3.6 Review 3.7 Moment of Momentum Equation (Conservation of ...

CEE 3310 – Control Volume Analysis 1 

3.6 Review 

• Conservation of Linear Momentum 

∫ 

∫ 

d 

⃗vρ d∀ + 

dt 

C.V. 

C.S. 

⃗vρ(⃗v · ⃗n) dA = ∑ ⃗ FC.V. 

3.7 Moment of Momentum Equation (Conservation 

of Angular Momentum) 

∑ ⃗F = m⃗a = 

d 

dt (m⃗v) 

Now, taking the moment of each side about some point A with position vector ⃗r 

And 

∑ 

⃗r × F ⃗ 

d = 

dt (⃗r × m⃗v) = d (⃗r × ρ∀⃗v) 

dt 

∑ 

(⃗r × ⃗ Fext ) = d dt 

∫ 

sys 

(⃗r × ⃗v)ρ d∀ 

Let H = ∫ (⃗r ×⃗v)ρd∀ be the moment-of-momentum of the system. Now we turn to the 

sys 

Reynolds Transport Theorem to find the conservation law for the moment-of-momentum 

with B = H and b = ⃗r × ⃗v 

dH sys 

dt 

= d [∫ 

] ∫ 

(⃗r × ⃗v)ρ d∀ + 

dt CV 

For a non-deformable inertial control volume we have 

∑ 

(⃗r × F ⃗ )ext = ∂ [∫ 

] ∫ 

(⃗r × ⃗v)ρ d∀ + 

∂t CV 

CS 

(⃗r × ⃗v)ρ(⃗v · ⃗n) dA 

CS 

(⃗r × ⃗v)ρ(⃗v · ⃗n) dA 

If we have 1-D inlets and outlets: 

∑ 

(⃗r × F ⃗ )ext = ∂ [∫ 

] 

(⃗r × ⃗v)ρ d∀ + ∑ (⃗r × ⃗v) out ṁ out − ∑ (⃗r × ⃗v) in ṁ in 

∂t CV 

Note that the (⃗v ·⃗n) term leads to a positive or negative flux of angular momentum, as we 

found with the conservation of mass and linear momentum. However, angular momentum

2 

itself has a sign (as does linear momentum), which arises from the ⃗r ×⃗v term so we must 

track two signs, which may cancel each other out. We note that our sign convention for 

the sign of angular momentum is the same as we used for torque, namely the right-handrule 

with a counter-clockwise sense yielding positive angular momentum and a clockwise 

sense yielding negative angular momentum. Hence the inward flux of clockwise angular 

momentum is positive and the outward flux of clockwise angular momentum is negative. 

3.8 Moment-of-Momentum Examples 

3.8.1 Example 1 – Single-Arm Lawn Sprinkler 

Consider the following single-arm lawn sprinkler 

What is the optimal coordinate system and control volume? 

A moving control volume ⇒ need absolute velocity, therefore 

V = velocity relative to nozzle + velocity of nozzle 

= Q A − ΩR


where A is the area of the pipe. Now, what is the solution? 

Ω = Q AR − T 0 

ρQR 2 

Aha! Even if T 0 , the retarding torque from friction, is zero Ω is finite. What is Ω when 

T 0 = 0? ⇒ Ω = V 0 /R, where V 0 is the nozzle exit velocity, why? 

velocity out of the nozzle is zero! 

⇒ The absolute 

3.8.2 Example 2 – Pipe Section and Bracket Torque 

Consider the following pipe section and supporting bracket: 

What is the reaction torque at the bracket wall mounting point (A)? You may assume the 

fluid in the pipe is constant density and the surrounding environment is at atmospheric 

pressure. 

How do we draw the control volume?

4 

What is the equation? 

What is ∑ (⃗r × ⃗ F ) CV ? 

What are the pressure forces? 

T A = h 2 (P 2 A 2 +ṁV 2 )−h 1 (P 1 A 1 +ṁV 1 ) or since ṁ = ρQ = ρA 1 V 1 = ρA 2 V 2 

T A = h 2 A 2 (P 2 +ρV 2 

2 )−h 1 A 1 (P 1 +ρV 2 

1 )


3.9 Review 

• Moment of momentum equation example problems 

3.10 The First Law of Thermodynamics and the Energy 

Equation 

Time rate of change of 

total system energy 

= 

Time rate of change by 

heat transfer 

+ 

Time rate of change by 

work transfer 

or 

( ) dE 

= 

dt 

˙Q + Ẇ 

sys 

where E is energy, Q is heat, and W is work. Recall that ( ˙ ) indicates time rate of 

change. 

