7 Head Loss 2 1

ANALYSIS OF VISCOUS FLOW IN PIPES The viscous flow in pipe systems may be analyzed by classifiying them in the following manner: i) Series system of pipes, ii) Parallel system of pipes, iii) Pipe networks, iv) Interconnected reservoir systems.

In designing or analysis of pipe flow systems, there are six primary parameters involved, which are 1) the total head loss of the system, hf, 2) the volumetric flow arte of the fluid, Q 3) the diameter of the pipe, d 4) the length of the pipe, L 5) the roughness of the pipe, ef 6) the fluid density r and viscosity, m

Usually, one of the first three parameters is to be determined, while the reamaining ones are either known or can be specified by the designer. The method of performing the design or the analysis is different depending on what is unknown.

Series Sytem of Pipes

Parallel Sytem of Pipes

Pipe networks

Interconnected reservoir systems

7 Head Loss 2 2

i) Series Pipe System

If the pipe system is arranged such that the fluid flows through a single path without branching, it is referred as a series pipe system. In a series pipe system, the total head loss is the sum of the head losses in each serially connected pipe, and the volumetric flow rate in pipes should be the same.

l la lb lch h h h a b cQ Q Q Q

Series pipe system analysis and design problems can be classified into three classes as follows:

Class I system: The total head loss, hl, of the system is to be determined.

Class II system: The volumetric flow rate , Q, of the system is to be determined.

Class III system: The diameter of the pipes, d, of the system is to be determined.

7 Head Loss 2 3

The procedure for solving Class I problems

1) Determine the relative roughness, es/d, for each pipe.

2) Calculate the average velocity in each pipe, V=4Q/pd2

3) Calculate Reynolds number, Re=rVd/m

4) Determine the friction factor f

5) Calculate the major head loss, hl=f(L/d)(V2/2g)

6) Determine minor head loss coefficients K and then calculate the minor head

losses for each variable are portions, hl=K(V2/2g).

7) Evaluate the total head loss of the system by adding the major and minor losses

calculated in stepes 5 and 6.

CLASS I SYSTEMS In class I systems, the volumetric flow rate Q, the diameter of the pipes d, the length of the pipes L, the roughness of the pipe es, and fluid properties density r and viscosity m are known and the total head loss hl is to be determined. Total hand loss between two points of the system is sum of the major and all local (minor) losses, i.e.

, ,minl l major l orh h h 2 2

2 2l

L V Vh f K

d g g

To complete these calculations, friction factor for each pipe and loss coefficient for all fittings need to be determined. The following procedure may be followed:

7 Head Loss 2 4

Solution:

Example: In the system shown, water at 20 oC is flowing at the volumetric flow rate of 0.015 m3/s. The suction line is made from commercial steel pipe of diameter 4 inch and length is 15 m. The discharge line is also commercial steel pipe and its diameter is 2 inch and length is 200 m. The entrance to the suction line from the reservoir is through a square edged inlet. The elbow is 90o standad elbow and the globe valve is fully open. Calculate the power supplied to the pump if its efficiency is 76%.

The required pump head for transporting water between the two reservoirs may be determined by applying the extended Bernoulli equation between points 2 and 1 along the streamline, which is shown in Figure.

1 2

2 2

1 1 2 21 2

2 2s f

P V P Vz h z h

g g g gr r

2 1 1 2t t s fh h h h

or solving for the pump head, we obtain, 2 1 1 2s t t fh h h h

Since the cross sectional areas of the two reservoirs are very large, when compared to the cross-sectional areas of the suction and the discharge pipes, the velocities at the free surfaces of the reservoirs are negligible, that is V1=0, and V2=0. Also, the free surfaces of the reservoirs are exposed to the atmosphere, so that P1=P2=Patm.

2

1 11 1

2

atmt

PP Vh z

g g gr r

2

2 22 2 2

2

atmt

PP Vh z z

g g gr r

Therefore, the required pump head may be expressed as, 1 2 2s fh h z

7 Head Loss 2 5

The frictional head loss, hf1-2, may now be evaluated by following the procedure which is presented in this section. i) The relative roughness for the 4" commercial steel suction pipe and the 2" commercial steel discharge pipe may be obtained from Figure as follows: (ϵs/d)s=0.00044 (ϵs/d)d=0.0009

ii) The average velocity in the suction and the discharge pipes may then be evaluated as follows:

3

2 2

4 (4)(0.015m /s)1.85 m/s

(0.1016m)s

s

QV

dp p

3

2 2

4 (4)(0.015m /s)7.4 m/s

(0.0508m)d

d

QV

dp p

iii) The kinematic viscosity of water at 20 0C is 1.003x10-6 m2/s. Therefore, the Reynolds number for the suction and the discharge pipes are as follows:

5

6

(1.85)(0.1016)Re (1.87)(10 )

(1.003)(10 )

s ss

V d

5

6

(7.4)(0.0508)Re (3.75)(10 )

(1.003)(10 )

d dd

V d

iv) Now, the friction factor for the suction pipe corresponding to a relative roughness of (ϵs/d)s=0.00044 and a Reynolds number of Res = 1.87x105 from the Moody diagram is, fs=0.019 Similarly, the friction factor for the discharge pipe corresponding to relative roughness of (ϵs/d)d=0.0009 and a Reynolds number of Red = 3.75x105 is, fd=0.020

v) The major head losses in the suction and the discharge pipes may be evaluated by using the Darcy-Weisbach equation as

2 2

2

(15m) (1.85m/s)(0.019) 0.49m

2 (0.1016m) (2)(9.81m/s )

s sfs s

s

L Vh f

d g

2 2

2

(200m) (7.4m/s)(0.020) 219.77m

2 (0.0508m) (2)(9.81m/s )

d dfd d

d

L Vh f

d g

7 Head Loss 2 6

vi) For the square edged inlet, the head loss coefficient, ki, may be obtained from the Figure as 0.5. Then the minor head loss through the inlet is

2 2

2

(1.85m/s)(0.5) 0.09 m

2 (2)(9.81m/s )

sfi i

Vh k

g

The head loss coefficient, ko, for the outlet from the discharge pipe into the reservoir may be obtained from figure as 1.0, then the minor head loss is

2 2

2

(7.4m/s)(1.0) 2.79m

2 (2)(9.81m/s )

dfo o

Vh k

g

For full open globe valve loss coefficient kv = 10

2 2(7.4)10 27.91m

2 (2)(9.81)

dfv v

Vh k

g

For 90o standard elbow loss coefficient k=0.3,

2 2(7.4)0.3 0.83 m

2 (2)(9.81)

dfe e

Vh k

g

vii) The total head loss is the sum of the major and minor losses, then

1 2f fs fd fi fo fv feh h h h h h h

1 20.49 219.77 0.09 2.79 27.91 0.83fh

1 2251.89mfh

The power required by the pump is

3 2 3(998.2kg/m )(9.81m/s )(0.015m /s)(251.89m)

(0.76)

sp

p

gQhr

50.62kW

7 Head Loss 2 7

CLASS II SYSTEMS

In class II systems, the total head loss of the system hl, the diameter of the pipes d, the length of the pipes L, the roughness of the pipe es, and fluid properties density r and viscosity m are known and the volumetric flow rate of flow Q is to be determined. Since the volumetric flow rate is unknown, the fluid velocity and hence Reynolds number can not be calculated directly. Thus, the friction coefficient can not be determine directly. Hence an iterative solution procedure is applied to determine the volumetric flow rate of the fluid. The main steps of this iterative solution procedure is given below:

The procedure for solving Class II problems

1) Determine the relative roughness, es/d, for each pipe.

2) Assume a friction factor f for each pipe from the Moody diagram using the relative

roughness es/d in fully rough region. The reason for using the fully rough region in

Moody diagram is that in this region friction factor is independent from the

Reynolds number. If this assumption is true no iteration is necessary.

3) Calculate the major head loss, hl=f(L/d)(V2/2g) in terms of the unknown velocity.

4) Determine minor head loss coefficients K and then calculate the minor head losses

for each variable are portions, hl=K(V2/2g) in terms of the unknown velocity.

