Ch 4.
Continuity, Energy, and Momentum Equation
Chapter 4 Continuity, Energy, and Momentum Equations
4.1 Conservation of Matter in Homogeneous Fluids
• Conservation of matter in homogeneous (single species) fluid → continuity equation
positive
dA1 - perpendicular
to the control
surface
negative
4.1.1 Finite control volume method-arbitrary control volume
- Although control volume remains fixed, mass of fluid originally enclosed (regions A+B)
occupies the volume within the dashed line (regions B+C).
Since mass m is conserved:
  mB t   mB t  dt   mC t  dt
 m A t
 mB t dt
  mB t
dt

 m A t
  mC t  dt
dt
(4.1)
(4.2)
•LHS of Eq. (4.2) = time rate of change of mass in the original control volume in the limit
 mB t dt
  mB  t
dt

  mB 


  dV 
t
t CV
4−1
(4.3)
Ch 4.
Continuity, Energy, and Momentum Equation
where dV = volume element
•RHS of Eq. (4.2)
= net flux of matter through the control surface
= flux in – flux out
=
 q
n
dA1 
 q
n
dA2

where qn = component of velocity vector normal to the surface of CV  q cos 


  dV  
t CV

CS
 qn dA1 

CS
 qn dA2
(4.4)
※ Flux (= mass/time) is due to velocity of the flow.
• Vector form

 



q

dV


 CS  dA
t CV

(4.5)
where dA = vector differential area pointing in the outward direction over an enclosed
control surface
 
 
 q  dA  q dA cos 
 positive for an outflow from cv,   90
 


 negative for inflow into cv, 90    180
If fluid continues to occupy the entire control volume at subsequent times
→ time independent
4−2
Ch 4. Continuity, Energy, and Momentum Equation

  dV  
t CV
LHS:

CV t dV
(4.5a)
Eq. (4.4) becomes

 

dV

q
CV t
 CS  dA  0
(4.6)
→ General form of continuity equation - integral form
[Re] Differential form
Use Gauss divergence theorem
F
V xi dV 

A
FdAi
Transform surface integral of Eq. (4.6) into volume integral

CS



 q  dA       q  dV
CV
Then, Eq. (4.6) becomes
 
 




q

 dV  0
CV  t



(4.6a)
Eq. (4.6a) holds for any volume only if the integrand vanishes at every point.


  q   0
t
(4.6b)
→ Differential form
◈ Simplified form of continuity equation
(1) For a steady flow of a compressible fluid
4−3
Ch 4.
Continuity, Energy, and Momentum Equation
V

CV t dv  0
Therefore, Eq. (4.6) becomes

CS


 q  dA  0
(4.7)
(2) For incompressible fluid (for both steady and unsteady conditions)
  const. 

d
 0,
 0
dt
t
Therefore, Eq. (4.6) becomes

CS
 
q  dA  0
(4.8)
[Cf] Non-homogeneous fluid mixture
→ conservation of mass equations for the individual species
→ advection - diffusion equation
= conservation of mass equation
q
 c


 0

x
 t



+ mass flux equation due to advection and diffusion  q  uc  D
c
 
c 

 uc  D   0
t
x 
x 

c
uc
  c 


D 
t
x
x  x 
4−4
c 

x 
Ch 4.
Continuity, Energy, and Momentum Equation
4.1.2 Stream - tube control volume analysis for steady flow
There is no flow across the longitudinal boundary
∴ Eq. (4.7) becomes


  q  dA   1q1dA1   2 q2 dA2  0
(4.9)
 qdA  const.
If density = const.
q1dA1  q2 dA2  dQ
(4.10)
where dQ = volume rate of flow
(1) For flow in conduit with variable density
V 
 qdA
A
→ average velocity
4−5
Ch 4.
 
