The objective of the lab is to analyze the performance of a hydraulic pump, responsible for the transfer of fluid between two tanks at a constant flow, in function of its rotational speed. As the RPM vary from 0 to 4000 we are mainly interested in studying the speed, flow rate and pressures when entering and exiting the pump, the coefficient of head losses associated with the delivery duct, the required hydraulic power and the hydraulic power generated. As the lab progresses, we find ourselves needing to solve the problem of cavitation that manifests itself in the aspiration duct, and are asked to calculate the plate angle of orientation when the cylinders are placed along a circumference with diameter of 60mm. The objective of the lab is to analyze the operativity of an actuator used to control the movements of an Airbus A320 rudder. The Airbus A320 uses three actuators with double redundancy, each of which is designed to control the mobile surface independently. Given the opposing moment that must be overcome we can calculate the muscular force required to control the mobile surface, from which we can determine the dimensional specifics for the actuator that will be introduced, the equations of operation and the approximate time required to complete the movement.

1. Polytechnic University of Milan
Department of Aerospace Engineering
Aerospace Systems

2. Andre Odu
Student ID : 789645
Professor : Paolo Astori
September 2014
Laboratory N° 1:
Analysis of a Hydraulic Pump

3. 1.Preface
The objective of the lab is to analyze the performance of a hydraulic pump
responsible for the transfer of fluid between two tanks at a constant flow, in
function of its rotational speed.
As the rotations per minute (rpm) vary from 0 to 4000, we are mainly interested in
studying the speed, flow rate and pressures when entering and exiting the pump,
the coefficient of head losses associated with the delivery duct, the required
hydraulic power and the hydraulic power generated.
As the lab progresses, we find ourselves needing to solve the problem of cavitation
that manifests itself it the aspiration duct, and are the asked to calculate the plate
angle of orientation when the cylinders are placed along a circumference with
diameter of 60 mm.

4. 2. Index
1. Preface
2. Index
3. List of Symbols
4. Description of the Problem and Method of Resolution
4.1 Aspiration Duct
4.2 Delivery Duct
4.3 Pressures
4.4 Coefficient of head losses
4.5 Generated and Required Power
4.6 Verifying Cavitation
4.7 Plate Angle of Orientation
5. Given Data
6. Calculations and Results
6.1 Flow Rates
6.2 Speeds
6.3 Pressures
6.4 Coefficient of head losses
6.5 Generated and Required Power
6.6 Verifying Cavitation
6.7 Plate Angle of Orientation
7. Final Considerations

5. 3.List Of Symbols
PA Pressure in tank A [Pa]
PB Pressure in tank B [Pa]
Np Number of pistons in the pump [ ]
Dp Diameter of the pistons in the pump [m]
Cp Stroke of the pistons [m]
ηv Volumetric Efficiency of the pump [ ]
ηm Mechanical Efficiency of the pump [ ]
L0 Length of the Aspiration Duct [m]
D0 Diameter of the Aspiration Duct [m]
L Length of the Delivery Duct [m]
D Diameter of the Delivery Duct [m]
k Coefficient of Minor Losses [ ]
Keq Coefficient of Head Losses [Pa/(m3/s)2]
λ Coefficient of Distributed Losses [ ]
e Average Roughness [m]
ν Cinematic Viscosity of the liquid [St]
ρ Density of the Liquid [Kg/m3]

6. Pc Vapor Tension of the Liquid [Pa]
V cilindrata [m3]
Qin Flow Rate in the Aspiration Duct [m3/s]
Qout Flow Rate in the Delivery Duct [m3/s]
vin Speed in the Aspiration Duct [m/s]
vout Speed in the Delivery Duct [m/s]
Pin Pressure at the Pump Entrance [Pa]
Pout Pressure at the Pump Exit [Pa]
Ain Section of the Aspiration Duct [m2]
Aout Section of the Delivery Duct [m2]
ein Relative Roughness of the Aspiration Duct [ ]
eout Relative Roughness of the Delivery Duct [ ]
WHyd Hydraulic Power Generated [W]
WMec Mechanical Power Necessary [W]
ΔP Variation of Pressure between pump exit and Tank B [Pa]
ΔPp Variation of Pressure between pump entrance and exit [Pa]
DT Diameter of the Rotating Drum [m]
α Swash Plate Angle of Orientation [rad]

