Thermal Fluid System Design

Design Problem #2

Introduction

The size of a piping network is a crucial aspect in designing a water supply system. The

network itself can contribute to 15-20% of the total cost of the system1, thus making it an

important parameter to optimize. This problem entails designing a piping system to supply water

to three distinct points (A, B, and C) in a residential area, as depicted in Figure 1. Complicating

the design, is the fact that the pressure drop, height of the storage tank, dimensions of the storage

tank, and pump size are all unknown, making this a realistic problem. With only the flow rates

known for each point, as shown in Table 1, along with the knowledge that points A and C run

(intermittently) during an 8 hour working day, while point B runs (intermittently) during non-

working hours, is the only information given regarding the requirements.

Further stipulations include that the piping must be within a foot border surrounding the

building and that the piping cannot go over, through or under the building. With these guidelines

known, it is possible to perform a mathematical analysis to aid in calculating each requirement,

as well as providing the necessary assumptions to complete the design.

Figure 1: Schematic for Design

Location A B C

gpm 40 55 25

hrs of

operation 8 16 8

Table 1: Flow Rates for Each Consumption Point

Analysis

Due to the comprehensive nature of this problem, and the fact that multiple parameters

rely on subsequent calculations, the analysis will be broken up into three sections.

Section 1: Discharge piping size

Section 2: Tank height

Section 3: Tank dimensions, supply piping size, and pump selection

To be able to size the piping (Section 1) that supplies water to the 3 points, as well as the tank

height (Section 2) it is necessary to have certain assumptions;

Pressure in the storage tank is = psig

Pressure at point C is = psig

To determine the range of diameters for each pipe section, Table 6.41 will be

used, in order to choose an economically sized pipe.

Schedule 40 pipe with be implemented due to the fact that the pressure in the

system doesn’t necessitate high pressure piping (such as XS or XXS wall), also

schedule 40 pipe is substantially cheaper than other alternatives.

Flanged fittings will be used due to the ease of removal. For example; repairing a

single pipe section can simply be unbolted, while threaded fittings make it

necessary to removal an entire section.

The system contains:

1. Four- 90° elbows

2. Two- T-joints

3. One- reducer

4. Well-rounded outlet, due to the lower minor loss

5. Commercial steel finish, due to cost savings

6. Three- Gate valves. Two of which are not in piping system, simply serving

as water flow controllers at points A and B. And one in the piping system

after the T-joint at point B, to channel flow into point B if point C isn’t

active.

Using Table 6.41, it is apparent that water’s economic velocity range is 4.4 ft/s – 8.8 ft/s. As

shown in Table 1, points A and C can be running simultaneously. Thus, the system will be

designed for this maximum flow rate of 65 gpm (simply combining the two flow rates). Then

points B and C will be analyzed to obtain a range of pipe sizes for each section of piping. In the

Results section, the schematics of the piping network serves as a visual aid for how the piping

system will be ran.

Equation (1) yields the formula for the maximum and minimum diameters associated with the

economic velocity range for each section.

(1)

Where Q is volume flow rate in ft3/s, and V is flow velocity in ft/s. With each range, it is

possible to use Table D.11 to find the nominal pipe sizes associated with each diameter.

Referencing the schematic once again shows that there are 4 possible cases, as seen in Table 2.

Case Pipe Size for each section

1 2-nom to 2-nom to 11/4

-nom

2 21/2

-nom to 11/2

-nom to 11/4

-nom

3 21/2

-nom to 2-nom to 11/4

-nom

4 2-nom to 11/2

-nom to 11/4

-nom

Table 2: Cases Associated With Each Possible Pipe Size

The detailed calculations for each case are seen in Appendix A. Nevertheless, each subsystem

consists of nearly identical equations.

