Assignment 6

Two similar centrifugal pumps 3 (Fig.6.) (their characteristics are given below in tables 6,7) are simultaneously supplying the fuel to tanks 6 and 7. Pipelines branching from the pumps are connected in A knot. The pressure at A knot is P. Pipeline bifurcation is accomplished in B knot. 6 and 7 tanks filling is finished simultaneously.

Determine:

1.Delivery, pressure head and consumed power for each of the pumps.

2.Diameters d₃ and d₄ of the pipelines connected to 6 and 7 fuel tanks.

Note: The fuel level alteration in the tanks is ignored.

Solution technique:

1. Draw characteristics of both pumps using the table data.

2. Set-up characteristic equation for each of the pipelines up to knot A and sketch-graphs having combined them with the respective pump characteristics.

3. Draw pressure versus flow rate curve in knot A separately for each of the pipelines and compose summary pressure – flow rate characteristic in the knot A.

Fig.6. Schematic diagram of the emergency fuel drain: 1,2,6,7 – fuel tanks; 3 – centrifugal pump; 4 – non-return valve; 5 – filter

4. Determine the total flow rate in the knot A and delivery at each of the pumps.

5. Calculate the pressure head and consumed power for each of the pumps.

6. Determine fuel flow in the pipelines connected to 6 and 7 tanks.

7. Set-up characteristic equations for the pointed pipelines and draw pressure versus diameter curve for each of the pipelines from point B to the vessels.

8. Determine pipelines diameters d₃ and d₄, using available pressure at point B.

Table 6

Data	Variant
l1, m	0,6	1,0	1,4	1,8	2,2	2,6	3,0	3,4	3,8	4,2
D1, m	0,020	0,022	0,034	0,025	0,028	0,030	0,032	0,035	0,038	0,040
L2, m
D2, m	0,040	0,042	0,045	0,048	0,050	0,055	0,060	0,065	0,070	0,075
P, kPa
L3, m	0,5	0,6	0,7	0,8	0,9	1,0	1,1	1,2	1,3	1,4
L4, m	3,5	4,0	4,5	5,6	5,8	6,0	6,5	7,0	7,5	8,0
H1, m	0,3	0,7	1,1	1,5	1,9	2,3	2,7	3,1	3,5	3,0
H2, m	3,3	3,7	4,1	4,5	4,9	5,3	5,7	6,1	6,5	6,9
h 3, m	0,7	0,8	0,9	1,0	1,1	1,2	1,3	1,4	1,5	1,6
H4, m		2,5	3,0	3,5	4,0	4,5	5,0	5,5	6,0	6,5
W1, m3	5,5
W2, m3

Table 7

Data
Qp, m3/s		0,0004	0,0008	0,0012	0,0014	0,0016	0,0018	0,002	0,0022
Hp, m
η %

ASSIGNMENT 7

Emergency fuel drain is accomplished (Fig.7.) in case of emergency aircraft landing. The principle scheme of the emergency drain system is shown on the lay-out. Kerosene is supplied to the drain pipeline by two groups of pumps, installed in the 1st and 2d fuel tanks. Kerosene viscosity is ν. The volume of drained fuel is W. The pipeline diameter is d. The lengths of the pipes are l₁ and l₂. Pipelines local resistance coefficients are ζ₁ and ζ₂, number of the pumps in the tanks are z₁ and z₂. Pump efficiency is η.

Basic calculation data are given in table 8.

There are additional data for estimating the H = f (Q) characteristic for centrifugal pump in tables 9.

Determine:

1. Emergency fuel drain duration.

Q, l/s		0,1	0,2	0,3	0,4	0,5	0,6
H, m	18,0	17,0	15,5	13,8	11,4	8,2	3,0

2. Volume of fuel drain from tank 2.

3. Total power of the pumps.

4. Emergency fuel drain duration in the case of 50% pumps failure.

Fig.7. Schematic diagram of the emergency fuel drain: 1, 2 – tanks; 3 – non- return valve; 4 – pumps; 5 – drain pipelines; 6 – tap

Note: The calculus should be performed on graphic and analytic method only for one drain pipeline branch.

Solution technique:

1. Plot up the pumps characteristics, according to table 9 and taking into consideration their parallel operation for two pipelines simultaneously. Subtract the value of pressure loss in pipeline 2 from point A to B from curve 1 to receive curve 2.

