TURBOMACHINES
A device that transfers energy between the moving fluid and rotating parts, rotating vanes, a
change in the flow direction, velocity, and/or pressure of the fluid and the work is done either by
the system or on the system. They may be classified as
1) Pumps
2) Turbines
PUMPS
Pump Classification
Piston - Pistons of varying stroke lengths
Constant and Variable Volume
pump fluid through the check valve
Reciprocating
Positive Displacement
(negative)
Diaphragm-Pulse one or two flexible diaphragm
to displace liquid, ideal with particulate filled fluid
Radial Flow
Single Suction
Centrifugal
Mixed Flow
Double Suction
Peripheral
Special
Axial Flow
Single Suction
Pumps
Kinetic
Rotary
Single Stage
Multistage
Single Stage
Multistage
a) Displacement
i) Positive:
Examples:
This type of pump provides a fixed amount of fluid to a system for each
revolution. The fluid is forced along by successive volume change or fluid
displacements as with a piston and cylinder. They are either reciprocating or
rotary pump. They can be used in fluid power, machining, tool manufacturing,
engine superchargers, and lubrication applications. They provide high capacity
but low quantity, they cannot sustain over pressurization therefore installation of
safety valves are necessary.
External/Internal gear pump; hydraulic jack, fluid power applications at 3000
psi (pounds per square inch), 150 gpm (gallons per minute) that provides low
cost, long life, and high efficiency
Vane pump - lubrication of machine tools
Lobe pump - engine superchargers, cooling, 1500 psi, low quality but very
durable.
ii) Negative: Kinetic, they provide large quantity of fluid at relatively low pressure compared
to positive displacement. Use: commercial and industrial applications.
Examples;
Centrifugal pumps (Diffuser type or Radial flow) - series of fixed diffusing
vanes surrounding the impeller the rotating element of the pump. Reduction in
velocity and increase in pressure occurs because of diffusion. Residential
purposes, relatively low in cost, highly reliable and they can handle almost all
types of fluids
Volute type – non-diffusing, Low cost and can handle almost any fluid
Supplementary notes on Turbomachines/Dr. Kalim/Page #1
Propeller type (Axial flow) - very large discharge at low head, they can be used
in extremely hostile environments, such as movement of coal slurries, crude oil,
and sewerage etc.
b) Delivery
i) Constant volume } Self explanatory
ii) Variable volume }
“
c) Motion
i)Rotary
:
Rotating elements fit the casing closely and as they rotate in opposite directions
in the casing, fluid is trapped in the space between two or three gear teeth and
the casing and is moved from inlet to outlet. Flow from rotary pump is steady
compared to reciprocating pumps. Such a pump shown in Figure 1 is suited for
viscous liquids and as the viscosity of the fluid increases the speed decreases
and power required increases.
ii) Reciprocating: In this pump (shown in figure 2) the circular motion of the crank at the left
causes, via the connecting rod, back and forth movement of the piston within the
cylinder. As the piston moves creating vacuum once and sucking water in the
cylinder, then pressurizing next and forcing the fluid out.
outlet
gear
casing
Inlet
(a)
(b)
(c)
Figure 1(a) Rotary Pump Schematics, (b) Rotary Vane Oil-less Vacuum Pump 0.5 Hp, (c) Gear Pump 0.25 hp
connecting rod
Piston
cylinder
outlet
Inlet
crank
reservoir
Figure 3a,b and c - Reciprocating Pump Schematics
Supplementary notes on Turbomachines/Dr. Kalim/Page #2
Industry wide common definitions
Max. Fluid Temperature
Max. Fluid Viscosity
Max. Flow Rate
Max Suction lift
Max Head
Max System Pressure
Motor Type
Duty Cycle
Motor Phase
Power Requirements
Wetted Materials
Self-Priming
Washdown Capable
Continuous Duty
Runs Dry
Submersible
Leak-Proof
Explosion Proof
Air-Powered
The maximum temperature that the wetted materials of the pump will be able to safely
handle
The “thickness” of a liquid or its ability to flow.
The largest amount of fluid that can be moved per unit of time
The maximum height the pump can “lift” the fluid, disregarding friction loss from
below the center of the pump.
Indicates the maximum height of a column of water being moved by the pump.
The maximum amount of pressure that the pump can withstand.
The type of motor the pump has- TENV stands for Totally-enclosed, non-ventilated;
TEFC stands for Totally-enclosed, fan-cooled; ODP stand for open- drip proof, XPRF
indicates explosion proof
The amount of time the pump can be run per day. Generally “continuous” indicates the
ability of the pump to withstand running 16 hours per day and “intermittent” generally
indicates the ability to withstand less than 8 hours per day.
Motor phase determines the ability to vary the speed through a controller. The speed of
a three-phase motor can be varied through the correct controller whereas the speed of a
single-phase motor cannot be varied.
The type of voltage that the unit requires to operate properly. The United States
generally uses 120 VAC as its line voltage. Much of the rest of the world uses 220
VAC as its line voltage.
These types of materials come in contact with the fluid in the normal operation of the
pump.
Pumps that draw liquid up from below pump inlet, as opposed to pumps requiring
flooded suction
Pumps with an enclosure that protects the motor from water and the external
environment. This enclosure allows the pump to be cleaned with water spray or be
placed in various weather conditions.
Able to operate with no off or rest periods.
The ability of pump to resist damage when air, rather than liquid, is being pumped
through the pump cavity.
The ability of a pump to be completely submersed with the fluid that it is pumping.
Contains no seals that would normally cause leakage if contaminated.
The ability of a pump to resist explosion when exposed to flammable gases. Generally
the wiring for these units must be done by a licensed electrician.
The ability of the pump to be run by air-power. Excellent for environments where
electric power may be hazardous or unpractical.