  dQ
Q
Continuity, Energy, and Momentum Equation
→ average density
1V1 A1   2 V2 A2
(4.11)
(2) For a branching conduit


  q  dA  0
  1q1dA1 
A1

A2
 2 q2 dA2 

A3
 3q3dA3  0
1V1 A1   2 V2 A2  3 V3 A3
4−6
Ch 4.
Continuity, Energy, and Momentum Equation
[Appendix 4.1] Equation of Continuity
→ Infinitesimal (differential) control volume method
At the centroid of the control volume:
 , u, v, w
Rate of mass flux across the surface perpendicular to x is


   u  dx 
 dydz
x 2 


   u  dx 
 dydz
x 2 
flux in    u 
flux out    u 
net flux = flux in  flux out  
  u 
dxdydz
x
net mass flux across the surface perpendicular to y  
  v 
dydxdz
y
net mass flux across the surface perpendicular to z  
   w
dzdxdy
z
Time rate of change of mass inside the c.v. =
4−7
   dxdydz 
t
Ch 4.
Continuity, Energy, and Momentum Equation
Time rate of change of mass inside = The sum of three net rates
   dxdydz 
 v
   w 
   u
 


 dxdydz
t



x
y
z


By taking limit dV  dxdydz

 u
 v
   w






    q  div   q 
t
x
y
z


   q  0
t
(A1)
→ general point (differential) form of Continuity Equation
[Re]
  u
 v
   w




    q  div   q 
x
y
z
By the way,
 

  q  q    q
Thus, (A1) becomes



 q     q  0
t
(A2)
1) For incompressible fluid
d
 0 (   const.)
dt

 
d
 q  
 0
t
dt
4−8
Ch 4.
Continuity, Energy, and Momentum Equation
Therefore Eq. (A2) becomes

  q  0


 q  0
(A3)
In scalar form,
u
v
w


 0
x
y
z
(A4)
→ Continuity Eq. for 3D incompressible fluid
For 2D incompressible fluid,
u
v

 0
x
y
2) For steady flow,

 0
t
Thus, (A1) becomes
 

   q  q    q  0
4−9
Ch 4.
Continuity, Energy, and Momentum Equation
4.2 The General Energy Equation
4.2.1 The 1st law of thermodynamics
The 1st law of thermodynamics: combine continuity and conservation of energy → energy
equation
– property of a system: location, velocity, pressure, temperature, mass, volume
– state of a system: condition as identified through properties of the system
The difference between the heat added to a system of masses and the work done by the
system depends only on the initial and final states of the system (→ change in energy).
→ Conservation of energy
Q
W
E
 Q   W  dE
where
(4.14)
 Q = heat added to the system from surroundings
 W = work done by the system on its surroundings
dE = increase in energy of the system
Consider time rate of change
Q
dt

W
dt

dE
dt
(4.15)
(1) Work
Wpressure = work of normal stresses acting on the system boundary
Wshear
= work of tangential stresses done at the system boundary
on adjacent external fluid in motion
4−10
Ch 4.
Continuity, Energy, and Momentum Equation
= shaft work done on a rotating element in the system
Wshaft
(2) Energy
Consider e = energy per unit mass = E
mass
eu = internal energy associated with fluid temperature = u
e p = potential energy per unit mass = gh
where h = local elevation of the fluid
q2
eq = kinetic energy per unit mass =
2
u 
p

 enthalpy
e  eu  e p  eq  u  gh 
•
q2
2
(4.16)
Internal energy
= activity of the molecules comprising the substance
= force existing between the molecules
~ depend on temperature and change in phase
4.2.2 General energy equation
Q
dt

W
dt

dE
dt
(4.15)
4−11
Ch 4.
Continuity, Energy, and Momentum Equation
Consider work done
W

dt
 W pressure
dt

 Wshaft
dt

 Wshear
dt
(4.15a)
outward normal
vector

d A2

q2
 W pressure
dt
= net rate at which work of pressure is done by the fluid on the surroundings
= work fluxout – work fluxin
=

