Handbook of Turbomachinery Second Edition Revised - Ventech!

We can then state (correctly) that, ‘‘A torque applied to a body will produce
a change in its angular momentum.’’ In the case of steady flow through a
control volume that consists of our rotor, this can be written as
T ¼ _mðr2Cu2 r1Cu1Þ= g ð3Þ
We also know that the applied torque, T, multiplied by the rotational speed
o gives the required power:
P ¼ To ¼ _moðr2Cu2 r1Cu1Þ= g ð4Þ
Notice above that the quantities or2 and or1 are the tangential velocities
(normally designated as ‘‘u’’) of the rotor at the respective stations.
Substituting the local rotor speeds u1 and u2 into Eq. (4) gives us
P ¼ _mðu2Cu2 u1Cu1Þ ð5Þ
We should also note that the delivered power P is the rate at which work
enters the fluid in its travel through the rotor. More specifically, a suitable
statement of the First Law of Thermodynamics says that the rate at which
work enters the fluid can be equated to a change in a fluid called enthalpy:
P
¼ D21
_m h0
where h0 is the quantity called stagnation enthalpy, and the right-hand side
of the above equation depicts a change from station 1 (rotor inlet) to station
2 (rotor outlet). For working fluids that are gases, this is a conveniently
calculable (from measurements) property. However, for pumps that move
liquids, enthalpy changes are not so easily measurable. For this reason, the
energy equation is rearranged slightly so as to substitute for enthalpy the
sum of internal energy (denoted as u, and not to be confused with rotor
speed) and the ‘‘pressure energy,’’ or ‘head,’ denoted as P=r where P and r
represent static pressure and density, respectively. For a liquid pump, if the
pumped flow can be approximated as incompressible (as is the case with
most rocket pumps to a fair approximation), the change in internal energy
between inlet and exit states can be neglected, and h0 becomes simply the
change in stagnation energy, or ‘‘head.’’
If we now combine Eq. (5) and (6), we can write
Dh0 ¼ðu2Cu2 u1Cu1Þ= g ð7Þ
Equation (7) is known as Euler’s equation. Unless stated otherwise, this
chapter assumes flow to be incompressible, and therefore h0 is taken to be
stagnation ‘‘head.’’ It should be pointed out, however, that for any given
example this limitation could be relieved by simply interpreting h0 as
Copyright © 2003 Marcel Dekker, Inc.
ð6Þ