5) Evaluate the total head loss of the system by adding the major and minor losses

calculated in steps 4 and 5.

6) Relate the unknown velocities in each pipe by using the continuity equation.

7) Express the total head loss hl of the system in terms of one of the unknown

velocities, and then solve the unknown velocities.

8) Calculate Reynolds number, Re=rVd/m for each pipe.

9) Determine the improved friction factor f for each pipe from the Moody diagram

using Reynolds number and es/d.

10) Compare the asumed and the improved values of the friction factor and repeat

steps 3 through 9 as many times as needed in order to obtain desired accuracy in

the friction factor.

11) Evaluate volumetric flow rate of the fluid, Q=pd2V/4

7 Head Loss 2 8

Example The piping system, which is shown in the figure is used to transfer water at 25oC from one storage tank to the other. The larger pipe is a 6 inch commercial steel pipe having a total length of 30 m. The smaller pipe is a 2 inch commercial steel pipe having a total length of 15 m. The entrance to the larger pipe from the storage tank is through an inward projecting pipe. 90o standard elbows are used, and the gate valve is half open. Determine the volumetric flow rate through the system.

Solution:

The total head loss during the transfer of water from one storage tank to the other may be determined by applying the extended Bernoulli equation between points 2 and 1 along the streamline which is shown in Figure. Then

2 1 1 2t t fh h h

or solving for the total head loss 1 2 1 2f t th h h

Since the areas of the two reservoirs are very large when compared to the cross-sectional areas of the larger and the smaller pipes, then the velocities at the free surfaces of the storage tanks are negligible that is, V1=0 and V2 =0. Also the free surfaces of both storage tanks are exposed to the atmosphere, so that P1 =P 2 = Patm. Therefore,

2

1 11 1 10

2

atmt

PP Vh z

g g gr r

2

2 22 2

2

atmt

PP Vh z

g g gr r

Therefore, the total head loss of the system is 1 2 10 10 matm atmf

P Ph

g gr r

7 Head Loss 2 9

The volumetric flow rate of water may now be evaluated by following the procedure described above,

i) The relative roughness of 6 and 2 inch commercial steel pipes are, (ϵs/d)l = 0.00028 (ϵs/d)s = 0.0009 ii) The friction factors corresponding to the relative roughnesses can be obtained from Moody Chart as, for (ϵs/d)l = 0.00028 → fl = 0.015 and for (ϵs/d)s = 0.0009 → fs = 0.019 iii) The major head losses in the larger and the smaller pipes may be evaluated by using the Darcy-Weisbach equation as

2 22

2

( )(30)(0.015) 0.151V

2 (0.1524m) (2)(9.81m/s )l

l l lf l l

l

L V Vh f

d g

2 22

2

( )(15m)(0.019) 0.286V

2 (0.1524m) (2)(9.81m/s )

s s Sfs s s

s

L V Vh f

d g

iv) For an inlet projecting pipe inlet, the head loss coefficient ki = 1.0 and for the reducer, kr=0.42. Using these coefficients head losses for the inlet and the reducer can be calculated as follows, For the inlet,

2 22

l

( )(1) 0.051V

2 (2)(9.81)

l lfi i

V Vh k

g

2 22

s

( )(0.42) 0.021V

2 (2)(9.81)

s sfr r

V Vh k

g For reducer,

The head loss coefficient for the outlet ko=1.0, then the minor head loss for the outlet is

2 22

s

( )(1.0) 0.051V

2 (2)(9.81)

s sfo o

V Vh k

g

7 Head Loss 2 10

Loss coefficient for half open gate valve is kv=2.1,

2 22

s

( )(2.1) 0.11V

2 (2)(9.81)

s sfv v

V Vh k

g

Loss coefficient for 90° standard elbow is 0.3 2 2

2( )2 (0.3) 0.035V

2 (2)(9.81)

l lfe e l

V Vh k

g

Total head loss can be calculated from the sum of the major and minor losses as,

1 2f fl fs fi fr fo fv feh h h h h h h f

1 2

2 2 2 2 2 2 20.151 0.286 0.051 0.021 0.051 0.11 0.035f l s l s s s lh V V V V V V V

2 20.237 0.468l sV V

v) The velocities in the larger and the smaller pipes may now be related by using continuity equation as,

2 2/ 4 / 4l l s sd V d Vp p

2( / )s l s lV d d V

2(0.1524 / 0.0508)s lV V

9s lV V

Substituting into total head loss equation,

2 210 0.237 0.468(9 )l lV V

Solving for Vl and Vs, Vl = 0.51 m/s Vs = 4.60 m/s ix) The kinematic viscosity of water at 25 oC is 0.893x106 m2/s

7 Head Loss 2 11

Reynolds number for the larger and smaller pipes,

5

6 2

(0.49m/s)(0.1524m)Re (8.36)(10 )

0.893(10 )m /s

l ll

water

V d

5

6 2

(4.60m/s)(0.0508m)Re (2.54)(10 )

0.893(10 )m /s

s ss

water

V d

x) Now for both large and small pipes, the improved value of friction factor can be read from Moody chart using for relative roughness values and the Reynolds numbers calculated. For larger pipes, (ϵs/d)s = 0.00028 and → fl = 0.02 For smaller pipes, (ϵs/d)s = 0.0009 and → fs= 0.0205 xi) These steps given above should be repeated until the friction factors converge. The iteration results are given in the table below. The volumetric flow rate can be obtained using the velocity value obtained in the current step as,

2 2 3 3/ 4 (0.473m/s)( )(0.1524) / 4 8.63(10 )m /sl lQ V dp p

Iteration 1 2

fl (Assumed) 0.015 0.02

fs (Assumed) 0.019 0.0205

hfl (m) 0.151 Vl2 0.201 Vl

2

hfs (m) 0.286 Vs2 0.309 Vs

2

hfi (m) 0.051 Vl2 0.051 Vl

2

hfr (m) 0.021 Vs2 0.020 Vs

2

hfo (m) 0.051 Vs2 0.051 Vs

2

hfv (m) 0.11 Vs2 0.11 Vs

2

hfe (m) 0.035 Vl2 0.035 Vl

2

hf1-2 (m) 0.237 Vl2 + 0.468 Vs

2 0.313 Vl2 + 0.547 Vs

2

Vl (m/s) 0.49 0.473

Vs (m/s) 4.60 4.26

Rel 8.36x104 8.07x104

Res 2.51x105 2.42x105

fl (Improved) 0.02 0.02

fs (Improved) 0.0205 0.0205

7 Head Loss 2 12

CLASS III SYSTEMS Class III system are true design problems. In class III systems, the total head loss of the

system hl, the volumetric flow rate Q, the length of the pipes L, the roughness of the pipes es, and fluid properties density r and viscosity m are known and the pipe diameter d is to be determined. Since the diameter is unknown, the fluid velocity and hence the Reynolds number can not be calculated directly. Thus, the friction coefficient can not be determine directly. Hence, an iterative solution procedure is applied to determine the pipe diameter. The main steps of this iterative solution procedure is given below:

The procedure for solving Class III problems

1) Assume the diameter d of each pipe

2) Determine the relative roughness, es/d, for each pipe.

3) Calculate the average velocity V=4Q/pd2 for each pipe.

4) Calculate Reynolds number, Re=rVd/m for each pipe.

5) Determine the friction factor f for each pipe from the Moody diagram using

Reynolds number and es/d.

6) Calculate the major head loss for each pipe using hl=f(L/d)(V2/2g).

7) Determine minor head loss coefficients K and then calculate the minor head loss for

each variable are portions using hl=K(V2/2g).

8) Evaluate the total head loss of the system by adding the major and minor losses

calculated in stepes 6 and 7.

9) Compare the calculated total head loss with the given head loss. If the calculated

head loss is less than the given head loss, then decrease pipe diameter. If the

calculated head loss is more than the given head loss, then increase the pipe

diameters. Repeat steps (2) through (8) until the calculated and the given head

losses are close enough.