CS
 
p  q  dA
p = pressure acting on the surroundings = F A  F / L2
 Positive for outflow into CV
 
q  dA  

 Negative for inflow
 
q  dA  Q  L3 / t
4−12
Ch 4.
Continuity, Energy, and Momentum Equation
F L3
 
p  q  dA  2
 FL / t  E / t
L t
Thus, (4.15a) becomes
W
dt

   Wshaft  Wshear
p
q

 CS  dA  dt  dt
(4.15b)
Consider energy change
dE
= total rate change of stored energy
dt
= net rate of energy flux through C.V.
(energy @ t  t  t  energy @ t  t )
+ time rate of change inside C.V.
=

 

e

 e dV 
q
dA



 CS
t CV
(4.15c)
 
e  E / mass;   q  dA  mass / time
 
e  q  dA  E / t
Substituting (4.15b) and (4.15c) into Eq. (4.15) yields
Q
dt


 Wshaft
dt

 Wshear
dt
  
p
 
 dA 

q
  CS

 
e


 e dV 


q

dA
 CS
t CV
Q
dt

 Wshaft
dt

 Wshear
dt
4−13
Ch 4.

Continuity, Energy, and Momentum Equation

p

 

e

q

CS     dA  t CV  e dV 
(4.17)
Assume potential energy e p  gh (due to gravitational field of the earth)
q2
Then e  u  gh 
2
Then, Eq. (4.17) becomes
Q
dt

 Wshaft
dt

 Wshear
dt
p

q2   
 

e dV




u
gh



q
dA

 CS  
t CV
2
◈ Application: generalized apparatus
At boundaries normal to flow lines → no shear
→ Wshear  0
4−14
(4.17)
Ch 4.
Q

dt
 Wshaft
dt
Continuity, Energy, and Momentum Equation
p

q2   
 
 CS    u  gh  2    q  dA  t CV e dV
(4.19)
For steady motion,
Q

dt
 Wshaft
p
q2   
 
 CS    u  gh  2    q  dA
dt
(4.20)
◈ Effect of friction
~ This effect is accounted for implicitly.
~ This results in a degradation of mechanical energy into heat which may be transferred away
( Q , heat transfer), or may cause a temperature change
→ modification of internal energy.
→ Thus, Eq. (4.20) can be applied to both viscous fluids and non-viscous fluids (ideal
frictionless processes).
4.2.3 1 D Steady flow equations
For flow through conduits, properties are uniform normal to the flow direction.
→ one - dimensional flow
1
2
V1
V2
Integrated form of Eq. (4.20) = ② - ①
Q
dt

 Wshaft
dt


p
V2
p
V2
 u   gh 
 Q  u   gh    Q
2  ②
2 ①




4−15
Ch 4. Continuity, Energy, and Momentum Equation
where
Divide by
V2
 average kinetic energy per unit mass
2


Section 1:
1   q  dA   Q  mass flow rate into CV
Section 2:
2   q  dA  Q  mass flow from CV


 Q (mass/time)
Wshaft

heat transfer
p
V2

 u   gh 
 
mass
mass

2

②

p
V2
u    gh  2 

①
Divide by g
Wshaft
u
u
heat transfer
p
V2 
p
V2 

   h
   g    h  2g 
weight
weight
g
g

2

② 
①
(4.21)
◈ Energy Equation for 1-D steady flow: Eq. (4.21)
~ use average values for p ,  , h , u and V at each flow section
~ use K e (energy correction coeff.) to account for non-uniform velocity distribution over
flow cross section


1 mV 2
K e V Q   q dQ ---- kinetic energy/time 
2
2
2 t
Ke 
2

2

2
 2 q dQ
2
2
1
V Q
4−16
Ch 4.
Continuity, Energy, and Momentum Equation
Wshaft
p
p
heat transfer
u  u1
V2 
V2 

   h  Ke
   h  Ke
 2


weight
weight
g
2 g ②
2 g ①


(4.23)
Ke =
2, for laminar flow (parabolic velocity distribution)
1.06, for turbulent flow (smooth pipe)
For a fluid of uniform density

