This preview shows page 1. Sign up to view the full content.
Unformatted text preview: XI. Turbomachinery
This chapter considers the theory and performance characteristics of the
mechanical devices associated with the fluid circulation.
General Classification:
Turbomachine  A device which adds or extracts energy from a fluid.
Adds energy:
Extracts energy: Pump
Turbine In this context, a pump is a generic classification that includes any device
that adds energy to a fluid, e.g. fans, blowers, compressors.
We can classify pumps by operating concept:
1. Positive displacement
2. Dynamic (momentum change)
General Performance Characteristics
Positive Displacement Pumps
1. Delivers pulsating or periodic flow (cavity opens, fluid enters, cavity
closes, decreasing volume forces fluid out exit opening.
2. Not sensitive to wide viscosity changes.
3. Delivers a moderate flow rate.
4. Produces a high pressure rise.
5. Small range of flow rate operation (fixed pump speed).
Dynamic Pumps
1. Typically higher flow rates than PD’s.
2. Comparatively steady discharge.
3. Moderate to low pressure rise.
4. Large range of flow rate operation.
5. Very sensitive to fluid viscosity.  v v XI  1  eText Main Menu  Textbook Table of Contents  Study Guide Typical Performance Curves (at fixed impeller speed) Fig. 11.2 Performance curves for dynamic and positive
displacement pumps
Centrifugal Pumps
Most common turbomachine used in industry. Includes the general
categories of (a) liquid pumps, (b) fans, (c) blowers, etc.
They are momentum change devices and thus fall within the dynamic
classification.
Typical schematic shown as Fig. 11.3 Cutaway schematic of a typical centrifugal pump  v v XI  2  eText Main Menu  Textbook Table of Contents  Study Guide Writing the energy equation across the device and solving for hp – hf ,we
have P − P1 V2 − V1
H = hp − hf = 2
+
+ Z 2 − Z1
ρg
2g
2 2 where H is the net useful head delivered to the fluid, the head that results in
pressure, velocity, and static elevation change.
Since for most pumps (not all), V1 = V2 and ∆Z is small, we can write H≅ ∆P
ρg Since friction losses have already been subtracted, this is
the ideal head delivered to the fluid. Note that velocity
head has been neglected and can be significant at large
flow rates where pressure head is small.
Pw = ρ Q g H The ideal power to the fluid is given
by The pump efficiency is given by η= ρQgH ρQgH
Pw
=
=
BHP
BHP
ωT where BHP = shaft power necessary to drive the pump
ω = angular speed of shaft
T = torque delivered to pump shaft
Note that from the efficiency equation, pump efficiency is zero at zero flow
rate Q and at zero pump head,H.  v v XI  3  eText Main Menu  Textbook Table of Contents  Study Guide Basic Pump Theory
Development of basic pump theory begins with application of the integral
conservation equation for momentofmomentum previously presented in Ch. III.
