S.F.E.E. per unit mass basis
(i) 1
+ C
12+
1+ d Q =
2+ C
22+
2+ d W
h g z h g z
2 d m 2 d m
[h, W, Q should be in J/Kg and C in m/s and g in m/s2]
(ii) 1
+
12+ gZ
1+ Q =
2+
22+ gZ
2+ W
h h
2000 1000 dm 2000 1000 dm
C d C d
[h, W, Q should be in KJ/Kg and C in m/s and g in m/s2]
S.F.E.E. per unit time basis
⎛ ⎞
+ + +
⎜ ⎟ τ
⎝ ⎠
⎛ ⎞
= ⎜ ⎝ + + ⎟ ⎠ + τ
2
1 1 1 1
2
2 2 2 2
w z
2
w z
2
x
C dQ
h g
d dW
h C g
d
Where, w = mass flow rate (kg/s)
Steady Flow Process Involving Two Fluid Streams at the Inlet and Exit of the Control Volume
S
he following e gineering sys ozzle and Di nozzle is a d op, whereas low a nozzle cific volume ( ncele of steady examples illu
stems.
iffuser:
device which a diffuser in which is insu 1 1
y flow proce ustrate the a
increases th ncreases the
ulated. The s
2
he velocity o pressure of a steady flow en
2 ntrol surface
apter 2
some of the
its pressure ure show in gives
First Law of Thermodynamics
S K Mondal’s Chapter 2
35 Here dQ 0;dWx 0,
dm= dm = and the change in potential energy is zero. The equation reduces to
2 2
1 2
1 2 2 2
C C
h + =h + (a)
The continuity equation gives
1 1 2 2
1 2
A A
w = C = C
v v (b)
When the inlet velocity or the ‘velocity of approach’ V1 is small compared to the exit velocity V2, Equation (a) becomes
2 2
1 2
2 1 2
2
2( ) /
h h C
or C h h m s
= +
= −
where (h1 – h2) is in J/kg.
Equations (a) and (b) hold good for a diffuser as well.
Throttling Device:
When a fluid flows through a constricted passage, like a partially opened value, an orifice, or a porous plug, there is an appreciable drop in pressure, and the flow is said to be throttled. Figure shown in below, the process of throttling by a prettily opened value on a fluid flowing in an insulated pipe. In the steady-flow energy equation-
0, Wx 0 dQ
dm= ddm =
And the changes in P. E. are very small and ignored. Thus, the S.F.E.E. reduces to
2 2
1 2
1 2 2 2
C C
h + =h +
(Fig.- Flow Through a Valve)
Often the pipe velocities in throttling are so low that the K. E. terms are also negligible. So
h
1= h
2or the enthalpy of the fluid before throttling is equal to the enthalpy of the fluid after throttling.
Turbine and Compressor:
Turbines and engines give positive power output, whereas compressors and pumps require power input.
For a turbine (Fig. below) which is well insulated, the flow velocities are often small, and the K.E.
terms can be neglected. The S.F.E.E. then becomes
First Law of Thermodynamics
S K Mondal’s Chapter 2
36 (Fig.-. Flow through a Turbine)
1 2
1 2
x
x
h h dW dm
or W h h
m
= +
= −
The enthalpy of the fluid increase by the amount of work input.
Heat Exchanger:
A heat exchanger is a device in which heat is transferred from one fluid to another, Figure shown in below a steam condenser where steam condense outside the tubes and cooling water flows through the tubes. The S.F.E.E for the C.S. gives
c 1 s 2 c 3 s 4
s 2 4 c 3 1
w w w w
, w ( ) w ( )
h h h h
or h h h h
+ = +
− = −
Here the K.E. and P.E. terms are considered small, there is no external work done, and energy exchange in the form of heat is confined only between the two fluids, i.e. there is no external heat interaction or heat loss.
Fig. -
Figure (shows in below) a steam desuperheater where the temperature of the superheated steam is reduced by spraying water. If w1, w2, and w3 are the mass flow rates of the injected water, of the steam entering, and of the steam leaving, respectively, and h1, h2, and h3 are the corresponding enthalpies, and if K.E. and P.E. terms are neglected as before, the S.F.E.E. becomes
1 1 2 2 3 3
w h + w h = w h and the mass balance gives
w1 + w2 = w3
S
he above law actical situat nity.
