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Solution Manual, Chapter 14 – Turbomachinery
Chapter 14
Turbomachinery
General Problems
14-1C
Solution We are to discuss energy producing and energy absorbing devices.
Analysis A more common term for an energy producing turbomachine is a
turbine. Turbines extract energy from the moving fluid, and convert that energy into
useful mechanical energy in the surroundings, usually in the form of a rotating shaft.
Thus, the phrase “energy producing” is from a frame of reference of the fluid – the
fluid loses energy as it drives the turbine, producing energy to the surroundings. On
the other hand, a more common term for an energy absorbing turbomachine is a
pump. Pumps absorb mechanical energy from the surroundings, usually in the form
of a rotating shaft, and increase the energy of the moving fluid. Thus, the phrase
“energy absorbing” is from a frame of reference of the fluid – the fluid gains or
absorbs energy as it flows through the pump.
Discussion From the frame of reference of the surroundings, a pump absorbs
energy from the surroundings, while a turbine produces energy to the surroundings.
Thus, you may argue that the terminology also holds for the frame of reference of the
surroundings. This alternative explanation is also acceptable.
14-2C
Solution We are to discuss the differences between fans, blowers, and
compressors.
Analysis A fan is a gas pump with relatively low pressure rise and high flow
rate. A blower is a gas pump with relatively moderate to high pressure rise and
moderate to high flow rate. A compressor is a gas pump designed to deliver a very
high pressure rise, typically at low to moderate flow rates.
Discussion The boundaries between these three types of pump are not always
clearly defined.
14-1
Solution Manual, Chapter 14 – Turbomachinery
14-3C
Solution We are to list examples of fans, blowers, and compressors.
Analysis Common examples of fans are window fans, ceiling fans, fans in
computers and other electronics equipment, radiator fans in cars, etc. Common
examples of blowers are leaf blowers, hair dryers, air blowers in furnaces and
automobile ventilation systems. Common examples of compressors are tire pumps,
refrigerator and air conditioner compressors.
Discussion Students should come up with a diverse variety of examples.
14-4C
Solution We are to discuss the difference between a positive-displacement
turbomachine and a dynamic turbomachine.
Analysis A positive-displacement turbomachine is a device that contains a
closed volume; energy is transferred to the fluid (pump) or from the fluid
(turbine) via movement of the boundaries of the closed volume. On the other
hand, a dynamic turbomachine has no closed volume; instead, energy is
transferred to the fluid (pump) or from the fluid (turbine) via rotating blades.
Examples of positive-displacement pumps include well pumps, hearts, some
aquarium pumps, and pumps designed to release precise volumes of medicine.
Examples of positive-displacement turbines include water meters and gas meters in
the home. Examples of dynamic pumps include fans, centrifugal blowers, airplane
propellers, centrifugal water pumps (like in a car engine), etc. Examples of
dynamic turbines include windmills, wind turbines, turbine flow meters, etc.
Discussion Students should come up with a diverse variety of examples.
14-5C
Solution We are to discuss the difference between brake horsepower and water
horsepower, and then discuss pump efficiency.
Analysis In turbomachinery terminology, brake horsepower is the power
actually delivered to the pump through the shaft. (One may also call it “shaft
power”.) On the other hand, water horsepower is the useful portion of the brake
horsepower that is actually delivered to the fluid. Water horsepower is always less
than brake horsepower; hence pump efficiency is defined as the ratio of water
horsepower to brake horsepower.
Discussion For a turbine, efficiency is defined in the opposite way, since brake
horsepower is less than water horsepower.
14-2
Solution Manual, Chapter 14 – Turbomachinery
14-6C
Solution We are to discuss the difference between brake horsepower and water
horsepower, and then discuss turbine efficiency.
Analysis In turbomachinery terminology, brake horsepower is the power
actually delivered by the turbine to the shaft. (One may also call it “shaft power”.)
On the other hand, water horsepower is the power extracted from the water
flowing through the turbine. Water horsepower is always greater than brake
horsepower; because of inefficiencies; hence turbine efficiency is defined as the
ratio of brake horsepower to water horsepower.
Discussion For a pump, efficiency is defined in the opposite way, since brake
horsepower is greater than water horsepower.
14-7C
Solution We are to explain the “extra” term in the Bernoulli equation in a
rotating reference frame.
Analysis A rotating reference frame is not an inertial reference frame. When we
move outward in the radial direction, the absolute velocity at this location is faster
due to the rotating body, since v
θ
is equal to
ω
r. When solving a turbomachinery
problem in a rotating reference frame, we use the relative fluid velocity (velocity
relative to the rotating reference frame). Thus, in order for the Bernoulli equation
to be physically correct, we must subtract the absolute velocity of the rotating
body so that the equation applies to an inertial reference frame. This accounts
for the “extra” term.
Discussion The Bernoulli equation is the same physical equation in either the
absolute or the rotating reference frame, but it is more convenient to use the form with
the extra term in turbomachinery applications.
14-3
Solution Manual, Chapter 14 – Turbomachinery
14-8
Solution We are to determine how the average speed at the outlet compares to
the average speed at the inlet of a water pump.
Assumptions 1 The flow is steady (in the mean). 2 The water is incompressible.
Analysis Conservation of mass requires that the mass flow rate in equals the
mass flow rate out. Thus,
Conservation of mass: in in in in out out out out
mVAm VA
ρ
ρ
=
==
or, since the cross-sectional area is proportional to the square of diameter,
22
in in in
out in in
out out out
D
VV V
DD
ρ
ρ
==
D
(1)
(a) For the case where Dout < Din, Vout must be greater than Vin.
(b) For the case where Dout = Din, Vout must be equal to Vin.
(c) For the case where Dout > Din, Vout must be less than Vin.
Discussion A pump does not necessarily increase the speed of the fluid passing
through it. In fact, the average speed through the pump can actually decrease, as it
does here in part (c).
14-9
Solution For an air compressor with equal inlet and outlet areas, and with both
density and pressure increasing, we are to determine how the average speed at the
outlet compares to the average speed at the inlet.
Assumptions 1 The flow is steady.
Analysis Conservation of mass requires that the mass flow rate in equals the
mass flow rate out. The cross-sectional areas of the inlet and outlet are the same.
Thus,
Conservation of mass: in in in in out out out out
mVAm VA
ρ
ρ
=
==
or
in
out in
out
VV
ρ
ρ
= (1)
Since
ρ
in <
ρ
out, Vout must be less than Vin.
Discussion A compressor does not necessarily increase the speed of the fluid
passing through it. In fact, the average speed through the pump can actually decrease,
as it does here.
14-4
