EAT223 Thermofluids and Engines
Turbomachinery
Summary
Motivation
Principles of operation
Dimensional analysis
Pump selection
Motivation
The most common practical engineering application for fluid mechanics is the design of fluid machinery.
This includes machines that add energy to the fluid (pumps), and machines that extract energy (turbines).
Both types are connected to a rotating shaft i.e. Both are turbmachines.
(Turbo Latin for spin, or whirl).
Whilst machines that deliver liquids are called pumps, when gases are involved different terms are used:
Fan: small pressure rises
Blower: up to 1 atm. Pressure rise
Compressor: > 1 atm. Pressure rise
We will study a device that is commonly used to move liquids through a piping system: the centrifugal pump.
This is a rotodynamic pump that uses a rotating impeller to increase the pressure of a fluid.
Principles of operation
Fluid enters the pump impeller along or near to the rotating axis and is accelerated by the impeller, flowing radially outward into a diffuser or volute chamber (casing)
From here it exits into the downstream piping system.
Dimensional analysis I
The power P of any rotary hydraulic pump typically depends upon the following quantities:
N: speed
D: impeller diameter
Q: volumetric flow rate (discharge)
: density
H: change in head
g: gravitational constant
The general equation is thus: gQHDNP ,,,,,
However, since if is normal to consider gH as one quantity, we may write this as:
QgHDNP ,,,,
Dimensional analysis II
Exercise:
Use Buckinghams theory to find 3 dimensionless groups.
Assume the repeating variables are , N and D.
Thus, the groups can be found by expressing the remaining variables in terms of a Nb Dc and finding a, b, and c.
E.g. P = 1 a Nb Dc solve to get 1 etc.
You should end up with three Pi groups, such that: 1 = [2, 3]
QgHDNP ,,,,
Solution:
22353321,,
DN
gH
ND
Q
DN
P
Power group
Flow (or discharge)
group
Head group
Dimensional analysis III
We can plot non-dimensional performance characteristics as follows:
22353,
DN
gH
ND
Q
DN
P
3ND
Q
53DN
P
22DN
gH
Power
Efficiency
Head
Worked Example
A pump draws water from one tank and delivers it to another at a higher elevation. The pump impeller is 500 mm diameter and revolves at 600 rev/min.
The pump is geometrically and dynamically similar to another pump with an impeller 550 mm diameter, for which the following data was acquired when the pump was running at 900 rev/min:
Determine the flow rate and developed head for the pump used, if the frictional resistance characteristic (i.e. the system characteristic) is given by the following expression:
238008 QH f
H (m) 37 41 44 45 42 36 28
Q (m3/s) 0 0.016 0.032 0.048 0.064 0.08 0.096
Expressions of this form come from the static head + the turbulent head loss term in Bernoullis equation, covered in EAT106
Solution to worked example I
First determine the head-flow characteristics for the pump actually used. To do this we know that the head and flow coefficients (calculated using the dimensionless groups) are constant:
1
3
1
2
1
2123
1
2
1
2
2
1
21222
5.0constant
367.0constant
QD
D
N
NQQ
ND
Q
HD
D
N
NHH
DN
gH
From these relationships, produce a table for the pump you are using:
H2 (m) 13.58 15.05 16.15 16.52 15.41 13.21 10.28
Q2 (m3/s) 0 0.008 0.016 0.024 0.032 0.04 0.048
H (m) 37 41 44 45 42 36 28
Q (m3/s) 0 0.016 0.032 0.048 0.064 0.08 0.096
Solution to worked example II
Now produce a table for frictional resistance Hf versus Q.
Choose Q values 0, 0.02, 0.04, 0.06, 0.08, 0.1
Hf (m) 8 9.52 14.08 21.68 32.32 46
Q (m3/s) 0 0.02 0.04 0.06 0.08 0.1
238008 QH f
Now plot H2 and Hf versus Q.
The operating point is where the two curves intersect.
Curves produced using Excel are on the next slide.
However, in an exam you will have to plot them by hand!
Solution to worked example III
At the matching point: Q = 0.038 m3/s Head = 13.5 m
Question 1
A centrifugal pump has an impellor diameter of 0.5 m. When running at 550 rev/min the following data was obtained:
Q (m3/min)
0 7 14 21 28 35 42 49 56
H (m) 40 40.6 40.4 39.3 38 33.6 25.6 14.5 0
(%) 0 41 60 74 83 83 74 51 0
Predict the performance of a geometrically similar pump having an impeller diameter 0.35 m, running at 1450 rev/min. Hence determine the efficiency of the pump at the operating point and the power required to drive the pump. The system characteristic is given by the relationship h = 20 + 0.01Q2. [41%; 720.5 kW]
The following non-dimensional relationships may be assumed without proof:
22353,
DN
gH
ND
Q
DN
P
Remember: Assume data is unchanged. Pump power P = QgH /
Question 2
Tests on a centrifugal pump having an impellor diameter of 8 cm and running at a constant speed of 500 rev/min gave the following results for water::
Q (m3/s) 0 0.5 1.0 1.5 2.0 2.5
H (m) 4.0 4.1 3.9 3.4 2.4 1.2
(%) 0 48 68 76 70 50
It is proposed to use a similar pump having an impellor diameter of 4 cm and running at a higher speed of 800 rev/min, in an application where the system resistance is given by the equation h = 10(1 + Q).
Determine the new flow rate and delivery head, the efficiency at which the pump operates, and the power required to drive the pump at the new operating point. [0.416 m3/s; 14.2 m; 67%; 86.5 kW]
22353,
DN
gH
ND
Q
DN
P
Remember: Assume data is unchanged. Pump power P = QgH /