diameter for the network.
In Section 4.3, these three types of problems have been applied to straight pipe lengths. The problems are no different in a complex pipe network. In addition to straight runs of pipe, there are valves and fittings as well.
EXAMPLE 4.8
Liquid cyclohexane at 35°C is pumped at a volumetric flow rate of 10 L/s through a pipe network to a large storage tank as shown in Figure E4.8.
Determine the pressure drop (kPa) from the pump discharge (A) to the pipe exit (B).
Threaded Flanged Welded Soldered Bell & spigot
FIGURe 4.5
Symbols used to indicate joining methods in pipe network drawings.
SOLUTION: The pressure drop from A to B is equal to the pressure loss in the pipe network including the head loss due to friction and the minor losses due to the valves and fittings. Applying the conservation of energy equation between points A and B results in
γ + =
∑
Rearranging this equation to solve for the pressure drop that the pump must overcome results in The properties of the cyclohexane can be found from Appendix B.1
γ = ρ =
3600 s 0.0007502 kg m s
3 2
2
g 3
The inside diameter of the pipe can be found in Appendix C and the roughness of the pipe can be found in Table 4.3.
D = 7.7927 cm = 0.077927 m
cast iron steel pipe
All elbows are standard 10 m
B
FIGURe e4.8
From the drawing, the elevations of points A and B as well as the length of straight pipe in the network are given as follows:
= = =
zA 0 m zB 10 m L 50 m
To determine the friction factor, the Reynolds number and relative roughness are required. To determine the Reynolds number, the velocity of the cyclohex-ane in the pipe is needed. This velocity can be found by,
4 4 10L s
m 1000 L
0.077927 m 2.097m s
The Reynolds number can now be found.
( )
The relative roughness of the pipe can also be found.
ε = 0.0026 m = 0.077927 m 0.03336 D
Using the Reynolds number and the relative roughness, the friction factor can be found using any of the correlations in Table 4.2. Using the Swamee–Jain correlation,
The only task left is to determine the total effect of the minor losses due to the valves and fittings. The check valve size is not given. However, the size can be specified by ensuring that the cyclohexane velocity in the pipe completely lifts the disc from its seat inside the valve. The minimum velocity required to fully lift the disc (Appendix E) is
= β
ρ 40
min
V 2
For a 3-in. check valve, β = 1. Therefore,
Notice that this calculation had to be done in English units before converting the velocity to the SI system. This is because expression used in Appendix E to calculate the minimum velocity is empirical. It was developed to use density in lbm/ft3 and the velocity in ft/s. This result indicates that the pipe velocity is higher than the minimum velocity required to completely lift the disc in the check valve. Therefore, a 3-in. check valve will work in this application, and no reducers are needed to connect the check valve to the pipe. Therefore, the loss coefficients required are
∑
K K= check+Kgate+4Kelbow+KexitThe loss coefficients for the individual valves and fittings can be found in Appendix E. For the check valve,
= 600
check
K fT
The pipe is made of cast iron. Therefore, the value of fT is computed using the modified Colebrook correlation shown in Appendix E (the table of fT values in Appendix E, is for commercial steel pipe).
= ε
3.7 0.25 log 0.03336
3.7 0.05978
2 2
f D
T
Using the calculated value of fT, the loss coefficient for the check valve can be found.
( )
=600 =600 0.05978 =35.87
check
K fT
The loss coefficients for the gate valve, elbows, and pipe exit can be determined from the appropriate loss coefficient expressions and the calcu-lated value of fT. 30 30 0.05978 1.79 1.00
Often pipe or tube networks have parallel runs. Figure 4.6 shows a sketch of a pipe system with two parallel flow paths, labeled 1 and 2. The label “c”
indicates combined flow. To analyze a system like this, a system boundary is sketched around the individual pipes in the network. Then the conservation of mass and conservation of energy equations are used. In Figure 4.6, if a sys-tem boundary was sketched around pipe 1 from A to B and a second syssys-tem boundary around pipe 2 from A to B, it must be true that the pressure drop from A to B is the same through each parallel leg.
