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Chapter 12

Q. 12.1

In Fig. 12.2, 100 gal/min of water at 60°F is flowing in a 2-in Schedule 40 steel pipe at section 1. The heat exchanger in branch a has a loss coefficient of K = 7.5 based on the velocity head in the pipe. All three valves are wide open. Branch b is a bypass line composed of 1¼-in Schedule 40 steel pipe. The elbows are standard. The length of pipe between points 1 and 2 in branch b is 20 ft. Because of the size of the heat exchanger, the length of pipe in branch a is very short and friction losses can be neglected. For this arrangement, determine (a) the volume flow rate of water in each branch and (b) the pressure drop between points 1 and 2.

Step-by-Step

Verified Solution

If we apply Step 1 of the solution method, Eq. (12–1) relates the two volume flow rates. How many quantities are unknown in this equation?


The two velocities v_a and v_b are unknown. Because Q = A_v , Eq. (12–1) can be expressed as

Q_1 = Q_2 = Q_a + Q_b     (12–1)

Q_1 = A_v v_a + A_b v_b     (12–3)

From the given data, A_a = 0.02333ft^2,A_b = 0.1039ft^2 , and Q_1 = 100gal/min. Expressing Q_1  in the units of ft^3/s gives

Q_1 = 100gal/min \times \frac{1ft^3/s}{449gal/min}=0.223ft^3/s

Generate another equation that also relates v_a and v_b , using Step 2.


Equation (12–2) states that the head losses in the two branches are equal. Because the head losses h_a and h_b are dependent on the velocities v_a and v_b, respectively, this equation can be used in conjunction with Eq. (12–3) to solve for the velocities. Now, express the head losses in terms of the velocities for each branch.


You should have something similar to this for branch a:

h_a = 2K_1 \left(v^2_a/2g\right) +K_2 \left(v^2_a/2g\right)

where

K_1 = f_{aT} \left(L_e/D\right) = Resistance coefficient for each gate valve

K_2 = Resistance coefficient for the heat exchanger = 7.5 (given in problem statement)

The following data are known:

f_{aT} = 0.019 for a 2–in Schedule 40 pipe (Table 10.5)

L_e/D = 8 for a fully open gate valve (Table 10.4)

Then,

K_1 = \left(0.019\right) \left(8\right) =0.152

Then,

h_a = \left(2\right) \left(0.152\right) \left(v^2_a/2g\right) +7.5 \left(v^2_a/2g\right)=7.80 \left(v^2_a/2g\right)       (12–4)

For branch b:

h_b = 2K_3 \left(v^2_b/2g\right) +K_4 \left(v^2_b/2g\right)+K_5 \left(v^2_b/2g\right)

where

K_3 = f_{bT}\left(L_e/D\right) = Resistance coefficient for each elbow

K_4 = f_{bT}\left(L_e/D\right) = Resistance coefficient for the globe valve

K_5 = f_{b}\left(L_b/D\right) = Friction loss in the pipe of branch b for a pipe length of L_b = 20ft.

The value of f_b  is not known and will be determined through iteration. The known data are

f_{bT}=0.021 for a 1¼-in Schedule 40 pipe (Table 10.5)

L_e/D = 30 for each elbow (Table 10.4)

L_e/D = 340 for a fully open globe valve (Table 10.4)

Then,

K_3 = \left(0.021\right) \left(30\right) =0.63

K_4 = \left(0.021\right) \left(340\right) =7.14

K_{5}=f_{b}(20 / 0.1150)=173.9 f_{b}

Then,

h_b = \left(2\right) \left(0.63\right) \left(v^2_b/2g\right) + \left(7.14\right) \left(v^2_b/2g\right) +f_b \left(173.9\right) \left(v^2_b/2g\right)

h_b = \left(8.40+173.9f_b\right) \left(v^2_b/2g\right)

This equation introduces the additional unknown, f_b. We can use an iteration procedure similar to that used for Class II series pipeline systems in Chapter 11. The relative roughness for branch b will aid in the estimation of the first trial value for f_b:

D/\epsilon = \left(0.1150/1.5 \times 10^{-4}\right)=767

From the Moody diagram in Fig. 8.7, a logical estimate for the friction factor is f_b= 0.023 . Substituting this into the equation for h_b gives

h_b = \left[8.40+173.9\left(0.023\right) \right] \left(v^2_b/2g\right) =12.40\left(v^2_b/2g\right)      (12-5)

We now have completed Step 3 of the solution procedure. Steps 4 and 5 can be done now to obtain an expression for v_a in terms of v_b.


