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1 Dipartimento di Ingegneria Aerospaziale Corso di Impianti e Sistemi aerospaziali Fluid Mechanics Tommaso Negri, Academic ID 830049 February 2017
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Dipartimento di Ingegneria Aerospaziale
Corso di Impianti e Sistemi aerospaziali
Fluid Mechanics
Tommaso Negri, Academic ID 830049
February 2017
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1. Introduction The purpose of the following report is to analyze the following hydraulic system: a constant pressure displacement pump displaces a fluid from the tank A through a suction line and forces it to the tank B through three delivery lines. Losses in pipe joints are ignored.
2. Contents
1. Introduction……………………………………………………………………………………………………………………2
2. Contents………………………………………………………………………………………………………………………….2
3. List of symbols………………………………………………………………………………………………………………...3
4. Description of the problem and solving method………………………………………………………………..4
4.1……………………………………………………………………………………………………………………………..4
4.2……………………………………………………………………………………………………………………………..4
4.3……………………………………………………………………………………………………………………………..4
4.4……………………………………………………………………………………………………………………………..4
4.5……………………………………………………………………………………………………………………………..5
4.6……………………………………………………………………………………………………………………………..5
4.7……………………………………………………………………………………………………………………………..5
5. Problem data……………………………………………………………………………………………………………………6
6. Calculations……………………...……………………………………………………………………………………..……….7
6.1……………………………………………………………………………………………………………………………..7
6.2……………………………………………………………………………………………………………………………..8
6.3……………………………………………………………………………………………………………………………..8
6.4……………………………………………………………………………………………………………………………..8
6.5……………………………………………………………………………………………………………………………..9
6.6……………………………………………………………………………………………………………………………..9
6.7…………………………………………………………………………………………………………………………...10
7. Summary results……………………………………………………………………………………………………………..12
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3. List of symbols pA Pressure of the tank A [MPa]
pB Pressure of the tank B [MPa]
pp Pressure of the pump [MPa]
p0 Pressure in suction [MPa]
ηv Volumetric efficiency of the pump
ηm Mechanical efficiency of the pump
L0 Length of suction line [m]
L1 Length of delivery line 1 [m]
L2 Length of delivery line 2 [m]
L3 Length of delivery line 3 [m]
D0 Diameter of suction line [mm]
D Diameter of delivery lines [mm]
k Coefficient of minor losses
e Average roughness of suction line [mm]
ν Kinematic viscosity of the fluid [cSt]
ρ Density of the fluid [kg/m3]
pc Vapor pressure of the fluid [kPa]
β Bulk modulus of the fluid [MPa]
v0 Speed in suction line [m/s]
v1 Speed in delivery line 1 [m/s]
v2 Speed in delivery line 2 [m/s]
v3 Speed in delivery line 3 [m/s]
λ0 Head loss coefficient in suction line
λ Head loss coefficient in delivery lines
Re0 Reynolds number in suction line
Re1 Reynolds number in delivery line 1
Re2 Reynolds number in delivery line 2
Re3 Reynolds number in delivery line 3
A0 Suction pipe area [m2]
A Delivery lines area [m2]
Q0 Flow in suction line [l/min]
Q1 Flow in delivery line 1 [l/min]
