R. Shanthini 05 Feb 2010. Whole System Design: An integrated Approach to Sustainable Engineering...

R. Shanthini 05 Feb 2010


Whole System Design:

An integrated Approach to Sustainable Engineering

Units 6 – 10: Worked Examples

July 2007

Citation: Stasinopoulos, P., Smith, M., Hargroves, K. and Desha, C. (2009) Whole System Design: An Integrated

Approach to Sustainable Engineering, Earthscan, London, and The Natural Edge Project, Australia


Whole System Design:

An integrated Approach to Sustainable Engineering

Unit 6: Industrial Pumping Systems


Pumps move liquid from one place to another against pipe friction and against changes in height and direction.

Industrial pumping systems account for 20% of the world’s industrial electrical energy demand,

and 12% of world’s electricity.

(Motors use 60% of the world’s electricity.)

Improving the efficiency of industrial pumping systems can reduce industrial energy consumption and hence greenhouse

gas emissions.

Significance of Pumping Systems and Design


Power plant loss 70%

Transmission & distribution loss 9%

Motor loss 10%

Pump loss 25%

Throttle loss 33%

Pipe loss 20%

9.5 units of energy output

100

units

of

fuel

ene

rgy

inpu

t

Drivetrain loss 2%

Saving a single unit of pumping energy can save more than ten times that energy in fuel.


Large pumping system (in the order of kW and MW) design is disciplined considering minimum velocities, thermal

expansion, pipe work and maintenance.

Small pumping system design is not disciplined since it accounts for a small fraction of the total cost of an

industrial operation.

However, it is likely that there are a lot more small pumping systems than large pumping systems.

Both small and large pumping systems design could benefit from Whole System Design.

Significance of Pumping Systems and Design (continued)


1. Ask the right questions

2. Benchmark against the optimal system

3. Design and optimise the whole system

4. Account for all measurable impacts

5. Design and optimise subsystems in the right sequence

6. Design and optimise subsystems to achieve

compounding resource savings

7. Review the system for potential improvements

8. Model the system

9. Track technology innovation

10.Design to create future options

10 Elements of applying a Whole System Design (WSD) approach


Worked example overview

A typical production plant scenario

Design Challenge: Consider water at 20ºC flowing from reservoir A, through the system in the figure above to a tap with a target exit volumetric flow rate of Q = 0.001 m3/s. Select suitable pipes based on pipe diameter, D, and a suitable pump based on pump power, P, and calculate the cost of the system.

Window(fixed into wall)

Machine press(movable)

(2) Elevation (Z2 = 10 m)

Elevation (Z1 = 0 m)

(1)

A

Q


Conventional Design solution: Conventional system with limited application of the

Elements of Whole System Design

Whole System Design solution: Improved system using the Elements of Whole

System Design

Performance comparison: Comparison of the economic and environmental

costs and benefit

Design Process


Conventional Design Solution


The system accommodates the pre-existing floor plan (window) and

equipment (machine press) in the plant. Reservoir A exit is very well rounded. The diameter of every pipe is D. A globe valve, which acts as an emergency cut off and stops the flow for

maintenance purposes, is fully open during operation. The existing tap is replaced by a tap with an exit diameter of D.

Conventional Design Solution (continued)


The energy balance between point 1 and point 2 in thesystem is given by Bernoulli’s Equation:

Σ Pi/ρgAiVi

= (p2/ρg - p1/ρg) + (α2V2

2/2g - α1V12/2g) + (z2 - z1)

+ Σ fi (Li/Di)(Vi2/2g) + Σ

KLiVi2/2g

Friction head losses

Head losses in pipe contractions,

expansions, bends, joints and valves

Pumping gains

Step 8 of WSD: Model the system

Pressure, kinetic energy and potential energy changes

Ref: Appendix 6A


Assumptions:

• p1 = p2 = atmospheric pressure; V1 = 0; z1 = 0

• The diameter of every pipe is D, and therefore the cross sectional area of every pipe is A.

• The average velocity of the fluid in the downstream of the pump is constant and equal to V2.

• The pipes are considered to be a single pipe of length L.

• Assume that head losses through reservoir A exit (well rounded), pump connectors and tap connectors are negligible.