You should have v_a = 1.261v_b,obtained as follows:

h_a = h_b

7.80 \left(v^2_a/2g\right) = 12.40\left(v^2_b/2g\right)

Solving for v_a gives

v_a =1.261v_b        (12–6)

At this time, you can combine Eqs. (12–3) and (12–6) to calculate the velocities (Steps 6 and 7).


The solutions are v_a = 5.60 ft/s and v_b = 7.06 ft/s. Here are the details:

Q_1 = A_a v_a+A_b v_b   (12-3)

v_a =1.261 v_b    (12-6)

Then, we have

Q_1 = A_a \left(1.261v_b\right) + A_b v_b =v_b \left(1.261A_a +A_b\right)

Solving for v_b, we get

v_b = \frac{Q_1}{1.261A_a+A_b}=\frac{0.223ft^3/s}{\left[\left(1.261\right) \left(0.02333\right)+0.01039 \right]ft^3 }

v_b = 5.60ft/s

v_a = \left(1.261\right)\left( 5.60\right)ft/s = 7.06ft/s

Because we made these calculations using an assumed value for f_b, we should check the accuracy of the assumption.
We can evaluate the Reynolds number for branch b:

N_{Rb} = v_b D_b/\nu

From Appendix A, Table A.2, we find \nu = 1.21 \times 10^{-5} ft^2/s . Then,

N_{Rb} = \left(5.60\right) \left(0.1150\right)/\left(1.21 \times 10^{-5}\right) =5.32 \times 10^4

Using this value and the relative roughness of 767 from before, in the Moody diagram, yields a new value, f_b = 0.0248. Because this is significantly different from the assumed value of 0.023, we can repeat the calculations for Steps 3–8. The results are summarized as follows:

h_b = \left[8.40+173.9\left(0.0248\right) \right] \left(v^2_b/2g\right) = 12.71\left(v^2_b/2g\right)

h_a = 7.80\left(v^2_a/2g\right) (same as for first trial)

Equating the head losses in the two branches gives

h_a = h_b

7.80\left(v^2_a/2g\right) = 12.71\left(v^2_b/2g\right)

Solving for the velocities gives

v_a = 1.277v_b

Substituting this into the equation for v_b  used before gives

v_b = \frac{0.223ft^3/s}{\left[\left(1.277\right)\left(0.02333\right)+0.01039 \right] ft^2} = 5.55ft/s

v_a = 1.277v_b = 1.277\left(5.55\right) =7.09ft/s

Recomputing the Reynolds number for branch b gives

N_{Rb} = v_b D_b/\nu

N_{Rb} = \left(5.55\right) \left(0.1150\right) /\left(1.21 \times 10^{-5}\right) =5.27 \times 10^4

There is no significant change in the value of f_b. Therefore, the values of the two velocities computed above are correct. We can now complete Steps 10 and 11 of the procedure to find the volume flow rate in each branch and the head loss and the pressure drop across the entire system.

Now calculate the volume flow rates Q_a  and Q_b (Step 10).


You should have

Q_a = A_a v_a = \left(0.02333ft^2\right)\left(7.09ft/s\right) =0.165ft^3/s

Q_b = A_b v_b = \left(0.01039ft^2\right)\left(5.55ft/s\right) =0.0577ft^3/s

Converting these values to the units of gal/min gives Q_a = 74.1 gal/min and Q_b = 25.9 gal/min.

We are also asked to calculate the pressure drop. How can this be done?


We can write the energy equation using points 1 and 2 as reference points. Because the velocities and elevations are the same at these points, the energy equation is simply

\frac{p_1}{\gamma} -h_L = \frac{p_2}{\gamma}

Solving for the pressure drop, we get

p_1 – p_2 = \gamma h_L     (12–7)

What can be used to calculate h_L ?