A B 1
)
A B 2)
(
P −P =(
P −P (4.26) These values can be used to determine the total loss coefficient for the minor losses in the system.∑
K=35.87 0.48 4 1.79+ + ( )+1.00 44.51=Now, everything is known except the pressure drop from point A to B. Using the conservation of energy equation developed above,
1000 Pa 214.6 kPa
B A
It is interesting to determine the magnitude of the head loss due to the straight pipe compared to the head loss due to the valves and fittings. For the system analyzed in this problem,
( ) ∑
In this case, the minor losses are larger than the head loss due to friction in the straight pipe! This is due to the check valve. Notice that its K value is much larger than the other gate valve, elbows, and exit.
If the diameters of each leg are constant, the conservation of energy between points A and B for each leg of the network can be written as follows:
2 2
branch,A branch,B c2
P P z z f L
Notice that the K values for the tees are separated out from the other fit-tings. This is because the minor losses of the branch and straight runs of the tee are associated with the combined velocity of the flow entering (or leaving) the tee. In Equations 4.27 and 4.28, Vc is the combined flow velocity. Since the pressure drop through each parallel leg is the same,
2 2
branch,A branch,B c2
f L
The following example illustrates how to apply these equations to a net-work with parallel flow paths.
c
A parallel pipe network (plan view).
EXAMPLE 4.9
Figure E4.9 shows a parallel flow copper tube network. Water at 70°F is flowing through the network such that the pressure drop from A to B is 0.2 psi.
Determine the volumetric flow rate (gpm) through each leg of the paral-lel network and also determine the total volumetric flow (gpm) through the network.
SOLUTION: The following data pertinent to the system can be found immediately:
3600 s 0.0006552lbm ft s
0.06758 ft 0.000005 ft 6 ft 6 2 2 ft 10 ft
Appendix D Table 4.3
1 2
D
L L
The conservation of energy equations for each tube in the system are shown as follows:
branch,A branch,B c2
(Parallel run 2)
The friction factors in each tube require the calculation of the Reynolds num-ber and the relative roughness. Since the volumetric flow rate in each tube is not known, this becomes a type 2 iterative solution.
2 (Top view)
In tube 1, there are two minor losses associated with the tees. The tee at A is a diverging tee and the tee at B is converging. The losses are those associ-ated with the value of Krun for each tee. In tube 2, the minor losses include a diverging tee branch (A), two elbows, a gate valve, and a converging tee branch (B). The K values for the elbows and gate valve can be easily determined from Appendix E. However, when calculating the K values for the tees, the value of yb is unknown since the volumetric flow through each leg is to be determined.
This compounds the iterative nature of this calculation.
Because of the iterative nature of this problem, it is most convenient to solve it using an equation solver such as EES. The EES code for this problem will be developed step by step. The preliminary information given in the problem is stated first, along with the calculated properties of the water.
"GIVEN: A parallel flow system as shown"
f$ = ‘steam_iapws’
T = 70[F]
rho = density(f$,T=T,x=0) gamma = rho*g/g_c
mu = viscosity(f$,T=T,x=0)*convert(lbm/ft-hr,lbm/ft-s)
$IFNOT Parametric Study DELTAP_AB = 0.2[psi]
$ENDIF
"Copper tubing"
D = 0.06758[ft] "Appendix D"
epsilon = 0.000005[ft] "Table 4.3"
L_1 = 6[ft]
L_2 = 2*2[ft] + 6[ft]
"Constants"
g = 32.174[ft/s^2]
g_c = 32.174[lbm-ft/lbf-s^2]
"FIND: Volume flow rate through each leg"
Notice that the pressure drop (DELTAP_AB) is inside of a parametric study directive. There are several directive commands that can be used in EES. The parametric study directive allows the user to build a Parametric Table and vary the variable(s) included inside the parametric study directive. When a normal
“Solve” command is issued, the solver uses the value within the directive.
However, when the “Solve Table” command is executed, the solver ignores the value within the directive and uses the value(s) listed in the Parametric Table.
Use of this directive eliminates the need to comment out variables used in a parametric study.