You can visit the website referenced in this document, other websites listed on my webpage, or any other website of
your choice to learn about various pumps, its characteristic curves. Etc. Today ITT A-C and Gould pumps are
installed in pump stations and water treatment plants in many parts of the world [ http://www.acpump.com/ ]
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Non-clog centrifugal pumps are available in horizontal or vertical mounting up to 54" or 110,000 GPM.
Axially splitcase pumps, again, horizontally or vertically mounted, range up to 72" discharge with
capacities to 110,000 GPM.
Self-priming, solids handling Trash Hog? Pumps provide lifts to 28 feet and capacities to 6,000 GPM.
Vortex type centrifugal Whirl-Flo pumps can pass port size solids for unsurpassed clog-free pumping to
2,400 GPM.
Vertical column and axial flow pumps for water supply, flood control, and irrigation exceeding a half
million gallons per minute.
Gould’s Brand ANSI chemical pumps, both conventionally sealed and magnetically driven for chemical
feed applications.
Hard metal slurry pumps for extreme grit problems.
Shearpeller impellers are designed to handle those applications that are too difficult for conventional solids
handling centrifugal pumps. The Shearpeller impeller design enables pumps to handle large, stringy or
fibrous solids, up to 26% entrained air in water by volume, and highly viscous as well as many other hard to
pump slurries.
Supplementary notes on Turbomachines/Dr. Kalim/Page #3
Centrifugal Pumps
Two most used pumps are the Centrifugal and Axial pumps. The centrifugal pumps are most
widely used in domestic and small commercial applications whereas the axial pumps are used
for large and commercial applications.
The horizontal axial flow pump is designed for pumping large capacities with low lift
requirements. It is designed for reliable, efficient pumping operation and for ease of
maintenance. The WCXH pump is available in sizes up to 132" (3355mm) with capacities up to
600,000 GPM (136,500m³/hr) and heads to 25 ft (7.5m) [3].
The centrifugal pump is one of the most widely used pumps for transferring liquids. The basic
components of a centrifugal pump are depicted in the figure 3. The principle operation of a
centrifugal pump is to convert fluid velocity into pressure energy. The pump consists of three
components, an inlet duct, an impeller, and a volute. Fluid enters the inlet duct (D). As the shaft
(A) rotates, the impeller (B), which is connected to the shaft, also rotates. The impeller consists
of a number of blades that project the fluid outward when rotating. This centrifugal force gives
the fluid a high velocity. Next, the moving fluid passes through the pump case (C) and then into
the volute (E). A cutout view of a single stage centrifugal pump is shown in the figure 3a.
outlet or
discharge nozzle
Volute or
Diffuser (E)
Shaft (A)
Casing (C)
Impeller
Suction (D)
Impeller Eye
Figure 3– Centrifugal Pump
Figure 3a- Centrifugal pump
SINGLE STAGE PUMPS (API 610 type OH2) as shown in the figure 3a are used for petrochemical and
hydrocarbon processing, transfer charge and hot services- Capacities to 11,000 US gpm or 2,500 m3/hr,
Heads to 1,310 feet or 400 m, Maximum Temperature 800°F or 427°C, Maximum Working Pressure
1,000 psi or 75 bar. Another cutout view of a two-stage centrifugal pump is shown in the figure 5a.
Sample characteristics curve for a centrifugal pump is shown in the figure 5b.
Centrifugal pumps are very quiet in comparison to other pumps. They have a relatively low
operating and maintenance costs. Centrifugal pumps take up little floor space and create a
uniform, non-pulsating flow. Pumps are used in almost all aspects of industry and engineering
from feeds to reactors or distillation columns in chemical engineering to pumping storm water in
civil and environmental. They are an integral part of engineering knowledge and an
understanding of how they work is important for all engineers. The volute chamber has a
uniformly increasing area. This increasing area decreases the fluids velocity, which converts the
velocity energy into pressure energy. The heart of the pump is the impeller, or impellers in the
case of multistage pump. The impeller is fastened to the shaft, which transmits energy to the
Supplementary notes on Turbomachines/Dr. Kalim/Page #4
pump. It is surrounded by a casing, which contains the suction and discharge nozzles. The fluid
flows from the suction nozzle to the eye of the impeller.
The impeller vanes impart velocity or kinetic energy to the fluid. This high velocity fluid is then
captured by the pump casing, which turns high velocity to pressure as it guides the fluid through
the discharge nozzle. All attachments are sealed to prevent any fluid or pressure leakage. Two
characteristics a pump produces are pressure head and volumetric flow. Different types of
impellers are shown in the figures 4a, b and c below.
Figure 3b- Centrifugal Pump
(a)
(b)
Figure 3c- Mixed Flow Vertical Wastewater Pump [3]
(c)
Figure 4. Impellers: (a) open, (b) semi-open, and (c) closed
Figure 5b- Centrifugal Pump Characteristics Curve [1]
Figure 5a- Centrifugal Two-Stage Pump
Figure 5c - Centrifugal Pump Characteristics Curve [2]
Supplementary notes on Turbomachines/Dr. Kalim/Page #5
Total head
Total head
Discharge
head
Suction head
Figure – 6a (hs is positive)
•
Suction lift
Discharge
head
Figure – 6b (hs is negative)
Dead head
It is the ability of a pump to continue running without damage when discharge is closed off. It is
only recommended for Centrifugal pump.
The centrifugal pump is basically employed to perform one of three functions (a) to raise fluid
from one level to another; (b) force fluid into high pressure zone such as pressure vessel; or (c) to
overcome the friction loss in the pipe and fittings. To perform these functions the pump
generates what is termed as total head,
Total Pump Head = Discharge Head ± Suction Head or Lift or alternatively
Hp = Hd ± Hs
Where Hd is the discharge pressure head, Hs is positive with suction head shown in Figure 6a, and
Hs is negative with suction lift shown in Figure 6b. Head is measured in units of length. Due to
the fact that head is a way of denoting pressure, it can be easily determined using a pressure
gauge, as long as the pressure taps are located at the pump suction and discharge ports. They
offer low to high flow rates at low pressures and pulse-less flow at low cost. The range of flow
rates for Low flow is (1-20 gpm), Moderate flow is (20-50 gpm) and High flow is >50 gpm. Also
some centrifugal pumps are Submersible. Most centrifugal pumps need priming, which means
one has to make sure to position the pump at a location such that liquid is always retained at a
level higher than the pump inlet port.
