Engineering Tripos 1B

Paper 4

Fluid Mechanics

Lecture 6 - One dimensional pipe flow

• Static pressure and stagnation pressure

• Stagnation pressure loss across an orifice plate

• Stagnation pressure loss along a pipe

• Stagnation pressure changes across pipe components

• Stagnation pressure and mechanical work

• Network analysis

Internal Flow Systems by D. S. Miller is an excellent source of practical informationon internal flow (ISBN 0-947711-77-5 and classmark TA 379/ TA 293 in the CUEDlibrary)

Matthew Juniper mpj1001@eng.cam.ac.uk

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6.1 Static pressure and stagnation pressure

In 1A you met static and stagnation pressures. At a point in a moving fluid thestatic pressure, p, is the pressure measured by a probe if it does not change thevelocity of the flow, e.g. at point A on the diagram below. The stagnation pressure,p0, is that measured by a probe that lets the flow come to rest without loss of me-chanical energy, e.g. at point B on the pitot tube (there will be more detail on thisin lecture 10). The two pressures are related by Bernoulli’s equation.

The stagnation pressure is the pressure measured at a stagnation point. Some booksadd the term ρgh because this gives the Bernoulli constant, which is conserved ifthere is no loss of mechanical energy in the flow. This, however, causes confusionbecause p + ρv2/2 + ρgh is not the pressure measured at a stagnation point. In anetwork of pipes we find it useful to follow the stagnation pressure, p+ρv2/2, ratherlike following the voltage in an electrical network. When divided by ρg, this is alsoknown as the head. The stagnation pressure drops as a fluid flows through pipes,bends, orifice plates and other components. It rises as it goes through pumps. Atsufficiently high Reynolds number the stagnation pressure loss in each componentis proportional to ρV 2/2.

In a network, I find it easiest to evaluate the stagnation pressure change across eachcomponent, ignoring gravity, and then compare this with the required network gain,including gravity. In the above example, this means that the slope of the pipe doesnot matter. Only the difference between initial and final water levels matters.

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6.2 Example - stagnation pressure loss across an orifice plate

In the pipe flow experiment an orifice plate is used to measure the flowrate. Theflowrate is calculated from the static pressure drop across the plate. The pressuredrop depends on the size of orifice, the sharpness of the edges and where the pressuretappings are placed relative to the plate. Orifice plates are calibrated experimen-tally but here we use a simple model to estimate the pressure drop. We assume thatthe velocity is, on average, uniform and steady across section 1, section 3 and thecentral jet in section 2.

The area of the central jet adjusts until the static pressure is constant across thewhole of section 2. If we assume that there are no viscous losses between section 1and section 2 then Bernoulli can be applied along a streamline.

p1 +1

2ρV 2

1 = p2 +1

2ρV 2

2

⇒ p1 − p2 =1

2ρV 2

1

(V 2

2

V 21

− 1

)

If we knew A2/A1 we could calculate V2/V1 from conservation of mass between sec-tions 1 and 2:

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In most situations we will not know A2/A1. However, the velocity ratio V2/V1

remains a function only of the orifice diameter and shape so the term in brackets isa constant that can be determined experimentally. Thus V 2

1 and the flowrate canbe found by measuring the static pressure drop p1 − p2. This is the main functionof orifice plates.

Between section 2 and section 3 the jet mixes turbulently. Turbulent eddies decayto smaller and smaller eddies, which quickly lose their mechanical energy throughviscous dissipation, so Bernoulli cannot be applied. However, the steady flow mo-mentum equation and conservation of mass can be used between these two sections∗ .

From the steady flow momentum equation, the net momentum flux equals the pres-sure difference multiplied by the area on each side of the control volume:

p2 − p3 =m

A3

(V3 − V2) = ρV3(V3 − V2) = ρV 23

(1− V2

V3

)

By conservation of mass, V3 is equal to V1 so the total static pressure drop is:

p1 − p3 =1

2ρV 2

1

(1− V2

V1

)2

=1

2ρV 2

1 K

The stagnation pressure drop is the same because V3 equals V1:

p01 − p03 = p1 +1

2ρV 2

1 − p3 − 1

2ρV 2

3 = p1 − p3 =1

2ρV 2

1 K

The loss coefficient, K, is a function of the orifice diameter and shape. K is aconstant for flows at high Reynolds number.