Assuming that the water is at 60° F;

ρ = 62.4 lbm/ft3

μ = lb-s/ft2

Apply the Modified Bernoulli Equation, recognizing the end conditions due to the selected

control volume (Figure 3) are;

z2 = 0

V2 = Vavg (as pipe diameter is constant in each section)

P1 = 0 psig in the storage tank

V1 = 0 in the storage tank

Yields Equation (2), which is valid for the tower to point A:

(2)

Where z is in ft, P2 is the gage pressure at point A, gc = 32.2 lbm-ft/lbf-s2, ρ is density in lbm/ft3,

g is 32.2 ft/s2, V velocity in ft/s, f is the friction factor, L is the length of the pipe section in ft, D

is the inner diameter for the given nominal size in ft, and is the sum of minor losses due to

fittings, bends, valves, etc.

Next, Equation (3) is valid for point A to point B:

(3)

Recognizing that P3 = pressure at point B, and that there is no elevation change between these

consumption points. Important to note is the fact that Case 1 is the only subsystem which

doesn’t require a reducer from point A to point B, thus altering the value.

Lastly, Equation (4) is valid for point B to point C:

(4)

Recalling that the pressure at point C is atmospheric, Equations (2)-(4) allow the pressures at

point A and B to be solved for, working backwards from point C. Important to note is that

Equations (2)-(4) are at unique velocities and diameters. Thus Equation (5), coupled with the

relative roughness ratio , is necessary to be able to obtain the friction factor, f.

(5)

This allows the Reynolds number to be computed (Note: V is in ft/s), where ε is the roughness

associated with commercial steel1 (.00015 ft). Now it is possible to have the piping network

sized as well as the storage tank height. These values are in the Results section and the

subsequent calculations can be viewed in Appendix A.

Now it is feasible to size the pump and tank (Section 3) using the results from Sections 1 and 2.

Making certain assumptions:

Flanged fittings for ease of repair and removal

Square inlet, as this is the norm for flanged assemblies1

Diameter into pump = Diameter out of pump

One-Globe valve, to throttle flow

One-Check valve, so that water doesn’t flow back into pump once it is off.

Three-90° elbows

Pressure of the source =

Pressure of storage tank =

Referring to Table 1, it is possible to calculate the necessary volume in the tank to supply one

day’s worth of water.

This is shown in Equation (6).

(6)

Where Q is in gpm. This volume will be the minimum level in the tank at which the pump

switches on. Thus, in case of pump failure, there will be at least a day’s worth of water

available.

A reasonable maximum volume for the tank is found by adding 50% of the minimum value, as

depicted in Equation (7).

(7)

By multiplying by a factor of 1.5, the tank will hold a day and half of water.

Using this maximum volume, it is possible to size the tank. This is shown in Equation (8), where

the radius of the tank is solved for as a function of h.

(8)

Where is in ft3. By arbitrarily selecting a radius and height value, a proper sized tank can

be calculated, which is not aesthetically unpleasing, or obtrusive to the residential area.

With the tank dimensioned, the supply line piping size is next. Using the economic velocity

range once again, it is possible to calculate the upper and lower flow areas, as shown in Equation

(9).

(9)

With the range of areas, Table D.11 lists nominal pipe sizes that are adequate. The smaller of the

pipes will be selected in an attempt to lower flow velocity, thus lowering friction, and ultimately

hindering any attempt for cavitations in the line. Now it is applicable to use the Bernoulli

equation for a pump, as shown in Equation (10).

(10)

Where the only new term is

, is the total head associated with the system. Due to the

control volume (Figure 4);

= 0

z2 = 30 ft

V1 = V2 = 0

P1 = 0 psig (assuming pressure in a well is atmospheric)

Equation (10) simplifies to:

(11)

This allows the total head to be computed.

Using Equation (5) to compute Reynolds number for the given flow rate, and combining that

value with , various friction factors can be found. It is now possible to solve Equation (11)

and develop the system curve for a range of flow rates.

Due to the fact that this system is in a residential area, it is important to test for excessive

vibrations in the piping system that could result in unwanted noise. Equation (12) is derived

from Equation (10). Assuming that over 30 psi drop per 1000 ft constitutes a noise problem:

(12)

Where L is 1000 ft.

Next to be computed is the theoretical pump characteristics. The specific speed ratio (ωs) is

computed via Equation (13).