•
Net Positive Suction Head (NPSH)
NPSH is the total suction head required to avoid cavitation problems in pumps. Inhibitors
are used to yield corrosion and to stop erosion. In other words, NPSH is the minimum fluid
energy required at the pump inlet for satisfactory operations. The NPSH required (NPSHR) must
be positive to avoid cavitation. In general a higher NPSH is desirable. It is usual for a
manufacturer to run a test to determine the NPSH for the machine that will permit it to operate
efficiently and without noise and damage. As long as the machine is operated above the NPSH
value the operation will be satisfactory. NPSH is also regarded as the pump head required at zero
flow. NPSHA is the net positive suction head available or actual energy available at the pump
inlet, which is calculated using equation A. To avoid cavitation the NPSHA must be greater than
the NPSHR. NPSHR of a pump is determined by test for various flows and pump speeds. The
Supplementary notes on Turbomachines/Dr. Kalim/Page #6
manufacturer for all pumps provides NPSHR. . The NPSHA is calculated using equation 1 given
below.
NPSHA = (P/γ) + (V2/2g) - hvp ± (hs) - hL (On suction side only)
(1)
Where, hvp = vapor pressure of the fluid corresponding to the liquid temperature (m or ft),
Table- 1. Water Vapor Pressure Corresponding to Temperature
Water Temp. oC
Vapor Pressure, hvp (m)
10
0.125
20
0.239
30
0.435
40
0.759
50
1.275
60
2.068
100
10.785
If the temperature is in degrees Celsius and the pressure p is in Pascal,
the vapor pressure can be found at ant temperature by solving the following equation.
Tv = 100 + 0.0002772 (p - 101000) - 1.24 10-9 (p - 101000)2
P/γ = pressure head at water level on suction side, V2/2g = velocity of water in the suction pipe at the entrance, p =
static pressure on the fluid surface (m or ft), γ = specific weight of the fluid (N/m3 or lb/ft3), hs = Suction head or
suction lift shown below, the distance from center line of pump to fluid level (m or ft); note that if the reservoir is
above the center line of the pump (Figure 6b with suction head) positive sign is considered, while for a reservoir
below the pump (Figure 6c with suction lift) negative sign is be considered. The hL = head loss due to friction
considered in the suction line only (m or ft).
Table- 2. Vapor Pressure of Gasses
Gas
Helium
Propane
Butane
Carbonyl sulfide
Acetaldehyde
Freon 113
Methylisobutylketone
Tungsten
•
Pumps in Series or Parallel Setting
vapor pressure
(bar)
1
22
2.2
12.55
0.987
0.379
0.02648
0.001
notes
@ -269.15 °C
@ 55 °C
@ 20 °C
@ 25 °C
@ 20 °C
@ 20 °C
@ 25 °C
@ 3203 °C
ΔP = ΔP1+ ΔP2
and flow rate of Q
Often in many engineering applications a single pump
Pump
Pump
cannot deliver the head necessary for a particular need, and Q
2
1
two (or more in practice) can be combined in series to
Assumed
ΔP2
increase the height to which the fluid can be pumped. A
ΔP1
No Loss
schematic representation of two pumps in series is shown
below. Ignoring any losses that occur between the two pumps,
Figure 6c. Pump in Series
the flow rate through both is the same, but the overall pressure
rise is the sum of the two individual values. If both pumps are
Pump
Q1
identical then for a given flow rate, the pressure rise is
1
Q1+ Q2
doubled normally. The pressure characteristics for two pumps
at ΔP
in series are shown in the figure 6a. This is useful when a
Pump
Q2
2
high pressure is needed but the same flow rate as of a single
pump is sufficient. In this case however the second pump in
ΔP
the series must have the ability to operate at a higher suction
pressure, which is produced by the first pump.
Q
Figure 6d. Pump in Parallel
Supplementary notes on Turbomachines/Dr. Kalim/Page #7
Similarly when the identical pumps are in parallel the flow rate is doubled but the pressure
head remains the same as shown in the figure 6d. When pumps run in parallel the flow is
increased and the pressure head produced is around the same as a single pump (these are for
identical pumps run in parallel.
•
Dissimilar Pumps in Series or Parallel Setting
It is possible to combine two or more pumps, in either series or parallel, which have
dissimilar characteristics. However the overall system characteristics, as shown below, can be the
same as that for one pump at certain conditions, therefore no benefit would be achieved.
Cavitation is the local vaporization of a liquid. This
occurs when the absolute pressure falls to less than (or equal
to) the liquids vapor pressure at a given temperature,
resulting in small pockets of vapor and boiling occur.
Vapor
Pressure
Cavitation
As most liquids contain air, this lowering of pressure
causes the release of air, and when combined with the
vaporization, cavitation occurs. As the vapor pressure is
Temperature
temperature dependent as shown in the figure 7a, a system
that may appear to problem free could have cavitation
Figure – 7a Vapor Pressure Vs Temperature
when the temperature rises.
Vapor bubbles generated at the inlet to the pump are
carried along until a region of higher pressure is reached and
then they suddenly collapse. If the vapor bubbles come in
contact with the walls at the time they collapse, pitting of
the surface can occur due to the high local pressures
generated. The entire process of the formation, growth, and
collapse of a bubble can occur in milliseconds in a turbomachine. Experimental data indicate that the pressures
occurring from the collapse of vapor bubbles are in the
order of 109 Pa, which is consistent with the damage
observed due to cavitation. To protect against cavitation the
system should be designed to avoid pressures that approach
the vapor pressure of the fluid. Effects of cavitation are
noise, and damage to the pump which results pump failure.
