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 P

_{h }= ρ. Q. g. h/ (3.6 x 10^{6}) KW
Efficiency
of new pump = 75 %, hence

Pump power
rating, P= P

_{h }/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 u

^{2}/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 m

Velocity in selected pipe = Q/A, where A is cross sectional area of pipe and Q is flow

Pressure head loss due to friction= 0.0015Lu

Flow of pump = 40 m

^{3}/hrVelocity in selected pipe = Q/A, where A is cross sectional area of pipe and Q is flow

Pressure head loss due to friction= 0.0015Lu

^{2}/D
Total head =
40 + h

_{F,}where h_{F}loss of head due to friction
Cost of pump
($) = 2[12 + 250 P

^{0.6}], where P in KW
Cost of pipe
per meter of length ($/m) = 9400 D

^{2}+ 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) / (πD

^{2}/4) = 1.98 m/s
Frictional
loss of head = 0.0015 L u

^{2}/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 P

_{h }= ρ. Q. g. h/ (3.6 x 10^{6}) KW
P

_{h }= 0.109 x h = 4.928
Efficiency
of new pump = 75 %, hence

Pump power
rating, P= P

_{h }/0.75
P=
4.928/0.75= 6.57 KW

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