7. 4.Description of the Problemand Method of Resolution
The instructions ask to calculate the following parameters, in function of the range
of values in which the rotational speed of the volumetric pump may vary:
· Speed and Flow Rate in the Aspiration and Delivery Ducts
· Pressure Entering and Exiting the Pump
· Coefficient of Head Loss Associated with the Delivery Duct
· Generated Hydraulic Power and Required Mechanical Power
The instructions also ask to verify the occurrenceof cavitation in the aspiration
duct and to determine the swash plate angle of orientation, knowing that its
diameter is of 60 mm.
4.1 Aspiration Duct
For starters, in order to calculate the Flow Rate in the aspiration duct, I need to
determine the displacement of the pump.
Given the piston diameter, the number of pistons and their stroke, the displacement
of the pump is calculated through the following equation:
V = Np * pi * Cp * ( Dp / 2 )2
Once the displacement of the pump is know, the flow rate of the aspiration duct is
easily obtainable, in function of the rotational speed (n):
Qin = n * V
The final step is finding the fluid's speed in the aspiration duct, which depends on
the section of the aspiration duct:
Ain = pi * (D0 / 2 )2

8. vin = Qin / Ain
4.2 Delivery Duct
The Flow Rate in the delivery duct depends on the Flow Rate in the aspiration duct
and on the volumetric efficiency of the pump through the following equation:
Qout = ηv * Qin
Having determined the Flow Rate in the delivery duct, the speed is obtainable
through the same procedure used for the aspiration duct
Aout = pi * (D / 2 )2
vout = Qout / Aout
4.3 Pressures
To calculate the pressures at the entrance and exit of the pump we need to
determine the coefficient of distributed losses ( λ ) which varies in function of the
Reynolds numbers of the aspiration and delivery ducts:
Rein = ( d0 * vin ) / v
Reout = ( d * vout ) / v
For Reynolds numbers below 2000, the flow is considered laminar and the
coefficient of distributed losses is approximated as:
λ = 64 / Re
For Reynolds number between 2000 and 4000 the flow is considered transitional,
and above 4000 the flow is considered completely turbulent.
Remembering that the two speeds depend of the rotational speed ( n ) we find that
the coefficient of distributed losses ( λ ) depends on the number of rotations per
minute.

9. In our case the Reynolds numbers become greater that 2000 in both the aspiration
duct ( at approximately 3000 rpm ) and the delivery duct ( at approximately 3800
rpm) ,while always remaining below the 4000 mark.
The problem now becomes that of a transitional phase with rough ducts, which is
solved by determining the coefficient of distributed losses in function of the
relative roughness of the ducts and their respective Reynolds numbers by
consulting the Moody diagram.
ein = e / D0 = 0.0062
eout = e / D = 0.0080
With these values of relative roughness, and low Reynolds numbers, the coefficient
of distributed losses appears as a curve with equation:
λ = ( 1.8log(Re) - 1.64 )-2
Given the equation for the coefficient of distributed losses, the next step is
calculating the pressures entering and exiting the pump:
λin = ( 1.8log(Rein) - 1.64 )-2
λout = ( 1.8log(Reout) - 1.64)-2
Pin = Pa - ( λin * ( L0 / D0 ) * 0.5 * ρ * vin
2
Pout = Pb + ( k + λout * ( L / D ) * 0.5 * ρ * vout
2
4.4 Coefficient of head losses
When considering the delivery duct, the coefficient of head losses is easily
calculated through the following equation:
Keq = ΔP / Qout

10. 4.5 Generated and Required Power
The generated Hydraulic Power depends strictly on the variation in pressure
between pump entrance and exit, and on the flow rate in the delivery duct( which
also depends on the flow rate in the aspiration duct )
Whyd = ΔP * Qout
The Mechanical Power required to generate this Hydraulic Power depends on both
the Mechanical Efficiency and the Volumetric Efficiency
Wmec = Whyd / ( ηm * ηv)
4.6 Verifying Cavitation
With the numerical data obtained we are now able to prevent cavitation, by
ensuring that the pressures at the entrance and exit of the pump stay above the
vapor tension of the fluid ( Pc ) for every rotation speed taken into consideration.
From the equations obtained for the entrance and exit pressures we notice that
above a certain number or rotations per minute the entrance pressure ( Pin ) falls
below the vapor tension ( Pc ) , compromising the pump.
4.7 Plate Angle of Orientation
The equation which puts in relation the piston stroke with the diameter of the
rotating drum is as follows:
Cp = DT * tan(α )
From this equation the plate angle of orientation is then:
α = tan-1( Cp / DT )