Collapse
of Bubble
Fluid
Flow
Figure – 7b- Cavitation
However in practice there are several reasons for cavitation such as, high velocity flow,
higher temperature flow. Often it appears that cavitation occurs a higher pressure than the vapor
pressure of the liquid, due to the generation of a vortex, say at an inlet valve as shown in the
figure 7c. To avoid cavitation NPSHavailable must be greater than NPSHrequired OR NPSHA >
NPSHR.
Cavitation can cause erosion within the pipes, the pumps impeller, vibration and thus noise,
performance fluctuations, and will probably decrease the performance, and therefore the
efficiency of the pump, as shown typically in the figure 7b.
Supplementary notes on Turbomachines/Dr. Kalim/Page #8
Pumps are used to transfer fluid in a system, either at the same level or to a new height. The
flow rate depends on the height to which the fluid is pumped, and the relationship between head
and flow rate is called the pump characteristic, which has to be determined experimentally.
Although it is difficult to predict when cavitation will occur, but another parameter which is use
by the pump manufacturers is the cavitation number (σ). It may be used in modeling and
comparing experimental results. The cavitation number σ is often compared with a known
parameter called critical cavitation parameter σc (a dimensionless factor given by manufacturer
and/or found by experiments).
Valve
Low Pressure
leads to
cavitation
Head H
high velocities
low pressures
Normal
Performance
Performance due
to Cavitation
Eddy shedding low periodic pressures
Quantity
Figure7c – Valve Flow
Figure 7d. Cavitation Performance
If σ<σc then cavitation may be avoided, σc is usually provided by the manufacturer.
Where
σ=
Pps − Pvp
γ
V2
2g
And Pps = Pump Suction Pressure
=
Pps − Pvp
ρV 2
2
Pps
=
γ
− h vp
V2
2g
(2)
Note that above conditions may not be the only reasons at which cavitation may result, but in real situations many
factors may affect such development such as suction specific speed described later.
•
Specific Speed
It is a performance factor. It is a non-dimensional speed of the pump at which discharge and
head can be estimated at best efficiency. Specific speed can be thought of as the speed required
for a pump to produce unit head at unit volume flow rate. It can be useful in optimizing a pump
application for a given capacity and head. The manufacturers sometimes offer Ns vs. Cost and
Efficiency performance curves that are very useful in selection of pumps.
Any constant value of specific speed describes all operating conditions of geometrically
similar machines with similar flow conditions. Specific speed is not really a speed in the ordinary
sense. Rather it is pump selection parameter that includes the effect of discharge, head and
relative impeller speed that will be best suited for a given application. It is calculated as
for US units, where, N = rpm, Q = gpm, H = feet
Ns = N Q1/2/ (H)3/4 = N P1/2/(H)5/4
(3)
for SI units, where, n = rps, Q = m3/sec, H = m
Supplementary notes on Turbomachines/Dr. Kalim/Page #9
Ns = n Q1/2/ (g H)3/4 = n P1/2/(g H)5/4
(4)
For multistage pumps replace H by (H/n), where n is the number of stages.
Here H can be replaced by NPSH. Ns can help in predicting or designing the combinations
of factors such as discharge, head and speed that are both possible and desirable. See figure
14.14 (text) for selection of pumps on the basis of the specific speed. This parameter can also be
used to figure out the optimum efficiency of the commercial pumps as given by various
characteristic curves shown in figures in chapter 14.
Figure 14.14, text [5]
Sometimes the suction performance is categorized by Suction Specific Speed Nss. It is a
dimensionless parameter that defines impeller eye geometry, and is an indicator of the cavitation
susceptibility. The Hydraulic Institute recommends that for cold-water flow in a pump the critical
suction speed Nss should be less than 8,500 for cavitation free operation. For cold water Nss
varies in the range of 0-8500, for boiler feed the range is 8,500-13,000 and for special
applications such as hydrocarbon flow Nss can be over 13,000.
The Suction Specific Speed Nss =
Nss =
RPM Q per eye
NPSHR 3/ 4
rps Q
(g * NPSHR)3 / 4
(US sys.)
(5)
(SI sys)
(6)
Supplementary notes on Turbomachines/Dr. Kalim/Page #10
•
Energy To The Pump
Total head or dynamic head = Ep = hp = the energy added to the pump per unit weight given
by equation 7. If it is desired to know the operating point of a pump in a given system, we must
consider that the pump will be required to deliver the fluid against a static discharge head and to
overcome all the flow losses in the system.
hp = (p2 - p1) / γ + (V22 - V12)/ 2*g + (Z2 - Z1) + hL
Ppower = Q * hp * γ = γ * Q * hp/η
Where η is the pump efficiency
•
(7a)
(7b)
Affinity Laws
Often it is important to operate a pump at a speed or head other than the published speed or
head. To determine how pump will perform at speeds other than the published speed we can
either look at the manufacturer’s published data or use affinity laws. These laws are used to
design actual machines (pumps) on the basis of models developed in the laboratories, Q =flow
rate, N =speed in rpm, D = diameter, P =Power, H =head such as;
Model (1)
Prototype (2)
Q1/(N1* D13)
H1/(N12 * D12)
Q1/(D12 * H11/2)
P1/(N13 * D15)
NPSHR2/NPSHR1
•
=
=
=
=
=
Q2/(N2*D23)
H2/(N22 * D22)
Q2/(D22 * H21/2)
P2/(N23 * D25)
(Q2/Q1)2
So using these laws, for same diameter or homologous pumps (D1=D2) we get
Q2 N 2
=
, Capacity varies directly as the speed ratio
Q1 N 1
(8)
(9)
(10)
(11)
(12)
(8a)
2
H2 ⎛ N2 ⎞
=⎜
⎟ , Head varies as the square of the speed ratio
H 1 ⎝ N1 ⎠
(9a)
3
HP2 ⎛ N 2 ⎞
=⎜
⎟ , HP varies as the cube of the speed ratio.
