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What is Pumping Head?
Dr. Faith A. Morrison
Professor of Chemical Engineering
Michigan Technological University
September 7, 1999
Copyright 1999 All rights Reserved
There is often considerable confusion among students when it comes
to
the definition and use of pumping head. This is understandable since
head
is a concept that was developed many years ago as a shortcut concept
for
field engineers when American engineering units (then called the
English
system) was the only set of units presented to students. American
engineering
units are still the dominant units used in the U.S. chemical
engineering
field, but for the most part, the science education of engineers in the
U.S. has moved over to the metric system. The concept of head can be a
stumbling block in a student's efforts to navigate the waters between
these
two systems.
The concept of head is related to the mechanical energy balance[Felder
& Rousseau]:
where D P=P_{2}P_{1}
is the pressure difference between two points, r
is the density of the fluid being pumped,
D(v^{2})=v_{2}^{2}
 v_{1}^{2} is the
difference
of the square of velocity at two points, g is the acceleration
due
to gravity, D z=z_{2}z_{1}
is the height difference between two points, F is the friction
in
the systems (expressed in consistent units), W_{s}
is the shaft work done on the system (again, expressed
in
consistent units), and m is the mass flow rate of the fluid
being
pumped. Note also that point 1 is an upstream point and point 2 is a
downstream
point.
Let us begin by examining the units of terms in equation 1 using the
American engineering system ([=] means, has units of).


[=] 

(2) 

g D
z 
[=] 

(3) 


[=] 
æ
ç
ç
è 



ö
÷
÷
ø 
æ
ç
ç
è 


ö
÷
÷
ø 
= 



(4) 
The units of expressions 2 and 3 are not the same as expression 4, and
the units must be reconciled before equation 1 may be used. The
solution
is to apply the definition of pounds force to the last expression.
This is the equivalent of dividing the velocity and height terms by g_{c}[Felder
& Rousseau]. It is not necessary to remember which terms to
divide by what; the units tell you where this conversion factor goes.
If we convert terms above using the lb_{f}
definition, we obtain three terms with common units of ft lb_{f}/lb_{m}.
These are not units of head and it is not permissible
to
``cancel '' lb_{f} with
lb_{m}.
What, then, is head? Head is mechanical energy per unit
weight.
What is weight? Weight is the force due to a mass in a gravity field,
and
it has units of force which are newtons (N) in the metric or SI system,
or lb_{f} in the American
engineering
system.
<>weight = mg [=] N or lb_{f}
(6)
The distinction between weight and mass is often forgotten since we
all live in the Earth's gravitational field. We say, ``That box weighs
90 kilograms," even though kilograms are units of mass rather than
weight.
The correct unit of weight in the metric system, the newton, is not
used
in common language (Even in countries where they use the metric system
they do not use the newton; they say ``I weigh 50 kilos.").
In the American engineering system the situation is at once simpler
and more complicated. In this system the units of both mass and weight
are the pound. The lb_{f}, in
fact,
is defined with the Earth's gravity field in mind since 1
lb_{m}
weighs 1 lb_{f} on Earth. It
becomes
tempting, however, to confuse the two types of pounds with each other
and
to cancel them in equations, and this can lead to errors.
If we examine equation 1 above, we see that it is written in terms
of
mechanical energy per unit mass. To rewrite it in terms of
mechanical
energy per unit weight, or head, we must divide by the
acceleration
due to gravity, g
term above which accounts for the heightdifference contribution to the
mechanical energy (the elevation head) now clearly has units of length
(ft or m). Examining the other two terms we see


[=] 

(8) 


[=] 
æ
ç
ç
è 



ö
÷
÷
ø 
æ
ç
ç
è 



ö
÷
÷
ø 
æ
ç
ç
è 



ö
÷
÷
ø 
= 


(9) 
The last expression still needs to be converted using the definition of
lb_{f}:


[=] 
æ
ç
ç
è 


ö
÷
÷
ø 
æ
ç
ç
è 

32.174 ft lb_{m} 

s^{2} lb_{f} 

ö
÷
÷
ø 
= ft (10) 
We thus see that head, mechanical energy per unit weight, has units
of ft which is a particularly easy unit to remember.
Is that the whole point? Why bother with all of this mess just to
use
a simple unit? Engineers deal with many types of units all the time and
could certainly learn to deal with other units such as ft lb_{f}/lb_{m},
or N, or whatever.
The desirability of using head in the mechanical energy expression
is
obscured in the calculations shown above by the fact that no numbers
have
been used. Let us examine the head produced by a pressure difference of
50 lb_{f}/ft^{2}
in a system pumping water.

ì
í
î 
mechanical energy 
per unit mass 
due to pressure 


ü
ý
þ 

= 

(11) 


= 
æ
ç
ç
è 
50 


ö
÷
÷
ø 
æ
ç
ç
è 



ö
÷
÷
ø 

(12) 


= 

(13) 

ì
í
î 
mechanical energy 
per unit weight 
due to pressure 


ü
ý
þ 

= 

= 
æ
ç
ç
è 



ö
÷
÷
ø 
æ
ç
ç
è 



ö
÷
÷
ø 

(14) 


= 
æ
ç
ç
è 
0.80 


ö
÷
÷
ø 
æ
ç
ç
è 



ö
÷
÷
ø 
æ
ç
ç
è 

32.174 ft lb_{m} 

s^{2} lb_{f} 


ö
÷
÷
ø 

(15) 


= 
0.80 ft 
(16) 
As we see in the final equation, the number 32.174 appears in two
places,
once in the numerator as part of the unit conversion from lb_{f}
to ft lb_{m}/s^{2},
and secondly in the denominator as g, the acceleration due to
gravity.
Because these two factors are numerically the same, they cancel (the
numbers
cancel, not the units), and expressions having units of ft lb_{f}/lb_{m}
and ft of head are
numerically the same.
Thinking in terms of head allows an engineer to have some intuition
about the amount of mechanical energy per unit weight under
consideration.
Let us examine elevation head. The value in ft of the elevation
head is the height difference between two reference points. Now if we
express
the velocity term in head units (velocity head ºD
(v^{2})/2
g) we can understand
what, for example, 10 ft of velocity head means. It is the
equivalent
amount of mechanical energy per unit weight to that represented by 10 ft
of water. Likewise if the pressure term is expressed in head units
(pressure
head ºD P/rg),
a similar understanding is obtained. In pumping this is particularly
useful
since we can get a handle on the characteristics of a pump by taking
about
the developed head of the pump. The developed head of a pump represents
the height to which a fluid may be lifted by the pump in a frictionless
piping system. If the pump is not used to lift fluid, we still can
understand
that the pump can use its ability to do work (produce mechanical energy
per unit weight) to produce elevation, velocity, or pressure head or
any
combination of these.
In summary, ft of head is a unit of length. It expresses the
mechanical energy per unit weight of a flowing system. It may be
converted
to other systems by using the conversion factors for length. When you
do
this you must remember that you are keeping track of a quantity which
expresses
mechanical energy per unit weight. In the SI system, m of head
has
exactly that meaning. Note that in the SI system N m/kg
(the
analogous expression to ft lb_{f}/lb_{m})
is not numerically equal to m of head. This is because
the
conversion factor for kg m/s^{2}
to
N is one, rather than being numerically equal to g in
the SI unit system (check this to convince yourself).
References
 [Felder &
Rousseau]
 Felder, Richard M. and Ronald W. Rousseau, Elementary
Principles
of Chemical Processes, 2nd Edition (John Wiley, & Sons: New
York,
1986).
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This document was translated from L^{A}T_{E}X
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