The conservation of energy equations for each leg of the parallel network are shown previously. In EES format, these equations are written as follows:
SOLUTION:"
"The pressure drop from A to B is the same for both circuits"
"Tube 1"
DELTAP_AB*convert(psi,lbf/ft^2)/gamma = f_1*(L_1/D)*(V_1^2/
(2*g))+ (K_run_A + K_run_B)*(V_c^2/(2*g))
&
"Tube 2"
DELTAP_AB*convert(psi,lbf/ft^2)/gamma = (f_2*(L_2/D) + K_2) &
(V_2^2/(2*g)) + (K_branch_A + K_branch_B)*(V_c^2/(2*g)) The friction factors for each leg of the network can be calculated using the built-in Moody Chart function in EES, or any of the formulations shown in Table 4.2. Here, the Swamee–Jain correlation is used.
"Friction factor in each tube,"
f_1 = 0.25*log10((epsilon/D)/3.7 + 5.74/Re_1^0.9)^(-2) Re_1 = rho*V_1*D/mu
Vol_dot_1*convert(gpm,ft^3/s) = A*V_1
f_2 = 0.25*log10((epsilon/D)/3.7 + 5.74/Re_2^0.9)^(-2) Re_2 = rho*V_2*D/mu
Vol_dot_2*convert(gpm,ft^3/s) = A*V_2 A = pi*D^2/4
The loss coefficients need to be found. The most complex calculation here is associated with the tees. The data for the tees are as follows:
"Loss coefficieints"
"Tee data"
beta = 1
alpha = 90[deg]
y_b = Vol_dot_2/Vol_dot_c
In the straight tube, the loss coefficients can be found by the following:
"Straight leg (subscript 1)"
K_run_A = BigM*y_b^2
BigM = IF(y_b,0.5,2,2,0.3)*(2*y_b - 1) K_run_B = 1.55*y_b - y_b^2
Notice in this set of EES equations, M in Equation 4.25 is assigned the EES variable BigM. This is done because EES variables are not case sensitive and m is often used to represent a mass (although it is not in this particular problem).
Tricks such as this help avoid unintentional variable overlapping, which can make debugging difficult. Also used in this set of equation is the IF statement.
The format of the IF statement is
( )
=
Q IF A,B,x,y,z
Having a program such as EES to solve Example 4.9 eliminates the tedious manual iterative calculations. However, another feature of most equation solvers is their ability to conduct and plot the results of parametric studies.
Consider a case where you are interested in how the flow splits between the two tubes of the parallel system shown in Example 4.9 as the pressure drop between points A and B varies. This study can be done by setting up a Parametric Table in EES. The results can then be plotted for a visual display of the behavior of the system, as shown in Figure 4.7.
EES compares A and B within the IF statement and assigns the following actions: if A < B, then “x” is executed, if A = B, then “y” is executed, or if A > B, then “z” is executed. In the IF statement, A,B,x,y,z do not need to be constant values. They can also be expressions.
The minor losses in tube 2 of the parallel system can be calculated as follows:
"Tube with elbows and valve (Tube 2)"
K _ 2 = 2*K _ el + K _ gv K _ el = 30*f _ T
K _ gv = 8*f _ T
f _ T = 0.25*(log10((epsilon/D)/3.7))^(-2)
"In leg 2, the tee flow is a branch flow,"
K _ branch _ A = BigG*(1 + BigH*(y _ b/beta^2)^2 - &
BigJ*(y _ b/beta^2))*cos(alpha) BigG = 1 + 0.3*y _ b^2
BigH = 0.3 BigJ = 0
K _ branch _ B = BigC*(1 + BigD*(y _ b/beta^2)^2 - &
BigE*(1 - y _ b)^2 - BigF*(y _ b/beta)^2) BigC = 0.55
BigD = 1 BigE = 1 BigF = 0
Two more equations are needed: one that allows for the solution of the com-bined flow rate and one that allows for the calculation of the comcom-bined flow velocity. Both of these values are needed in the calculation of the loss coef-ficients for the tees.