Applying this equation to a centrifugal pump with one inlet, one exit, and uniform
properties at each inlet and exit, we obtain
&
&
T = me r x Ve − mi r x Vi where T is the shaft torque needed to drive the pump Vi , Ve are the absolute velocities at the inlet and exit of the pump
This is used to determine the change of angular momentum across the device. Fig. 11.4 Inlet and exit velocity diagrams for an idealized impeller
Since the velocity diagram is key to the analysis of the device, we will discuss the
elements in detail.  v v XI  4  eText Main Menu  Textbook Table of Contents  Study Guide 1. At the inner radius r1 have two velocity components:
a. the circumferential velocity due to the
impeller rotation u1 = r1 ω V
1 w1 blade tip speed at inner radius
β1 b. relative flow velocity tangent to the
blade w1 tangent to the blade angle α1 Vn 1 u1 Vt β1 1 These combine to yield the absolute inlet
velocity V1 at angle α1
The absolute velocity can be resolved into two absolute velocity components:
1. Normal ( radial ) component: Vn1 = V1 sin α1 = w1 sin β1 Note that for ideal pump design, Vn1 = V1 and α 1 = 90 o 2. Absolute tangential velocity: Vt1 = V1 cos α 1 = u 1  w1 cos β1 again, ideally Vt1 = 0 It is also important to note that Vn1 is use to determine the inlet flow rate, i.e., Q = A1 Vn1 = 2 π r1 b1 Vn1
where b1 is the inlet blade width  v v XI  5  eText Main Menu  Textbook Table of Contents  Study Guide Likewise for the outer radius r2 we have the following:
a. the circumferential velocity due to the
impeller rotation u 2 = r2 ω blade tip speed at outer radius
b. relative flow velocity tangent to the
blade w2 tangent to the blade angle V2 w2
β2 Vn2 α2 Vt u2 2 β2 These again combine to yield the absolute
outlet velocity V2 at angle α2
The exit absolute velocity can also be resolved into two absolute velocity
components:
1. Normal ( radial ) component: Vn2 = V2 sin α 2 = w2 sin β2 = Q
2 π r2 b2 Note that Q is the same as for the
inlet flow rate 2. Absolute tangential velocity: Vt 2 = V2 cos α 2 = u 2  w2 cos β2 Vt 2 = u 2  Vn2 tan β 2 = u2  Q
2 π r2 b 2 tan β2 Q = A1 Vn1 = 2 π r1 b1 Vn1 = A 2 Vn2 = 2 π r2 b 2 Vn 2 where Again, each of the above expressions follows easily from the velocity diagram,
and the student should draw and use the diagram with each pump theory
problem.  v v XI  6  eText Main Menu  Textbook Table of Contents  Study Guide We can now apply moment  of – momentum equation. { T = ρ Q r2 * Vt 2 − r1 * Vt1 } (again Vt1 is zero for the ideal design) For a sign convention, we have assumed that Vt1 and Vt2 are positive in the
direction of impeller rotation.
The “ ideal” power supplied to the fluid is given by { Pw = ω T = ρ Q ω r2 Vt2 − ω r1 Vt1
or { } } Pw = ω T = ρ Q u 2 Vt 2 − u1 Vt1 = ρ Qg H
Since these are ideal values, the shaft power required to drive a nonideal pump
is given by BHP = Pw ηp The head delivered to the fluid is H= ρ Q{u 2 Vt2 − u1 Vt1 }
ρQg {u
= 2 Vt 2 − u 1 Vt1 } g For the special case of purely radial inlet flow H* = u 2 Vt 2
g  v v XI  7  eText Main Menu  Textbook Table of Contents  Study Guide From the exit velocity diagram, substituting for Vt2 we can show that u
ωQ
H= 2 −
g 2 π b 2 g tan β 2
2 2 has the form C1  C2 Q shutoff head, the head produced at zero flow, Q = 0 u2
where: C1 =
g
Example: A centrifugal water pump operates at the following conditions:
speed = 1440 rpm, r1 = 4 in, r2 = 7 in, β1 = 30o, β2 = 20o, b1 = b2 = 1.75 in
Assuming the inlet flow enters normal to the impeller (zero absolute tangential
velocity):
find: (a) Q, (b) T, (c) Wp, (d) hp, (e) ∆P ω = 1440 rev 2 π
rad
= 150.8
min 60