7 Head Loss 2 13

Example: In a chemical processing system, benzene at 20 oC is taken from the bottom of a larger tank and transferred by gravity to another part of the system, as shown in the figure. The length of the pipe between two tanks is 7 m and it is commercial steel pipe. A filter is installed in the line, and is known to have a loss coefficient of 8.5. The gate valve is fully open. Determine the diameter of the pipe which would allow a volumetric flow rate of 0.0025 m3/s through this system.

Solution: The total head loss during the transportation may be determined by applying extended Bernoulli Eq. between points 2 and 1 along the streamline shown.

1 2 1 2t t fh h h

Solving for the total head loss,

1 2 1 2f t th h h

Since the surface areas of the reservoirs are much larger than the cross-section area of the piping, velocities at the free surfaces are negligible, that is V1=0 and V2=0. Also, P1=P2=Patm.

2

1 11 1 8

2

atmt

PP Vh z

g g gr r

2

2 22 2

2

atmt

PP Vh z

g g gr r

The total head loss of the system is

1 2 8 8atm atmf

P Ph m

g gr r

7 Head Loss 2 14

Step 1 In order to start iteration procedure, a pipe diameter of 2” may be assumed. Step 2 The relative roughness for 2” commercial steel pipe is (ϵs/d) = 0.0009 Step 3 The average velocity in the pipeline may be calculated as

3

2 2

4 (4)(0.0025m /s)1.23m/s

(0.0508m)s

s

QV

dp p

Step 4 Calculation of the Reynolds number

3

4

34

4

895kg/m

6.5(10 )Pa s

(895kg/m )(1.23m/s)(0.0508m)Re 8.6(10 )

6.5(10 )Pa s

Vd

r

m

r

m

Step 5 Now the friction factor corresponding to a relative roughness of (ϵs/d) = 0.0009 and a Reynolds number of 8.6x104 can be read from Moody diagram as f=0.022 Step 6 The major head loss in pipeline may now be evaluated using the Darcy-Weisbach equation as

2 2

2

(7) (1.23)(0.022) 0.234m

2 (0.0508m) (2)(9.81m/s )fp

L Vh f

d g

Step 7 For an inlet through an inward projecting pipe, the head loss coefficient ki is 1.0.

2 2

2

(1.23)(1.0) 0.077m

2 (2)(9.81m/s )fi i

Vh k

g

7 Head Loss 2 15

For the filter, loss coefficient kf for the filter was given as 8.5.

2 2

2

(1.23)(8.5) 0.655m

2 (2)(9.81m/s )fi f

Vh k

g

For the outlet, loss coefficient ko for the discharge pipe is 1.0. 2 2

2

(1.23)(1.0) 0.077m

2 (2)(9.81m/s )fo o

Vh k

g

Loss coefficient for the fully open gate-valve is 0.15.

2 2

2

(1.23)(0.15) 0.011m

2 (2)(9.81m/s )fv v

Vh k

g

Step 8 Total head loss is the sum of the major and minor losses.

1 2

0.234 0.077 0.655 0.077 0.011

1.154m

f fp fi ff fo fvh h h h h h

Step 9: Compare the calculated head loss with the given head loss. Calculations from Step 2 to Step 8 must be repeated until the total head loss converges. The summary of the calculation from the iterations are given in the table below.

Iteration 1st 2nd 3rd 4th

d (Assumed) 2” 1 ½” 1” 1 ¼”

ϵs/d 0.0009 0.0012 0.0018 0.0015

V (m/s) 1.23 2.19 4.93 3.16

Re 8.16x104 11.49x104 17.24x104 13.82x104

f 0.022 0.0225 0.024 0.023

hfp (m) 0.234 1.011 8.194 2.581

hfi (m) 0.077 0.245 1.239 0.509

hff (m) 0.655 2.078 10.530 4.326

hfo (m) 0.077 0.245 1.239 0.509

hfv (m) 0.011 0.72 0.387 0.152

hf1-2 (m) 1.054 3.651 21.589 8.077

Change in d Decrease Decrease Increase O.K.

7 Head Loss 2 16

Parallel System of Pipes

If the pipe system causes the flow to branch into two or more lines, then it is referred as

a parallel system of pipes. A typical parallel system of pipes is shown in Figure. The flow

in the main line splits into three branches at section 1, and then rejoins at section 2.

The head loss in each branch between sections 1 and 2 must be equal, that is

ℎ𝑓1−2= ℎ𝑓𝑎 = ℎ𝑓𝑏 = ℎ𝑓𝑐

Also, one should observe that the volumetric flow rate in the main line is equal to the

sum of the volumetric flow rates through each branch. Hence

𝑄 = 𝑄𝑎 + 𝑄𝑏 + 𝑄𝑐

Usually, two types of problems occur in parallel system of pipes. These are:

i) Class I Systems for which the head loss in each branch is known, and the

volumetric flow rate in each branch and in the main line are to be determined.

ii) Class II Systems for which the total volumetric flow rate is known, and the

volumetric flow rate and the head loss in each branch are to be determined.

Parallel system of pipes

7 Head Loss 2 17

Class I Systems In Class I problems, the pressure drop or the head loss across the parallel branches is known, and it is desirable to determine the volumetric flow rates in each branch and in the main line. The procedure for solving Class I problems may be given as follows: 1) Determine the relative roughness, es/d, for each branch by using the pipe

diameter, d, and the pipe material.

2) Assume a friction factor, f, for each branch from the Moody diagram by using the relative roughness, ɛs/d, in the fully rough region. The reason for the assumption of, the flow in the fully rough region is that no iteration is necessary, if this assumption is true.

3) Calculate the major head losses for each branch by using the Darcy-Weisbach equation, hf = f (L/d)(V2/2g) in terms of the unknown branch velocities.

4) Determine the minor head loss coefficients K and then calculate the minor head losses for the variable area portions in each branch by using hf= kV2/(2g) in terms of the unknown branch velocities.

5) Evaluate the total head loss ,hf, of each branch by adding up the major and the minor head losses from steps (3) and (4).

6) Equate this head loss to the given head loss across the parallel branches and solve for the unknown branch velocities.

7) Calculate the Reynolds number, Re = ρVd/μ for each branch,

8) Determine the improved value of the friction factor, f, for each branch from the Moody diagram by using the Reynolds number, Re, and the relative roughness, ɛs/d.

9) Compare the assumed and the improved values of the friction factor for each branch and repeat steps (3) through (8) as many times as needed in order to obtain the desired accuracy in the friction factor.

10) Evaluate the volumetric flow rate of the fluid, Q = πd2 V/4, in each branch and in the main line.

7 Head Loss 2 18

Example 10.4 The arrangement, which is shown in Figure, is used to supply lubricating oil to the bearing of a large machine. The bearings act as restrictions to the flow. The head loss coefficients for the journal bearings in branches x and y are 11 and 4, respectively. The length of 1" commercial steel branch x is 10 m, while the length of 1 ½ inch commercial steel branch y is 5 m. Each of the four bends in the tubing has a radius of 100 mm. The pressures before and after the branching are 275 kPa and 195 kPa respectively. The density and the absolute viscosity of the lubricating oil are 881 kg/m3 and 2.2x10-3 Pa.s. Determine: a) the volumetric flow rate through each bearing, and b) the total volumetric flow rate.