"The combined flow and combined velocity are"
Vol_dot_c = Vol_dot_1 + Vol_dot_2 Vol_dot_c*convert(gpm,ft^3/s) = A*V_c
The solution to this iterative set of equations isV 1= 4.747 gpm, V 2= 3.129 gpm, andV c= 7.876 gpm. This result indicates that tube 1, the straight run, carries more of the flow. This is to be expected because tube 1 has a lower head loss due to friction and fittings. In addition, the length of straight tube in tube 1 is smaller than in tube 2, resulting in a lower friction loss.
4.6 Economic Pipe Diameter
Many possible pipe diameters can be used to transport a fixed flow rate through a pipe. A small pipe has a low initial cost, but the pressure drop may be quite high resulting in a significant pumping cost to move the fluid through the pipe. On the other hand, a large pipe will have a significant first cost, but the pressure drop is much lower resulting in a low pumping cost.
This thinking leads one to believe that there must be a pipe diameter that minimizes the total cost of the pipe system. This diameter is known as the economic pipe diameter or optimum pipe diameter.
The expressions developed in this section are based on the work of Darby and Melson (1982) and further refined by Janna (2015). Even though many simplifying assumptions are made as the method is developed, the resulting expressions provide a very reasonable estimate to the economic diameter.
4.6.1 Cost of a Pipe System
There are many costs that are incurred in a pipe system. However, these costs can be grouped into two categories: capital (i.e., initial) and annual costs. The capital cost of a pipe system includes the cost of the pipe itself and costs associated with installation, valves, fittings, pumps, insulation, hangers, and supports. The costs incurred annually are the cost of energy required to move the fluid through the pipe system and any maintenance costs required.
Since capital and annual costs occur at different times, economic interest factors must be considered to align all costs to a common point on the cash
00 4 8 12 16 20 24 28 32 36 40
1 2 3 4
ΔPAB (psi)
5 6 7
Flow through tube 2 Flow through tube 1
8 9 10
V (gpm).
FIGURe 4.7
Result of a parametric study for example problem 4.9.
flow diagram for the pipe system. In the method presented here, costs are converted to annual costs. Interest is assumed to be compounded annually.
4.6.2 Determination of the economic Diameter
The strategy used to determine the economic diameter is to develop a cost function that represents the total annual costs of the pipe system as a function of the pipe diameter.
ACT= f D (4.30)(
)
The diameter that results in the lowest annual cost is the economic diameter.
This diameter can be found by taking the derivative of Equation 4.30, setting it equal to zero, and solving for D. To ensure that this value is indeed a mini-mum, the second derivative can be used.
The total annual cost of the pipe system is defined as
ACT=ACP+ACH+ACM+ACE (4.31) In this equation, ACP is the annual cost of the pipe in the system, including installation; ACH the annual cost of the hardware including valves, fittings, pumps, insulation, supports, and so on; ACM the annual cost of maintenance for the system; and ACE the annual cost of the energy required to move the fluid through the system.
The initial cost of installed pipe per foot is determined from cost data col-lected in 1980. These data can be fit to the following equation:
ICP = +(1
)
(yr 1980− ) 1L e C Ds (4.32)
In this equation, C1 and s are constants. C1 represents the installed cost per foot of a 12-nom pipe based on 1980 dollars. The exponent s corresponds to various pipe classes. The values of C1 and s are given in Table 4.10. The term in parenthesis involving the variables, e and yr, is a way to convert the constant C1 from 1980 dollars to the current dollars. The variable e is an average yearly escalation rate (expressed in decimal form) from 1980 to present, which takes into account inflation. The variable, yr, is the year of installation of the fluid transport system. Equation 4.32 is valid for pipe diameters up to 36 in.
Table 4.10
Constants C1 and s in Equation 4.32
Pipe Class (max working pressure) C1($/fts+1) s
300 (720 psig) 22.50 1.14
400 (960 psig) 23.14 1.20
600 (1440 psig) 28.28 1.29
900 (2160 psig) 36.10 1.32
1500 (3600 psig) 53.39 1.35
Source: Janna, W. S., Design of Fluid Thermal Systems, Stamford, CT, Cengage Learning, 2015.