Q2 Flow in delivery line 2 [l/min]
Q3 Flow in delivery line 3 [l/min]
Qtot Total flow in delivery [l/min]
ΔQ Total loss of flow in the pump [l/min]
ΔQCOMP Loss of flow in the pump by compressibility [l/min]
ΔQLEA Loss of flow in the pump by leakage [l/min]
Keq Equivalent coefficient of minor losses
Wh Hydraulic power generated [kW]
Wm Mechanical power requested [kW]
Dp Diameter of the pistons [mm]
d Diameter of the circumference described by the pistons [mm]
Np Number of pistons
Ap Pistons area [mm2]
V Total volume displaced per evolution [cm3]
c Pistons stroke [mm]
α Incline angle of the swash plate [°]
n Number of revolutions per minute [rpm]
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4. Description of the problem and solving method 4.1 Calculation of speed and flow in the delivery pipes and check of laminarity condition hypothesis
Assuming the flow laminar, λ is calculated with the formula: 𝜆 =64
𝑅𝑒=
64 𝜈
𝑣 𝐷
Knowing the pump’s output pressure pp and the tank’s pressure pb, the fluid’s density ρ and the geometrical characteristics of each pipe, it is possible to find the speed in delivery pipes from
Darcy-Weisbach law: 𝑝𝑝 − 𝑝𝑏 = 𝜆 𝐿
𝐷
1
2 𝜌𝑣2; adding minor losses in the first pipe: 𝑘
1
2𝜌𝑣2
Once derived the three values of speed, it is necessary to verify the laminarity condition by checking the Reynolds number is lower than 2000. Then it’s possible to calculate the flow in each pipe by the formula: 𝑄𝑖 = 𝑣𝑖𝐴
4.2 Calculation of the equivalent coefficient of minor losses in delivery lines The equivalent coefficient of minor losses can be calculated by the formula: 𝑝𝑝 − 𝑝𝑏 = 𝑘𝑒𝑞𝑄𝑡𝑜𝑡
2 ,
where 𝑄𝑡𝑜𝑡 is the total flow in delivery lines, obtained summing the flows of each delivery line. Another way to calculate the equivalent coefficient is to calculate the coefficient of minor losses of each delivery line and then, considering that are three lines in parallel, derive the equivalent
from the formula: 1
√𝑘𝑒𝑞=
1
√𝑘1+
1
√𝑘2+
1
√𝑘3
4.3 Calculation of speed and flow in suction line It is possible to derive the suction flow through the volumetric efficiency of the pump and the
total delivery flow: 𝑄0 = 𝑄𝑡𝑜𝑡
𝜂𝑣
Now it’s possible to calculate the suction speed by the area of the suction pipe 𝑣0 =𝑄0
𝐴0
4.4 Verification of the existence of the cavitation problem in suction line and eventual resolution Cavitation problem exists if the pressure in the suction pipe is lower than the vapor pressure of
the fluid. Suction pressure is calculated with the formula 𝑝𝑎 − 𝑝0 = 𝜆𝐿0
𝐷0
1
2𝜌𝑣2. λ is find into the
Moody diagram given in the data, considering the relative roughness 𝜖 =𝑒
𝐷0 of the pipe and the
Reynolds number of the fluid in the suction line. Now is checked if the suction pressure is lower than the vapor pressure. In case it is, it is possible to overcome cavitation problem with different solutions:
Increase the pressure of tank A Enlarge the diameter of the suction pipe Change the roughness of the pipe
The first solution would require a considerable increase of pressure, so it would be impractical; The second would be a very expensive solution; The better way to solve the problem is to enlarge the suction pipe’s diameter, and to do it is possible to use a Matlab script with the function [Head_loss].
