1. Abstract
The objective from this experiment was determining the efficiency of a centrifugal
pump for half and fully opened diffuser and determining the head loss for system of
pumps in series. Two graphs were obtained for five cases the performance graph and
head graph.
2. Introduction
Centrifugal pumps are a sub-class of dynamic axisymmetric work-absorbing
turbo machinery. Centrifugal pumps are used to transport fluids by the conversion of
rotational kinetic energy to the hydrodynamic energy of the fluid flow. The rotational
energy typically comes from an engine or electric motor. The 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 where it exits.
Common uses include water, sewage, petroleum and petrochemical pumping. The
reverse function of the centrifugal pump is a water turbine converting potential energy
of water pressure into mechanical rotational energy. like most pumps, a centrifugal
pump converts mechanical energy from a motor to energy of a moving fluid. A
portion of the energy goes into kinetic energy of the fluid motion, and some into
potential energy, represented by fluid pressure (hydraulic head) or by lifting the fluid,
against gravity, to a higher altitude.
The transfer of energy from the mechanical rotation of the impeller to the motion and
pressure of the fluid is usually described in terms of centrifugal force, especially in
older sources written before the modern concept of centrifugal force as a fictitious
force in a rotating reference frame was well articulated. The concept of centrifugal
force is not actually required to describe the action of the centrifugal pump.
The outlet pressure is a reflection of the pressure that applies the centripetal force that
curves the path of the water to move circularly inside the pump. On the other hand,
the statement that the "outward force generated within the wheel is to be understood
as being produced entirely by the medium of centrifugal force" is best understood in
terms of centrifugal force as a fictional force in the frame of reference of the rotating
impeller; the actual forces on the water are inward, or centripetal, since that is the
direction of force needed to make the water move in circles.
This force is supplied by a pressure gradient that is set up by the rotation, where the
pressure at the outside, at the wall of the volute, can be taken as a reactive centrifugal
force. This was typical of nineteenth and early twentieth century writings, mixing the
concepts of centrifugal force in informal descriptions of effects, such as those in the
centrifugal pump.
3. Theory
3.1 Head loss can be calcualted by
∆𝑯 =
∆𝑷
𝝆𝒈
Where
ΔP: Pressure difference is taking from the pump machine (pa)
ρ: density of the fluid (Water) (kg/m3)
g: The gravitational acceleration (m/s2)
3.2 The hydraulic power Ph can be calculated from
𝑷𝒉 = ∆𝑷 × 𝑸
Where
ΔP: Pressure difference is taking from the pump machine (pa)
Q: Volumetric flow rate (m3/s)
3.3 The mechanical power Pm can be calculated from
𝟐𝝅𝑵
𝑷𝒎 = 𝑻 × 𝝎 = 𝑭𝒓(
)
𝟔𝟎
Where
T: Torque of the pump force equal F.r (N.m)
ω: Angular velocity of the impeller (rad/s)
3.4 The efficiency ƞ of the pump can be found by
Ƞ=
𝑯𝒚𝒅𝒓𝒖𝒂𝒍𝒊𝒄 𝑷𝒐𝒘𝒆𝒓
𝑴𝒆𝒄𝒉𝒂𝒏𝒊𝒄𝒂𝒍 𝑷𝒐𝒘𝒆𝒓
4. Apparatus & Procedures
Fig 1: Cenrifugal Pump apparatus
Fig 2: Sketch of the pump
Experimental procedure
1. The pump has to be primed before been able to operate.
2. The diffuser vanes have to be set at certain opening, start with fully open vanes.
3. The pump is started and the speed increases to 1800 rpm.
4. The flow rate is varied from zero to maximum possible. This is done using the
throttling valve from fully closed until fully open.
5. At each valve opening, the reading of Hg-manometer, ΔHm [mmHg], torque-meter
spring force, F [N], rotating speed, N [rpm], suction and discharge pressure, Ps and Pd
[bar], voltage and current, V and I [volts, and ampere]
6. Repeat the experiment for different diffuser vane angle (half Open).
7. Repeat the above experiment at different speed, 2000 rpm, [at diffuser vane angle
full open]
For the system of pumps in series
1. Establish the characteristic curves for each pump separately at certain operating
speed.