HP1 ⎝ N 1 ⎠
•
Similarly, for constant speed, N1=N2; we get
Q1/D13 = Q2/D23 or D2 = ((Q1/D1)*Q2)(1/3)
2
•
(11a)
2
(8b)
H1/D1 = H2/D2 ; and
(9b)
P1/ D15
(11b)
= P2/ D2
5
The critical speed of a pump is any RPM, which corresponds to the natural frequency of the
pump rotor. Such a critical speed is experienced when an imbalance creates a vibration
whose frequency is equal to a natural frequency of the pump rotor. If shafts and rotors were
in perfect balance, critical speed would be of no importance. However, it is never possible,
because even the perfectly balanced rotor has some imbalance. Therefore, the lubrication
system, bearings, couplings, and sealing play very important role in the smooth and efficient
operation and design of the pump. We want to install the pumps as vibration free as possible.
Supplementary notes on Turbomachines/Dr. Kalim/Page #11
•
The Torque on a pump is calculated by T =
5250xHP
RPM
Pressure Head
System Curve Components
(12)
Pump
Curve
Liquid
Pump
Static
Head
B
Friction
head (valve
System
partially
Curve
closed)
Valve
Friction
head (valve
open)
Head
ft
Liquid
A
Pressure
Head
Figure 9-Pump system with Static Head
Static
Head
•
Usually in any given pump system shown in
GPM Q
figure 9, there is a control valve, which adds
additional friction head. This control valve will
Figure –10. Pump system curve
be adjusted to add or remove frictional head to control pump flow rate. A pump curve can be
selected to interact with the system curve to yield optimum performance.
•
All pumps operate on a system curve. A system curve is made up of three basic components;
Static Head, Pressure Head, Friction Head (imparted by the suction loss and valve loss (if a
valve exist in the system, the loss can be calculated by
(Q =
Cv
(S / ΔP)1 / 3
S is the specific gravity)
(13)
•
•
The pump always operates at the intersection of the curve and the system curve.
Pump characteristics or performance curve usually provides relationship between Efficiency,
Power, and Specific speed etc.
•
For a given head, capacity and size, the cost of the pump decreases as the specific speed
increases. The specific speed is a powerful tool in making right selection for the pump.
The cost of a pump increases as the number of pumping stage increases.
Efficiency generally increases as the pump size increases. Therefore, optimum efficiency can
be graphically represented as a function of both specific speed and pump size.
•
•
The electrical power input to the pump is also measured
in Watts (W) and with pump power P, the overall pump
efficiency ηp is,
ηp =
P
W
=
γ Q hp
W
or W =
Head H
The increase in head H between the inlet and outlet of a pump is a function of the flow rate Q
and rotational speed N. This pump characteristic is
expressed graphically, as shown in the figure 11.
N2
N1
N3
Increasing
Speed
γ Q hp
ηp
Mass flow rate ρQ
Supplementary notes on Turbomachines/Dr. Kalim/Page #12
Δp Q
W
Figure 11. Pump Speed Characteristics
In terms of pressure rise Δp across the pump, the efficiency is written as, ηp =
Often Pumps are available in base mounted units with pump, base, motor and coupling as
shown in figure 11. Inquire if you are interested in a base mounted unit.
MAXIMUM PUMPING SPEED
MAXIMUM PUMPING TEMPERATURE
MAXIMUM INLET PRESSURE
MAXIMUM CASE WORKING PRESSURE
3500 RPM
500 F
100 PSI
200 PSI
Sample characteristics or performance curves are shown in
the figure 12a and b.
Figure 11a. Base Mounted Pump
Figure –12a. Typical Pump Characteristics Curve
Figure 12b. Sample Pump Characteristics Curve
Pump Example:
A prototype test of a mixed flow pump with a 72 in diameter discharge operating at 225 rpm
resulted in the characteristics shown in table 1 and plotted in the figure. What size of homologous
pump should be used to produce 200 cfs at 60 ft head at best efficiency? Draw the characteristics
curve for the new pump homologous to the model.
Examining Table 1, for best efficiency of the pump is η1= η2= 88 %. At this point, the pump
has the characteristics of; H1= 45 ft, Q1 = 345 cfs, N1=225 rpm and D1=72 inch. We need to find
the characteristic of the homologous pump at H = 60 ft and Q = 200 cfs at best efficiency. The
characteristic curve of the given pump is shown in the figure 13a.
H1 / N12 * D12 = Hn / Nn2 * Dn2 ⇒ 60 = 45 * (Nn* Dn)2/(225* 72)2
⇒ 200 = 345 * (Nn* Dn3)/(N1* D13)
Q1 /N1* D13
= Qn /Nn*Dn3
Supplementary notes on Turbomachines/Dr. Kalim/Page #13
Using above equations (2 equations, two unknown), the characteristics for the new pump is,
Nn = 366.7 rpm and Dn = 51 inch as shown below. Since N= 367 is not a round number for
practical purposes (for 60 Hz cycles, multiple of 60 is better), therefore assume that N = 360 rpm.
For this new rpm calculate new D. The new value of D is 52 inch and Q = 208 cfs. Note this is
the characteristic of the pump at best efficiency.