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4.5 Calculation of the flow lost in the pump by leakage and compressibility The total flow loss percentage can be calculated using the definition of volumetric efficiency:
𝜂𝑣 =𝑄𝑡𝑜𝑡
𝑄0=
𝑄0 − Δ𝑄
𝑄0= 1 −
Δ𝑄𝐶𝑂𝑀𝑃 + Δ𝑄𝐿𝐸𝐴
𝑄0
The flow lost by compressibility can be derived from the compressibility relation Δ𝑄𝐶𝑂𝑀𝑃
𝑄0=
1
𝛽(𝑝𝑝 − 𝑝0)
Finally, the flow lost by leakage is obtained subtracting the part lost by compressibility from the total loss of flow in the pump. 4.6 Calculation of the hydraulic power generated and the mechanic power requested by the pump The hydraulic power generated is given by the formula: 𝑊ℎ
𝑔𝑒𝑛= 𝑄𝑡𝑜𝑡(𝑝𝑝 − 𝑝0)
The mechanic power requested is obtained using the volumetric and the mechanical efficiency
of the pump: 𝑊𝑚𝑟𝑒𝑞 =
𝑊ℎ𝑔𝑒𝑛
𝜂𝑣𝜂𝑚
4.7 Definition of the geometrical and mechanical characteristics of an axial pump that can provide the requested flow An axial displacement pump like the one in figure will be used in the system. Values of number of revolutions per minute, number of pistons, diameter of the pistons and of the circumference described by the pistons will be chosen arbitrarily in order to derive the stroke of the pistons and the incline angle of the swash plate necessary to provide the requested flow:
𝑄0 = 𝑛𝑉; 𝑉 = 𝑁𝑝𝑐𝐴𝑝; 𝑐 = 𝑑 tan 𝛼
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5. Data Pressure of the tank A pA 1 MPa Pressure of the tank B pB 15 MPa Pressure of the pump pp 21 MPa Volumetric efficiency of the pump ηv 0.96 Mechanical efficiency of the pump ηm 0.90 Length of suction line L0 1 m Length of delivery line 1 L1 20 m Length of delivery line 2 L2 20 m Length of delivery line 3 L3 25 m Diameter of suction line D0 6 mm Diameter of delivery lines D 10 mm Coefficient of minor losses k 20 Average roughness of suction line e 18/100 mm Kinematic viscosity of the fluid ν 95 cSt Density of the fluid ρ 950 kg/m3
Vapor pressure of the fluid pc 15 kPa Bulk modulus of the fluid β 1500 MPa
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6. Calculations
6.1 Calculation of speed and flow in the delivery pipes and check of laminarity condition hypothesis
1. 𝑝𝑝 − 𝑝𝑏 = 𝜆𝐿1
𝐷
1
2𝜌𝑣1
2 + 𝑘1
2𝜌𝑣1
2
𝑣12 + 𝑣164
𝐿1
𝐷
𝜈
𝑘𝐷+ 2
(𝑝𝑏 − 𝑝𝑝)
𝑘𝜌= 0
𝑣1 = −32𝐿1𝜈
𝑘𝐷2± √
(32𝐿1𝜈)2
𝑘2𝐷4−
2(𝑝𝑏 − 𝑝𝑝)
𝑘𝜌= 9,04
𝑚
𝑠
𝑅𝑒1 =𝑣1𝐷
𝜈= 951,6
The laminarity condition is verified.
𝑄1 = 𝑣1𝐴 = 𝑣1
𝐷2
4𝜋 = 710,23 ∙ 10−6
𝑚3
𝑠= 42,61
𝑙
𝑚𝑖𝑛
2. 𝑝𝑝 − 𝑝𝑏 = 𝜆𝐿2
𝐷
1
2𝜌𝑣2
2
𝑣2 =(𝑝𝑏 − 𝑝𝑝)𝐷2
32𝐿2𝜈𝜌= 10,39
𝑚
𝑠
𝑅𝑒2 =𝑣2𝐷
𝜈= 1093,7
The laminarity condition is verified.
𝑄2 = 𝑣2𝐴 = 𝑣2
𝐷2
4𝜋 = 815,24 ∙ 10−6
𝑚3
𝑠= 48,92
𝑙
𝑚𝑖𝑛
3. 𝑝𝑝 − 𝑝𝑏 = 𝜆𝐿3
𝐷
1
2𝜌𝑣3
2
𝑣3 =(𝑝𝑏 − 𝑝𝑝)𝐷2
32𝐿3𝜈𝜌= 8,31
𝑚
𝑠
𝑅𝑒3 =𝑣3𝐷
𝜈= 874,7
The laminarity condition is verified.
𝑄3 = 𝑣3𝐴 = 𝑣3
𝐷2
4𝜋 = 6,53 ∙ 10−4
𝑚3
𝑠= 39,18
𝑙
𝑚𝑖𝑛
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6.2 Calculation of the equivalent of minor losses in delivery lines 𝑝𝑝 − 𝑝𝑏 = 𝑘𝑒𝑞𝑄𝑡𝑜𝑡
2
𝑘𝑒𝑞 =𝑝𝑝 − 𝑝𝑏
𝑄𝑡𝑜𝑡2 = 1,26 ∙ 1012
𝑃𝑎 ∙ 𝑠2
𝑚6
Using the other method, it is necessary first to calculate the coefficient of each delivery line.