Wshaft
V12
p2
V2 2
heat transfer
u  u1
 h1  K e1

 h2  K e2


 2
weight
weight
g

2g

2g
p1
(4.24)
→ unit:
m (energy per unit weight)
For viscous fluid;

heat transfer
u  u1
 2
 H L12
weight
g
→ loss of mechanical energy
~ irreversible in liquid
Then, Eq. (4.24) becomes
p1

 h1  K e1
V12
p
V2
 2  h2  K e2 2  H M  H L12
2g

2g
(4.24a)
where H M = shaft work transmitted from the system to the outside
H1  H 2  H M  H L12
(4.24b)
where H1 , H 2 = weight flow rate average values of total head
4−17
Ch 4.
Continuity, Energy, and Momentum Equation
◈ Bernoulli Equation
Assume
①
ideal fluid → friction losses are negligible
② no shaft work → H M  0
③ no heat transfer and internal energy is constant → H L12  0
V12
p2
V2 2
 h1  K e1

 h2  K e2

2g

2g
p1
H1  H 2
(4.25)
Potential head
Pressure head
If K e1  K e2  1 , then Eq. (4.25) reduces to
V12
p2
V2 2
 h1 

 h2 
H 

2g

2g
p1
~ total head along a conduct is constant
◈ Grade lines
4−18
Velocity head
(4.26)
Ch 4.
Continuity, Energy, and Momentum Equation
1) Energy (total head) line (E.L) ~ H above datum
2) Hydraulic (piezometric head) grade line (H.G.L.)
~
p

   h  above datum


For flow through a pipe with a constant diameter
V1  V2

V12
V2
 2
2g
2g
1) If the fluid is real (viscous fluid) and if no energy is being added, then the energy line may
never be horizontal or slope upward in the direction of flow.
2) Vertical drop in energy line represents the head loss or energy dissipation.
4−19
Ch 4.
Continuity, Energy, and Momentum Equation
4.3 Linear Momentum Equation for Finite Control Volumes
4.3.1 Momentum Principle
Newton's 2nd law of motion




dq d  mq  dM

F  ma  m


dt
dt
dt


M  linear momentum vector = mq
(4.27)

F  external force

 boundary (surface) forces: normal to boundary - pressure, Fp

tangential to boundary - shear, Fs

body forces - force due to gravitational or magnetic fields, Fb
External forces cause momentum change




dM
Fp  Fs  Fb 
dt

Fb 

CV
(4.28)
fb   dv  , where fb = body force per unit mass
4.3.2 The general linear momentum equation

dA1

dA2
positive
negative
4−20
Ch 4.
Continuity, Energy, and Momentum Equation
RHS of Eq. (4.28)
Consider change of momentum

dM
= total rate of change of momentum
dt
momentum@t2-momentum@t1
= net momentum flux across the CV boundaries
+ time rate of increase of momentum within CV


  

q
 CS   q  dA  t CV q  dV



where q  q  dA

(4.29)
= flux of momentum = velocity  mass per time

dA = vector unit area pointing outward over the control surface
Substitute (4.29) into (4.28)



Fp  Fs  Fb 

  

q


q  dV



q
dA

 CS
t CV
(4.30)
For steady flow and negligible body forces


Fp  Fs 

CS
  
q   q  dA
(4.30a)
• Eq. (4.30)
1) It is applicable to both ideal fluid systems and viscous fluid systems involving friction and
energy dissipation.
2) It is applicable to both compressible fluid and incompressible fluid.
3) Combined effects of friction, energy loss, and heat transfer appear implicitly in the
magnitude of the external forces, with corresponding effects on the local flow velocities.
• Eq. (4.30a)
1) Knowledge of the internal conditions is not necessary.
2) We can consider only external conditions.
4−21
Ch 4.
Continuity, Energy, and Momentum Equation
4.3.3 Inertial control volume for a generalized apparatus
• Three components of the forces