Sign conventions: 

˙Q 

→ is the transfer by radiation, conduction, convection of heat. Transfer into the 

control volume is positive. 

W > 0 is work done on the system by the surroundings and W < 0 is work done by the 

system on the surroundings. 

Ẇ = Ẇshaft + Ẇpressure + Ẇviscous stress 

Note that gravitational work will enter our energy budget through potential energy. 

Ẇ shaft = T shaft Ω where T shaft is the torque and Ω is the angular velocity.

6 

3.10.1 Stress Induced Work per Time (Power) 

Work occurs by applying a force over a distance. Therefore pressure (normal stress) and 

shear (tangential stress) can produce work. If we look at the rate of work, or work per 

unit time, we have power and we see that we can think of this as applying a force on a 

system with a given velocity. 

Therefore for pressure we have 

δẆ = δ ⃗ F · ⃗v 

δẆpres = −P⃗n δA · ⃗v = −P⃗v · ⃗n δA 

Therefore 

∫ 

Ẇ pres = − P⃗v · ⃗n dA 

CS 

This term is usually moved to the right-hand-side flux term of the energy equation as it 

is a flux, which is how we will treat it. 

For shear stress we have 

δẆshear = ⃗τδA · ⃗v = ⃗τ · ⃗v δA 

Therefore 

∫ 

Ẇ shear = ⃗τ · ⃗v dA 

But shear is internally self-canceling. On the control surface ⃗v = 0 at solid boundaries 

(no-slip boundary condition), if the control surface is normal to the flow then ⃗τ ⊥ ⃗v and 

hence ⃗τ ·⃗v = 0. Hence it is often reasonable to assume that the shear stress induced work 

is small. 

What is a good environmental example of when this assumption breaks down? 

CS 

3.11 The Energy Equation 

With our definitions of work and energy we are now ready to apply the Reynolds 

Transport Theorem to produce the conservation of energy equation. Let B = E and


b = e = E/m, the energy per unit mass. We can decompose e into the following components 

e = ŭ + v2 

2 

+ gz + other 

where ŭ is the internal energy per unit mass, v 2 /2 is the kinetic energy per unit mass, 

and gz is the potential energy per unit mass. We will ignore other sources of internal 

energy but from the definition they are easily included. 

We can write the Reynolds Transport Theorem flux term (including the pressure work 

term) 

∫ 

CS 

(ŭ + v2 

2 + gz + P ) 

ρ(⃗v · ⃗n) dA 

ρ 

Therefore 

( ) dE 

= 

dt 

˙Q+Ẇshaft = d ∫ 

sys 

dt 

CV 

) ∫ 

(ŭ + v2 

2 + gz ρ d∀+ 

(ŭ + v2 

CS 2 + gz + P ) 

ρ(⃗v·⃗n) dA 

ρ 

The 1-D form is 

˙Q + Ẇshaft = d dt 

∫ 

CV 

(ŭ + v2 

2 + gz ) 

ρ d∀+ 

∑ 

outflows 

(ŭ + v2 

2 + gz + P ρ 

) 

ṁ − ∑ 

inflows 

We sometimes will choose to write (ŭ + P/ρ) = ˘h = enthalpy. 

(ŭ + v2 

2 + gz + P ) 

ṁ 

ρ 

3.11.1 Forms of the Energy Equation 

We often find it convenient to cast the energy equation in alternate forms. 

Velocity Squared form: 

If the flow is at steady state then conservation of mass gives us that ṁ out = ṁ in = ṁ 

and we can normalized all of our quantities by ṁ which leaves our homogeneous equation 

with terms having units of velocity squared 

˙Q 

ṁ + Ẇshaft 

ṁ 

= ŭ out + v2 out 

2 + gz out + P out 

ρ 

− ŭ in − v2 in 

2 − gz in − P in 

ρ

8 

Defining 

q = 

˙Q 

ṁ 

= 

heat transfer 

unit mass 

and w s = Ẇshaft 

ṁ 

shaft work 

= 

unit mass 

we can write the above as 

Head form: 

ŭ out + v2 out 

2 + gz out + P out 

ρ 

= ŭ in + v2 in 

2 + gz in + P in 

ρ + q + w s 

Manometers have historically lead to a desire to think of energy in units of length, or 

head. We see that we can arrive at units of length by dividing our velocity squared form 

of the energy equation by gravity. Hence we have 

ŭ out 

g 

+ v2 out 

2g + z out + P out 

γ 

= ŭin 

g + v2 in 

2g + z in + P in 

γ 

+ h q + h s 

where h q and h s are q/g and w s /g, respectively – the head forms of the heat transfer 

power and shaft power. 