Solution

The head loss across the parallel branches may be determined by applying the extended Bernoulli Equation between sections 2 and 1 in Figure as

𝑃1𝜌𝑔+𝑉12

2𝑔+ 𝑧1 =

𝑃2𝜌𝑔

+𝑉22

2𝑔+ 𝑧2 + ℎ𝑓1−2

Note that d1=d2 hence V1=V2, and 𝑧1 = 𝑧2, so that the head loss between sections 1 and 2 may be evaluated as

ℎ𝑓1−2 =𝑃1 − 𝑃2𝜌𝑔

=275000

𝑁𝑚2 − 195000

𝑁𝑚2

(881 𝑘𝑔/𝑚3)(9.81 𝑚/𝑠2)= 9.26 𝑚

a) The volumetric flow rate through each branch may now be determined by applying

the procedure which is presented above.

i) The relative roughness for the 1″ and 1/2″ commercial steel pipes may be obtained from as follows:

(𝜀𝑠/𝑑)𝑥= 0.0018 (𝜀𝑠/𝑑)𝑦= 0.0012

7 Head Loss 2 19

ii) The friction factors in branches x and y corresponding to relative roughness values of (𝜀𝑠/𝑑)𝑥= 0.0018 and (𝜀𝑠/𝑑)𝑦= 0.0012 in the fully flow region may be obtained from

the Moody diagram as 𝑓𝑥 = 0.0225 𝑓𝑦 = 0.02

iii) The major head loss in branches x and y may be evaluated by using the Darcy-Weisbach equation as

ℎ𝑓𝑝𝑥 = 𝑓𝑥𝐿𝑥𝑑𝑥

𝑉𝑥2

2𝑔= (0.0225)

(10 𝑚)

(0.0254 𝑚)

𝑉𝑥2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑝𝑥 = 0.452 𝑉𝑥2

ℎ𝑓𝑝𝑦 = 𝑓𝑦𝐿𝑦𝑑𝑦

𝑉𝑦2

2𝑔= (0.0225)

(10 𝑚)

(0.0254 𝑚)

𝑉𝑦2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑝𝑦 = 0.134 𝑉𝑦

2

iv) The minor head losses in the journal bearings are

ℎ𝑓𝑗𝑥 = 𝑘𝑗𝑥𝑉𝑥2

2𝑔= 11

𝑉𝑥2

2 9.81𝑚𝑠2

= 0.561 𝑉𝑥2

ℎ𝑓𝑗𝑦 = 𝑘𝑗𝑦𝑉𝑦2

2𝑔= 4

𝑉𝑦2

2 9.81𝑚𝑠2

= 0.204 𝑉𝑦2

For standard tees with flow through branch, loss coefficient from Table is obtained as k=1.0

ℎ𝑓𝑡𝑥 = 2𝑘𝑉𝑥2

2𝑔= 2 (1)

𝑉𝑥2

2 9.81𝑚𝑠2

= 0.102 𝑉𝑥2

ℎ𝑓𝑡𝑦 = 2𝑘𝑉𝑦2

2𝑔= 2 (1)

𝑉𝑦2

2 9.81𝑚𝑠2

= 0.102 𝑉𝑦2

The minor head loss coefficient for bends 𝒌𝒃 = 𝟏. 𝟑

ℎ𝑓𝑏𝑥 = 2𝑘𝑏𝑉𝑥2

2𝑔= 2 (1.3)

𝑉𝑥2

2 9.81𝑚𝑠2

= 0.13 𝑉𝑥2

ℎ𝑓𝑏𝑦 = 2𝑘𝑏𝑉𝑦2

2𝑔= 2 (1.3)

𝑉𝑦2

2 9.81𝑚𝑠2

= 0.13 𝑉𝑦2

7 Head Loss 2 20

v) Then the total head losses in branches x and y may be evaluated as

ℎ𝑓𝑥 = ℎ𝑓𝑝𝑥 + ℎ𝑓𝑗𝑥 + ℎ𝑓𝑡𝑥 + ℎ𝑓𝑏𝑥

ℎ𝑓𝑥 = 0.452 𝑉𝑥2 + 0.561 𝑉𝑥

2 + 0.102 𝑉𝑥2 + 0.13 𝑉𝑥

2 = 1.245 𝑉𝑥2

ℎ𝑓𝑦 = ℎ𝑓𝑝𝑦 + ℎ𝑓𝑗𝑦 + ℎ𝑓𝑡𝑦 + ℎ𝑓𝑏𝑦

ℎ𝑓𝑦 = 0.134 𝑉𝑦2 + 0.204 𝑉𝑦

2 + 0.102 𝑉𝑦2 + 0.13 𝑉𝑦

2 = 0.57 𝑉𝑦2

vi) The velocity in branch x is

𝑉𝑥 = (ℎ𝑓𝑥1.245

)1/2= (9.26 𝑚

1.245)1/2= 2.72 𝑚/𝑠

and the velocity in branch y is

𝑉𝑦 = (ℎ𝑓𝑦0.57

)1/2= (9.26 𝑚

0.57)1/2= 4.04 𝑚/𝑠

vii) Now, the Reynolds number in branches x and y may be calculated as

𝑅𝑒𝑥 =𝜌𝑉𝑥𝑑𝑥𝜇

=(881

𝑘𝑔𝑚3)(2.80

𝑚𝑠)(0.0254 𝑚)

(2.2𝑥10−3 𝑁. 𝑠/𝑚2)= 2.85𝑥104

𝑅𝑒𝑦 =𝜌𝑉𝑦𝑑𝑦𝜇

=881

𝑘𝑔𝑚3 4.04

𝑚𝑠

0.0381 𝑚

2.2𝑥10−3 𝑁.𝑠𝑚2

= 6.67𝑥104

viii) Now, the improved value of the friction factor for branch x corresponding to a relative roughness of (𝜀𝑠/𝑑)𝑥= 0.0018 and a Reynolds number of 𝑅𝑒𝑥 = 2.81𝑥10

4 from the Moody diagram is

𝑓𝑥 = 0.028 Similarly, the improved value of the friction factor for branch y corresponding to a relative roughness of (𝜀𝑠/𝑑)𝑦= 0.0012 and a Reynolds of 𝑅𝑒𝑦 = 6.67𝑥10

4 is

𝑓𝑦 = 0.024

ix) As long as the assumed and the improved values of the friction factors are not the same, then the calculations in steps (iii) through (vii) must be repeated. The summary of these iterations is presented in the Table.

7 Head Loss 2 21

x) The volumetric flow rate through branches x and y may now be evaluated as

𝑄𝑥 =𝜋𝑑𝑥

2𝑉𝑥4

=𝜋 0.0254 𝑚 2(2.64 𝑚/𝑠)

4= 1.34𝑥10−3 𝑚3/𝑠

𝑄𝑦 =𝜋𝑑𝑦

2𝑉𝑦4

=𝜋 0.0381 𝑚 2(4.14 𝑚/𝑠)

4= 4.72𝑥10−3 𝑚3/𝑠

b) The volumetric flow rate through the main line is the sum of the volumetric flow rates in branches x and y. that is

𝑄 = 𝑄𝑥 + 𝑄𝑦 = 1.34𝑥10−3𝑚3

𝑠+ 4.72𝑥10−3

𝑚3

𝑠

𝑄 = 6.06𝑥10−3 𝑚3

𝑠

Iteration 1 2

𝑓𝑥 (assumed) 0.0225 0.028

𝑓𝑦 (assumed) 0.02 0.024

ℎ𝑓𝑝𝑥 0.452 𝑉𝑥2 0.562 𝑉𝑥

2

ℎ𝑓𝑝𝑦 0.134 𝑉𝑦2 0.161 𝑉𝑦

2

ℎ𝑓𝑗𝑥 0.561 𝑉𝑥2 0.561 𝑉𝑥

2

ℎ𝑓𝑗𝑦 0.204 𝑉𝑦2 0.204 𝑉𝑦

2

ℎ𝑓𝑡𝑥 0.102 𝑉𝑥2 0.171 𝑉𝑥

2

ℎ𝑓𝑡𝑦 0.102 𝑉𝑦2 0.147 𝑉𝑦

2

ℎ𝑓𝑏𝑥 0.13 𝑉𝑥2 0.040 𝑉𝑥

2

ℎ𝑓𝑏𝑦 0.13 𝑉𝑦2 0.028 𝑉𝑦

2

ℎ𝑓𝑥 1.183 𝑉𝑥2 1.334 𝑉𝑥

2

ℎ𝑓𝑦 0.484 𝑉𝑦2 0.540 𝑉𝑦

2

𝑉𝑥 2.80 2.64

𝑉𝑦 4.37 4.14

𝑅𝑒𝑥 2.65x104 2.69x104

𝑅𝑒𝑦 6.67x104 6.32x104

𝑓𝑥 (improved) 0.028 0.028

𝑓𝑦 (improved) 0.024 0.024

7 Head Loss 2 22

Class II Systems In Class II problems, the volumetric flow rate in the main line is known, and it is desirable to determine the volumetric flow rates in each branch and the head loss across the parallel branches. The procedure for solving Class II problems may be given as follows 1) Determine the relative roughness, ɛs/d, for each branch by using the pipe

diameter, d, and the pipe material.