x  dir. : Fpx  Fsx  Fbx 

 
u

 dA  CV u  dV

q
 CS
t



  
y  dir. : Fp y  Fs y  Fby  
v

 dA  CV v  dV

q
 CS
t



  
z  dir. : Fpz  Fsz  Fbz  
 CS w  q  dA  t CV w dV
• For flow through generalized apparatus



x  dir. : Fpx  Fsx  Fbx   u  dQ 
2
For 1 - D steady flow,
 u dQ 
1

u  dV
t CV

q dV  0
t CV
~ Velocity and density are constant normal to the flow direction.



x  dir.: Fpx  Fsx  Fbx   Fx  Vx  Q 2  Vx  Q 1
4−22
(4.32)
Ch 4.
Continuity, Energy, and Momentum Equation
 Vx2  2 Q2  Vx1 1 Q1  Q  Vx2  Vx1   Q  Vxout  Vxin 
y  dir.:
 F  V  Q   V  Q 
z  dir.:
 F  V  Q   V  Q 
y
z
y
z
y
2
z
2
1 Q1   2 Q2  Q  (4.12)
1
1
(4.33)
where V = average velocity in flow direction
If velocity varies over the cross section, then introduce momentum flux coefficient

V

q

  
q


K
V

q
dA


m  VA 



q

dQ

K
V
m Q


Km
 q dQ
 
V Q
where
V = magnitude of average velocity over cross section = Q A

V = average velocity vector
K m = momentum flux coefficient  1
=
1.33 for laminar flow (pipe flow)
1.03-1.04 for turbulent flow (smooth pipe)
4−23
Ch 4.
Continuity, Energy, and Momentum Equation
[Cf] Energy correction coeff.

Ke 
 2 q dQ
2
 
2
VQ
F
  K mVx  Q 2   K mVx  Q 1
F
  K mVy  Q 2   K mVy  Q 1
F
  K mVz  Q 2   K mVz Q 1
x
y
z
(4.34)
[Example 4-4] Continuity, energy, and linear momentum with unsteady flow
A1  20 ft 2 ,
A2  0.1 ft 2 , h0  16
At time t  0 valve on the discharge nozzle is opened. Determine depth h , discharge rate
Q , and force F necessary to keep the tank stationary at t  50 sec.
4−24
Ch 4.
Continuity, Energy, and Momentum Equation
i) Continuity equation, Eq. (4.4)

 dV    qn dA1    qn dA2
t CV
 qn dA1  0 (because no inflow across the section (1))
dV  A1dh ,
  A1
A1
 h
dh   V2 A2
t 0
dh
  V2 A2
dt
(A)
ii) Energy equation, Eq. (4.19)
~ no shaft work
~ heat transfer and temperature changes due to friction are negligible
Q
dt

 Wshaft
dt

 Wshear
dt
p

q2   

 CS    u  gh  2    q  dA  t CV e dV
I
II
q2
e = energy per unit mass = u  gh 
2

p
q2   
I= 
 CS  u    gh  2    q  dA


p
q2 
p
q2 
  u   gh   V2 A2   u   gh   V1 A1
2 2
2 1





p
q2 
  u   gh   V2 A2

2 2

4−25
V1  0 
Ch 4.
II =
Continuity, Energy, and Momentum Equation



q2 


e
dV
 u  gh 
  dV
t CV
t CV 
2 
A1 dh
∵ nearly constant in the
tank except near the nozzle
 A1 
 h
 u  gh  dh
t 0

 h
p
q2 
V
A
A
 0   u   gh 



 u  gh  dh
2 2
1


0
t


2

2
Assume
  const. , p2  patm  0 , h2  0 (datum)
V22
dh
dh
0  uV2 A2 
V2 A2  uA1
 A1 gh
dt
dt
2
(B)
Substitute (A) into (B)
V22
0  uV2 A2 
V2 A2  u  V2 A2   gh  V2 A2 
2