1. 𝑝𝑝 − 𝑝𝑏 = 𝑘1𝑄12
𝑘1 =𝑝𝑝 − 𝑝𝑏
𝑄12 = 11,9 ∙ 1012
𝑃𝑎 ∙ 𝑠2
𝑚6
2. 𝑝𝑝 − 𝑝𝑏 = 𝑘2𝑄22
𝑘2 =𝑝𝑝 − 𝑝𝑏
𝑄22 = 9,10 ∙ 1012
𝑃𝑎 ∙ 𝑠2
𝑚6
3. 𝑝𝑝 − 𝑝𝑏 = 𝑘3𝑄3
2
𝑘3 =𝑝𝑝 − 𝑝𝑏
𝑄32 = 14,1 ∙ 1012
𝑃𝑎 ∙ 𝑠2
𝑚6
1
√𝑘𝑒𝑞
=1
√𝑘1
+1
√𝑘2
+1
√𝑘3
⇒ 𝑘𝑒𝑞 = 1,26 ∙ 1012𝑃𝑎 ∙ 𝑠2
𝑚6
6.3 Calculation of speed and flow in suction line
𝑄0 = 𝑄𝑡𝑜𝑡
𝜂𝑣=
𝑄1 + 𝑄2 + 𝑄3
𝜂𝑣= 2,27 ∙ 10−3
𝑚3
𝑠= 136,13
𝑙
𝑚𝑖𝑛
𝑣0 =𝑄0
𝐴0=
𝑄0
𝐷2
4 𝜋= 80,28
𝑚
𝑠
6.4 Verification of the existence of the cavitation problem in suction line and eventual resolution
𝑅𝑒0 =𝑣0𝐷0
𝜈= 5070,3
From the Moody diagram, we find a value of 𝜆 = 5070,3 (that means the fluid is turbulent).
𝑝𝑎 − 𝑝0 = 𝜆𝐿0
𝐷0
1
2𝜌𝑣2 ⇒ 𝑝0 = 𝑝𝑎 − 𝜆
𝐿0
𝐷0
1
2𝜌 𝑣0
2 = −22,96 𝑀𝑃𝑎 ≪ 15 𝑘𝑃𝑎 = 𝑝𝑐
This means that there is the problem of cavitation in the suction pipe. It is possible also to know the point in the pipe where the bubbles begin to form.
𝑝𝑎 − 𝑝𝑐 = 𝜆x
𝐷0
1
2𝜌𝑣0
2 ⇒ x =2𝐷0(𝑝𝑎 − 𝑝𝑐)
𝜆𝜌𝑣02 = 0,043 𝑚
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The following Matlab script finds a new value of suction’s pipe diameter that avoids cavitation and also the values of pressure, speed, Reynolds number and friction factor
%D0 needed to solve the cavitation problem
d0=[]; p0=[];
while(P0<Pc) D0=D0+0.001; v0=Q0/(pi*(D0^2)/4);
Re0=v0*D0/(ni*1e-4); eps=e/D0;
lambda=Head_loss(Re0,eps); P0=Pa-0.5*lambda*L0*rho*(v0^2)/D0;
d0=[d0;D0]; p0=[p0;P0];
end
The results are: 𝐷0 = 11 𝑚𝑚
𝜆0 ≅ 0,034
𝑝0 ≅ 0,146 𝑀𝑃𝑎
𝑣0 = 23,88 𝑚
𝑠
6.5 Calculation of the flow lost in the pump by leakage and compressibility
𝜂𝑣 =𝑄𝑡𝑜𝑡
𝑄0=
𝑄0 − Δ𝑄
𝑄0= 1 −
Δ𝑄𝐶𝑂𝑀𝑃 + Δ𝑄𝐿𝐸𝐴
𝑄0= 0,96 ⇒
Δ𝑄
𝑄0= 0,04 = 4%
Δ𝑄𝐶𝑂𝑀𝑃
𝑄0=
1
𝛽(𝑝𝑝 − 𝑝0) = 0,014 = 1,4%
Δ𝑄𝐶𝑂𝑀𝑃 = 𝑄0 ∙ 0,014 = 1,91 𝑙
𝑚𝑖𝑛
Δ𝑄𝐿𝐸𝐴
𝑄0= Δ𝑄 − Δ𝑄𝐶𝑂𝑀𝑃 = 0,026 = 2,6%
Δ𝑄𝐿𝐸𝐴 = 𝑄0 ∙ 0,026 = 3,54 𝑙
𝑚𝑖𝑛
6.6 Calculation of the hydraulic power generated and the mechanic power requested by the pump
𝑊ℎ𝑔𝑒𝑛
= 𝑄𝑡𝑜𝑡(𝑝𝑝 − 𝑝0) = 45,4 𝑘𝑊
𝑊𝑚𝑟𝑒𝑞 =
𝑊ℎ𝑔𝑒𝑛
𝜂𝑣𝜂𝑚= 52,6 𝑘𝑊
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6.7 Definition of the geometrical and mechanical characteristics of an axial pump that can provide the requested flow
The values of revolution per minute and diameter of the circumference described by the pistons are fixed arbitrarily: n=3500 rpm, d=60 mm. With a Matlab script will be confronted pumps with such characteristics and different numbers of pistons and different diameters, and the relative strokes and incline angles. Np=[5,7,9]; Dp=[0.011:0.001:0.02] [m]
%Sizing of the axial piston pump
rpm=3500; V=Q0/(rpm/60); Dp=[0.011:0.001:0.02];