11. 5.Given Data
PA = 0.25 MPa
PB = 0.80 MPa
Np = 9
Dp = 0.5 in
Cp = 22.5 mm
ηv = 96 %
ηm = 88 %
L0 = 1 m
D0 = 13 mm
L = 4 m
D = 10 mm
k = 4
e = 0.08 mm
ν = 80 cSt
ρ = 950 kg/m3
Pc = 15 kPa
n = 0÷4000 rpm

12. 6.CalculationsAnd Results
All the following results and the associated graph were calculated and plotted with
matlab. The required codewill be displayed
6.1 Flow Rates
Vtot = Np * (pi*Cp*(Dp /2)2 (m3)
Vtot_l = Vtot * 1000 (liters)
Qin = n*Vtot_l (liters / minute )
ηv = 0.96
Qout = Qin*ηv (liters / minute )

13. 6.2 Speeds
d0 = 0.013 (m)
Ain = pi * (d0/2)2 (m2)
vin = Qin/(Ain*60000) (m/s)
d = 0.01 (m)
Aout = pi * (d/2)2 (m2)
vout = Qout/(Aout*60000) (m/s)

14. 6.3 Pressures
Rein = (d0 * vin) / v
Reout = (d * vout) / v

15. lambdain(1:3820) = 64 ./ Rein(1:3820)
lambdain(3821:4000) = ( 1.8 .* log10(Rein(3821:4000)) - 1.64 )-2
lambdaout(1:3061) = 64 ./ Reout(1:3061) ;
lambdaout(3062:4000) = ( 1.8 .* log10(Reout(3062:4000)) - 1.64 )-2
Pin = Pa - (lambdain * (l0 / d0) * 0.5 * rho * (vin)2)
Pout = Pb + (k + lambdaout * (l/d) ) * 0.5 * rho * (vout)2

16. 6.4 Coefficient of head losses
ΔP = Pout - Pb
Keq = ΔP / ( (Qout/60000)2 )

17. 6.5 Generated and Required Power
Whyd = (Pout - Pin) * (Qout/60000)
Wmec = Whyd / (ηm*ηv)

18. 6.6 Solving Cavitation
d1 = 0.015
Ain = pi * (d1/2)2
vin = Qin/(Ain*60000)
Rein = (d1 * vin) / v
lambdain(1:3820) = 64 / Rein(1:3820) ;
lambdain(3821:4000) = ( 1.8 * log10(Rein(3821:4000)) - 1.64 )-2

19. Pin = Pa - ( lambdain * (l0/d0) * 0.5 * rho * (vin
2 )
6.7 Plate Angle of Orientation
α = atan(Cp / d) * (180/pi) = 20.556

20. 7.Final Considerations
After thoroughly analyzing the pump we can deducethat, becauseof how the plant
was structured , the best parameters are obtained when working at a low number of
rotations per minute. At approximately 3800 rpm the Reynolds number in the
aspiration duct becomes greater than 2000, and in the transition to a turbulent flow
the pressure falls below that of cavitation, compromising the pump. However,
when considering a constant flow rate, the loss of pressure along the ducts depends
exclusively on the width of the duct. Hypothetically speaking, by increasing the
diameter of the aspiration duct by few millimeters one could increase the pressure
enough to avoid cavitation in the entire range of rotation speeds considered. In this
particular case, increasing the diameter to 15 millimeters ( increasing by merely 2
millimeters ) is enough to avoid cavitation.