V22
V2 A2  ghV2 A2
2
V2 
2 gh
(C)
Substitute (C) into (A)
A2 2 gh   A1
dh
dt
dh
A
  2 2 gdt
A1
h
4−26
Ch 4. Continuity, Energy, and Momentum Equation
Integrate

h
h0
dh

h
 12
A
h   h0  2
A1

2g
2

h   16  0.1
20

  4  0.0201t 
At
t  50sec ,
V2 
 h  12
1 h 



h
dh
 h
 2h 2  h0 
0


A
0  A12 2 gdt
t

t

2
2  32.2  
t
2

2
2
h   4  0.0201 50   8.98 ft
2
2 gh 
2  32.2  8.98   24.05 fps
Q2  VA 2  24.05  0.1  2.405 cfs
iii) Momentum equation, Eq. (4.30)



Fp  Fs  Fb 
I
II

  

q


q
 dV
q

dA


 CS
t CV
II = Time rate of change of momentum inside CV is negligible
if tank area
I=

CS
 A1  is large compared to the nozzle area  A2  .
  
q   q  dA   qn  qn dA2   qn  qn dA1  V2 V2 A2
 Fpx  V2 V2 A2  V2  Q2
Fpx  1.94  24.05  2.405   112 lb
4−27
Ch 4. Continuity, Energy, and Momentum Equation
4.4 The Moment of Momentum Equation for Finite Control Volumes
4.4.1 The Moment of momentum principle for inertial reference systems
Apply Newton's 2nd law to rotating fluid masses
→ The vector sum

 r  F 
of all the external moments acting on a fluid mass equals the
time rate of change of the moment of momentum (angular momentum) vector

 r  M 
the fluid mass.
d  
 
(4.35)
T  r  F  r M 
dt

where r = position vector of a mass in an arbitrary curvilinear motion

M = linear momentum
[Re] Derivation of (4.35)
Eq. (4.27):


dM
F 
dt

Take the vector cross product of r
4−28
of
Ch 4. Continuity, Energy, and Momentum Equation

 
 dM
rF  r
dt
By the way,

 
d  
dr
 dM
r M    M  r 
dt
dt
dt
I
 
dr


I 
 M  q  mass  q  0
dt

 
 
q  q  q q sin 0  0 


d  
  dM 

r M 

r


dt  dt



 dr  
 q

 dt

d  
 
r  F  r M 
dt


where r  M = angular momentum (moment of momentum)
[Re] Torque

 
T  rF
translational motion → Force – linear acceleration
rotational motion
→ Torque – angular acceleration
[Re] Vector Product

 
V  ab



Magnitude = V  a  b sin  = area of parallelogram


direction = perpendicular to plane of a and b → right-handed triple
4−29
Ch 4. Continuity, Energy, and Momentum Equation
 
 
b  a   a  b 


 ka   b  k  a  b 
 

 
 
a  b  c    a  b    a  b 
• External moments arise from external forces
Forces
Moment
Boundary (surface) force
 
Fp , Fs
 
Tp , Ts
Body force

Fb

Tb




 r  F    r  F    r  F   dtd  r  M 
p

Tp
s
b

Tb

Ts




d 
Tp  Ts  Tb 
r  M 
dt


(4.36)

where Tp , Ts , Tb = external torque
4.4.2 The general moment of momentum equation

dM

dt

Eq. (4.30)

  

q
 CS   q  dA  t CV q  dV
d  
r  M  
dt
 



r
q



q
 dA 

 CS



T  T  T 
p
s
b


 
r
 q   dV

t CV

 