Step 8 of WSD: Model the system (continued)


P/ρgAV2 = α2V22/2g + z2 + f (L/D)(V2

2/2g) + V22/2g (Σ KLi)

V2 = Q / A = Q / (πD2/4)

P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2

Design equation relating the pump power (P) and pipe diameter (D)

Under the assumptions, energy balance reduces to

Velocity is related to volumetric flow rate by

Combining them, we get

Step 8 of WSD: Model the system (continued)


The known variables are: ρ = 998.2 at 20oC (Table 6A.3 in Appendix 6A) Q = 0.001 m3/s (design criterion)

α2 = 1 for uniform velocity profile

> 1 for non-uniform velocity profile f depends on the Reynolds number of the flow L = 30 m (from system plan)KLi for 90o threaded elbows = 1.5 (Table 6A.2)

KLV for fully open glove valve = 10 (Table 6A.2)

KLT for tab = 2

g = 9.81 m/s2 z2 = 10 m (from system plan)




Re = ρV2D/μ = 4ρQ/ πDμ

= 4(998.2 kg/m3)(0.001 m3/s)/ πD(1.002 x 10-3 Ns/m2)

(μ is obtained from Table 6A.3 of Appendix 6A)

Re = 1268.411 / D

Re Type of fluid flow f

Re < 2100 Laminar 64 / Re

2100 < Re < 4000 Transitional

Re > 4000 Turbulent function of Re and ε/D,where ε is the equivalent roughness.

The friction factor, f, is dependent on the Reynolds number, Re

For Re < 2100, D > 0.604 m = 60.4 cm

For Re > 4000, D < 0.317 m = 31.7 cm




A pipe of diameter D = 0.317m is much larger than what would even be suitable for the system.

Therefore, Re > 4000 and the flow is turbulent.

Turbulent velocity profile can be assumed to be uniform, therefore α2 = 1.

P = (8.0911x10-7/D4)[f (30/D) + 19] + 97.923

Design equation therefore becomes

For Re < 2100, D > 0.604 m = 60.4 cm

For Re > 4000, D < 0.317 m = 31.7 cm



P = (8.0911x10-7/D4)[f (30/D) + 19] + 97.923

Suppose drawn copper tubing of diameter D = 0.015 m was selected for the pipes.

Then, Re = 1268.411/(0.015 m) = 84561

For drawn tubing, ε = 0.0015 mm (Table 6A.1 of Appendix 6A)

Thus, ε/D = 0.0015/15 = 0.0001

Therefore, f = 0.0195 (from the Moody chart in Figure 6A.1 of Appendix 6A

at Re = 84533 and ε/D = 0.0001)



P = (8.0911x10-7/D4)[f (30/D) + 19] + 97.923

Substituting D = 0.015 m and f = 0.0195 into the above equation, we get

P = (8.0911x10-7/(0.015 m)4)[0.0195 (30/(0.015 m)) + 19] + 97.923

= 1025 W

To generate an exit volumetric flow rate of Q = 0.001 m3/s, drawn copper tubing of 0.015 m diameter (D) pipe and

1025 W power pump (P) are required.


From ‘Water pumps pricelist’ (Appendix 6B) we can select pump model:

Waterco Hydrostorm Plus 150 at P = 1119 W (1.5 hp)

Pump cost = $616

From ‘Hard drawn copper tube (6M length)’ in ‘Kirby copper pricelist’ (Appendix 6C) we can select pipe:

T24937 at D = 15 mm (5/8 in)

cost is $57.12 per 6m

Pipe cost = ($57.12 per 6m)(30 m)/6 = $285.60

Total Cost of the Conventional Design Solution



Elbow cost = ($2.34)(4) = $9.36(For standard radius 90º elbows of 15mm (5/8 in) diameter J00231 ‘Copper

fittings’ in Appendix 6C gives a cost of $2.34 each.)

Estimated globe valve cost = $13 (US$10)(For a globe valve of diameter 15mm (5/8 in), by interpolating a

‘Components pricelist’, Appendix 6D)

Tap cost = $6.70(For a tap of exit diameter 0.015 m in ‘Components pricelist’, Appendix 6D)

Installation costs = ($65/hr)(8 hrs) = $520(Installation costs for 8hrs at $65/hr)

Total capital cost of the system = $616 + $285.60 + $9.36 + $13 + $6.70 + $520

= $1451



For the pump running at output power P = 1025 W, the monthly pump running costs for 12 hrs/day, 26 days/month are:

Running cost = ($0.1/kWh)(1.025 kW)(12 hrs/day)(26 day/mth)/(0.47) = $68/month

cost of electricity(2006 price for large energy users)

pump efficiency for an electrical pump


Whole System Design Solution

Step 1 of WSD: Ask the right questions

• Could we change the pipe configuration so as to reduce the head losses?

• Did the selection procedure for pipe diameter, D, and pump power, P, address the whole system?


Whole System Design Solution (continued)

Step 7 of WSD: Review the system for potential improvements

• From Bernoulli’s equation,

we can see that increasing diameter dramatically reduces the power required

• Can the system be designed with less bends?

