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## 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. ## 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 1¼ DN 32 0.021 1½ DN 40 0.02 2 DN 50 0.019 2½ 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 