3.12 Incompressible 1-D Flow With No Shaft Work 

Rearranging our 1-D head form and setting w s = 0 we have 

( ( P 

γ + v2 

P 

2g 

)out 

+ z = 

γ + v2 

2g 

)in 

+ z 

− ŭout − ŭ in − q 

g 

Now, defining 

ŭ out − ŭ in − q 

g 

= h loss = h f 

where we can think of h f as the friction losses and we see that h f > 0 (note that 

in adiabatic flow, say a perfectly insulated pipe, friction will heat the flow, therefore 

ŭ out > ŭ in and hence h f > 0. Thus we write 

( ( P 

γ + v2 

P 

2g 

)out 

+ z = 

γ + v2 

2g 

)in 

+ z − h f



• Work and Power 

• The Energy Equation 

˙Q + Ẇshaft = d dt 

∫ 

CV 

) ∫ 

(ŭ + v2 

2 + gz ρ d∀ + 

(ŭ + v2 

CS 2 + gz + P ) 

ρ(⃗v · ⃗n) dA 

ρ 

where ˙Q is the heat energy transfer rate, Ẇ shaft is the shaft power (work rate), ŭ 

is the internal energy per unit mass, v 2 /2 is the kinetic energy per unit mass, and 

gz is the potential energy per unit mass. 

3.13.1 Example – Gas Pipeline 

Consider the following pipe flow: 

If Q = 75 m 3 /s, the pipe radius is r = 6 cm, the inlet pressure is maintained at 24 atm 

by a pump, the outlet vents to the atmosphere, the pipe rises 150 m from inlet to outlet 

and the pipe length is 10 km, what is h f ? What is the velocity head? 

h f =198 m Therefore the friction loss is greater than the ∆z and the pump must drive 

against both!

10 

The velocity head is only 0.17 m! 

Note that the length did not come into our solution. h f includes the total losses along 

the pipe due to friction effects and hence includes the effect of length implicitly. We will 

see more about this a bit later in the course. 

3.14 Bernoulli Equation 

Bernoulli wrote down a verbal form of his famous equation in 1738 and Euler completed 

the analytic derivation in 1755. The differential form of the Bernoulli Equation is known 

as the Euler Equation. 

Consider our 1-D head form of the energy equation and let’s apply it along a streamline of 

a flow. If the flow is steady then the integral over the control volume vanishes. Further, 

since by definition there is no flow normal to the streamline we only have flux terms 

at the starting and ending points along the streamline (really we are talking about a 

volume and hence a streamtube – just a cylindrical volume element defined by a family 

of streamlines - a virtual pipe!). Clearly there is no shaft work along the streamline. If 

we assume the flow is frictionless (i.e., inviscid or ν = 0) h f =0 and we have 

( ( P 

γ + v2 

P 

2g 

)out 

+ z = 

γ + v2 

2g 

)in 

+ z 

This is the Bernoulli equation. Clearly anywhere along a streamline, as long as no work 

is done between analysis points and the assumption of frictionless flow is good, we can 

write 

P 

γ + v2 

2g + z = h 0


where the constant h 0 is referred to as the Bernoulli constant and varies across streamlines. 

The Bernoulli Equation can be derived by considering Newton’s second law F ⃗ = m⃗a 

along a streamline (conservation of linear momentum). This leads to the steady form of 

the Bernoulli equation. If we add conservation of mass we can derive the unsteady form. 

Bernoulli Equation Assumptions 

• Flow along single streamline ⇒ different streamlines, different h 0 . 

• Steady flow (can be generalized to unsteady flow). 

• Incompressible flow. 

• Inviscid or frictionless flow, very restrictive! 

• No w s between analysis points on streamline. 

• No q between points on streamline. 

3.14.1 Illustrations of Valid and Invalid Regions for the Application 

of the Bernoulli Equation

12 

3.15 Pressure form of Bernoulli Equation 

If we multiply our head form of the Bernoulli equation by the specific weight we arrive 

at the pressure form of the Bernoulli Equation: 

P + ρ v2 

2 + γz = P t 

where we call the first term the static pressure, the second the dynamic pressure, the third 

the hydrostatic pressure, and the right-hand-side the total pressure. Hence the Bernoulli 

Equation says that in inviscid flows the total pressure along a streamline is constant. 