(13)

Where ω is the RPM of the pump (supplied by the manufacturer), Q is flow rate in gpm, and ΔH

is the total head. Coupling the specific speed along with the flow rate allows, the theoretical

efficiency to be viewed in Table 6.31, along with the corresponding pump type.

The power from the pump to the fluid and power from the motor to the pump must also be

computed. This is seen in Equations (14) and (15).

(14)

Where Q is in ft3

/ s, to keep units consistent.

To calculate the motor power, simply divide Equation (14) by the efficiency:

(15)

Where η is the pump efficiency.

Due to the fact that these are the ideal pump characteristics, the pump performance curve (Figure

B-2) yields the actual efficiency and power ratings.

Lastly, it is imperative to ensure that the net positive suction head available (NPSHa) is greater

than the net positive suction head required (NPSHr). The pump manufacturer supplies the

NPSHr. Equation (16) allows NPSHa to be solved for.

(16)

Where is the distance from the well water level to the center of the impeller, in ft. Assuming

suction lift, that the pump is above the source, is 10 ft. Note Pv is the absolute vapor pressure

of water at 60°F. This corresponds to .256 lbf / in2.

For Equation (16) to be verified NPSHa must be greater than NPSHr. If so, the pump is deemed

adequate. With these equations denoted and the methodology for solving the problem laid out, it

is reasonable to move onto the results obtained via the calculations.

Results / Discussion

Firstly, it is vital to know what the entire system’s configuration is. Figure 1 shows the

schematic for Case 1, while Figure 2 and shows the schematic for Cases 2-4. Two different

schematics are necessary as Cases 2-4 require an extra reducer, as the pipe diameter changes

from point A to B, as well as point B to C. Figures 3, 4, and 5 show the different control volume

setups, as this affects which parameters are included in the Bernoulli equations.

Figure 2: Schematic for Cases 2-4

Figure 3: Control Volume for Discharge Side

1

2

3

4

Figure 4: Control Volume for Supply Side

Figure 5: Control Volume for Pressure at Point C

1

2

3

4

1

2

Next, with the system analyzed, tabulated results can be seen for the discharge side associated

with each case.

Case Tank Height

(ft)

1 39.3 ft

2 46.7 ft

3 28.2 ft

4 58.1 ft

Table 3: Tank Height and Associated Case

This table shows each case and how tank height is affected by different combinations of pipe

size. Table 4 shows the price breakdown of the different pipe sizes, while Table 5 shows the

subsequent cost to fabricate each case.

Pipe Size Cost / ft

21/2

-nominal $8.35

2-nominal $5.45

11/2

-nominal $4.75

1-nominal $4.75

Table 4: Cost per foot for piping

These prices were recorded from www.globaltecheng.com.

Applying these costs to each case yields:

Case System

Cost

1 $2630.40

2 $3283.30

3 $3225.30

4 $2641.10

Table 5: System Cost

At first glance it would appear that the obvious case to optimize the cost of the system

would be Case 1. But referring to Table 4 shows that the tank height would then have to be over

40 ft (assuming a factor of safety). Furthermore, Case 1 contains a 2-nominal pipe from the

tower to point A. While this may not pose a problem under short term use, as sedimentation and

corrosion occur in the system it would be best to have the slightly larger (21/2

-nominal pipe), to

accommodate for any possible pressure drops. Also, to spend an extra $600 on piping is more

economical than attempting to build the tank an extra 10 feet, keeping in mind that this is a

residential area and a visual deterrent would surely not make home owners happy.

With this being said, the optimal choice to balance the initial cost of piping, with the cost

of building the storage tank, with the future cost of operation, is Case 3. While not apparent

now, but will be seen later in sizing the pump, the lower tank height allows a smaller pump to be

selected, as the total head of the system will be less.

Having the supply line fully designed, along with the height of the tank, it is logical to

discuss the maximum volume of the tank. Knowing that the consumption points require 84,000

gal / day, applying Equation (7) yields that the maximum volume of the tank should be 126,000

gal / day. This is a reasonable value as the water level switch will be activated when the volume

is 84,000 gal (i.e. the minimum volume is 84,000 gal). Applying Equation (8), a diameter of 30

ft and height of 25 ft will satisfy the 126,000 gallon maximum volume. These dimensions are

neither obtrusive to the surrounding community, nor aesthetically unpleasing (i.e. too tall for its

width or too wide for its height).