Because h_{L_{1-2}} = h_a = h_b , we can use either Eq. (12–4) or (12–5). Using Eq. (12–4), we get

h_a = 7.80\left(v^2_a/2g\right) = \left(7.80\right)\left(7.09\right)^2/64.4ft = 6.09 ft

Note that this neglects the minor losses in the two tees. Then, we have

p_1 – p_2 = \gamma h_L = \frac{62.4lb}{ft^3} \times 6.09ft \times \frac{1ft^2}{144in^2} = 2.64psi

This example problem is concluded.

TABLE 10.5 Friction factor in zone of complete turbulence for new, clean, commercial Schedule 40 steel pipe

U.S. (in) Metric (mm) Friction factor, f_T
½ DN 15 0.026
¾ DN 20 0.024
1 DN 25 0.022
DN 32 0.021
DN 40 0.02
2 DN 50 0.019
DN 65 0.018
3, 3½ DN 80, DN 90 0.017
4 DN 100 0.016
5.6 DN 125, DN 150 0.015
8 DN 200 0.014
10–14 DN 250 to DN 350 0.013
16–22 DN 400 to DN 550 0.012
24–36 DN 600 to DN 900 0.011

TABLE 10.4 Resistance in valves and fittings expressed as equivalent length in pipe diameters, L_{e} ,/D

Type Equivalent Length in Pipe Diameters L_e/D
Globe valve—fully open 340
Angle valve—fully open 150
Gate valve—fully open 8
—¾ open 35
—½ open 160
—¼ open 900
Check valve—swing type 100
Check valve—ball type 150
Butterfly valve—fully open, 2–8 in 45
—10–14 in 35
—16–24 in 25
Foot valve—poppet disc type 420
Foot valve—hinged disc type 75
90° standard elbow 30
90° long radius elbow 20
90° street elbow 50
45° standard elbow 16
45° street elbow 26
Close return bend 50
Standard tee—with flow through run 20
—with flow through branch 60

(Reprinted with permission from “Flow of Fluids Through Valves, Fittings and Pipe, Technical Paper 410” 2011. Crane Co. All Rights Reserved.)

Table A.2 U.S. Customary System units (14.7 psia)

Kinematic Viscosity \nu \left(ft^2/s\right) Dynamic Viscosity \eta \left(lb–s/ft^2\right) Density \rho \left(slugs/ft^3\right) Specific Weight \gamma \left(lb/ft^3\right) Temperature (°F)
1.89 \times 10^{-5} 3.66 \times 10^{-5} 1.94 62.4 32
1.67 \times 10^{-5} 3.23 \times 10^{-5} 1.94 62.4 40
1.40 \times 10^{-5} 2.72 \times 10^{-5} 1.94 62.4 50
1.21 \times 10^{-5} 2.35 \times 10^{-5} 1.94 62.4 60
1.05 \times 10^{-5} 2.04 \times 10^{-5} 1.94 62.3 70
9.15 \times 10^{-6} 1.77 \times 10^{-5} 1.93 62.2 80
8.29 \times 10^{-6} 1.60 \times 10^{-5} 1.93 62.1 90
7.37 \times 10^{-6} 1.42 \times 10^{-5} 1.93 62 100
6.55 \times 10^{-6} 1.26 \times 10^{-5} 1.92 61.9 110
5.94 \times 10^{-6} 1.14 \times 10^{-5} 1.92 61.7 120
5.49 \times 10^{-6} 1.05 \times 10^{-5} 1.91 61.5 130
5.03 \times 10^{-6} 9.60 \times 10^{-6} 1.91 61.4 140
4.68 \times 10^{-6} 8.90 \times 10^{-6} 1.9 61.2 150
4.38 \times 10^{-6} 8.30 \times 10^{-6} 1.9 61 160
4.07 \times 10^{-6} 7.70 \times 10^{-6} 1.89 60.8 170
3.84 \times 10^{-6} 7.23 \times 10^{-6} 1.88 60.6 180
3.62 \times 10^{-6} 6.80 \times 10^{-6} 1.88 60.4 190
3.35 \times 10^{-6} 6.25 \times 10^{-6} 1.87 60.1 200
3.17\times 10^{-6} 5.89 \times 10^{-6} 1.86 59.8 212