all the energy terms must be the same
Steady Flow Processes Devices
V2
Q = m ∆ (h +
+ gz) + Wshaft Steady Flow Energy Equation
2
Turbine, Compressor, Pump
∆Velocity, ∆Elevation, Q = 0
W = ∆H = m∆h
W = m(h in − h out )
Boiler, Condenser, Heat Exchanger
∆Velocity ≅ 0, ∆Elevation ≅ 0, Work = 0
Q = ∆H = m∆ h
Q = m(h in − h out )
Diffuser, Nozzle
∆Elevation ≅ 0, Q = 0, W = 0
V22
V12
= h2 +
h1 +
2
2
Valve - throttling process
∆Velocity = 0, ∆Elevation = 0, Q = 0, W = 0
∆H = 0
H in = H out
h in = h out
What range of 850 kPa steam quality
can be meaaured with this device?
open thermodynamic system
Steady Flow Energy Equation
V2
Q = ∆(h +
+ zh) + Wshaft
2g
∆KE = 0, ∆PE = 0, W = 0, Q = 0
h1 = h 2 (T2, Pbarometer )
h 2 = h g @Pbarometer = 100. kPa
h 2@maximum mesurable quality = 2506.1 kJ/kg
@850 kPa
T
h v = 732.22 kJ/kg
h fg = 2039.4 kJ/kg
h − h1f 2506.1 kJ/kg − 732.22 kJ/kg
x= 2
=
h1fg
2039.4 kJ/kg
x = .87, 87% to 100% quality can be measured
1
2
850 kPa
1
100 kPa
2
v
500 kg/sec of 60 O C water is mixed with 200 kg/sec 60O C
saturated steam in a tank at a pressure of 15kPa.
What are the exit conditions?
open thermodynamic system
Mass Balance m c = m a + m b
m c = 500 kg/sec + 200 kg/sec
500 kg/sec
Steady Flow Energy Equation
60 O water
2
V
Q = m∆(h +
+ zh) + Wshaft
a
2g
Q = 0, W = 0, ∆KE = 0, ∆PE = 0,
b
mh = constant
200 kg/sec
ma h a + m bh b = mch c
h a = h f @60O C = 251.13 kJ/kg
60 O steam
h b = h v @60O C = 2373.1 kJ/kg
500 kg × 251.13 kJ/kg + 200 kg × 2373.1 kJ/kg = 700 kg × h c
h c = 924.98 kJ/kg
at 15 kPa h f = 225.94 kJ/kg, h g = 2373.1 kJ/kg
925.98 − 225.94
x=
= .29, 29%quality
2373.1
T = 53.97 O C
15 kPa
c
4.233 An adiabatic air compressor is to be powered by a direct coupled adiabatic steam
turbine that is also driving a generator. Steam enters the turbine at 12.5 MPa and 500 C at a
rate of 25 kg/sec and exits at 10 kPa and a quality of .92. Air enters the compressor at 98
kpa and 295 K at a rate of 10 kg/sec and exits at 1 MPa. Determine the net power delivered
to the generator by the turbine.
air
12.5 kPa
p
1 Mpa
steam
p
500 C
620 K
generator
Wnet
compressor
98 kPa
295 K
turbine
10 kPa
x=.92
v
v
h c1 = airtable@(T = 295) = 210.5kJ/kg
h c2 = airtable@(T = 620) = 628.1kJ/kg
h t1 = superheat@(T = 500, p = 12.5) = 3341.8
h t2 = h f @10kPa + xh fg @10kPa
h t2 = 191.83 + .92 × 2392.8 = 2393.2kJ/kg
Wnet = Wturbine − Wcompressor = m t (h t1 − h t2 ) − m c (h c1 − h c2 )
Wnet
Wnet
kg
kg
(628.1 − 210.5)
= 25
× (3341.8 − 2393.2) − 10
sec
sec
= 19539 kJ/sec
4.65
Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a
velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the exit
temperature and (b) the ratio of inlet to exit area.
Open thermodynamic system - a region in space
3 MPa
400C
40m/sec
2.5MPa
300m/sec
1
p
2
v
h1 & v1 = superheat @ (T = 400., P = 3.)
h1 = 3203.9 kJ/kg
v1 = .09936 m 3 /kg
V12
V22
h1 +
steady flow energy equation
= h2 +
2
2
 V12 V22 

h 2 = h1 + 
−
2
2


 40 2 300 2  m 2  1 kJ/kg 

h 2 = 3203.9 kJ/kgm + 
−
2 
2
2 
2
2
sec
1000
m
/sec




h 2 = 3203.9 − 44.2 = 3159.7 kJ/kg
a) T2 = superheat @ (h = 3159.7, p = 2.5)
T2 = 364.78 C
 p 
b) m = ρAV = 
AV
 RT 
 p 
 p1 
 A1 V1 =  2  A1 V2