) Work deve (a) Water In this cas
1 1v +z g 1
p
(b) Steam In this cas
(
h1W =
) Work abso (a) Centrifug
The system
In this sys
1 1v +z g 1 tions as work
eloping syst
gal water pum
m is shown in led as stead k developing
tems dy flow ene system and
equation bec
sumed to be
below
he energy equ
system Δz =
= 0, p1 1v =
Therm
ergy equati work absorp
omes
zero and the
uation now b
0 and the equ
e equation be
becomes,
uation becom
= 0; now the
amics
Cha
n be applied . Let the ma
comes
mes,
e energy simp
apter 2
d to various ass flow rate
plifies to
S
and hence th2 equation for Steam con are very sma (h1 – h2) and heat lost by Steam nozz
In this syste possible heat The ene
ii) Unsteady any flow proc n be analyzed nder non-stea
thin the cont rface, as give
Fir
fans the tem he energy equ2 all. Under ste d this heat is steam will be zle:
y Flow Anal cesses, such a d by the cont ady state cond
trol volume is en below:
rst Law
2 1
s C C mperature ris
uation for fan
essor – In a
g and absorb s system we omes Q = (h2
n this system eady conditio as filling up a trol volume te ditions (Figu ompressor is
bing system neglect ΔZ, Δ
2 – h1) m the work do
ons the chang o the change at gained by ure-shown in ed is equal to
Therm
all and heat
g compressor
ms
ΔKE and W
one is zero a ge in enthalp e in enthalpy the cooling w
zero and hea
ing gas cylind onsider a dev
below). The r
at transfer w
ders, are not vice through w
rate at which e of mass flow
amics
Cha
ected (i.e.) Δh
PE are neglig
KE = W = 0;
also assume Δ o heat lost by water circulat
which is noth
steady, Such which a fluid h the mass of w across the c
apter 2
h = 0, q = 0
gibly energy
; the energy
ΔZ and ΔKE y steam. Q = ted (i.e.) the
ing but any
h processes d is flowing
f fluid control
S
ross the cont ate of energyd the equatio dEv
ny finite peri
1 2
m – mΔ Δ cumulation of
rol surface. I increase = R
uid within th iod of time
f energy with If Ev is the en Rate of energy
1
these energy
2
eral energy e
For a closed s
he control vol
hin the contr nergy of fluid y inflow – Ra
trol volume a
2 y flux quantit
2
rol volume is d within the c
at any instant
dW
− d τ ties. For any
2 2
+Z g dm⎞
⎟⎠
Fig.
r steady flow
0, w2 = 0, the
y time interva
,
en from equat
amics
al, equation (
tion (A),
apter 2
energy flow tant,
(B) becomes
S
ariable flow p chnique, as il Conside e beginning td gas flows in pply to the pi
P P P
,T , v , h ,
ystem Techn hich would ev
Energy en omitted.
Now, ther lume. Then t
sing the first
Q = ΔE = m2
Fir
dal’s
s
variable flo processes may llustrated be er a process i the bottle con nto the bottle ipeline is ver
P P
u and v .
nique: Assum ventually ent terms are ne re is a chang the work don
W = y be analyzed
low.
in which a ga ntains gas of
e till the mas ry large so th
me an envelo ter the bottle efore filling.
)
2glected. The ge in the vo as bottle is fi mass m1 at s
the pipeline a
)
2he system tec lled from a p state p1, T1, v he bottle is m of gas in the
extensible) o n Figure abov
and tube whi
uP⎞ + ⎟
⎠ ottle is not in because of t
chnique or th pipeline (Figu
ich would en
n motion, and the collapse
amics
Cha
he control vol ure shown in The valve is , T2, v2, h2 an onstant at
pipeline and
ter the bottle
d so the K.E.
S
hich gives the ontrol Volumgure above, A ritten on a tim
nce hP and CP
t us consider plying first la
suming K.E.
gain
hich shows t asi-static.
e energy bala me Techniq Applying the me rate basis
dEv dQ
that the pro
arging the tan
rst Law
que: Assume energy equa s -nt, the equati
2
(
charging a flu ntrol volume,
cess is adiab
nk
ion is integra
)
case, the follated to give f
Tank
upply line (Fi
+ ⎞⎟
ng and Disc
amics
Cha
ol surface as y balance ma
process
e dW = 0 anx
charging a
apter 2
shown in ay be
nd dmin = 0,
Tank
First Law of Thermodynamics
S K Mondal’s Chapter 2
42
( )
= Δ = −= −
∫
in 2 2 1 12 2 1 1
V p p
hdm U m u m u m h m u m u
where the subscript p refers to the constant state of the fluid in the pipeline. If the tank is initially empty, m1 = 0.
= 2 2 m hp p m u Since
=
=
2
2