s Calculate blade tip velocities: u1 = r1 ω = 4
rad
ft
ft150.8
= 50.3
12
s
s u 2 = r2 ω = Since design is ideal, at inlet 7
rad
ft
ft150.8
= 88
12
s
s V = V1
n
1 w1 α1 = 90 , Vt1 = 0
o Vn1 = U1 tan 300 = 50.3 tan 30o = 29.04 ft/s 30Þ 90Þ Q = 2 π r1 b1 Vn1 r1
•  v v XI  8  eText Main Menu  Textbook Table of Contents  Study Guide 30Þ
u1 3 ft
ft
4
Q = 2 π ft1.75 ft 29.04 = 8.87
s
12
s
3 ft
gal
gal
s
Q = 8.87 60
7.48 3 = 3981
ft
min
s
min
Repeat for the outlet: ft 3
8.87
Q
s
Vn2 =
=
2 π r2 b 2 2 π 7 ft 1.75 ft
12 12
ft
Vn2 = 16.6
s
Vn2
16.6 ft/s
ft
w2 =
=
= 48.54
s
sin 20 o sin 20 o V
2 w2 α2 20Þ 20Þ
u2 r2
• Vt 2 = u 2  w 2 cos β2 = 88 − 48.54 cos 20 o = 42.4 ft
s We are now able to determine the pump performance parameters. Since for the
centrifugal pump, the moment arm r1 at the inlet is zero, the momentum equation
becomes
Ideal moment of momentum delivered to the fluid: { T = ρ Q r2 * Vt2 } 3 slug
ft 7
ft
= 1.938 3 8.87
ft 42.4 = 425.1ft − lbf
ft
s
s 12 Ideal power delivered to the fluid: P = ω T = 150.8 rad
ft − lbf
425.1ft − lbf = 64,103
= 116.5 hp
s
s  v v XI  9  eText Main Menu  Textbook Table of Contents  Study Guide Head produced by the pump (ideal): H= P
64,103 ft − lbf/s
=
= 115.9 ft
lbf
ft 3
ρ gQ
62.4 3 8.87
ft
s Pressure increase produced by the pump:
3 ft
∆ P = ρ g H = 62.4 115.9 ft = 7226 psf = 50.2 psi
s Pump Performance Curves and Similarity Laws Pump performance results are typically obtained from an experimental test of the
given pump and are presented graphically for each performance parameter.
• Basic independent variable  Q {usually gpm or cfm }
• Dependent variables typically
– head pressure rise, in some cases ∆P H BHP – input power requirements (motor size)
η – pump efficiency • These typically presented at fixed pump speed and impeller diameter
Typical performance curves appear as  v v XI  10  eText Main Menu  Textbook Table of Contents  Study Guide Fig. 11.6 Typical Centrifugal Pump Performance Curves
at Fixed Pump Speed and diameter
These curves are observed to have the following characteristics:
1. hp is approximately constant at low flow rate.
2. hp = 0 at Qmax.
3. BHP is not equal to 0 at Q = 0.
4. BHP increases monotonically with the increase in Q.
5. ηp = 0 at Q = 0 and at Qmax.
6. Maximum pump efficiency occurs at approximately Q* = 0.6 Qmax . This
is the best efficiency point BEP. At any other operating point, efficiency is
less, pump head can be higher or lower, and BHP can be higher or lower.
7. At the BEP, Q = Q*, hp = hp*, BHP = BHP*.
Measured Performance Data
Actual pump performance data will typically be presented graphically as shown in
Fig. 11.7. Each graph will usually have curves representing the pump head vs flow
rate for two or more impeller diameters for a given class/model of pumps having a
similar design. The graphs will also show curves of constant efficiency and
constant pump power (BHP) for the impeller diameters shown. All curves will be
for a fixed pump impeller speed.  v v XI  11  eText Main Menu  Textbook Table of Contents  Study Guide Fig. 11.7 Measured performance curves for two models of a centrifugal
water pump  v v XI  12  eText Main Menu  Textbook Table of Contents  Study Guide How to Read Pump Performance Curves
Care must be taken to correctly read the performance data from pump curves. This
should be done as follows:
(1) For a given flow rate Q
(2) Read vertically to a point on the pump head curve h for the impeller
diameter D of interest.
(3) All remaining parameters ( efficiency & BHP) are read at this point; i.e.,
graphically interpolate between adjacent curves for BHP to obtain the pump
power at this point.