TABLE –1 Pump data
GPM
EFFICIENCY
CHARACTERISTICS CURVE OF THE MODEL PUMP
500
90
450
85
400
80
EFFICIENCY %
350
cfs
-------------------------------------------------Pump head
cfs
Efficiency
-------------------------------------------------60.000
200.00
69.000
57.500
228.00
75.000
55.000
256.00
80.000
52.500
280.00
83.700
50.000
303.00
86.000
47.500
330.00
87.300
45.000
345.00
88.000
42.500
362.00
87.400
40.000
382.00
86.300
37.500
396.00
84.400
35.000
411.00
82.000
32.500
425.00
79.000
30.000
438.00
75.000
27.500
449.00
71.000
25.000
459.00
66.500
-------------------------------------------------
300
75
250
70
200
150
65
20
30
40
50
60
70
PUMP HEAD in ft
Figure. 13a- Characteristics curve for the given pump
With N= 360 and D = 52, equations for transforming the corresponding values of H &
Q at any efficiency given in table 1 can be obtained by the following relations:
N D3
Ηn = H1 * (360* 52)2/(225* 72)2 =1.335 H1} and Qn = Q1 * n n = 0.603 Q1 }------- (6)
N1D13
Plugging different values of head and flow rate, in the equation 6, the data for the
homologous pump is tabulated as follows in Table 2. The characteristic curve of this pump is
shown in the figure 13b.
The characteristic curve for a pump is for example, may be a plot of (Q/ND3) as abscissa against (γH/N2D2)
as ordinate. This curve obtained from actual tests on one unit of the series and applied to all homologous. Similarly
there are many other combinations of parameters (Q, N, h, and D etc.), which may be utilized to plot characteristics
curve that help find the performance of other similar pumps and hydraulic machines as shown in figures 1 and 2. A
number of characteristic curves are shown in the figures 14.6, 14.7, 14.9, 14.10, and 14.14 in the text.
We can define a quantity called Specific Diameter Ds =
DH1 / 4
Q
. Consider N in (rpm), Q in (gpm) and D in
inches. If Ds is plotted against Ns the characteristics curve obtained by this plot is sometimes more practical. The
radial flow pumps (centrifugal) is recommended in the specific speed range of 400-4000, mixed flow pump in the
range of 4000-7000, and axial flow pumps in the range of 7000-60,000 and up.
Supplementary notes on Turbomachines/Dr. Kalim/Page #14
Table 2- Pump Data
----------------------------------------------Pump head
cfs
Efficiency
----------------------------------------------80.000
121.00
69.000
76.700
138.00
75.000
73.400
155.00
80.000
70.000
169.00
83.700
183.00
86.000
66.700
63.500
200.00
87.300
60.000
208.00
88.000
56.700
219.00
87.400
53.500
231.00
86.300
50.000
239.00
84.400
46.700
248.00
82.000
43.400
257.00
79.000
40.000
264.00
75.000
36.700
271.00
71.000
33.400
277.00
66.500
-----------------------------------------------
Efficiency
cfs
Characteristic Curve of the Pump, Homologous to the Model Pump
300
90
85
255
Efficiency %
80
cfs
210
75
165
70
120
65
30
40
50
60
70
80
90
Pump Head in Feet
Figure 13b Characteristics curve for the given pump
4.0
Radial Flow
Mixed Flow
Ds- ft 1.5
80
Axial Flow
70 %
0.2
300
4000
7000
Conversion Factor:
3.785 liters = 1 gallon
1 gal = 0.1337 cu ft. = 0.003785 cu m.
60 %
50
60000
Ns
Figure 14 - Characteristics curve of a pump
Supplementary notes on Turbomachines/Dr. Kalim/Page #15
Factors that must be considered before the selecting the pump
•
•
•
•
•
•
•
•
•
Nature of liquid
Gpm
Conditions on suction/inlet side and NPSH
Conditions on discharge side
Power source (motor, engine, and turbine etc)
Space, weight, position limitation
Environmental conditions
Cost of pump, installation, operational, capital
Governing codes and standards
Factors that must be specified after the pump selection
•
•
•
•
•
•
•
•
•
Type of pump and manufacturer
Size of pump
Suction/discharge connections
Operational speed
Driver specifications (voltage, frequency, phase, frame size)
Coupling type manufacturer, model #
Mounting details (vibration)
Special materials and accessories required
Shaft seal design & seal materials
Example problems solved in class
1.
A pump runs at 1500 rpm with inlet area of 2 sq. ft. The pump is located 2 ft below the reservoir and moves
3 cusecs of water (20 oC) 50 ft above the pump inlet. The total head loss in suction side is 3 ft while in the
discharge side it is 4 ft. The water dumps water in the second reservoir. Find NPSHA, total pump head,
specific speed and Nss. What type of pump would you recommend? What is the operating efficiency of this
pump? Is this pump would be susceptible to cavitation at these conditions and when the pump speed is
increased to 20,000 rpm?
2.
An 80% efficient pump delivers 100 gpm with discharge head of 3 ft. Determine the pump power. The
suction and discharge pipe diameter is 3 in. and 1 in. respectively. The pressure difference between inlet
and outlet is 6 psi if the head loss in suction side is 2 ft.
3.
A 2-ft diameter pump produces 30 ft head and delivers 500 gpm at 1000 rpm. Determine the performance
parameters (gpm, head, Sp. Speed, Power etc.) at 1100 rpm. Determine the pump diameter at the newly
obtained data if the power required of two pumps are the same and select the pump.
4.
Pump example in the handout.
5.
If the manufacturer’s critical cavitation parameter for the pump in the problem 1 is “60”, whether this pump
would cavitate for the given operating conditions.
Supplementary notes on Turbomachines/Dr. Kalim/Page #16
Pump Problems - Assigned
1) A pump delivers water at 30 C from a lower reservoir to an upper reservoir. The pump outlet is 2 m above
the pump exit, and static pressure taps indicate an outlet pressure that is 100 kPa greater than the inlet pump
pressure. If the flow rate is 0.05 m3/sec and inlet diameter of the pipe leading to the pump is 100 mm while
the outlet diameter is 50 mm, calculate the power added to the water. If the lower reservoir is at the
atmospheric pressure equivalent to 1 bar and it is 1 m below the pump level calculate NPSH.
