ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES Some background information first: We have seen that a major limitation of the Bernoulli equation is that it does not account for friction. This is easily done for flow through pipes by adding some terms to the Bernoulli equation:

2

22

21

12

1

22z

g

V

g

Phhz

g

V

g

PLp �� ����

UU

Note that two terms have been added: hP and Lh .The first refers to any work put into the flow (by a pump) or work done by the flow ( by a turbine), and the second term Lh to frictional losses, or head losses between points 1 and 2. The head loss term Lh refers to the pressure drop experienced in a length L of pipe. Lh is expressed in meters or feet, in other words, the pressure lost is expressed as the lost height in a column of the fluid. In the drawing below, friction causes the pressure to drop downstream by 0.2 m.

If Lh is 0.2 m, as shown above then the pressure drop is easily calculated using the hydrostatic formula:

PamsmmkgghP 19602.0)2^/8.9(3^/1000 ' U

We will make use of a dimensionless parameter known as the friction factor, which indicates how rough a pipe’s walls are:

2

)2(

LV

gDhf L

------------------------------------------------------------------------------------------------------------------

h=0.5m h=0.3

m

Problem F1 The friction factor for a pipe is given as 0.1 from a chart. Calculate the head losses in a pipe if water is moving at 1 m/s, pipe diameter is 0.3 m, and the pipe length is 50 m. Solution: The solution is straight-forward. We rearrange the equation for the head loss hl:

2

2222

/)8.9(2)3.0(

)/1)(50(1.0

2 smm

smm

gD

fLVh L =.87 m

The head loss in Pa is (1000 kg/m^3)(9.8m/s^2)(0.87m) = 8526 Pa ----------------------------------------------------------------------------------------------------------------- Problem F2 Repeat the problem above, with the same friction factor, with the same numerical values, but using feet instead of meter. Solution:

ftsftft

sftft

gD

fLVh L 26.0

/)2.32(2)3.0(

)/1)(50(1.0

2 2

2222

The pressure drop in lbf/ft^2 is 62.4 (lbf/ft^3)(.26 ft)= 16.2 lbf/ft^2 ---------------------------------------------------------------------------------------------------------------- We will distinguish between laminar and turbulent flow because the friction in laminar flow can be easily calculated from first principles – that is, starting with F=ma one can calculate the head losses. Turbulent flow, on the other hand, usually requires the use of experimentally obtained data. Head losses/Friction in Laminar Flow: In class we derived:

)/(8 4RLQP SP ' This equation is known as the Hagen-Poiseuille Equation. Where delta P is the pressure drop due to laminar flow through a pipe, and Q is the volumetric flow rate. ---------------------------------------------------------------------------------------------------------------

Problem F3 From the expression for pressure drop for laminar flow, show that the friction factor in laminar flow is 64/Re Solution: We start with the definition of the head loss:

gD

fLVh L

2

2

Since head loss times rho g is the pressure drop, we can say:

)/(822

422

RLQD

fLV

gD

gfLVghP L SP

UUU '

We solve for the friction factor:

24

)8(2

LVR

LQDf

US

P

We recall that Q = VA, that A is pi D^2/4, and that Re = VD U�P, and we can easily manipulate the equation above to show f = 64/Re for laminar flow. ----------------------------------------------------------------------------------------------------------------- Problem F4 A 0.5 m diameter pipe has a volumetric flow rate of 0.01 kg/s, and is 300 m long. Calculate the pump head required to move the flow. The inlet and outlets of the pipe are the same diameter. The water is being pumped from a lake to a free jet 200 m above the lake. First determine if the flow is laminar.

Free jet

200m Pump

Solution: Calculate the velocity from the mass flow rate, then the Reynolds number. Re should be below 2000 for this analysis to be accurate, since the formulas developed apply only to laminar flow. You should first calculate the head loss term, then apply it to the modified Bernoulli equation at the beginning of this section (with the hp and hl terms). Solve the equation for pump head required. Make point 1 at the top of the water – this makes P1 = 0 gauge and V1 = 0. Making point 2 the free jet exit makes the exit pressure P = 0 gauge also (free jet). Calculate the head required from the pump in meters and Pascals.

----------------------------------------------------------------------------------------------------------------- Problem F5 Re-do the problem above instead assuming that the pipe exit has a nozzle with 0.01 m diameter. Solution: You must now find the flow velocity out the pipe, using mdot = rho V A, then include this term in the Bernoulli equation. We note that in reality this term will introduce additional head loss terms, but we will ignore these at this time.

------------------------------------------------------------------------------------------------------------------- PREDICTING HEAD LOSSES IN TURBULENT FLOW For laminar flow we were able to predict head losses from first principles, but turbulence does not yield to such a simple analysis – we’ll see that in the next chapter. In order to predict the head losses in turbulent flow, we will require the use of experimental data and dimensional analysis – we cannot easily predict head losses from first principles as we did with laminar flow. We reason that the head losses should be a function of: hl = f( V, D, U��P��H) where H is the height of roughness in the pipe over the surface of the pipe, measured in meters or feet.� There are 6 factors and 3 units, which results in 3 dimensionless groups required. These are the friction

factor f (already discussed above, where 2

)2(

LV

gDhf L ), the Reynolds number Re, and a new group

called the relative roughness: H /D. Notice that the relative roughness is dimensionless, as H is the size of the roughness on the pipe walls. Nikudrase did many experiments in the1930’s in which he varied flow rate, pipe diameter and pipe roughness. A few years later, Colebrook expressed Nikudrase’s experimental results in what is now known as the Colebrook equation:

