5.1-Rate of Flow, fluid mech
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Rate of Flow
5.1 Rate of Flow
In order to develop the Eulerian, or control volume approach, there is a need to be able to calculate the flow
rates through a control volume. Also, the capability to calculate flow rates is important in analyzing water supply
systems, natural gas distribution networks, and river flows. The equations for calculating flow rates are
developed in this section.
Discharge
The
discharge
,
Q
, often called the
volume flow rate
, is the volume of fluid that passes through an area per unit
time. For example, when filling the gas tank of an automobile, the discharge or volume flow rate would be the
gallons per minute flowing through the nozzle. Typical units for discharge are ft
3
/s (cfs), ft
3
/min (cfm), gpm,
m
3
/s, and L/s.
The discharge or volume flow rate in a pipe is related to the flow velocity and crosssectional area. Consider the
idealized flow of fluid in a pipe as shown in Fig. 5.3 in which the velocity is constant across the pipe section.
Suppose a marker is injected over the cross section at section
A
A
for a period of time
t
. The fluid that passes
A
A
in time
t
is represented by the marked volume. The length of the marked volume is
V
t
so the volume is
, where
A
is the crosssectional area of the pipe. The volume flow per unit time past
A
A
is
Taking the limit as
t
→ 0 gives
(5.1)
which will be referred to as the
discharge
or
volume flow rate equation
. It is important to realize that discharge
refers to a volume flow rate.
Figure 5.3
Volume of fluid in flow with uniform velocity distribution that passes section AA in time t.
The discharge given by Eq. (5.1) is based on a constant flow velocity over the crosssectional area. In general,
the velocity varies across the section such as shown in Fig. 5.4. The volume flow rate through a differential area
of the section is
V dA
, and the total volume flow rate is obtained by integration over the entire crosssection:
(5.2)
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Rate of Flow
Figure 5.4
Volume of fluid that passes section AA in time t.
In many problems—for example, those involving flow in pipes—one may know the discharge and need to find
the mean (average) velocity without knowing the actual velocity distribution across the pipe section. By
definition, the
mean velocity
is the discharge divided by the crosssectional area,
(5.3)
For laminar flows in circular pipes, the velocity profile is parabolic like the case shown in Fig. 5.4. In this case,
the mean velocity is half the centerline velocity. However, for turbulent pipe flow as shown in Fig. 4.7
b
, the
timeaveraged velocity profile is nearly uniformly distributed across the pipe, so the mean velocity is fairly close
to the velocity at the pipe center. It is customary to leave the bar off the velocity symbol and simply indicate the
mean velocity with
V
.
The volume flow rate equation can be generalized by using the concept of the dot product. In Fig. 5.5 the flow
velocity vector is not normal to the surface but is oriented at an angle θ with respect to the direction that is
normal to the surface. The only component of velocity that contributes to the flow through the differential area
dA
is the component normal to the area,
V
n
. The differential discharge through area
dA
is
If the vector,
d
A
, is defined with magnitude equal to the differential area,
dA
, and direction normal to the
surface, then
V
n
dA
= |
V
| cos θ
dA
=
V
·
d
A
where
V
·
d
A
is the dot product of the two vectors. Thus a more
general equation for the discharge or volume flow rate through a surface
A
is
(5.4)
If the velocity is constant over the area and the area is a planar surface, then the discharge is given as
If, in addition, the velocity and area vectors are aligned, then
which reverts to the original equation developed for discharge, Eq. (5.1).
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Rate of Flow
Figure 5.5
Velocity vector oriented at angle θ with respect to normal.
Mass Flow Rate
The
mass flow rate
, , is the mass of fluid passing through a crosssectional area per unit time. The common
units for mass flow rate are kg/s, lbm/s, and slugs/s. Using the same approach as for volume flow rate, the mass
of the fluid in the marked volume in Fig. 5.3 is
, where ρ is the average density. The
mass flow rate
equation
is
(5.5)
The generalized form of the mass flow equation corresponding to Eq. (5.4) is
(5.6)
where both the velocity and fluid density can vary over the crosssectional area. If the density is constant, then
Eq. (5.5) is recovered. Also if the velocity vector is aligned with the area vector, such as integrating over the
crosssectional area of a pipe, the mass flow equation reduces to
(5.7)
In summary, Eqs. 5.1 to 5.7 can be combined to create several useful formulas for volume flow rate (discharge):
(5.8)
Useful formulas for mass flow rate are:
(5.9)
The equations for discharge and mass flow rate are summarized in Table F.2.
Example 5.1 shows how to calculate the discharge and mean velocity using the mass flow rate, fluid density, and
pipe diameter.
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Rate of Flow
EXAMPLE 5.1 VOLUME FLOW RATE AD MEA
VELOCITY
Air that has a mass density of 1.24 kg/m
3
(0.00241 slugs/ft
3
) flows in a pipe with a diameter of 30 cm
(0.984 ft) at a mass rate of flow of 3 kg/s (0.206 slugs/s). What are the mean velocity and discharge in
this pipe for both systems of units?
PROBLEM DEFINITION
Situation:
Airflow in pipe with 30 cm diameter at 3 kg/s.
Find:
1.
Discharge (m
3
/s and ft
3
/s).
2. Mean velocity (m/s and ft/s).
Assumptions:
Properties are uniformly distributed across section.
Properties:
ρ = 1.24 kg/m
3
(0.00241 slugs/ft
3
).
PLAN
1. Find the volume flow rate using the volume flow rate equation, Eq. (5.5).
2. Calculate the mean velocity using Eq. (5.3).
SOLUTION
1. Discharge:
2. Mean velocity
Example 5.2 shows how to evaluate the discharge when the velocity vector is not normal to the cross section
area by using the dot product.
Example 5.3 illustrates how to evaluate the volume flow rate for a nonuniform velocity distribution by
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Rate of Flow
integration.
EXAMPLE 5.2 FLOW I SLOPIG CHAEL
Water flows in a channel that has a slope of 30°. If the velocity is assumed to be constant, 12 m/s, and
if a depth of 60 cm is measured along a vertical line, what is the discharge per meter of width of the
channel?
Sketch:
PROBLEM DEFINITION
Situation:
Channel slope of 30°. Velocity is 12 m/s and vertical depth is 60 cm.
Find:
Discharge per meter width (m
2
/s).
Assumptions:
Velocity is uniformly distributed across channel.
PLAN
Use Eq. (5.7) with area based on 1 meter width.
SOLUTION
REVIEW
The discharge per unit width is usually designated as
q
.
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