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Sunday, 14 December 2014

HOW TO WRITE TECHNICAL MEMORANDUM FOR A PUMPING SYSTEM

 In this blog we will consider how to write one technical memorandum for a pumping system.
Let us discuss how we may write in desired format.

TECHNICAL MEMORANDUM

Project:   Design of pumping system for filling a roof tank from a reservoir

Date     :       To     :

Back ground

A water pumping system will be used to fill a roof- tank from a large reservoir (lake). Pump should be located nearby reservoir and pumping system will run for time period of 3 hrs. Tank is under atmospheric pressure and reservoir is also under atmospheric pressure. It is decided that 120 m3 volume of water will be pumped by pumping system in 3 hrs to fill tank which is located at 40 m height from reservoir or lake.

Objectives

Objective of this technical memorandum involves….
  • Pumping system design to suit the above application
  • Preparation of preliminary cost estimation for recommended pumping system
  • Pumping system should be cost efficient  

Technical analysis

Pump size might be calculated on the basis of following formula
Power required lifting water for given height Ph = ρ. Q. g. h/ (3.6 x 106) KW
Efficiency of new pump = 75 %, hence
Pump power rating, P= Ph /0.75
During the calculation of pump power rating, we have to consider loss of head due to friction and which might be calculated by following way
 F=0.0015 L u2/D
Minimum pressure rating of pipe (excluding pressure head loss due to friction) = ρ g h
Minimum pressure rating of pipe (excluding pressure head loss due to friction) = 1000 x 9.81 x 40 = 392400 Pa = 56.91 PSI
As the condition of site is unknown, so we will take length of pipe as mentioned,
Hence length of pipe = 75 meters.     

Flow of pump = 40 m3/hr 
Velocity in selected pipe = Q/A, where A is cross sectional area of pipe and Q is flow 

Pressure head loss due to friction= 0.0015Lu2/D
Total head = 40 + h F, where h F loss of head due to friction
Cost of pump ($) = 2[12 + 250 P0.6], where P in KW
Cost of pipe per meter of length ($/m) = 9400 D2 + 1100 D + 50, Where D in meter
As we may see in above table , That for diameter of 0.048590 meter , costing will be low but in this case frictional head loss is much higher and hence will be costly

We will prefer pipe of diameter 0.084480 will be best to use
Pipe diameter (ID) = 0.084480 m= 3.326 inch
Pipe OD = 4 inch,                
Pipe pressure rating = 350 PSI (suitable for our application)
Pump size Capacity = 40 m3/ Hr. Power= 7.26 KW
Head = 50 meters (considering safety factor, hence 50 m taken instead of 45.22 meters)
Pumping system would be designed as above for better performance i.e. with positive suction
I will solve for one only one, for you reference,
Let us see for green one,
Pipe ID from table provided in assignment
ID=0.084480
Velocity = Q/A= (40/3600) / (πD2/4) = 1.98 m/s
Frictional loss of head = 0.0015 L u2/D = 5.22 m
Where, L is 75 meter,
Total head = 40 m+5.22 m = 45.22 m
Power required lifting water for given height Ph = ρ. Q. g. h/ (3.6 x 106) KW
Ph = 0.109 x h = 4.928
Efficiency of new pump = 75 %, hence
Pump power rating, P= Ph /0.75
P= 4.928/0.75= 6.57 KW

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