• Can the system be designed with more-shallow bends?

• Is it worthwhile moving the plant equipment (machine press)?

• Is an alternative pipe material more suitable?

• Is there a more suitable valve? Do we even need a valve?



Assumptions are the same as for the conventional system:

• p1 = p2 = atmospheric pressure; V1 = 0; z1 = 0

• The diameter of every pipe is D, and therefore the cross sectional area of every pipe is A.

• The average velocity of the fluid in the downstream of the pump is constant and equal to V2.

• The pipes are considered to be a single pipe of length L.

• Assume that head losses through reservoir A exit (well rounded), pump connectors and tap connectors are negligible.


P = (8ρQ3/π2D4)(α2 + f (L/D) + Σ KLi) + ρgQz2 Same design equation as before


The known variables are: ρ = 998.2 at 20oC Q = 0.001 m3/s

α2 = 1 (assuming turbulent velocity profile)

f depends on the Reynolds number of the flow L = 30 24 mKLi for 90o 45o threaded elbows = 1.5 0.4 (Table 6A.2)

KLV for fully open glove gate valve = 10 0.15 (Table 6A.2)

KLT for tab = 2

g = 9.81 m/s2 z2 = 10 m (from system plan)



P = (8.0911x10-7/D4)[f (24/D) + 3.95] + 97.923


P = (8.0911x10-7/D4)[f (24/D) + 3.95] + 97.923

Suppose drawn copper tubing of diameter D = 0.015 0.03 m was selected for the pipes.

Then, Re = 1268/(0.03 m) = 42280

For drawn tubing, ε = 0.0015 mm (Table 6A.1 of Appendix 6A)

Thus, ε/D = 0.0015/30 = 0.00005

Therefore, f = 0.0195 0.0215(from the Moody chart in Figure 6A.1 of Appendix 6A

at Re = 42267 and ε/D = 0.00005)



Substituting D = 0.03 m and f = 0.0215 into the above equation, we get

P = (8.0911x10-7/(0.03 m)4)[0.0215 (24/(0.03 m)) + 3.95] + 97.923

= 1025 119 W

P = (8.0911x10-7/D4)[f (24/D) + 3.95] + 97.923


To generate an exit volumetric flow rate of Q = 0.001 m3/s, drawn copper tubing of 0.03 m diameter (D) pipe and

119 W power pump (P) are required.


From ‘Water pumps pricelist’ (Appendix 6B) we can select pump model:

Monarch ESPA Whisper 500 at P = 370 W (0.5 hp)

From ‘Hard drawn copper tube (6M length)’ in ‘Kirby copper pricelist’ (Appendix 6C) we can select pipe:

T22039 at D = 31.75 mm (1¼ in)


Is this a optimal solution for the whole system?



Step 3 of WSD. Design and optimize the whole system

D (m) Re ε/D f P (W)

0.015 84561 0.0001 0.0195 660

0.02 63421 0.000075 0.0205 242

0.025 50736 0.00006 0.0210 148

0.03 42280 0.00005 0.0215 119

0.04 31710 0.0000375 0.0230 104

Pump power calculated for a spectrum of pipe diameters

The capital and running costs for each pipe and pump combination are calculated in a similar way as for the conventional solution, and is given in the next slide.



D (m)

Pipes and

components cost

P (W)Pump

selectedPump cost

Total capital cost

Running cost

Life cycle cost

(-NPV)

0.015 $602 660

Monarch ESPA Whisper 1000

$357 $959 $49/mth $10,227

0.02 $745 242Monarch ESPA Whisper 500

$331 $1076 $19/mth $4,670

0.025 $827 148 $331 $1158 $12/mth $3,428

0.03 $914 119 $331 $1245 $9/mth $2947

0.04 $1126 104 $331 $1457 $8/mth $2970



The efficiency of Monarch ESPA Whisper Model 1000 = 42% Model 500 = 40%

The life cycle economic cost of each solution is estimated as the net present value (NPV) calculated over a life of 50 years and at a discount rate of 6%.

Assumptions used for the cost calculation on the previous slide:


Comparing the cost of the two solutions

Solution D (m)Pipes and componen

ts costP (W)

Pump cost

Total capital cost

Running cost

Life cycle cost

(-NPV)

Conven-tional

0.015 $8351025

$616 $1451 $61/mth $12,989

WSD 0.03 $914 119 $331 $1245 $9/mth $2947

R. Shanthini 05 Feb 2010. Whole System Design: An integrated Approach to Sustainable Engineering...

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