21. Laboratorio N° 7:
Comandi di Volo
1. Premessa
L'obbiettivo del laboratorio e' di studiare la funzionalita' dell'attuatore usato per il
comando del timone in un Airbus A320. Nel caso dell'Airbus A320 sono presenti
tre attuatori con doppiaridondanza, ognuno e' quindi progettato per azionare in
modo autonomo la superficie di comando.
Dato il momento della cerniera da contrastare, possiamo ricavare la forza
muscolare necessaria per muovere la superficie di co. mando, possiamo
determinare le dimensioni dell'attuatore che verra' introdotto, e a sua volta ricavare
l'equazioni di funzionamento e stimare il tempo richiesto per completare il
movimento.

22. 2. Indice
1. Premessa
2. Indice
3. Lista dei Simboli Utilizzati
4. Descrizione del Problema e Metodo di Risoluzione
4.1 Stimare l'Entita' della Forza Muscolare Richiesta
4.2 Dimensionare l'Attuatore
4.3 Tracciare i Diagrammi Significativi del Sistema
4.4 Scrivere l'Equazione Dinamica dell'Attuatore
4.5 Stimare il Tempo di Attuazione
5. Dati del Problema
6. Calcoli e Risultati
6.1 Stimare l'Entita' della Forza Muscolare Richiesta
6.2 Dimensionare l'Attuatore
6.3 Disegnare lo Schema Idraulico Completo
6.4 Tracciare i Diagrammi Significativi del Sistema
6.5 Stimare il Tempo di Attuazione

23. 3. Lista Dei Simboli Utilizzati
LC lavoro fatto per ruotare il timone fino alla sua apertura massima [N*m]
LP lavoro fatto per far estendere completamente l'asta di comando [N*m]
M Momento d'Inerzia della cerniera [kg*m2]
J momento d'inerzia del timone [kg*m2]
S Corsadel pedale [m]
b Distanza tra cerniera e punto di attacco dell'attuatore [m]
dS Diametro stelo [m]
AC Area del cilindro [m2]
AS Area dello stelo [m2]
PP Pressione pilota [Pa]
PS Pressione serbatioio [Pa]
a Accellerazione [m/s2]
v Velocita' [m/s]
Q Flusso [m3/s]

24. 4. Descrizione del Problema e Metodo di Risoluzione
Le istruzioni chiedono di studiare il funzionamento di uno dei tre attuatori
indipendenti utilizzati per azionare il timone. In particolare e' richiesto di:
 Stimare l'Entita' della Forza Muscolare necessaria in Assenza di
Potenziamento Idraulico
 Dimensionare l'Attuatore
 Disegnare lo Schema Idraulico Completo del Servocomando (FBW)
 Tracciare i Diagrammi Significativi
 Scrivere l'Equazione Dinamica dell'Attuatore
 Stimare il Tempo di Attuazione

4.1 Stimare la Forza Muscolare Necessaria
Considerando lineare il momento di cerniera, possoricavare il lavoro fatto per
ruotare il timone fino alla sua apertura massima
Considerando poi la forza manuale applicata all'asta, possoricavare il lavoro fatto
per far estendere completamente l'asta di comando collegata alla cerniera
Del bilancio dei due lavori ottengo, in funzione del momento massimo della
cerniera, la forza massima che deve esercitare manualmente il pilota
Mmax * ϑmax = Fmusc * s
Fmusc = ( Mmax * ϑmax ) / s

25. 4.2 Dimensionamento dell'Attuatore
Una volta ricavata la forza muscolare necessaria, e' evidente il bisogno di
utilizzare un attuatore idraulico. Conoscendo la distanza tra la cerniera e il punto di
attacco dell'attuatore si ricava la corsa dell'attuatore e il braccio della forza
esercitata rispetto alla cerniera stessa
h = b * cos(ϑ )
x = b * sin(ϑ )
Siccome corsae braccio dipendono dalla variazione dell'angolo theta, per theta
massimo o minimo otteniamo dei valori significativi
hmin = b * cos(ϑmax )
xmax = b * sin(ϑmax )
Noto il valore minimo assunto dal braccio della forza, e conoscendo il valore
massimo assunto dal momento della cerniera, si ricava la forza massima necessaria
per muovere completamente il timone, ovvero la forza massima che dovra' poter
esercitare l'attuatore
Fmax = Mmax / hmin
Conoscendola forza che deve esercitare l'attuatore, lo si puo' dimensionare in
funzione delle pressioni esterne e dell'impianto. Sovradimensionando del 20%
ottengo una superficie utile dell'attuatore che mi garantisce il funzionamento
corretto
A = 1.2 * Fmax / ΔP
Dalla superficie utile dell'attuatore, avendo gia' scelto il diametro dello stelo ricavo
il diametro del cilindro richiesto
As = pi * ( ds / 2 )2
Ac = A + As
dc = (Ac / pi )0.5 * 2

