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
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
No comments:
Post a Comment