By making this assumption, the storage tank will hold at least a day’s worth of water.

This is imperative as the pump may fail, or the supply line may need to be altered, and the tank

must be able to supply water for a day.

Another assumption is that there is already 84,000 gallons of water in the tank. This is

reasonable as the pump could work intermittently for a few days prior to the consumption points

being active. This would have no extra stress on the pump. It can be thought of as “priming” the

system. Because of this, the pump would only need to supply 42,000 gal / day, as calculated by

subtracting from .

Now it is necessary to determine how many hours per day the pump will run. Obviously

it cannot run 24 hours due to excessive wear, along with the noise associated with a pump. Thus,

it will be assumed that the pump runs only during the 8 hour day, every other hour. This makes a

total of 4 hours per day. Dividing 42,000 gal / day by 4 hours yields a flow rate of 175 gpm, or

.390 ft/s. Equation (9) allows a choice between 3, 31/2

, and 4-nominal pipe. In an attempt to

hinder cavitation, the 3-nominal pipe will be used. The ideology here is that the smaller pipe

will reduce velocity, thus reducing friction and ultimately lowering the risk of cavitation in the

supply line.

The total head of the system is then calculated to be 49.6 ft at a flow rate of 175 gpm.

Table 6 shows a range of flow rates, and the subsequent parameters. From this data, the system

curve is created, as shown in Figure 6.

Q, gpm G,ft3/s

V = Q/A,

ft/s Re f ΔH, ft

35 0.08 1.52 39618.51 0.024 30.85

70 0.16 3.04 79237.03 0.023 33.35

105 0.23 4.56 118855.54 0.0215 37.35

140 0.31 6.08 158474.06 0.02 42.76

175 0.39 7.60 198092.57 0.019 49.61

210 0.47 9.11 237711.08 0.019 58.23

245 0.55 10.63 277329.60 0.0185 68.11

280 0.62 12.15 316948.11 0.0185 79.77

315 0.70 13.67 356566.63 0.018 92.46

Table 6: Various Flow Rates and Calculated Head

Figure 6: System Curve

25.00

35.00

45.00

55.00

65.00

75.00

85.00

95.00

105.00

0 50 100 150 200 250 300 350

ΔH

in f

t

Q in gpm

System Curve

Due to the fact that this system is in a residential area, it is preferred to test for excessive

vibrations in the system, to limit any potential noise created by the network. Equation (12) gives

a value of 66.4 lbf / ft2, which is .463 psi. Because this value is much less than the limit of 30 psi

drop per 1000 feet of pipe, the system is not subject to excessive vibrations.

Referencing the Bell and Gossett curve booklet, it is possible to determine that a potential

pump is the Series 1510, 2 BC, by using the basic performance curve (Figure B-1). This is a

base mounted centrifugal pump that operates at 1750 RPM. The performance map associated

with this pump yields an actual efficiency of 64% and a 31/2

hp motor. The theoretical efficiency

is found in Table 6.31. To use that table, it is necessary to use the result from Equation (13),

along with the flow rate of 175 gpm. This yields 67.8% efficiency, very comparable to the actual

64%. Furthermore, Equation (15) allows the motor power to be computed. This comes to a

value of approximately 3.25 hp, which is also comparable to the performance map.

With the theoretical efficiency and power compatible with the actual efficiency and

power, the pump seems like an adequate fit. Nevertheless, it is important to note that the desired

flow rate and total head falls in between impeller sizes, thus the larger (8” diameter) impeller

must be selected. By doing so, the flow rate increases to 190 gpm. To get back to the desired

175 gpm, it is possible to use the globe valve and throttle the flow after the pump, although this

essentially wastes energy, or it may be possible to alter the RPM of the pump slightly.

The last piece of data to ensure that the Series 1510, 2BC, 1750 RPM model is indeed an

adequate pump is the check that the NPSHa is greater than the NPSHr.