If we remain at a constant elevation the above equation reduces to 

P + ρ v2 

2 = P s 

where we refer to P s as the stagnation pressure. Thus by definition the stagnation pressure 

is the pressure along horizontal streamlines when the velocity is zero. 

3.15.1 Stagnation Point and Pressure 

Consider the flow around a circular cylinder: 

P + ρ v2 

2 + γz = P t 

We see that the stagnation pressure is simply the conversion of all kinetic energy to 

potential energy and hence there is a subsequent pressure rise. The elevation head simply 

accounts for any change in the potential energy due to vertical changes in elevation.


3.16 Pitot-Static Tube 

The static and stagnation pressures can be measured simultaneously using a Pitot-static 

tube. Consider the following geometry: 

Now, we see the streamlines around the tube, either at the tip or away from the tip (but 

not around the curved front end), are horizontal. If the tube is not very long it is very 

reasonable to assume friction is negligible for this analysis. Now the velocity at the tip of 

the Pitot-static tube is zero hence the pressure at this point (and hence along the entire 

horizontal leg of the Pitot tube as this portion of the device is known) is the stagnation 

pressure. The holes perpendicular to the flow are similar to piezometers - they simply 

measure the static pressure in the fluid flow. If we write the equation for the Pitot tube 

we have 

P 1 + ρ v2 1 

2 = P 2 = γH 

Now, at the static tube we have the free-stream pressure, P 2o but at this point, which is 

along the streamline from point 1 to point 2, the velocity is the same as it is for point 

1 (assuming the Pitot-static tube is small and does not effect the flow) and there is no 

elevation change so the Bernoulli Equation gives us: 

P 1 = P 2o = γh 

Substituting this expression into the equation for the Pitot tube we arrive at the Pitot 

formula 

or in terms of heads 

V 1 = 

√ 

2 P 2 − P 1 

ρ 

V = √ 2g(H − h)

14 

3.16.1 Example – Flow accelerating out of a reservoir 

⎡ 

⎤ 

2gh 

V 2 = ⎢ ( ) 

⎣ 

2 

⎥ 

A2 ⎦ 

1 − 

A 1 

1 

2 

and if A 1 ≫ A 2 

1 − 

( 

A2 

A 1 

) 2 

≈ 1 

⇒ V 2 ≈ √ 2gh, again! 

Let’s look at this a bit further by asking the question what speed will a parcel of fluid 

dropped a distance h be traveling at? 

Therefore 

v = 

∫ t 

0 

g dt = gt ⇒! h = 

t = 

√ 

2h 

g 

and 

∫ t 

0 

v dt = 

∫ t 

v = gt = √ 2gt aha! 

0 

gt dt = 1 2 gt2 

Thus we see that in an inviscid flow, which by definition has no frictional energy losses, we 

simply convert potential energy to kinetic energy and hence the same result, v = √ 2gh 

keeps showing up. This was first noted by Torricelli.



• Head Form of the Energy Equation 

( ( P 

γ + v2 

P 

2g 

)out 

+ z = 

γ + v2 

2g 

)in 

+ z − h f 

where h f is the friction head loss. 

• A special case of the Energy Equation – The Bernoulli Equation 

P 

γ + v2 

2g + z = h 0 

Can think of it as the inviscid (frictionless) conservation of energy equation (i.e., 

h f = 0 = h s ). 

• Stagnation pressure – the sum of the static pressure and the dynamic pressure 

P s = P + ρ v2 

2 

3.18 Irrotational Flow and Bernoulli 

Consider 

P 

γ + v2 

2g + z = h 0 

0 + v2 

2g + 0 = h 0 = v2 

2g 

h 2 γ 

γ 

+ v2 

2g − h √ 

2 = h 0 

h 3 γ 

γ 

+ v2 

2g − h √ 

3 = h 0 

hγ 

γ + 0 − h = 0 ≠ h 0

16 

Notice that in the unsheared regions (uniform flow) h 0 = a constant across streamlines 

while where shear exists (e.g., shear is non-zero), h 0 varies across the streamlines. More 

strictly speaking we actually want to know if the flow is rotational. Our test is if we stick 

a small neutrally buoyant + shaped probe in the flow and see if it will rotate. In uniform 

flow it will not, in a linear shear, like the shear profile shown here, it will. Hence we 

say that h 0 is constant in irrotational (non rotational) flows. This allows us to connect 

Bernoulli points that are not on the same streamline in flows that are irrotational, further 

expanding the power of the Bernoulli equation but also the opportunities to misuse it! 