 RT2 
 RT1 
3 A1 40
2.5 A 2 300
=
(400 + 273.15) (364.78 + 273.15)
A 2 .1783
=
= .1517
A1 1.1757
Steam at 3Mpa and 400 C enters an adiabatic nozzle steadily with a
velocity of 40 m/sec and leaves at 2.5 MPa and 300 m/sec. Determine (a) the
exit temperature and (b) the ratio of inlet to exit area.
Open thermodynamic system - a region in space
3 MPa
400C
40m/sec
2.5MPa
300m/sec
1
p
2
v
h1 & v1 = superheat @ (T = 400., P = 3. Mpa)
a) T2 @ (h = 3187.5, p = 2.5)
h1 = 3231.7 kJ/kg
T2 = 376.5 C,
v1 = .09938 m 3 /kg
v 2 @ (h = 3187.5, p = 2.5)
V12
V22
h1 +
steady flow energy equation
= h2 +
2
2
 V12 V22 

h 2 = h1 + 
−
2
2


 40 300
h 2 = 3231.71.9 kJ/kgm + 
−
2
2

h 2 = 3231.71 − 44.2 = 3187.5 kJ/kg
2
2
 m  1 kJ/kg 

2 
2
2 
sec
1000
m
/sec



2
v 2 = .11626
b) m = ρAV
A1 V1 A1 V2
=
v1
v2
3 A1 40 2.5 A 2 300
=
.09938
.11626
A 2 13.951
=
= .1872
A1 74.535
4.65
FIRST LAW IN UNSTEADY SYSTEMS
Wboundary
p1
h1,
u1,
T1
m 1 = (m f − m i )
Q = ∆E + W
Q = E f − E i + (Wflow + Wboundary )
E i = m i u i + (m f − m i )u1
Ef = mf u f
Wf = ∫ pdV = (m f − m i ) p1 (v end − v1 )
Wf = −(m f − m i )× p1v1
Q = m f u f − m i u i − (m f − m i )× (u1 + p1v1 ) + Wb
Q = m f u f − m i u i − (m f − m i )× h1 + Wb
FIRST LAW FOR UNSTEADY SYSTEMS
Q − Wb + H in = ∆U contents
mi
Q
For :
m i = 0, adiabatic vacuum
Q = 0, W b = 0
u f = h1
c v (Tf − To ) = c p (Ti − To )
 T − To
=  f
c v  Ti − T o
To is arbitrary,
cp
T f = k Ti