∗See Derivation 1 at the back of the handout for all the intermediate steps in this calculation

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The loss coefficient, K, is tabulated or plotted in books for different orifice shapes:

6.3 Stagnation pressure loss along a pipe

In lecture 5 we derived an expression for the pressure drop along a pipe in terms ofthe friction coefficient, cf :

dP

dx= −ρV 2

Rcf

If this is constant along a pipe then:

If the pipe has constant cross-sectional area and the flow inside is fully-developedthen the average velocity, V , is constant. Consequently, the stagnation pressuredrop is exactly equal to the static pressure drop:

p02 − p01 = p2 +1

2ρV 2

2 − p1 − 1

2ρV 2

1 = p2 − p1 =

There are actually two definitions of the friction coefficient. The other is denotedf and is equal to 4cf . In this course we call f the friction factor although in somebooks it too is called the friction coefficient.

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6.4 Stagnation pressure loss at a sudden expansion

There is a stagnation pressure loss when the cross-sectional area of a pipe suddenlyincreases. If we analyse∗ this in the same way as an orifice plate we calculate thatthe stagnation pressure drop is given by:

p02 − p03 =1

2ρV 2

2

(1− A2

A3

)2

=1

2ρV 2

2 K

where K is a new loss coefficient. We can compare this value of K with the experi-mental values that are shown on the right.

If a pipe discharges into a reservoir without a gradual expansion of the cross-sectionalarea, then A3 tends to infinity and the loss coefficient, K is equal to 1.

6.5 The pressure against a wall

Students are often surprised that the pressure against a wall is assumed to be thatof the fluid next to the wall. Some think that the pressure should be zero.

If the wall and the fluid are at thermal equilibrium, molecules hit the wall, stick toit for a while, and are released back into the fluid with the same energy with whichthey hit the wall. Statistically, this is equivalent to replacing the wall with a fluidat the same temperature and pressure as the fluid next to the wall. You can thinkof the wall as a ‘pressure mirror’. If, instead, we were to assume that the pressureis zero at the wall, it would be equivalent to replacing the wall with a vacuum. Inother words, all molecules would disappear on hitting the wall. This is evidentlynot a good model of a wall’s real behaviour.

∗See Derivation 2 in the back of this handout for the intermediate steps in this calculation

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6.6 Stagnation pressure loss at a pipe entrance

There is a similar stagnation pressure loss at the entrance to a pipe:

6.7 Stagnation pressure changes across general pipe com-ponents

In summary, all pipe components, such as valves, junctions and bends cause a stag-nation pressure loss. At high Reynolds number the flow is turbulent and this loss isequal to KρV 2/2 where K is the loss coefficient. K has been measured experimen-tally for all components and is tabulated in book such as Internal Flow Systems byD. S. Miller:

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6.8 Pumps and turbines

Pumps do mechanical work on a fluid and cause a stagnation pressure rise. Turbines,on the other hand, extract mechanical work from a fluid and cause a stagnationpressure loss. The exact mechanisms of this are described in the third year (3A3Compressible Flow) and the fourth year (Turbomachinery). There are always somelosses in such systems due to irreversibility.

6.9 Stagnation pressure and mechanical work

A drop in stagnation pressure in the fluid corresponds to a loss of mechanical energyby the fluid. The mechanical energy may have been converted to internal energythrough an irreversible thermodynamic process such as viscous dissipation (see lec-ture 10). Alternatively, it may have done shaft work, Wx, on its surroundings via adevice such as a turbine.

