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Two-Phase Flow Momentum and Energy Balances Revisited.

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Dev. Chem. Eng. Mineral Process., I I(1/2), pp. 121-126, 2003.
Two-Phase Flow, Momentum and Energy
Balances Revisited
R.S. Raghunathan, P.L. Spedding” and R.K. Cooper
School of Aeronautical Engineering, The Queen’s‘University of Belfast,
Stranmillis Road, Berfast BT9 5AG, Nort~ernIreland, UK
The apparent differences between the momentum balance and energy balance
derivation for the case of two-phase flow are resolved by examination of the liquid
holdup within the system.
Introduction
There has been a long running discussion over the apparent differences between the
pressure, momentum and energy balances for two-phase flow. The problem does not
arise with single-phase flow where consistency has been demonstrated between the
various balances [ 1,2].
Pressure Drop in Two-phase Flow
The total pressure drop in two-phase flow can be written as,
(dP/dlh = (dP/dlf,
+ (dpldl), + (dP/dl),
(1)
where (dP/dl)* and (dP/dl),, are due to acceleration and hydrostatic head respectively
and are commonly termed the reversible components, while the frictional term
(dP/dl)F is the irreversible component of the pressure loss being mainly dissipated as
heat within the system. Methods of calculating the acceleration term have been given
by a number of workers [3-111 as:
* Authorfor correspondence.
R.S. Raghunathan, P.L. Spedding and R.K. Cooper
while experimental measurement of this quantity have been reported by several
workers [12-151. Generally in the case of adiabatic, steady state flow in a uniform
duct at low gas turbulent Reynolds numbers the acceleration component of pressure
drop is small and there is little difference between the results obtained by the various
experimental methods.
There exists a divergence of views about the absolute value of the hydrostatic
head term with some workers indicating coincidence of the balances while others
showed a significant difference depending on whether the derivation was based on the
energy or momentum balance.
For the energy balance:
while the momentum balance yields:
Equations (3) and (4) yield vastly different results since in general the liquid hoIdup,
RL,is much greater than the corresponding (1 p) input condition [16]. In fact, it is
the difference between these two parameters that lies at the centre of the enigma
regarding the energy and momentum balances with two phase flow.
The frictional pressure loss components (dP/dl)FErefers to the total mechanical
energy dissipated into thermal energy, while (dP/dl)FMrefers to the wall shear stress.
These two frictional terms are identical irrespective of the basis of their derivations.
-
Two-phase Momentum Balance
Using the deviation of Harrison [17, 181 yields:
-42w
d
+G,-[xV,
dl
+(I-x)VL]+gpTPsinO
which agrees with Equation (2), and other derivations reported [5-11, 19-22].
122
Two-Phase Flow,Momentum and Energy Balances Revisited
Two-phase Energy Balance
By adding the pressure, potential, kinetic and internal energies Harrison [17, 181
derived:
dq
dl
dw - d [H,x+ H, (I - x)]+ d [-+xV:
dl dl
dl 2
(1-x)Vt ]+gsin,
where the enthalpy is given by:
dH =dq + dF+ dP/ p
(7)
and dF is the energy degraded into heat for an incompressible fluid or into both heat
and pressure energy in the case of a compressible fluid. Substitution of Equation (7),
taking dw = 0 and the pressure drop being equal in both phases as given by:
The terms in Equation (8) correspond to the frictional energy loss, the acceleration
and hydrostatic head. This latter will now be examined in detail.
Hydrostatic Head and Potential Energy
Many workers have used the two-phase density of Equation (4) in the potential energy
term with the energy balance [6, 8-1 1, 21, 23, 241. However, others have used the
homogeneous density of Equation (3) [25 - 281. As has been pointed out already the
effects of this variation in the choice of density are not trivial.
Within the energy balance the potential energy or head loss term represents the
energy necessary to bring the inlet flow of the phases to the elevation at the outlet of
the duct as shown in Figure 1. However also shown on this figure, there is an excess
of liquid holdup in the body of the duct which acts like a ladder allowing the outlet
flow of the liquid to take place. If it were not present then liquid outflow would not
occur. This excess of liquid regurgitates continuously within the body of the duct
adding both to the pressure energy and the degraded energy, i.e. the internal energy of
the whole system. Hence if the homogeneous density of Equation (3) is used to
calculate the potential energy, the effect of the liquid holdup in the conduit must also
be considered by increasing the pressure and internal energies of the system
accordingly in order to account for the energy needed to support the liquid
regurgitating within the duct. A simpler, but approximate, way is to use the twophase density of Equation (4) in the potential energy calculation.
123
R.S. Raghunathan, P.L. Spedding and R.K. Cooper
GAS FLOW
LIQUID FLOW
LIQUID HELD IN
Figure 1. Showing diagrammatically the diflerence between delivered liquid and gas
and in situ liquid and gas holdup.
Conclusions
The momentum balance and energy balance derivations for two-phase gas-liquid flow
are shown to give identical results. The potential energy calculations in the energy
balance reflect the energy expanded to raise the flowing fluids to the outlet elevation.
However in order to achieve this outflow, particularly of the liquid phase, an amount
of liquid must be maintained or held-up continuously regurgitating within the duct.
The energy necessary for the support of this internal movement of liquid will be
drawn from both the pressure energy and internal energy of the system. The
pragmatic way of calculating the resultant energy output is to use the two-phase
density in the potential energy calculation of the energy balance.
124
Two-Phase Flow, Momentum and Energy Balances Revisited
Nomenclature
D
F
g
G
1
P
9
Q
R
X
V
W
W
P
e
Diameter (m)
Energy degraded (J kgl)
Gravity (m ss2)
Mass flow rate (kg m-* s-I)
Length (m)
Pressure (kg m-1s-2)
Heat added (J kg-')
Volume flow (m3 s")
Holdup
Mass input ratio (WG/WT)
Velocity (m s-l)
Net work done by fluid (J kg-')
Mass flow (kg s-')
Volume input ratio (QG/QT)
Angle to horiztonal (degrees)
p
T
Density (kg m")
Shear stress (kg rn-' s - ~ )
Subscripts
Acceleration
E
Energy
F
Friction
G
Gas
H
Hydrostatic head
L
Liquid
M
Momentum
T
Total
T P Two phase
W
Wall
A
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5.
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R.S. Raghunathan, P.L. Spedding and R.K. Cooper
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126
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