To = 0
h1
A 200 cubic ft tank contains
u1
2. lbm carbon dioxide and .1
T1
mole helium at an initial
m 1 = (m i − m f )
temperature of 70 F. 3 lbm
p1
of air at 14.7 and 70 F are
admitted to the tank.
What is the final temperature of the tank?
mi
Q = m f u f − m i u i − (m f − m i )(h i )
m f u f = (3 × .174 + 2 × .1565 + .4 × .745 )Tf = 1.125Tf
m i u i = (2 × .1565 + .4 × .745 )× (460 + 70 ) = 323.83
(m i − m f )h1 = 3 × .24 × (460 + 70 ) = 381.6
Q = 1.125Tf − 323.83 − 381.6 = 0
Tf = 627 o R
Tf = 167 o F
Wboundary
8 kg liquid water and 2 kg vapor at 300 kPa are
contained in an insulated piston cylinder. Steam
at .5 MPa and 350 C are admitted until the piston
cylinder contains only vapor. Determine the final
temperature and the amount of steam admitted.
the system is the mass finally in the piston cylinder, m 2
Q − Wboundary + (m 2 − m1 ) h o = m 2 u 2 − m1u1
Wboundary = m 2 p 2 v 2 − m1p1v1
substituting for Wboundary ,
steam
ho
0 − (m 2 p 2 v 2 − m1p1v1 ) + (m 2 − m1 ) h o = m 2 u 2 − m1u1
0 = −m 2 h o + m1h o + m 2 u 2 − m1u1 + m 2 p 2 v 2 − m1p1v1
@ 300 kPa
0 = −m 2 h o + m 2 u 2 + m 2 p 2 v 2 + m1h o − m1u1 − m1p1v1
T2 = Tsaturation @300kPa = 133.55 C
since = u + pv,
0 = m 2 (h 2 − h o ) + m1 (h o − h1 )
m 2 = m1
(h o − h1 )
(h o − h1 )
h 2 = v g = 2725.3 kJ/kg
h1 = h f + x × h fg
h1 = 561.47 + .8 × 2163.8
h1 = 2292.51 kJ/kg
(3167.7 kJ/kg − 2292.51 kJ/kg ) = 19.78 kg
m 2 = 10kg
(3167.7 kJ/kg − 2725.3 kJ/kg )
@.5 MPa, 350 o C
m 2 − m1 = 19.78 kg − 10 kg = 9.78 kg
h o = 3167.7 kJ/kg
4-153
First Law
Energy defined , Energy conserved
E in − E out = ∆E (page 72)
E is all forms, Q, W, PE, KE, U
(Qin − Q out ) + (Win − Wout ) + (E massin − E massout ) = U 2 − U1
(2 - 32)
CLOSED SYSTEM a contained quantity of mass
(Qin − Q out ) + (Win − Wout ) + (E mass in − E mass out ) = U 2 − U1 (page 173)
0
(Qin − Q out ) + (Win − Wout ) = U 2 − U1
Q = U 2 − U1 + W
Q = ∆E + W
OPEN SYSTEM a region in space
(Qin − Q out ) + (Win − Wout ) + (E mass in − E mass out ) = U 2 − U1 (page 233)
0
W = Wshaft + Wflow
(Qin − Q out ) + (Win − Wout ) = (E mass out − E mass in )
for Wnet = 0, Q = H 2 − H1 = m(h 2 − h1 )
for Q net = 0, W = H 2 − H1 = m h 2 − h 1
(
)
UNSTEADY SYSTEM
quantity of mass, m 1 or m 2
(Qin − Q out ) + (Win − Wout ) + (E mass in − E mass out ) = U 2 − U1
0
(Qin − Q out ) + (Wb in − Wb out ) + (Wflow in − Wflow out ) = U 2 − U1
(Qin − Q out ) + (Wb in − Wb out ) + (p o Vo )in − (p o Vo )out = U 2 − U
(Qin − Q out ) + (Wb in − Wb out ) + (m 2 − m1 )(p o v o )in − (m 2 − m1 )(p o v o )out =
m 2 u 2 − m1u1 + (m 2 − m1 )u out − (m 2 − m1 )u in
(Qin − Q out ) + (Wb in − Wb out ) + (m 2 − m1 ) h in − (m 2 − m1 )h out =m 2 u 2 − m1u1
with Wout , Q in , +
Q - W + (m 2 − m1 ) h = m 2 u 2 − m 1u1
UNSTEADY SYSTEM
region in space
(Qin − Q out ) + (Win − Wout ) + (E mass in − E mass out ) = U 2 − U1
(Qin − Q out ) + (Wb in − Wb out ) + (E mass in − E mass out ) = U 2 − U1
(Qin − Q out ) + (Wb in − Wb out ) + (m 2 − m1 ) h in − (m 2 − m1 ) h out =m 2 u 2 − m1u1
Problem Set Up Strategies
System
- open, closed, unsteady
- schematic at each system state
State Point
- property diagram
- locate points
Properties
- property values – tables, Ideal Gas Law
- what properties remain constant ?
Cycle
- what remains constant ?
Mass Balance
Identify Energy
- Forms
- Sources
- Uses
Energy Balance
- general equation
- from schematic
- specific for open system
- specific for closed system
- specific for unsteady system
From Chapter 1
Concepts
System
Properties
State Point
Process
Cycle
V 2 m 2 /sec 2
1 kJ/kg
Kinetic Energy =
=
×
2
2
1000 m 2 /sec 2
V2
ft 2 /sec 2
1
Kinetic Eenergy =
=
×
= Btu/lb
2g 2 × 32.3 ft/sec 2 778 ft − lb/Btu
1 kJ
Work = ∫ pdV = kPa × m 3 ×
= kJ
3
1 kPam
lb 144 in 2
1 Btu
3
Work = ∫ pdV = 2 ×
×
ft
×
= Btu
2
in
ft
778ft − lb