Note that the resulting values are valid only for the conditions of these curves:
(1) pump model and design, (2) pump speed – N, (3) impeller size – D, (4) fluid
(typically water)
Thus for the pump shown in Fig. 11.7a with an impeller diameter D = 32 in, we
obtain the following performance at Q = 20,000 gpm:
Q = 20,000 gpm, D = 32 in, N = 1170 rpm
H ≅ 385 ft, BHP ≅ 2300 bhp, ηp ≅ 86.3 %
Note that points that are not on an h vs. Q curve are not valid operating points.
Thus for Fig. 11.7b, the conditions
Q = 22,000 gpm, BHP = 1500 bhp, hp = 250 ft
do not correspond to a valid operating point because they do not fall on one of the
given impeller diameter curves. However, for the same figure, the point
Q = 20,000 gpm, BHP = 1250 bhp
is a valid point because it coincidentally also falls on the D = 38 in impeller curve
at hp = 227 ft.  v v XI  13  eText Main Menu  Textbook Table of Contents  Study Guide Net Positive Suction Head  NPH
One additional parameter is typically shown on pump performance curves:
NPSH = head required at the pump inlet to keep the fluid from cavitating.
NPSH is defined as follows:
2 P
NPSH = i +
ρg Vi
P
−v
2g ρg where Pi = pump inlet pressure
Pv = vapor pressure of fluid
Pump inlet Considering the adjacent figure,
write the energy equation between
the fluid surface and the pump
inlet to obtain the following: P
NPSH = i +
ρg zi Pa Pi
z=0 2 Vi
P
P
P
− v = a − Z i − h f,a − i − v
2g ρg ρ g
ρg For a pump installation with this configuration to operate as intended, the righthandside of the above equation must be > the NPSH value for the operating
flow rate for the pump.
Example:
A water supply tank and pump are connected
as shown. Pa = 13.6 psia and the water is at
20 o C with Pv = 0.34 psia. The system has a
friction loss of 4.34 ft. Will the NPSH of the
pump of Fig. 11.7a at 20,000 gpm work? a
10 ft
i  v v XI  14  eText Main Menu  Textbook Table of Contents  Study Guide Applying the previous equation we obtain NPSH = Pa
P
− Z i − h f,a − i − v
ρg
ρg (13.6 − 0.34) lbf/in2 *144 in2 /ft 2 − (−10 ft) − 4.34 ft
NPSH =
3
62.4 lbf/ft NPSH = 36.26 ft The pump will work because the system NPSH as shown in
Fig. 11.7a is 30 ft which provides a 6.3 ft safety margin.
Conversely, the pump could be located as close as 3.7 ft
below the water surface and meet NPSH requirements. Pump Similarity Laws
Application of the dimensional analysis procedures of Ch. V will yield the
following three dimensionless performance parameters:
Dimensionless flow coefficient: CQ = Q
ω D3 Dimensionless head coefficient: CH = gH
ω 2 D2 Dimensionless power coefficient: CP = BHP
ρω 3 D5 where ω is the pump speed in radians/time and other symbols are standard design
and operating parameters with units that make the coefficients dimensionless.
How are these used?
These terms can be used to estimate design and performance changes between two
pumps of similar design.  v v XI  15  eText Main Menu  Textbook Table of Contents  Study Guide Stated in another way:
If pumps 1 and 2 are from the same geometric design family and are operating at
similar operating conditions, the flow rates, pump head, and pump power for the
two pumps will be related according to the following expressions: Q2 N2
=
Q1 N1 D2 D1 2 3 H 2 N2 D2 = H1 N1 D1 Use to predict the new flow rate for a design
change in pump speed N and impeller
diameter D.
2 Used to predict the new pump head H for a
design change in pump speed, N and
impeller diameter D.