2) Assume a friction factor, f, for each branch from the Moody diagram by using the relative roughness, ɛs/d in the fully rough region. The reason for the assumption of the flow in the fully rough region is that no iteration is necessary, if this assumption is true.

3) Calculate the major head losses for each branch by using the Darcy-Weisbach equation, hf= f(L/d)(V2/2g) in terms of the unknown branch velocities.

4) Determine minor loss coefficients k and then calculate the minor head losses for the variable area portions in each branch by using hf = kV2/2g in terms of the unknown branch velocities.

5) Evaluate the total head loss, hf, of each branch by adding up the major and the minor head losses from steps (3) and (4).

6) Equate the the total head loss of each branch and express all unknown branch velocities in terms of one unknown branch velocity.

7) As long as the volumetric flow rate in the main line is the sum of the volumetric flow rates of each branch, then express it in terms of the unknown branch velocities.

8) Solve for the unknown branch velocities.

9) Calculate the Reynolds number, Re = ρVd/μ, for each branch.

10) Determine the improved value of the friction factor, f, for each branch from the Moody diagram by using the Reynolds number, Re, and the relative roughness, ɛs/d.

11) Compare the assumed and the improved values of the friction factor for each branch and repeat steps (2) through (10) as many times as needed in order to obtain the desired accuracy in the friction factor.

12) Evaluate the volumetric flow rate of the fluid, Q = πd2V/4 in each branch.

7 Head Loss 2 23

Example 0.0063 m3/s of water at 15°C is flowing in a 2" commercial steel pipe at section 1, as shown in Figure. The heat exchanger in branch x has a head loss coefficient of 12.5. Branch y is a bypass line, which is composed of 1 1/4" commercial steel pipe with a length of 6 m. All three valves are wide open, and the elbows are standard. Determine; a) the volumetric flow rate of water in each branch and b) the pressure drop between points 1 and 2.

Solution

a) To determine the volumetric flow rate of water in each branch one may apply the procedure which is presented in this section

i) The relative roughness for the 2″ and 1 1/4″ commercial steel pipes may be obtained from Figure 10.4 as follows:

(𝜀𝑠/𝑑)𝑥= 0.0009 (𝜀𝑠/𝑑)𝑦= 0.0014

ii) The friction factors in branches x and y corresponding to relative roughness values of (𝜀𝑠/𝑑)𝑥= 0.0009 and (𝜀𝑠/𝑑)𝑦= 0.0014 in the fully flow region may be

obtained from the Moody diagram in Figure 10.5 as 𝑓𝑥 = 0.019 𝑓𝑦 = 0.021

iii) The major head loss in branches y may be evaluated by using the Darcy-Weisbach equation as

ℎ𝑓𝑝𝑦 = 𝑓𝑦𝐿𝑦𝑑𝑦

𝑉𝑦2

2𝑔= (0.021)

(6 𝑚)

(0.03175𝑚)

𝑉𝑦2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑝𝑦 = 0.202 𝑉𝑦2

7 Head Loss 2 24

iv) The minor head losses in the journal bearings are

ℎ𝑓ℎ𝑥 = 𝑘ℎ𝑉𝑥2

2𝑔= 12.5

𝑉𝑥2

2 9.81𝑚𝑠2

= 0.637 𝑉𝑥2

For fully open gate valve; loss coefficient 𝑘𝑣 = 0.15. Then, the minor head loss in branches x and y are;

ℎ𝑓𝑣𝑥 = 2𝑘𝑣𝑉𝑥2

2𝑔= 2 0.15

𝑉𝑥2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑣𝑥 = 0.015 𝑉𝑥2

ℎ𝑓𝑣𝑦 = 𝑘𝑉𝑦2

2𝑔= (0.15)

𝑉𝑦2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑣𝑦 = 0.0076 𝑉𝑦2

For tee with flow through loss coefficient 𝑘𝑡 = 0.2;

ℎ𝑓𝑡𝑥 = 𝑘𝑡𝑉𝑥2

2𝑔= 2 0.2

𝑉𝑥2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑡𝑥 = 0.020 𝑉𝑥2

For tee with flow through branch, the loss coefficient 𝑘𝑡 = 1.0

ℎ𝑓𝑡𝑦 = 2 𝑘𝑡𝑉𝑦2

2𝑔= (2) 1

𝑉𝑦2

2 9.81𝑚𝑠2

= 0.102 𝑉𝑦2

Loss coefficient for standard elbow is 𝑘𝑒 = 0.3

ℎ𝑓𝑒𝑦 = 2 𝑘𝑒𝑉𝑦2

2𝑔= (2)(0.3)

𝑉𝑦2

(2)(9.81 𝑚/𝑠2)

ℎ𝑓𝑣𝑦 = 0.031 𝑉𝑦2

v) Now, the total head losses in branches x and y are

ℎ𝑓𝑥 = ℎ𝑓ℎ𝑥 + ℎ𝑓𝑣𝑥 + ℎ𝑓𝑡𝑥

ℎ𝑓𝑥 = 0.637 𝑉𝑥2 + 0.015 𝑉𝑥

2 + 0.020 𝑉𝑥2 = 0.682 𝑉𝑥

2

ℎ𝑓𝑦 = ℎ𝑓𝑝𝑦 + ℎ𝑓𝑣𝑦 + ℎ𝑓𝑡𝑦 + ℎ𝑓𝑒𝑦

ℎ𝑓𝑦 = 0.202 𝑉𝑦2 + 0.0076 𝑉𝑦

2 + 0.102 𝑉𝑦2 + 0.031 𝑉𝑦

2 = 0.349 𝑉𝑦2

7 Head Loss 2 25

vi) As long as the total head loss in each branch must be equal then ℎ𝑓𝑥 = ℎ𝑓𝑦 = 0.682 𝑉𝑥

2 = 0.349 𝑉𝑦2

or 𝑉𝑥2 = 0.715 𝑉𝑦

2

vii) The volumetric flow rate in the main line may now be expressed as

𝑄 =𝜋𝑑𝑥

2𝑉𝑥4

=𝜋𝑑𝑦

2𝑉𝑦4

Now, the numerical valves may now be substituted to obtain

𝑄 = 𝜋(0.0508 𝑚)2𝑉𝑥

4+𝜋(0.03175 𝑚)2𝑉𝑦

4= 0.0063

𝑚3

𝑠

or 2.027 𝑉𝑥 + 0.792 𝑉𝑦 = 6.3

viii) The velocity in branches x and y may now be evaluated as

𝑉𝑥 = 2.05 𝑚/𝑠 𝑉𝑦 = 2.69 𝑚/𝑠

ix) The kinematic viscosity of water at 15˚C is 1.139x10-6 m2/s from Table A.2. Then the Reynolds number in branches x and y may now be evaluated as

𝑅𝑒𝑥 =𝑉𝑥𝑑𝑥𝜈

=(2.05

𝑚𝑠)(0.0508 𝑚)

(1.139𝑥10−6 𝑚2/𝑠)= 9.14𝑥104

𝑅𝑒𝑦 =𝑉𝑦𝑑𝑦𝜈

=2.69

𝑚𝑠

0.03175 𝑚

1.139𝑥10−6 𝑚2/𝑠= 7.50𝑥104

x) Now, the improved value of the friction factor for branch x corresponding to a relative roughness of (𝜀𝑠/𝑑)𝑥= 0.0009 and a Reynolds number of 𝑅𝑒𝑥 = 9.19𝑥10

4 from Moody diagram in Figure 10.5 is

𝑓𝑥 = 0.022 Similarly, the improved value of the friction factor for branch y corresponding to a relative roughness of (𝜀𝑠/𝑑)𝑦= 0.0014 and a Reynolds of 𝑅𝑒𝑦 = 7.50𝑥10