 CS  r  q    q  dA
4−30


 
r  q   dV


t CV
(4.37)
Ch 4. Continuity, Energy, and Momentum Equation
x  dir. :


 r  q  yz

 ryz q yz sin    yz    rq cos  
2

angle between q yz and ryz



 Tpx  Tsx  Tbx 


 CS  rq cos  yz   q  dA

 rq cos   yz  dV
t CV

(4.39a)



y  dir. : Tpy  Tsy  Tby 

  rq cos    q  dA 



z  dir. : Tpz  Tsz  Tbz 

 CS  rq cos  xy   q  dA

zx
CS



 rq cos   zx  dV
t CV

 rq cos   xy  dV
t CV
4.4.3 Steady-flow equations for Turbomachinery
■ Turbomachine
~ pumps, turbines, fans, blows, compressors
~ dynamic reaction occurs between a rotating vaned element (runner) and fluid passing
through the element
If rotating speed is constant → Fig. 4.14
Tr 
Where r1 , r2

r V cos  2  dQ 
A2 2 2

rV cos 1  dQ
A1 1 1
– Euler's Eq.
= radii of fluid elements at the entrance and exist of the runner
V1 , V2 = velocities relative to the fixed reference frame (absolute)
1 ,  2 = angles of absolute velocities with tangential direction
A1 , A2 = entrance and exit area of CV which encloses the runner
4−31
(4.40)
Ch 4. Continuity, Energy, and Momentum Equation
Tr
• Tr
= torque exerted on fluid by the runner
positive for pump, compressor
negative for turbine
If r2V2 cos  2 and rV
1 1 cos 1 are constant,
 is constant
Tr   Q  r2V2 cos  2  rV
1 1 cos 1 
If r2V2 cos  2 = rV
1 1 cos 1
 QrV cos   Qrv  const.
~ zero torque, constant angular momentum
If
  0 → free vortex flow
4−32
(4.41)
Ch 4. Continuity, Energy, and Momentum Equation
Homework Assignment # 2
Due: 1 week from today
4-11. Derive the equation for the volume rate of flow per unit width for the sluice gate
shown in Fig. 4-20 in terms of the geometric variable b , y1 , and cc . Assume the pressure
in hydrostatic at y1 and ccb and the velocity is constant over the depth at each of these
sections.
4-12. Derive the expression for the total force per unit width exerted by the above sluice gate
on the fluid in terms of vertical distances shown in Fig. 4-20.
4-14. Consider the flow of an incompressible fluid through the Venturi meter shown in Fig. 422. Assuming uniform flow at sections (1) and (2) neglecting all losses, find the pressure
difference between these sections as a function of the flow rate Q , the diameters of the
sections, and the density of the fluid,  . Note that for a given configuration, Q is a
function of only the pressure drop and fluid density. The meter is named for Venturi, who
investigated its principle in about 1791. However, in 1886 Clemens Herschel first used the
meter to measure discharge, and he is usually credited with its invention.
4−33
Ch 4. Continuity, Energy, and Momentum Equation
4-15. Water flows into a tank from a supply line and out of the tank through a horizontal
pipe as shown in Fig. 4-23. The rates of inflow and outflow are the same, and the water
surface in the tank remains a distance h above the discharge pipe centerline. All velocities
in the tank are negligible compared to those in the pipe. The head loss between the tank and
the pipe exit is H L . (a) Find the discharge Q in terms of h , A , and H L . (b) What is the
horizontal force, Fx , required to keep the tank from moving?
(c) If the supply line has an
area A' , what is the vertical force exerted on the water in the tank by the vertical jet?
4-28. Derive the one-dimensional continuity equation for the unsteady, nonuniform flow of
an incompressible liquid in a horizontal open channel as shown in Fig. 4-29. The channel
has a rectangular cross section of a constant width, b. Both the depth, y0 and the mean
4−34
Ch 4. Continuity, Energy, and Momentum Equation
velocity, V are functions of x and t.
4−35