The stagnation pressure change across a control volume is related to the shaftwork transferred from the fluid by:

m

ρ(p0out − p0in

) = −Wx

In lecture 10 we will relate this equation, which deals with mechanical energy, tothe Steady Flow Energy Equation, which deals with both mechanical and thermalenergies. For now will note that if there are any irreversibilities within the controlvolume, mechanical energy is lost to thermal energy and the equation becomes:

Similarly, if mechanical shaft work is done on the fluid then the stagnation pres-sure rises. Again, the work done is equal to the volumetric flowrate multiplied bythe stagnation pressure change.

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6.10 Network analysis

The velocities of a fluid in a network of pipes are related through mass conservation.We can analyse the flowrates in a manner analogous to electrical circuits by keepingtrack of the stagnation pressure. This can be done by hand but is easier with aspreadsheet or a dedicated software package. Here is an example:

Component diameter density velocity Re K ∆p0 head(m) (kgm−3) (ms−1) (Nm−2) (m)

pipe inlet 1.0 1000 3.50 3.2 ×106 0.1 613 0.06pipe 1.0 1000 3.50 3.2 ×106 1.35 8270 0.84valve 0.8 1000 5.47 4.0 ×106 0.5 7480 0.76pipe 0.8 1000 5.47 4.0 ×106 12 179500 18.30orifice 0.8 1000 5.47 4.0 ×106 0.5 7480 0.76pipe outlet 0.8 1000 5.47 4.0 ×106 1.0 14960 1.53total losses 218300 22.26requirednetwork gain 68700 7.00

pump gain 287000 29.26

(for pipes, K = 4CfL/D with cf = 3.75× 10−3)

Therefore the pump must supply a stagnation pressure rise of 287 KPa. The volu-metric flowrate is 2.75 m3s−1 so the required work by the pump on the fluid is:

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Derivation 1 - Orifice plate

From the steady flow momentum equation, the net momentum flux equals the pres-sure difference:

A3p2 − A3p3 = mV3 − mV2

⇒ p2 − p3 =m

A3

(V3 − V2) =ρA3V3

A3

(V3 − V2) = ρV 23

(1− V2

V3

)

V3 is equal to V1 because of conservation of mass so the total static pressure drop is:

p1 − p3 = (p1 − p2) + (p2 − p3)

=1

2ρV 2

1

([V2

V1

]2

− 1

)+ ρV 2

1

(1− V2

V1

)

=1

2ρV 2

1

(1− 2

V2

V1

+

[V2

V1

]2)

=1

2ρV 2

1

(1− V2

V1

)2

⇒ p1 − p3 =1

2ρV 2

1 K

where the loss coefficient, K, is (1− V2/V1)2.

The total stagnation pressure drop is:

p01 − p03 = p1 +1

2ρV 2

1 − p3 − 1

2ρV 2

3

= p1 − p3

=1

2ρV 2

1 K

because V1 is equal to V3.

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Derivation 2 - Abrupt expansion

There is a stagnation pressure loss when the cross-sectional area of a pipe suddenlyincreases.

Conservation of mass:

V2A2 = V3A3 (1)

Steady flow momentum equation:

p2A3 + mV2 = p3A3 + mV3 (2)

Re-arranging equation (2) gives:

p2 − p3 =m

A3

(V3 − V2) =ρA2V2

A3

(V3 − V2) = ρV 22

(A2V3

A3V2

− A2

A3

)

Substituting equation (1) into this expression gives:

p2 − p3 = ρV 22

([A2

A3

]2

− A2

A3

)(3)

The stagnation pressure drop is given by:

p02 − p03 = p2 +1

2ρV 2

2 − p3 − 1

2ρV 2

3

= p2 − p3 +1

2ρ(V 2

2 − V 23 )

= p2 − p3 +1

2ρV 2

2

(1−

[V3

V2

]2)

Substituting equation (1) and equation (3) into this expression gives:

p02 − p03 = ρV 22

([A2

A3

]2

− A2

A3

)+

1

2ρV 2

2

(1−

[A2

A3

]2)

=1

2ρV 2

2

(1− 2

A2

A3

+

[A2

A3

]2)

=1

2ρV 2

2

(1− A2

A3

)2

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