Linköping University
IEI – Fluid and Mechanical Engineering Systems
Laboratory exercise
Orifices and pressure measurement
Laboratory exercise in Hydraulic servo system (TMHP 51) for M4
Target for the laboratory experiment
In this laboratory experiment shall:
9 The flow coefficient, Cq, be decided for a sharp edged orifice and for a pipe orifice in
two different ways, measurements and calculations.
9 The critical pressure drop be decided at cavitation as function of the pressure before
the orifice
9 Investigate the effect losses at the orifice through temperature measurements.
9 Calibration of pressure transducers be carried out.
Test and measurement equipment
Orifice holder
Figure 1: Hydraulic diagram for the test equipment.
Seals
Orifice washer
Figure 2: The orifice holder for the exchangeable orifices.
2
The pressure p1 and p2 are measured with electric pressure transducers. The flow through the
orifice is measured with a flow meter of gear wheel type and is read direct in litres/minute.
The pressure p1 is varied with an electric hydraulic pressure relief valve and p2 with a manual
valve. The orifice holder can be opened with the six hex bolts. The orifices can hereby easily
be shifted. Note the two seals which can easily come off when the holder is apart.
Cavitation in orifices
The figure below shows the static pressure as function of the position for an orifice.
Position
Figure 3: The static pressure for an orifice.
At constant pressure p1 decreases the pressure in vena contracta linear with the increased Δp
(p2 decreases), i.e.
(1)
pvc = p1 − KΔp
At a certain Δp decreases however pvc immediately down to the oil’s evaporation pressure
(results in cavitation), i.e.
(2)
Δp = Δpcav when pvc = pvc min
Δp1 = constant
Δpcav
Δpvap
Figure 4: Pressure when cavitation is apparent.
When pvc decreased to pvap can the pressure drop p1-pvc not increase more and consequently
the flow can not increase either, even if p2 decreases and consequently Δp increases.
3
Equation (1) and (2) gives
pvc min = p1 − KΔpcav ⇒ Δpcav = −
Δpcav
pvc min p1
+
= C1 + C 2 p1
K
K
(3)
Cavitation
Figure 5: p1 against Δp
The critical pressure drop where cavition occurs is thus linear dependent of the inlet pressure
p1.
4
Exercise
You shall measure the pressure/flow characteristic (Δp as function of q) for a sharp edged
orifice and a pipe orifice.
• Identify the equipment, both hardware and software.
• Place a sharp edged orifice in the orifice holder. Check the seals.
• Open the Argus valves and the orifice valves.
• Start the pump.
• Test the equipment through rise the pressure in the system slowly to max 15 MPa.
Measurement
Record the pressure/flow characteristic for the different orifices. Don’t forget to record the oil
temperature for each orifice.
To decide the critical pressure drop that causes cavitation (Δpcav) as function of inlet pressure
p1 is the test repeated with the pipe orifice at different inlet pressures p1. Δp and p1 are
recoded when cavitation noise is occurred.
Flow coefficient
Calculate Cq from the measurement results for pressure/flow characteristic for the orifices.
Determine also Cq with help of the Handbook and compare the results.
Cavitation
Sketch Δpcav as function of p1. Calculate C1 and C2 (see page 4).
Effect losses in the orifice
Compare the effect losses in the orifice with the increased oil temperature after the orifice at
some working conditions (see page 8). Discuss the results.
5
Figure 6: Specific gravity against temperature and pressure for Shell Tellus Oil 27
6
Figure 7: Kinematic viscosity against temperature and pressure for Shell Tellus Oil 27
7
Temperature rise for the oil passing through an orifice
Q − L t = m 1 i 1 − m 2 i 2
(4)
The technical work Lt = 0 for an orifice.
Adiabatic process gives that Q = 0 and m1 = m2 .
This gives i1 − i2 = 0 .
The enthalpy for an arbitrary medium can be expressed as
di = c p dT + v(1 − αT )dp
where
(5)
v=spec. volume
α=thermic expansion coefficient
Integration of (5) when cp, v and α are constant and v =
1
ρ
i1 − i2 = c p (T2 − T1 ) + (1 − αT )
gives
p2 − p1
(6)
ρ
i1 − i2 = i2 − i1 = 0 gives then
T2 − T = ΔT = (1 − αT )
where
α = 7E − 4
1
K
8
p 2 − p1
cpρ
(7)
9