4 is

𝑓𝑦 = 0.024

7 Head Loss 2 26

xi) As long as the assumed and the improved values of the friction factors are not the same, then the calculations in steps (iii) through (xii) must be repeated. The summary of these iterations is presented in Table 10.9 xii) The volumetric flow rate through branches x and y may now be evaluated as

𝑄𝑥 =𝜋𝑑𝑥

2𝑉𝑥4

=𝜋 0.0508 𝑚 2(2.10𝑚/𝑠)

4= 4.26𝑥10−3 𝑚3/𝑠

𝑄𝑦 =𝜋𝑑𝑦

2𝑉𝑦4

=𝜋 0.03175 𝑚 2(2.59 𝑚/𝑠)

4= 2.05𝑥10−3 𝑚3/𝑠

b) The pressure drop between sections 1 and 2 may be evaluated by using the extended Bernoulli Equation as

𝑃2𝜌𝑔+𝑉22

2𝑔+ 𝑧2 =

𝑃1𝜌𝑔+𝑉12

2𝑔+ 𝑧1 − ℎ𝑓1−2

At this point, one should note that V1=V2 and z1=z2. Also the frictional head loss between sections 1 and 2 is

ℎ𝑓1−2 = ℎ𝑓𝑥 = ℎ𝑓𝑦 = 0.711 𝑉𝑥2 = 0.711 (2.10

𝑚

𝑠)2= 3.14 𝑚

The density of water is 999.1 kg/m3 from Table, therefore

𝑝2 − 𝑝1 = 𝜌 𝑔 ℎ𝑓1−2 = (999.1 𝑘𝑔/𝑚3)(9.81 𝑚/𝑠2)(3.14 𝑚)

𝑝2 − 𝑝1 = 30.78 𝑘𝑃𝑎

Iteration 1 2

𝑓𝑥 (assumed) 0.019 0.028

𝑓𝑦 (assumed) 0.021 0.024

ℎ𝑓𝑝𝑦 0.202 𝑉𝑦2 0.231 𝑉𝑦

2

ℎ𝑓ℎ𝑥 0.637 𝑉𝑥2 0.637 𝑉𝑥

2

ℎ𝑓𝑣𝑥 0.015 𝑉𝑥2 0.029 𝑉𝑥

2

ℎ𝑓𝑣𝑦 0.0076 𝑉𝑦2 0.016 𝑉𝑦

2

ℎ𝑓𝑡𝑥 0.020 𝑉𝑥2 0.045 𝑉𝑥

2

ℎ𝑓𝑡𝑦 0.102 𝑉𝑦2 0.147 𝑉𝑦

2

ℎ𝑓𝑒𝑦 0.031 𝑉𝑦2 0.073 𝑉𝑦

2

ℎ𝑓𝑥 0.682 𝑉𝑥2 0.711 𝑉𝑥

2

ℎ𝑓𝑦 0.349 𝑉𝑦2 0.467 𝑉𝑦

2

𝑉𝑥 2.05 2.10

𝑉𝑦 2.69 2.59

𝑅𝑒𝑥 9.14x104 9.37x104

𝑅𝑒𝑦 7.50x104 7.22x104

𝑓𝑥 (improved) 0.022 0.022

𝑓𝑦 (improved) 0.024 0.024

7 Head Loss 2 27

Pipe Networks In municipal water distribution systems, the pipelines are frequently branching and looping in a complex manner which is often referred as a pipe network. A simple pipe network is shown in Figure. The most important factor which makes the analysis of pipe networks difficult, is the uncertainty about the direction of the flow in different pipes one should note that, it is almost impossible to tell the direction of the flow, such as in pipe c of Figure.

The fluid flow in a pipe network must satisfy the basic principles of the conservation of mass and the conservation of energy. These principles can be summarized as follows

i) At each junction of the pipe network, the sum of the volumetric flow rates into the junction must be equal to the sum of the volumetric flow rates out of the junction.

ii) The algebraic sum of the head losses around any closed loop must be zero.

The pipe network problems are generally solved by a trial-and-error procedure. The most practical and widely used method of flow analysis in pipe networks is that of successive approximations, which is developed by Hardy Cross. In the solution procedure, which is outlined below, the pipe network should be divided into several closed loops. 1) Determine the relative roughness, ɛs/d, for each pipe from Figure 10.4 by using the

pipe diameter, d, and the pipe material.

2) Assume a friction factor, f, for each pipe from the Moody diagram by using the relative roughness, ɛs/d in the fully rough flow region.

3) Express the total head loss, hf, for each pipe in the form of

ℎ𝑓 = 𝐾𝑄2

where K is the equivalent resistance to the flow in the pipe, and Q is the unknown volumetric flow rate through the pipe.

7 Head Loss 2 28

4) Assume a value for the volumetric flow rate in each pipe such that the sum of the volumetric flow rates into each junction is equal to the sum of the volumetric flow rates out of the junction. Note that the fluid tends to follow the path of least resistance through the pipe network, so that the pipes having high values of K have relatively low volumetric flow rate.

5) Determine the algebraic sum of total head losses, 𝒉𝒇, for each closed loop by

using the following sign convention. If the flow is clockwise, then the total head loss and the volumetric flow rate are positive. However, if the flow is counterclockwise, then the total head loss and the volumetric flow rate are negative.

6) Determine the arithmetic sum of 𝟐𝑲𝑸, that is 𝟐𝑲𝑸 for each closed loop. During this summation, consider all 𝟐𝑲𝑸 values as positive.

7) Calculate the correction for the volumetric flow rate, ΔQ, for each loop from

∆𝑄 =𝑆ℎ𝑓

(2𝑄𝐾)

8) Calculate the improved values of the volumetric flow rate, Q', by using

𝑄′ = 𝑄 − ∆𝑄

9) Repeat calculations in steps (5) through (8) until ΔQ from step (7) becomes negligibly small. The Q' value is used for the next cycle of iteration.

10) Calculate the average velocity, V = 4Q/(πd2) for each pipe.

11) Calculate the Reynolds number, Re = ρVd/μ for each pipe.

12) Determine the improved value of the friction factor, f, for each pipe from the Moody diagram in Figure 10.5 by using the Reynolds number, Re, and the relative roughness, ɛs/d.

13) Compare the assumed and the improved values of the friction factor for each pipe and repeat steps (lii) through (xii) as many times as needed in order to obtain the desired accuracy in the friction factor.

7 Head Loss 2 29

Example Water at 20°C is flowing through the pipe network, as shown in Figure. All pipes are 1" cast iron, and the elbows are standard. Determine the volumetric flow rate of water in pipes a, b, c, d and e.

Solution:

The volumetric flow rate in each pipe of the network may be determined by applying the procedure, which is presented in this section: i) Relative roughness for the 1” cast iron pipes may be obtained from Figure as ii) The friction factor corresponding to this relative roughness in the fully rough flow region may be obtained from the Moody diagram as f=0.038 iii) For standard elbow, the standard tee with flow through run and the standard tee with flow through branch loss coefficient can be obtained from the table. Tee through branch- Kt1=1 Tee with line flow- Kt2=0.2 Elbow- Ke=0.3

0.01D

d

e

2

1 2 2

2

2 2

6 2

41

2

46 1(0.038) 1 0.3 0.2

0.0254 (2)(9.81 / ) (0.0254 )

2.079 10

af a t e t

a

a

QLh f K K K

D g D

Q

m m s m

Q

p

p

7 Head Loss 2 30

2

2 1 2

2

2 2

6 2

41

2

45 1(0.038) 1 0.3 0.2

0.0254 (2)(9.81 / ) (0.0254 )

2.32 10

b bfb t e t

b

b

L Qh f K K K

D g D

Q

m m s m

Q

p

p

2

2 2

2

2 2

6 2

412

2

44 1(0.038) 2(1)

0.0254 (2)(9.81 / ) (0.0254 )

2.09 10

c cfc t

c

c

L Qh f K

D g D

Q

m m s m

Q

p

p

2

2 1 2

2

2 2

6 2

41

2

44.5 1(0.038) 1 0.3 0.2

0.0254 (2)(9.81 / ) (0.0254 )