3.19 Energy and the Hydraulic Grade Line 

As we have seen we can write the head form of the Energy equation as 

P 

γ + v2 

+ z = H = Energy Grade Line (EGL) 

2g 

In the case of Bernoulli flows the energy grade line is simply a constant since by assumption 

energy is conserved (there is no mechanism to gain/lose energy). For other flows 

it will drop due to frictional losses or work done on the surroundings (e.g., a turbine) 

or increase due to work input (e.g., a pump). Note that this is the head that would be 

measured by a Pitot tube. 

We can also write 

P 

γ 

+ z = Hydraluic Grade Line (HGL) 

and we see that the HGL is due to static pressure – the height a column of fluid would 

rise due to pressure at a given elevation or in other words the head measured by a static 

pressure tap or the piezometric head.


Example – Venturi Flow Meter 

Consider 

⎡ 

⎤ 

2g∆h 

Q = A 2 V 2 = A 2 ⎢ ( ) 

⎣ 

2 

⎥ 

A2 ⎦ 

1 − 

A 1 

1 

2 

3.20 Variations From Uniform Flow 

As we have discussed we frequently assume that a flow is 1-D while we know in actuality 

it is not. Often this is an excellent assumption but sometimes the assumption is not as 

good and we may wish to correct for the effects of the dependence of the velocity on 

position. The term that is effected in the energy equation is the flux term. If we wish 

to use the average velocity, 〈V 〉, as representative of a 1-D velocity equivalent to the 2-D 

velocity then we have 

∫ 

CS 

v 2 

( 

2 ρ(⃗v · ⃗n) dA = ṁ αout 

2 〈V 〉2 out − α ) 

in 

2 〈V 〉2 in 

where α is known as the kinetic constant and it accounts for the effect of the non-uniform 

velocity profile on the surface flux of energy. The definition of the mean velocity is 

〈V 〉 = 

∫ 

ρ(⃗v · ⃗n) dA 

ρA 

which for incompressible flows with the velocity vector normal to the control surface 

reduces to simply 〈V 〉 = Q/A. Hence 

ṁ α out 

2 〈V 〉2 = 

∫ 

CS 

v 2 

ρ(⃗v · ⃗n) dA 

2

18 

and 

α = 

∫ 

CS v2 ρ(⃗v · ⃗n) dA 

ṁ 〈V 〉 2 

Example 

Consider the Laminar flow through a pipe sitting in a uniform velocity field in water: 

What is the head loss? 

h L = V 2 

2 

3.20.1 Frictional Effects 

If we have abrupt losses, say at a contraction, a simple way of accounting for this is 

through a discharge coefficient. We can write a modified form of Torricelli’s formula for 

incompressible flow 

〈V 〉 = Q A = C d√ 

2gh 

where C d is the discharge coefficient and is 1 for frictionless (inviscid) flow and can range 

down to about 0.6 for flows strongly effected by friction. Note we can handle non-uniform 

(violation of 1-D assumption) flow effects with a C d as well.


3.20.2 Vena Contracta Effect 

For a flow to get around a sharp corner there would need to be an infinite pressure 

gradient, which of course does not happen. Hence if the boundary changes directions too 

rapidly at an exit, the flow separates from the exit and forms what is known as a vena 

contracta 

Clearly A j /A ≤ 1. For a round sharp-cornered exit the coefficient is C c = A j /A = 0.61 

and typical values of the coefficient fall in the range 0.5 ≤ C c ≤ 1.0. 

3.21 Cavitation and the Bernoulli Equation 

Consider the following flow geometry: 

If we assume over such a short section friction is negligible then, given the constant z, 

the Bernoulli equation reduces to 

P 0 = P + ρ v2 

2 

Therefore high pressure occurs when the velocity is low and as the velocity increases the

20 

pressure drops. Plotting the variation of the dynamic and static pressure: 

we find that the pressure might fall below the vapor pressure of the fluid and hence 

cavitate. Thus in general we should be concerned about cavitation any time we are 

dealing with relatively high velocities.

3.6 Review 3.7 Moment of Momentum Equation (Conservation of ...

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