2) A test is conducted on a pump and test data show the pump capable of developing a total head of 15 m
while delivering 20 liters/sec when operated at 1000 rpm. Estimate the head and capacity when the pump is
operated at 800 rpm. Determine the initial specific speed of the pump.
3) A pump is to deliver 4000 liters/min against a head of 20 m. If the pump is operated at 3600 rpm, select the
type of pump to be used.
4) What is the specific speed of the pump of example solved in the handout at its point of best
efficiency?
a. Determine the size and synchronous speed of a pump homologous to the 72 inch pump in the
example that will produce 3 m3/sec at 100 m head, at its point of best efficiency?
b.
Plot the characteristic curve of this pump.
5) A pump has an impeller diameter of 1.5 meters and operates at 1200 rpm. If the speed is increased to 1400
rpm, what diameter impeller should be used to maintain a constant power input to the pump?
6) What type of pump should be selected that will deliver 2000 gpm against a head of 25 ft while operating at
3600 rpm?
7) A pump delivers 95 liters/min of water at 70 oC (μ = 4.04*10-4 N-s/m2, hvp = 31.2 Kpa, ρ = 978 Kg/m3)
through 1.5 inch steel pipe (Ks = 0.046 mm). The inlet pressure is 20 Kpa vacuum. The atmospheric
pressure is 1 bar. The pump is operating at a suction head of 2.5 m. Consider that there is a wide open globe
valve (Kv = 10) and one bend (Kb=0.9) in the system. Determine the NPSHA. If the manufacturer’s NPSHR
is 6 m, should this pump be recommended for the intended operation? If the pump operates at 2500 rpm
would it be prone to cavitation.
References
1.
2.
3.
4.
5.
http://www.albanypump.com/burks/centrifugal-pumps.html
http://www.pumpgroup.com/
http://www.acpump.com/ModelWCXH.html
http://www.goulds.com/pdf/GLICS375.pdf
Text Book
Supplementary notes on Turbomachines/Dr. Kalim/Page #17
Hydraulic Turbines
Classification
Hydraulic action: Impulse or Reaction.
Impulse:
total head is converted to kinetic energy through a series of nozzles. Flow on the
bucket is at atmospheric pressure. Example; Pelton wheel
Reaction: partly head and partly kinetic energy. Flow in the turbine is under pressure.
Flow completely fills the impeller chamber. Examples are Francis and Kaplan
turbine.
Direction of water:
Tangential flow Axial flow Radial flow Mixed flow-
water runs tangent to the wheel, Example; Kaplan
water flows parallel to the axis of the shaft, Example; Kaplan type
water runs toward center and then runs from center to periphery,
Example; Francis type, Figure 1a
Example; radial/axial, Francis, Figure 1b
Disposition of shaft:
Horizontal - Pelton turbine (Impulse type)
Vertical - Francis/Kaplan turbine (Reaction type)
Figure 1a. Radial type
Figure 1b.Mixed Flow type
Vt 2
Ts
ω
n2
Vt1
R1
Vn
1
V t = Positive in direction of rotation,
tangential velocity
n1
R2
Vn 2
u = R ω = pheripheral rotor velocity
V = Absolute velocity
w = angular velocity
Vn = fluid velocity
Figure 2. A general turbomachine
Q=A1 Vn1=A2 Vn2, Ts = r Q (R2 Vt2 - R1 Vt1) Euler equation for Turbomachine,
Power = Rate of work =- dws/dt = Ts ω = ρQ (R2 Vt2 - R1 Vt1) ω= mflowrate(u2 Vt2 - u1 Vt1). Similar to pumps the
turbine performance or characteristics curve are shown in the figures 3a,b
Supplementary notes on Turbomachines/Dr. Kalim/Page #18
8
Sp. Speed N(1)
Sp. Speed N(2)
7
6
5
4
3
Sp. Speed N(2) using
ft,hp, and rpm
Sp. Speed N(1) using m,kw, and rpm
30
28
26
24
22
20
18
16
14
12
10
8
6
4
2
2
0
0.42
0.43
0.44
0.45
0.46
1
0.47
φ
Figure 3a. Turbine Performance or Characteristics Curve
800
100
90
80
600
70
50
400
Whp
efficiency %
60
40
30
200
20
10
0
0
0
100
200
300
400
500
600
700
Bhp
Figure 3b. Turbine Performance or Characteristics Curve
Supplementary notes on Turbomachines/Dr. Kalim/Page #19
Turbine formulae:
Elevation H
DAM
Lake
concrete
friction loss in pipe
& penstock = h L
Buckets
Shaftwork
to generator
Turbine
Jet or Nozzle
d
jet
Pipe Diameter (D)
Vp
Pipe and penstock
length in thousands
of feet or meters.
H
Vj
hn
Pj
η
η
Pe
t
P
t
h
P
t
WHP
ηm
ηo
SHP
SHP
Figure: A Pelton (horizontal shaft) turbine setup
•
•
HP = Head at the top of the Penstock = PP/γ + VP2 /2g + ZP
H = HP - hLP = Effective Head at bottom of the Penstock
= PP/γ + VP2 /2g + ZP, hLP is loss in the penstock.