2.17 10

d dfd t e t

d

d

L Qh f K K K

D g D

Q

m m s m

Q

p

p

2

2 1 2

2

2 2

6 2

41

2

43.5 1(0.038) 1 0.3 0.2

0.0254 (2)(9.81 / ) (0.0254 )

1.87 10

e efe t e t

e

e

L Qh f K K K

D g D

Q

m m s m

Q

p

p

iv) As long as the K value for pipe is a larger than the K value for pipe B, then the volumetric flow rate in pipe will be larger. The continuity equation at junction J1 yields

3( ) 0.01 m /a bQ Q s

Therefore it is possible to assume Qa and Qb as Qa=0.004 m3/s and Qb=0.006 m3/s Now the continuity equation may be applied to junction J2 as (Qc+Qd) = Qa- 0.01 m3/s=0.003 m3/s

7 Head Loss 2 31

When the flow in pipe C is assumed to proceed from junction J21 to junction J3. Since the K value for pipe C is smaller than the K value in pipe d, then it is probable that the volumetric flow rate in pipe C will be larger. Hence Qc and Qd may be assumed Qc=0.002 m3/s and Qd=0.001 m3/s Finally, the continuity equation for junction J3 is Qe=(Qc+Qb) - 0.004 m3/s=0.004 m3/s v) The sum of the total heat losses in loop 1 may be now be calculated as

2 2 2 6 2 6 2 6 2

1 (2.61 10 )(0.004) (2.09 10 )(0.002) (2.32 10 )(0.006) 33.4f a c b a a c c b bh h h h K Q K Q K Q m

Similarly, the sum of the head losses in loop 2 is 2 2 2 6 2 6 2 6 2

2 (2.17 10 )(0.001) (1.87 10 )(0.004) (2.09 10 )(0.002) 36.11f d e c d d e e c ch h h h K Q K Q K Q m

vi) The sum of the values 2KQ for loop 1 is

6 6 6 2

1(2KQ) 2 2 (2.61 10 )(0.004) (2.609 10 )(0.002) (2.32 10 )(0.006) 57080 /a a c c b bK Q K Q K Q s m

Similarly, the sum of the values of 2KQ for loop 2 is 6 6 6 2

2(2KQ) 2 2 (2.617 10 )(0.001) (1.87 10 )(0.004) (2.09 10 )(0.002) 27640 /cd d e e cK Q K Q K Q s m

vii) The estimates of error in the assumed values of the volumetric flow rates for loop 1 and 2 are:

1 3 3

1 2

1

2 3 3

2 2

2

33.40.59 10 /

(2 ) 57080 /

36.111.31 10 /

(2 ) 27640 /

p

f

hQ m s

KQ s m

hQ m s

KQ s m

viii) The volumetric flow rate in pipes a, b and c of loop 1 may now be corrected as ' 3 3 3 3 3 3

1

' 3 3 3 3 3 3

1

' 3 3 3 3 3 3

1

4 10 / ( 0.59 10 / ) 4.59 10 /

6 10 / ( 0.59 10 / ) 5.41 10 /

2 10 / ( 0.59 10 / ) 2.59 10 /

a a

b b

c c

Q Q Q m s m s m s

Q Q Q m s m s m s

Q Q Q m s m s m s

Similarly, the correction may be applied to the volumetric flow rates of pipes c, d and e as '' ' 3 3 3 3 3 3

2

'' 3 3 3 3 3 3

2

'' 3 3 3 3 3 3

2

2.59 10 / ( 1.31 10 / ) 1.28 10 /

1 10 / ( 1.31 10 / ) 2.31 10 /

4.10 10 / ( 1.31 10 / ) 2.69 10 /

c c

d d

e e

Q Q Q m s m s m s

Q Q Q m s m s m s

Q Q Q m s m s m s

7 Head Loss 2 32

ix) In order to make Q1 and Q2 negligibility small, it is necessary to repeat the calculations. These iterations are presented in Table. x) The velocity in each pipe may now be evaluated as

3 3

2 2

3 3

2 2

3 3

2 2

3 3

2 2

2

4 (4)(4.79 10 / )9.45 /

(0.0254m)

4 (4)(5.21 10 / )10.28 /

(0.0254m)

4 (4)(1.23 10 / )12.43 /

(0.0254m)

4 (4)(2.56 10 / )5.05 /

(0.0254m)

4

aa

bb

cc

dd

ee

Q m sV m s

D

Q m sV m s

D

Q m sV m s

D

Q m sV m s

D

QV

D

p p

p p

p p

p p

p

3 3

2

(4)(2.44 10 / )4.82 /

(0.0254m)

m sm s

p

xi) The kinematic viscosity of water at 20°C is 1.003x10-6 m2/s from Table A.2, so that the Reynolds number for each pipe may be calculated as

5

6 2

5

6 2

5

6 2

6 2

(9.45 / )(0.0254 )Re 2.39 10

(1.003 10 / )

(10.28 / )(0.0254 )Re 2.60 10

(1.003 10 / )

(2.43 / )(0.0254 )Re 6.15 10

(1.003 10 / )

(5.05 / )(0.0254 )Re

(1.003 10 / )

aa

ab

ac

ad

d m s m

m s

d m s m

m s

d m s m

m s

d m s m

m s

m

m

m

m

5

5

6 2

1.28 10

(4.82 / )(0.0254 )Re 1.22 10

(1.003 10 / )

ae

d m s m

m s

m

xii) The friction factor corresponding to a relative roughness values of e/D = 0.01 and Reynolds numbers in the range 6.15x104 to 2.6x105 may be determined from the Moody diagram as f = 0.028

xiii) As long as the assumed and the improved values of the friction factor are the same, then there is no need to iterate. Consequently, the volumetric flow rate in each pipe may be given as follows:

3 3

3 3

3 3

3 3

3 3

4.79 10 /

5.21 10 /

4.23 10 /

2.56 10 /

4.44 10 /

a

b

c

d

e

Q m s

Q m s

Q m s

Q m s

Q m s

7 Head Loss 2 33

Iteration 1 2 3 4

Qa (m3/s) 4.00 ×10-3 4.59 ×10-3 4.76 ×10-3 4.79 ×10-3

Qb (m3/s) -6.00 ×10-3 -5.41 ×10-3 -5.24 ×10-3 -5.21 ×10-3

Qc (m3/s) 2.00 ×10-3 1.28 ×10-3 1.24 ×10-3 1.23 ×10-3

Qc (m3/s) -2.00 ×10-3 -1.28 ×10-3 -1.24 ×10-3 -1.23 ×10-3

Qd (m3/s) 1.00 ×10-3 2.31 ×10-3 2.52 ×10-3 2.56 ×10-3

Qe (m3/s) -4.00 ×10-3 -2.69 ×10-3 -2.48 ×10-3 -2.44 ×10-3

Sh1 -33.40 -9.49 -1.35 -0.07

Sh2 -36.11 -5.38 -0.93 -0.07

S(2kQ)1 (s/m2) 57080 54413 54344 54320

S(2kQ)2 (s/m2) 27640 25436 25395 25377

Q1 (m3/s) -0.59 ×10-3 -0.17 ×10-3 -0.03 ×10-3 -0.001 ×10-3

Q2 (m3/s) -1.31 ×10-3 -0.21 ×10-3 -0.04 ×10-3 -0.003 ×10-3

Q’a (m3/s) 4.59 ×10-3 4.76 ×10-3 4.79 ×10-3 4.79 ×10-3

Q’b (m3/s) -5.41 ×10-3 -5.24 ×10-3 -5.21 ×10-3 -5.21 ×10-3

Q’c (m3/s) 2.59 ×10-3 1.45 ×10-3 1.27 ×10-3 1.23 ×10-3

Q’’c (m3/s) -1.28 ×10-3 -1.24 ×10-3 -1.23 ×10-3 -1.23 ×10-3

Q’d (m3/s) 2.31 ×10-3 2.52 ×10-3 2.56 ×10-3 2.56 ×10-3

Q’e (m3/s) -2.69 ×10-3 -2.48 ×10-3 -2.44 ×10-3 -2.44 ×10-3

7 Head Loss 2 34

Interconnected Reservoir Systems Another example of pipe systems, which has practical importance in water supply systems is when three or more reservoirs at various elevations are interconnected at a joint. Figure shows three interconnected reservoirs. Reservoir A is at the highest elevation, reservoir B is at the intermediate elevation and reservoir C is at the lowest elevation. Three pipes 1, 2 and 3 meet at a common junction J, and Q1 . Q2 and Q3 are the volumetric flow rates in the respective pipes. It is obvious that the flow takes place from the highest reservoir, that is reservoir A, towards the junction, and then from the junction to the lowermost reservoir, that is reservoir C. However, the direction of the flow in the second pipe is still uncertain. If the total head at the free surface of reservoir B is greater than the total head at junction J, then the flow takes place from reservoir B towards junction J. In this case, the continuity equation at junction J becomes