3 BHP2 ρ2 N 2 D2 = BHP1 ρ1 N1 D1 5 Used to predict the new pump power BHP
for a design change in fluid, ρ, pump speed
N and impeller diameter D. Example
It is desired to modify the operating
conditions for the 38 in diameter
impeller pump of Fig. 11.7b to a
new pump speed of 900 rpm and a
larger impeller diameter of 40 in. •
H(ft) •
BEP1 Determine the new pump head and
power for the new pump speed at
the BEP. Q(gpm)  v v XI  16  eText Main Menu  Textbook Table of Contents  Study Guide BEP 2 For the D = 38 in impeller of Fig. 11.7b operating at 710 rpm, we read the best
efficiency point (BEP) values as
Q* = 20,000 gpm, H* = 225 ft, BHP * = 1250 hp Applying the similarity laws for N2 = 900 rpm and D2 = D1 = 38 in, we obtain
3 3
Q2 N2 D2 900 40 =
= 1.478 =
710 38 Q1 N1 D1 Q2 = 20,000*1.478 = 29,570 gpm
2 ans. 2 2
2
H 2 N2 D2 900 40 = 1.78
= =
710 38 H1 N1 D1 H2 = 225*1.78 = 400.5 ft
3 ans. 5 3
5
BHP2 ρ2 N 2 D2 900 40 = 2.632
= = (1)
710 38 BHP1 ρ1 N1 D1 BHP2 = 3290 hp ans. Thus, even small changes in the speed and size of a pump can result in significant
changes in flow rate, head, and power.
It is noted that every point on the original 38 in diameter performance curve
exhibits a similar translation to a new operating condition.
The similarity laws are obviously useful to predict changes in the performance
characteristics of an existing pump or to estimate the performance of a modified
pump design prior to the construction of a prototype.  v v XI  17  eText Main Menu  Textbook Table of Contents  Study Guide Matching a Pump to System Characteristics
The typical design/sizing requirement for a pump is to select a pump which has a
pump head which matches the required system head at the design/operating flow
rate for the piping system. Key Point hp = hsys at Qdes. It is noted that pump selection should occur such that the operating point of the
selected pump should occur on the pump curve near or at the BEP.
From the energy equation in Ch. VI, the system head is typically expressed as
2 h sys 2 2
P2 − P1 V2 − V1
f L + ∑ K V
=
+
+ Z 2 − Z1 + i
D 2g
ρg
2g Thus the selection of a pump for a
piping system design should result in
a pump for which the pump head
hp at the design flow rate Qdes is
equal ( or very close) to the head η p hp
Hdes requirements hsys of the piping
system at the same flow rate, and this
should occur at or near the point of
maximum efficiency for the chosen
pump. •
hsys Q(gpm) Qdes Other operating and performance requirements (such as NPSH) are obviously also
a part of the selection criteria for a pump.  v v XI  18  eText Main Menu  Textbook Table of Contents  Study Guide Pumping Systems: Parallel and Series Configurations
For some piping system designs, it may be desirable to consider a multiple pump
system to meet the design requirements. Two typical options include parallel and
series configurations of pumps. Specific performance criteria must be met when
considering these options.
Given a piping system which has a known design flow rate and head requirements,
Qdes, hdes. The following pump selection criteria apply.
Pumps in Parallel:
Assuming that the pumps are identical,
each pump must provide the following:
Q(pump) = 0.5 Qdes
h(pump) = hdes Pumps in Series:
Assuming that the pumps are identical,
each pump must provide the following:
Q (pump) = Qdes
h(pump) = 0.5 hdes
For example, if the design point for a given piping system were Qdes = 600 gpm,
and hsys = 270 ft, the following pump selection criteria would apply:
1. Single pump system Q(pump) = 600 gpm, hp = 270 ft 2. Parallel pump system Q(pump) = 300 gpm, hp = 270 ft
for each of the two pumps 3. Series pump system Q(pump) = 600 gpm, hp = 135 ft
for each of the two pumps  v v XI  19  eText Main Menu  Textbook Table of Contents  Study Guide ...
View
Full
Document
This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

Click to edit the document details