FLUID FLOW
Mechanical Energy Balance
V
gz   

 2
2
dp
potential
expansion
energy change
work
losses

  Wo   F

kinetic
work added/ sum of
energy change subtracted by friction
pumps or
compressors
Note that the balance is per unit mass. In differential form
Rewrite as follows
dp    g  dz  V  dV  F  Wo 
divide by dL (L is the length of pipe)
dp
dz
dV
F
W
  g 
 V 

 o
dL Tot
dL
dL
L
L
or :
dp 
dp 
dp 
dp 







dL  Tot
dL  elev dL  accel dL  frict
(
Wo
is usually ignored, as the equation applies to a section of pipe )
L
The above equation is an alternative way of writing the mechanical energy
balance. It is not a different equation.
The differential form of the potential energy change is
dL
dZ

g
dZ
 g sin 
dL
What about the friction losses?
1) Fanning or Darcy-Weisbach equation (Often called Darcy equation)
2V 2 f
F 
dL
D
This equation applies for single phase fluids !!!
The friction factor is obtained from the Moody Diagram (see P&T page
482).
Friction factor equations. (Much needed in the era of computers and excel)
16
Re
0.046
f 
Re a
f 
Laminar Flow
smooth pipes: a=0.2
Iron or steel pipes a=0.16
 
1
2.51 

 2 log10 

f
 3.7 D Re f 
Colebrook equation for
turbulent flow.
Equivalent length of valves and fittings.
Pressure drop for valves and fittings is accounted for as equivalent length of
pipe. Please refer to P&T for a table containing these values (page 484).
SCENARIO I
Piping is known. Need pressure drop. (Pump or compressor is not present.)
Incompressible Flow
a) Isothermal (is constant)
dp
dV F 
 dZ
  g 
 V


 dL
dL Tot
dL L 
for a fixed 

V constant

dV = 0
 L
F  2 V 2  f   
 D

L
p     g  Z  2V 2  f  
D



F

b) Nonisothermal
It will not have a big error if you use (Taverage), v(Taverage)
Compressible Flow (Gasses)
a) Relatively small change in T (known)
For small pressure drop (something you can check after you are done) can
use Bernoulli and fanning equation as flows
 V2 
g  dz  v  dp  d   F
 2 
g
1
V
F
 dz   dp  2  dV   2
2
v
v
v
v
V  v
but
G
A
V = Velocity
v = Specific volume (m3/Kg)
G = Molar flow (Kg/hr)
A = Cross sectional area
g
1
F
 G  dV
 G  dL
 dz   dp    
  2  2  f   
2
 A v
 A D
v
v
v
2
Now put in integral form
dp  G 
g 2
 
v  A
v

dz

Assume
2

2
dV
 G 1
 2      f dL
 A D
V

Tin  Tout
2
fin  fout

2
(Tin , Pin )  (Tout , Pout )

2
f (Tin , Pin )  f (Tout , Pout )

2
Tav 
fav
fav
The integral form will be
  g  z  
2
av
Now use
Then
out
in
dp  G   Vout 
L
 G
   ln
  2     f av 
 A
v  A   Vin 
D
2
pv 
Z  RT
M
v  Z av 
RTav
pM
2
M; Molecular weight

dp
1

v
Zav RTav
 p  dp  2  Z

1
2
pout
 pin2
av RTav

Therefore ;
 2av g  z 


2
2
V 
M
L
 G
 G
2
2
pout
 pin
   ln out   2  f av
 A
 A
2  Zav RTav
D
 Vin 
but,
Vout  Zout  Tout  pin


Vin  Zin  Tin  pout

Z RT
2
pout   pin
 2 av av
M

1


 Zout Tout pin 
L  G
  G
 2
2
2

f


ln



g


z


 
  

av
av
 A
D  A
ZinTin pout 





2
2
This is an equation of the form p out  F( p out )
Algorithm
a) Assume p (out1)
b) Use formula to get a new value p (out2 )  F p (out1) 
c) Continue using p (outi 1)  F p (outi) 
until
i  1)
i)
p (out
 p (out

i)
p (out
OR BETTER: USE Solver in EXCEL, or even better use PRO II, or any
other fluid flow simulator.
CAN THIS BE APPLIED TO LONG PIPES. What is the error ?
===> If
pout  pin
 0.2  0.3 you will be OK. What to do if not. Use shorter
pin
sections of pipe.
What if temperature change is not known
Use total energy balance as your second equation
g  dz  d ( pv)  VdV du  q  w o
Potential
Energy
rate of work
done on fluid
element by
pressure forces
change of
internal
energy
Energy
Kinetic
Energy
work done on
fluid element
by external
forces
Heat
Note : friction is not present !
external temp
q  UTo  T  DdL 
}
dA
1
G
Flow rate
Flux
Then, (ignore wo ,will not use when pumps or compressors are not present)
UTo  TDdL
 V2 
gdz  dh  d  
G
 2 
Integrate and solve for hout (use Tav in the heat transfer equation)
hout  hin 
2
U To  Tav DL Vout
 Vin2 

  g z 2  z1 
G
2


But
Vout  v out 
RTav G
G
 Z av 
A
pout M A
Tout  Tout  p out , h out 
Procedure :
a) Assume Tout, pout
b) Use mechanical energy balance to obtain p (out1)
c) Use total energy balance to obtain h (out1)
(1)
d) get temperature Tout
e) continue until convergence
Heat Balance
Subtract mechanical energy balance from total energy balance to get
dh  vdp  q  F
Integrate to get the result (use averages as before)
h out  h in 
UTo  Tav D
G
2

1
G
1
L
f av 
 p out  p in   2   
 av
A  av
D
How is it done in simulators?
Pipe is divided in several "short" segments and either averaging is done, or
the inlet temperature is used.
SCENARIO II
Have turbine or Compressor/pump need Wo
Easy : use total energy with q = 0 and z = 0
dh  dw o 
dV 2
2
Joule/sec
wo
 V2 
W
 h    
G
 2 
Joule/g
g/sec

V2 
W  G  h  
2 

Positive for compressor
Negative for turbine
for turbines, this is about 0.
h is known for turbines but not for compressors.
Therefore we need to go back to the Mechanical Energy equation for
pumps/compressors. Indeed, the Bernoulli equation gives
neglect (?)

V 2 
W  G  vdp 
2 

Pumps ( is constant)
1
 vdp    p
d
 ps 
W
G
p

For compressors
pvn = constant (The evolution is nearly isentropic)
n = Cp/Cv (Ideal gas)
n  Cp/Cv (Real gas)
1
Substitute v  p s
n
 1
 vs   
 
1
n
integrate to get
n 1


n


 pd 
 n 
W  G
 1
 p s v s  
 n  1
 p s 