𝑄1 + 𝑄2 = 𝑄3 𝑤ℎ𝑒𝑛 ℎ𝑡𝐵 > ℎ𝑡𝐽

However, when the total head at the free surface of the reservoir B is smaller than the total head at junction J, then the flow takes place from junction J towards reservoir B. Hence, the continuity equation at junction J is 𝑄1= 𝑄2 + 𝑄3 𝑤ℎ𝑒𝑛 ℎ𝑡𝐵 < ℎ𝑡𝐽

7 Head Loss 2 35

The procedure for determining the volumetric flow rate for each pipe may be given as below: 1) Calculate the total head, ht, at the free surface of each reservoir.

2) Determine the relative roughness, ɛs/d, for each pipe by using the pipe diameter, d, and the pipe material.

3) Assume a friction factor, f, for each pipe from the Moody diagram by using the relative roughness, ɛs/d, in the fully rough flow region. The reason for the assumption of the flow in the fully rough region is that no iteration is necessary if this assumption is true.

4) Express the total head loss, hf, for each pipe in the form of hf = KQ2, where K is the equivalent resistance to the flow and Q is the unknown volumetric flow rate through the pipe.

5) Assume the total head for the junction, and determine the direction of flow in pipes which are connected to intermediate reservoirs. Write the continuity equation for the junction.

6) Write the extended Bernoulli equation along the streamlines between the free surface of each reservoir and the junction.

7) Determine the volumetric flow rate, Q, for each pipe,

8) Determine whether the continuity equation is satisfied at the junction or not. If the continuity equation is not satisfied, then repeat the calculations in steps (5) through (7) as many times as necessary.

9) Calculate the average velocity, V = 4Q/(πd2) for each pipe,

10) Calculate the Reynolds number, Re = ρVd/μ for each pipe,

11) Determine the improved value of the friction factor, f, for each pipe from the Moody diagram in Figure 10.5 by using the Reynolds number, Re, and the relative roughness, ɛs/d.

12) Compare the assumed and the improved values of the friction factor for each pipe and repeat steps (4) through (9) as many times as needed in order to obtain the desired accuracy in the friction factor.

7 Head Loss 2 36

Example: Water at 20°C is flowing through the three reservoir system, which is shown in Figure. All pipes are cast iron. Neglecting all minor head losses, determine the volumetric flow rate for each pipe.

Solution

The volumetric flow rate in each pipe may be determined by applying the procedure, which is presented in this section: 1) Since the cross-sectional area of the reservoirs are very large compared to the cross-sectional area of the pipes, then the velocities at the free surface of the reservoirs are negligible that is VA = VB = VC= 0 Therefore, the total head at the free surfaces of the reservoirs are

2 2

3 2

2 2

3 2

2 2

3 2

101325 /700 710.35

2 (998.2 / )(9.81 / )

101325 /450 460.35

2 (998.2 / )(9.81 / )

101325 /100 110.35

2 (998.2 / )(9.81 / )

A AtA A

B BtB B

C CtC C

P V N mh z m m

g g kg m m s

P V N mh z m m

g g kg m m s

P V N mh z m m

g g kg m m s

r

r

r

7 Head Loss 2 37

2) The relative roughness of cast iron pipes are obtained from the diagram: For d1 = 0.3 m (es/d)1 = 0.00084 For d2 = 0.35 m (es/d)2 = 0.00072 For d3 = 0.4 m (es/d)3 = 0.00062 3) The friction factors for corresponding relative roughness values are obtained from the fully rough region as f1 = 0.0185 f2 = 0.018 f3 = 0.0175 4) Since the minor head losses are neglected, the total head loss for each pipe may be evaluated from the Darcy-Weisbach equation as follows:

2 2

21 1 11 1 12 2 2

1 1

2 2

22 2 22 2 22 2 2

2 2

2

3 33 3 2

3 3

4 41 200 10.0185 125.81

2 0.3 2 9.81 / (0.3 )

4 41 300 10.018 84.95

2 0.35 2 9.81 / (0.35 )

41 4000.0175

2

f

f

f

L Q Qmh f Q

d g d m m s m

L Q Qmh f Q

d g d m m s m

L Q mh f

d g d

p p

p p

p

2

2332 2

4156.48

0.4 2 9.81 / (0.4 )

QQ

m m s mp

5) The total head loss at junction J may me assumed as htj = 400 m For this case htA > htB > htJ > htC Hence water is flowing from reservoir A and B into reservoir C. As a result, the continuity equation at junction J becomes Q1 + Q2 = Q3 6) The extended Bernoulli equation may now be applied along the streamlines between the free surfaces of the reservoirs and the junction J as

1

2

3

tA tJ f

tB tJ f

tJ tC f

h h h

h h h

h h h

7 Head Loss 2 38

7) The volumetric flow rate in each pipe may now be determined as

1/2 1/2

3

1

1/2 1/2

3

2

1/2 1/2

3

3

710.35 4001.57 /

125.81 125.81

460.35 4000.84 /

84.95 84.95

400 110.352.27 /

56.48 56.48

tA tJ

tB tJ

tJ tC

h hQ m s

h hQ m s

h hQ m s

8) Continuity equation at junction J Q1 + Q2 = Q3 1.57 + 0.84=2.27 2.41 > 2.27 Ass seen the continuity is not satisfied at junction J. Fluid flowing into the junction is greater than the fluid flowing out from the junction. To satisfy the continuity, flow rate from reservoirs A and B to junction J should decrease and flow rate from junction J to reservoir C should increase. For this to happen, the total head at junction J should be increased and the calculations in step 7 should be repeated. The summary of the iterations is presented in the table below. As a result of 4 iterations, the flow rates in pipes are obtained as, Q1 = 1.54 m3/s Q2 = 0.76 m3/s Q3 = 2.30 m3/s 9) Now the velocity in each pipe may be calculated as below:

11 2 2

1

22 2 2

2

33 2 2

3

4 4 1.5421.79 /

(0.3 )

4 4 0.767.90 /

(0.35 )

4 4 2.3018.30 /

(0.4 )

QV m s

d m

QV m s

d m

QV m s

d m

p p

p p

p p

7 Head Loss 2 39

10) The kinematic viscosity of water at 20 oC is 1.003×10-6 m2/s, then the Reynolds number for each pipe may be evaluated as

61 11 6

62 22 6

63 33 6

21.79 0.3Re 6.52 10

1.003 10

7.90 0.35Re 2.76 10

1.003 10

18.30 0.4Re 7.30 10

1.003 10

V d

V d

V d

11) Friction factors corresponding to the relative roughness and Reynolds numbers for each pipe are obtained from Moody diagram as follows: f1 = 0.0185 f2 = 0.018 f3 = 0.0175 12) As seen the assumed and the improved values of friction factors are the same for each pipe. Hence there is no need to iterate. Volumetric flow rate for each pipe will be as Q1 = 1.54 m3/s Q2 = 0.76 m3/s Q3 = 2.30 m3/s

Iteration 1 2 3 4 htJ (m) 400 420 410 411

Q1 (m3/s) 1.57 1.52 1.55 1.54

Q2 (m3/s) 0.84 0.69 0.77 0.76

Q3 (m3/s) 2.27 2.34 2.30 2.30

Q1+Q2+Q3 (m3/s) 0.14 -0.13 0.02 0.00

Change in htJ (m) increase Decrease Increase O.K.