Np=[5 7 9]; d=0.06; %Stroke in function of the diameter of the pistons
c=V./[Np(1)*pi*(Dp.^2)/4; Np(2)*pi*(Dp.^2)/4; Np(3)*pi*(Dp.^2)/4];
figure(1)
plot(Dp*1000,c(1,:)*1000,Dp*1000,c(2,:)*1000,Dp*1000,c(3,:)*1000)
title('Stroke in function of the diameter of the pistons’)
xlabel('diameter [mm]') ylabel('stroke [mm]') legend('5 pistons','7 pistons','9 pistons')
grid on %Incline angle in function of the diameter of the pistons
alpha=atan(c./d).*180./pi; figure(2) plot(Dp*1000,alpha(1,:),Dp*1000,alpha(2,:),Dp*1000,alpha(3,:))
title(‘Incline angle in function of the diameter of the pistons’)
xlabel('diameter [mm]') ylabel('alpha [°]') legend('5 pistons','7 pistons','9 pistons','Location','Best')
grid on
The script gives the following results:
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5 pistons
pistons diameter 11 12 13 14 15 16 17 18 19 20 stroke 82 69 59 51 44 39 34 31 27 25 incline angle 54 49 44 40 36 33 30 27 25 22 7 pistons
pistons diameter 11 12 13 14 15 16 17 18 19 20 stroke 59 49 42 36 32 28 25 22 20 18 incline angle 44 39 35 31 28 25 22 20 18 16 9 pistons
pistons diameter 11 12 13 14 15 16 17 18 19 20
stroke 46 38 33 28 25 22 19 17 15 14
incline angle 37 32 28 25 22 20 18 16 14 13
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7. Summary Results
Delivery line 1
Speed: 𝑣1 = 9,04 𝑚
𝑠
Flow: 𝑄1 = 42,61 𝑙
𝑚𝑖𝑛
Reynolds number: 𝑅𝑒1 = 951,6
Delivery line 2
Speed: 𝑣2 = 10,39 𝑚
𝑠
Flow: 𝑄2 = 48,92 𝑙
𝑚𝑖𝑛
Reynolds number: 𝑅𝑒2 = 1053,7
Delivery line 3
Speed: 𝑣3 = 8,31 𝑚
𝑠
Flow: 𝑄3 = 39,18 𝑙
𝑚𝑖𝑛
Reynolds number: 𝑅𝑒3 = 874,7
Equivalent coefficient of minor losses in delivery lines
𝑘𝑒𝑞 = 1,26 ∙ 1012𝑃𝑎 ∙ 𝑠2
𝑚6
Speed in suction pipe
𝑣0 = 80,28𝑚
𝑠
Flow in suction pipe
𝑄0 = 136,13𝑙
𝑚𝑖𝑛
Pressure in the suction pipe
𝑝0 = −22,96 𝑀𝑃𝑎 (≪ 15 𝑘𝑃𝑎 = 𝑝𝑐)
Increased diameter of the suction pipe that allows to avoid the problem of cavitation
𝐷0 = 11 𝑚𝑚
Pressure lost in the pomp by compressibility
Δ𝑄𝐶𝑂𝑀𝑃 = 1,91 𝑙
𝑚𝑖𝑛
Pressure lost in the pump by leakage
Δ𝑄𝐿𝐸𝐴 = 3,54𝑙
𝑚𝑖𝑛
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Hydraulic power generated in the pump
𝑊ℎ
𝑔𝑒𝑛= 45,4 𝑘𝑊
Mechanic power requested by the pump
𝑊𝑚
𝑟𝑒𝑞 = 52,6 𝑘𝑊
Characteristics of the chosen pump
Number of revolutions per minute: n=3500 rpm Diameter of the circumference described by the pump: d=60 mm Number of the pistons: Np=7 Diameter of the pistons: Dp=15 mm
Stroke of the pistons: c=32 mm Incline angle of the plate swash: α=28°