•
H = the Effective Head at bottom of the Penstock is =Vj2/2g+Zj+hn where, hnis the head loss
in the nozzle. If the head loss in nozzle hn is negligible and Zj = 0 (when the nozzle center
line is at datum) then the Effective Head at bottom of the Penstock H = Vj2/2g, Unless
otherwise mentioned the nozzle loss hn is considered negligible (for this course)
•
Vj = Nozzle Jet Velocity = Cv {2gH}1/2 = Qj/Aj where Cv is the nozzle coefficient
1/ 2
•
•
•
•
•
•
•
d = Diameter of nozzle jet =
⎞
⎛
4 Qone jet
⎟
⎜
⎜ πC 2g (H - h ) ⎟
v
LP
⎠
⎝
, Normally,
Cv = 0.98
60u
πN
u = peripheral speed = φ Cv {2gH}1/2 = φ *Vj
D = Diameter of Turbine Wheel =
N = Turbine speed = 60 ω/2π rpm, and
φ = ratio of wheel speed to jet speed = Vbucket/Vj
V
ω = Bucket angular speed = bucket rad/sec.
(D/2)
Jet ratio = m = D/d
Supplementary notes on Turbomachines/Dr. Kalim/Page #20
Power
•
•
Water Horse Power (WHP) = Power available at the nozzle inlet =[γQ(H-hL)]/550 *
Horsepower available at nozzle outlet = (m• Vj2) /2*550 = Pj
•
•
Horsepower developed by Turbine = Pt = m• (Vj2-Ve2) /2*550
Power lost in the stream (exit) =(m• Ve2 ) /2*550
•
•
•
Power developed by turbine = Pt = ηh * WHP
Power at nozzle outlet = Pj = Torque *ω or, Torque = Pj /ω
Actual power developed by Turbine = SHP =ηo * (WHP)
Efficiency
•
•
•
•
Nozzle efficiency ηn
Turbine efficiency ηt
Mechanical efficiency ηm
Overall efficiency ηo
=Pj /WHP If no loss in nozzle i.e. ηn = 100% , Pj = WHP
=Pt / Pj
=SHP / Pt
=SHP /WHP=SHP /Pt =ηm* ηh=SHP/(Pj+nozzle power loss)
=η m * η n * η t
•
Hydraulic efficiency ηh
•
Volumetric efficiency ηv
=Pt /WHP = ηn * ηt =2 * u ( Vj - u ) (1+ K cosα) /Vj2
where, α = deflection angle of the nozzle jet by the turbine blade
usually 180o−β, also K=0.9~1.0
= Q/(Q+Qloss), Qloss is the loss of fluid from the buckets in the turbine.
Miscellaneous
•
•
•
Size of Buckets; Length = 3 d, Width = 4 d, depth = 1.0 d
Number of Buckets; They should be as small as possible, to keep the frictional loss
minimum. But the number and the size must be sufficient enough so that the jet (water) is
always intercepted by bucket. Empirically, Nbucket = D/2d + 15 (15 is a constant )
# of Jets = Q/Qj= Q /Vj Aj
Turbine Problems
1)
In a laboratory test on an impulse turbine, the following data were obtained, if the total head is 20 meters, the
friction loss in the penstock is 5 meters of head, Nozzle discharge is 0.02 m3/sec, Diameter of the jet is 3.90
cm, the Shaft horsepower (SHP) is 3.2 with turbine mechanical efficiency of 90%. Determine the power lost
in mechanical friction and in the nozzle. (Note total head is the effective head at the nozzle inlet, and there is a
power loss in the nozzle)
2)
What Power in KW can be developed by the impulse turbine shown. If the turbine efficiency is 85%, its
diameter is 3 meters and the total head at the reservoir level in the penstock is 1000 meters. Assume that
friction factor f of the penstock is 0.015 and head loss in the nozzle is neglected. What will be the angular
speed of the turbine ω if the ratio of the wheel (turbine) velocity to jet velocity φ is 0.5. Also find the torque
that will be exerted on the turbine. The length and diameter of the penstock is 6 Km and 1 m respectively. The
diameter of the jet is 18 cm.
*(746
for SI system)
Supplementary notes on Turbomachines/Dr. Kalim/Page #21
Example problems solved in class
1. An impulse turbine is required to develop a shaft hp of 10,000 when operating under an
effective head of 250 ft at the nozzle and at a speed of 550 rpm. If the values of Cv, the
overall efficiency (ηo), jet ratio m, and the ratio of wheel speed to jet speed φ are 0.98, 85%,
12, and 0.45, respectively, determine i) the quantity (Q) of water required ii) the diameter of
the turbine, iii) the diameter of the jet iv) nozzle velocity
2. A multiple jet Pelton wheel is required to develop 12,500 hp. The turbine operates under an
effective head of 300 m. Assume values of the velocity coefficient Cv, speed ratio φ, and jet
ratio m as 0.98, 0.45, and 12 respectively. The overall efficiency of the turbine plant is 85%
and the wheel speed is 550 rpm. Determine the following:
a)
b)
c)
d)
The quantity of water required to develop turbine power
The wheel diameter
Diameter of the jet
Number of jets necessary to produce the power.
DIMENSIONAL HOMOGENEITY:
A physical equation is the relationship between two or more physical quantities. A
dimensional homogeneous equation is applicable to all system of units. On the other hand the
dimensional non-homogeneous equation is applicable to only the system of units for which it had
originally been derived.
Consider pressure and height relationship, P =γ h
Here, P = F/L2 and γ = F/L3, h = L and therefore, both RHS and LHS are dimensionally equal
to F/L2. Therefore P =γ h is a dimensionally homogeneous equation. Any system (SI, CGS,
FPS, or US) can be used to evaluate pressure from this equation without any error.
Now consider Manning Formula Q = (1.49/n) R2/3 S1/2
Here, Q = L3 /T and n is a constant, R = A/P = L2 /L = L, S = hf / L = L/L = 1, therefore, LHS
= L3T-1 and RHS=L2/3. Notice that LHS and RHS are not equal, so they are dimensionally
nonequivalent. The Manning equation is therefore dimensionally inhomogeneous. The
physical equations, which are dimensionally homogeneous, obey Fourier's Law of
dimensional homogeneity.
Supplementary notes on Turbomachines/Dr. Kalim/Page #22