26. Quindi ottengo la forza massima che puo' esercitare questo attuatore
sovradimensionato, che deve essere maggiore della forza massima necessaria per
muovere il timone
Fatt = A * ΔP
4.3 Tracciare i Diagrammi Significativi
Al variare dell'angolo theta studio la corsae il braccio tra forza e cerniera
x = b * sin(ϑ )
h = b * cos(ϑ )
Avendo interpolato il grafico del momento della cerniera in funzione dell'angolo
theta, possiamo studiare la forza richiesta per bilanciare il momento al variare
dell'angolo theta
F = M / h
4.4 Scrivere l'Equazione Dinamica dell'Attuatore
Bilanciando forze e pressioni applicate all'attuatore, possoscrivere la sua
equazione dinamica
F = ( P1 - P2 ) * A - F(x )
( m + ( J / b2 ) ) * a = ΔP * A - F(x )
P1 = PP - K * Q2 = PP - K * ( A * v )2
P2 = PR + K * Q2 = PR + K * ( A * v )2
( m + ( J / b2 ) ) * a = ( PP - PR ) * A - 2 * A * K * ( A * v )2 - F(x )

27. 4.5 Stimare il Tempo di Attuazione
Per semplificare anziche' considerare F(x) considero una forza media, e
considerando una velocita media (costante) ipotizzo accelerazione nulla
L'equazione dinamica si riduce quindi a
0 = ( PP - PR ) * A - 2 * K * A3 * ( vmed )2 - Fmed
vmed = ( ( ( PP - PR ) * A - Fmed ) / 2 * K * A3 )0.5
Avendo ricavato la velocita' media, banalmente si ottiene il tempo di attuazione
t = xmax / vmed
5. Dati del Problema
S = 0.2 m
b = 0.1 m
Mmax = 1500 kg*m2
K = 2.1600e+15 Pa/(m3/s)2
PP = 21000000 Pa
PS = 101300 Pa
ϑmax = 25 °
m = 2.5 kg

28. 6 Calcoli e Risultati
Tutti i risultati e i loro grafici sono stati calcolati tramite matlab. Il codiceassociato
ad ogni singola richiesta verra' fornito.
6.1 Stimare l'Entita' della forza muscolare richiesta
ϑmax = 25 * ( pi/180 )
Fmusc = ( Mmax * ϑmax ) / s = 3272.5 (N)
6.2 Dimensionamento dell'Attuatore
hmin = b * cos(ϑmax) = 0.0906 (m)
xmax = b * sin(ϑmax) = 0.0423 (m)
Fmax = Mmax / hmin = 16.551 (kN)
ΔP = PP - PR (Pa)
A = 1.2 * Fmax / ΔP = 9.5034e-04 (m2)
dS = 0.02 (m)
As = pi * ( ds / 2 )2 = 3.1416e-04 (m2)
Ac = A + As = 0.0013 (m2)
dc = (Ac / pi )0.5 * 2 = 0.0401 (m)
Fatt = A * ΔP = 19.861 (kN)

29. 6.4 Disegnare lo Schema Idraulico Completo
Nel caso di un controllo fly-by-wire consistema idraulico, il sistema viene
scematizzato come segue

30. 6.4 Tracciare i Diagrammi Significativi
x = b * sin(ϑ )

31. h = b * cos(ϑ)

32. F = M / h = M / ( b * cos(ϑ) )
6.5 Stimare il Tempo di Attuazione
Fmed = sum(abs(F))/ length(F) = 10.025 (kN)
Q = 5/60000 (m3/s)
vmed = ( ( ( PP - PR ) * A - Fmed ) / 2 * K * A3 )0.5 = 0.0515 (m/s)
t = xmax / vmed = 0.8205 (s)