Assuming that the source is a well, the pump will be in a suction lift setup, as depicted in Figure

7.

Figure 7: Suction Lift

Note: Suction Lift is denoted by the pump being above the source tank. By doing so, Equation

(16) can be used. The calculated NPSHa (if the pump is 10 ft above the well) is 13.9 ft, well

above the required 6 ft, as per the performance curve. Thus, cavitation is not going to be an issue

with this system.

With the system designed, a Summary of Specifications Sheet is made that lists every result of

the calculations.

Summary of Specifications

Economic line size (discharge side) = 226 ft of 21/2

-nominal, ~132 ft of 2-nominal, ~132ft of 11/4

-

nominal schedule 40 pipe

Economic line size (supply side) = ~94ft of 3-nominal schedule 40 pipe

Storage tank height = 30 ft

Storage tank dimensions = 30 ft diameter, 25 ft height

Layout = Figures 1 and 2

System Curve = Figure 6

Specific speed = 1238.6 RPM at 175 gpm

Expected pump efficiency ≈ 67.8%

Pump type = Centrifugal

Motor power 3.25 hp

Conclusion

With the system designed and built, as per the specification sheet, adequate water will be

supplied to each of the three consumption points. Nevertheless, it is important to look at the

system’s behavior. In the case that all three valves were opened, the system would observe a

pressure of 2.6 psig at point A and -6.7 psig at point B. The negative gage pressure indicates that

if all consumption points are active, the storage tank isn’t high enough to create enough pressure

drop to adequately supply water at each point. For this situation to not be a problem, tank height

would have to be 56 ft, with a minimum volume of 172,800 gallons, if all consumption points

were active for one day. This situation would overwork the previously designed pump, as the 8”

diameter impeller cannot handle the new head of 76.5 ft and flow rate of 247 gpm (assuming the

pump now ran 6 hours per day). Potential motor failure could occur. Nevertheless, the pump

could be equipped with the 91/2

“diameter impeller. While this would be overkill for the normal

operating conditions, it would suffice if all consumption points were active. A last option would

be to run two pumps in series. An alternate supply line with the extra pump could be activated

by a gate valve if this situation were to arise. By doing so, the head wouldn’t be a problem (as

head is added when pumps are in series) but if two Series 1510, 21/2

AB, 1750 RPM, with a 61/2

”

diameter impeller pumps were used, they would have to run for over 8 hours to achieve a flow

rate close to 175 gpm (Figure B-3). Thus, if this situation were a major concern, the system

should be redesigned for such a condition.

With this being said, it is equally interesting to observe the system if point C isn’t

assumed to be at atmospheric pressure. Analyzing the control volume shown in Figure 5, as well

as using the characteristics the system was designed for (point A and C operating together at 65

gpm, and B by itself at 25 gpm), computations yield that the pressure at point A is 9.7 psig, point

B is at 6.35 psig and point C is at 1.2 psig. Thus, the assumption that point C is atmospheric

isn’t unrealistic, as 1.2 psig is only slightly over 14.7 psia. This affirms that the discharge line

will have enough pressure in it to supply water to each point.

Having performed calculations for the system at specifications other than those designed

for, it allows the engineer to see the performance of the network and how it is altered. By doing

this, failure of the system can be avoided and knowledge regarding the system is gained. Instead

of just designing for the maximum possible flow rate and assuming that is adequate for each

situation, this system has been carefully inspected to know when water supply may be an issue.

References

1) Design of Fluid Thermal Systems. 3rd ed. Stamford, CT: Cengage Learning, 2010. Print.

2) Dr. Litkouhi. "Design Problem #2." Personal interview. 17th, 19th, 26th Feb. 2010.

3) Dr. Walker. "Design Problem #2." Personal interview. 24th Feb. 2010.

Appendix A: Detailed Calculations

(See Attached)

Appendix B: Bell and Gossett Performance Curves

Figure B-1: General Selection Curves for Series 1510 Pumps

Figure B-2: Series 1510, 2BC, 1750 RPM Performance Curve

Figure B-3: Series 1510, 21/2

AB, 1750 RPM Performance Curve