»»¼

º

««¬

ª��

f

D

f Re

51.2

7.3

/log)0.2(

1 H

We make some important observations about this equation:

a) the formula is not explicit in f – in other words, you use this formula to calculate the friction factor f, but f occurs on both sides of the equation, and it’s impossible (as best as the author can determine) to solve the equation for f (note the f on the right side of the equation is inside the log calculation). If you can live with 2% accuracy from what the Colebrook equation, you could use the Haaland equation, which does solve explicitly for f:

»»¼

º

««¬

ª¸¹

ᬩ

§��

11.1

7.3

/

Re

9.6log)8.1(

1 D

f

H

b) The Colebrook and Haaland equations only works for turbulent flow, for Re > 4000 or so. Below

Re = 2000 you can use f = 64/Re, and in-between you cannot be sure of what’s going to happen, as the flow will be transitioning from laminar to turbulent.

c) You need to know the roughness of the pipe H . This can be looked up easily. d) The Colebrook equation will give results with accuracy of about 10-15%. In addition, once the

pipe ages, scale may accumulate and change the roughness (as well as the diameter).

e) The equation has three dimensionless groups: Re, /D and friction factor f, as promised.

f) The friction factor can also be determined from the Moody diagram once Re and the roughness

are known. The Moody diagram was made from the Colebrook equation precisely because the Colebrook equation is difficult to solve. See the diagram below.

The Reynolds number is on the x-axis and the relative roughness H /D is selected from the right vertical axis. The friction factor f is then read from the left vertical axis. As an example: a cast iron pipe of 0.1 m diameter is exposed to flow with Re = 1e6. The friction factor is found by calculating H /D as 1.5e-4/0.1 = 1.5e-3 (see small box at lower left of Moody diagram for Hvalues). The value of friction factor f is then read as about 0.023. No need to strain your eyes getting ultra-accurate readings, since the f values from the Colebrook, Haaland, and Moody diagram are not that accurate anyway – probably errors in f as large as 20% are possible. The complete turbulent line marked on the Moody diagram stresses that f values to the right of the line are constant since the lines are almost horizontal. You should also note that the line furthest down labeled “Smooth Pipe” applies to PVC and glass pipes, with a relative roughness H /D of zero.

Since the Colebrook equation can’t be solved analytically, it must be solved either by iteration (keep guessing values of f until the right side of the equation equals the left side), or use a solver like EES (which iterates for you automatically). Or you can use the Moody diagram, which was made using the

Colebrook equation. ------------------------------------------------------------------------------------------------------------------- Problem F6 A 300 meter long, 2 cm pipe has a mass flow rate of 0.5 kg/s. The pipe is made of cast iron, with a roughness of 0.26 mm (see for example Table 14-2 in the text, which gives the roughness for pipe materials). The fluid is water with a density of 1000 kg/m^3, T = 20 C. Assume the water is at 101000 Pa for properties determination. Calculate the head losses a) by using the Moody diagram d) by using the Haaland equation: Solution: Results: f = .04781, Re = 31765, mu = 0.001, V = 1.592 m/s, headloss = 92.67 m, ed (relative roughness) = 0.013 Haaland equation: We achieve a very similar result of f = 0.043

---------------------------------------------------------------------------------------------------------------- Once you know how much pump head is required, you will need to calculate power. From thermodynamics, we know that the work required to pump an incompressible fluid reversibly (no friction) is: -w = v 'P, where v is the specific volume. The 'P is the pressure rise through the pump (calculated by 'P = Ugh), expressed in N/m^2. The resulting units are:

kg

mN

m

N

kg

mw

� �

2

3

The sign is negative to indicate work must be put into the system. This formula will give the minimum work required, since it assumes no friction. In reality you might expect to put in more than that minimum, or if you know the pump efficiency, divide the number calculated by the equation above by the pump efficiency. This relation expresses the work which must be done per kg of fluid flowing. To get power, we simply multiply by the mass flow rate in kg/s. The result is N-m/s or watts. In U.S. units:

lbm

lbfft

ft

lbf

lbm

ftw

� �

2

3

And to get power:

slbffts

lbm

lbm

lbfftPower /�

� �

-------------------------------------------------------------------------------------------------------------------- Problem P1 Using the Bernoulli equation with the head loss equations, you determine that the pump must provide 3 feet of head to the flow. a) explain what this means b) express the head in lbf/ft^2 c) calculate the power required to run the pump if the mass flow rate is 2 lbm/s. d) will this be a realistic figure? Solution:

a) The fact that the pump must provide three feet of head means that it must provide the pressure found under a 3 foot high column of water. Instead of using a pump, you could elevate the upstream point by three feet to achieve the same result.

b) We can easily calculate the pressure by 'P = Ugh = 62.4 lbf/ft^3 (3 ft) = 187.2 lbf/ft^2 c) The power by the pump is v'P(mdot) = 187 lbf/ft^2 (1/62.4 ft^3/lbm) (2 lbm/s) = 6 ft-lbf/s, or

about 0.01 HP. d) The equation used assumes frictionless operation, so the calculated power required will be less