2. Connect the two pumps in series
3. Operate the pumps at the same speeds as were in first step.
4. At different flow rates, adjusted using the discharge valve, measure the following
variables:
-The pressure developed by the (pumps) system
-The flow rate developed by (pumps) system
-The power input to each pump
Calculate and plot the characteristic curves for the (pumps) system
5. Imposed the characteristic curves of single pumps on the above plots.
5. Data and Calculations
I will perform calculations for the first row from the data on the next one page for
the small pump.
F=6N, r = 0.17m, ω=1472 rpm, Q= 0.002m3/s, ΔP=0.2 bar
1. T=F*r = 1.02 N.m
2. ω = (1472*2π)/60 = 154.1475 rad/s
3. Pmechanical = T*ω =157.23 W
4. Q = 0.002m3/s
5. ΔP = 0.2*100000 = 20000 pa
6. Phaydrulic = Q*ΔP = 0.002*20000 = 40 W
7. ƞ = Phaydrulic /Pmechanical = 40/157.23 = 0.2544
Table1: the data from small pipe
P[pa]
F[N]
Q[m3/s] radius[m] N
20000
70000
30000
6
6
6.2
0.002
0.0018
0.002
0.17
1472
H[m]
Phydro[w]
2.038736 40
7.135576 126
3.058104 60
5.1.2 Half Opened Flow:
F=14.9 N, r = 0.267 m, ω=1800 rpm, Q= 49m3/h, ΔP=0.14 bar
1. T=F*r = 14.9*0.267 = 3.97 N.m
2. ω = (1800*2π)/60 = 188.5 rad/s
3. Pmechanical = T*ω = 749.5 W
4. Q = 49/3600 = 0.014 m3/s
5. ΔP = 0.14*100000 = 14000 pa
6. Phaydrulic = Q*ΔP = 0.0136*14000 = 190.5 W
7. ƞ = Phaydrulic /Pmechanical = 190.5/749.5 = 0.25
5.2 Pumps connected in series
5.2.1 Pump A
F=12.7 N, r = 0.16 m, ω=1500 rpm, Q= 2.2 L/s, ΔP=0.21 bar
1. T=F*r = 12.7*0.16 = 2.032 N.m
2. ω = (1500*2π)/60 = 157 rad/s
P-in[w]
effecincy
157.2304 0.254404
157.2304 0.801372
162.4714 0.369296
3. Pmechanical = T*ω = 2.032*0.16 = 319.024 W
4. Q = 2.2/1000 = 0.0022 m3/s
5. ΔP = 0.21*100000 = 21000 pa
6. Phaydrulic = Q*ΔP = 0.0022*21000 = 46.2 W
7. ƞ = Phaydrulic /Pmechanical = 46.2/319.024 = 0.145
8. H= ΔP/ρg = 21000/ (1000*9.81) = 2.14 m
5.2.2 Pump B
F= 4.61 N, r = 0.16 m, ω=1500 rpm, Q= 0.0021 m3/s, ΔP=0.1 bar
1. T=F*r = 4.61*0.16 = 0.7376 N.m
2. ω = (1500*2π)/60 = 157 rad/s
3. Pmechanical = T*ω = 0.7376*0.16 = 115.8 W
4. ΔP = 0.1*100000 = 10000 pa
5. Phaydrulic = Q*ΔP = 0.0021*10000 = 21 W
6. ƞ = Phaydrulic /Pmechanical = 21/115.8 = 0.181
7. H= ΔP/ρg = 10000/ (1000*9.81) = 1.02 m
5.2.3 Pump A+B in Series
F= 9.3 N, r = 0.16 m, ω=1500 rpm, Q= 0.002 m3/s, ΔP=0.2 bar
1. T=F*r = 9.3*0.16 = 1.488 N.m
2. ω = (1500*2π)/60 = 157 rad/s
3. Pmechanical = T*ω = 0.7376*0.16 = 233.61 W
4. ΔP = 0.2*100000 = 20000 pa
5. Phaydrulic = Q*ΔP = 0.002*20000 = 40 W
6. ƞ = Phaydrulic /Pmechanical = 40/233.61 = 0.17
7. H= ΔP/ρg = 20000/ (1000*9.81) = 2.04 m