2. Fined the pressure flow rate characteristic for point B summing estimated characteristics.

3. Plot the drain pipeline characteristic from point B to C.

Table 8

Data	Variant
ν,m2 /s	5*10-6	4*10-6	3*10-6	1*10-6	2*10-6	1*10-6	3*10-6	4*10-6	6*10-6	5*10-6
W, m3
d, m	0,10	0,08	0,09	0,08	0,08	0,08	0,09	0,08	0,085	0,08
l1, m
l2, m
ζ1
ζ2
z1
z2
η	0,56	0,62	0,70

Table 9

Q, l/s		0,1	0,2	0,3	0,4	0,5	0,6
H, m	18,0	17,0	15,5	13,8	11,4	8,2	3,0

ASSIGNMENT 8

The aircraft engines (Fig.8.) are fed with kerosene by two centrifugal pumps 10 installed in service tanks 3 and 7 of the fuel system. The kerosene transferring is dosed by D₁ and D₂ throttles from tanks 1 and 8, emergency kerosene drains from tank 8 along pipeline 9 through the left and right detaching part of the wing. Kerosene viscosity is v, the volume of drained fuel is W, and emergency drain pipeline diameter and length are d₃ and l₃ respectively. Total head losses coefficient of drain pipeline is ζ₃. D₁ and D₂ are the throttles dosing the fuel under transferring. Q1 and Q2 are throttle D₁ and D₂ fuel discharge. Quantity of pumps equals z₃ and z₄. Calculation data are given in table 10.

There are additional data in table 2 which is necessary for plotting pump characteristic.

Determine:

1. Emergency fuel drain duration.

2. Cross section area of throttles feeding the fuel to tanks 1 and 2.

3. Total pumps power.

4. Fuel drain duration in case of 50% pumps failure.

Note: The calculi are carried out by graphic and analytic methods in complex.

Solution technique:

1. Plot the pump characteristic.

2. Plot the drain pipeline characteristic and find the stationary points on the graphs.

Fig.8. Schematic diagram of the engine fuel supply: 1 – tank; 2 – non-return valve; 3 – tank; 4 – feeding pipeline; 5 – throttle D₁; 6 – throttle D₂; 7 – tank; 8 – tank; 9 – emergency fuel drain pipeline; 10,12 – transferring pumps; 11 – feeding pumps; 13 – stopcock

3. Estimate the fuel emergency drain time.

4. Plot the pump characteristic for tank 5.

5. Find the necessary pressure for kerosene to escape from the throttles.

6. According to the estimated pressure find the throttles diameters D₁ and D₂.

Notes: 1. Head losses in pipes 3 and 8 should be ignored because of their insignificance in comparison with the throttles losses.

2. The explanation given for assignment 7 should be used while plotting the hydraulic characteristics.

Table 10

Data	Variant
ν,m2 /s	5*10-6	3*10-6	4*10-6	2*10-6	1*10-6	5*10-6	3*10-6	4*10-6	2*10-6	1*10-6
W, m3
Q1, m3/h	0,4	0,3	0,4	0,5	0,4	0,3	0,4	0,5	0,3	0,4
Q2, m3/h	1,2	1,4	1,5	1,0	1,6	1,2	1,4	1,5	1,6	1,7
l3, m
ζ3
d3, mm
z6
z5
η	0,65	0,55	0,58

Table 11

Q, l/s		0,001	0,002	0,003	0,004	0,005	0,006
H, m	18,0	17,0	15,5	13,8	11,4	8,2	3,0

ASSIGNMENT 9

Gear pump (Fig.9.) supplies АМG-10 oil into hydrocylinder 7, which is loaded by external force R. The area of the piston is F. Some liquid returns into the service tank along the pipeline l₁, d_1, through the throttle with coefficient of resistance ζ₄. Oil, running into the cylinder through the pipeline l₂, d₂ and leaving it through the pipeline l₃, d₃ passes two-time control valve 6. Resistance coefficient is equal to ζ₂ in both cases. Piston flow velocity is V.

Calculation data are given in table 12.

Fig.9. Schematic diagram of the hydraulic system: 1 – tank; 2 – pump; 3 – non-return valve; 4 – throttle; 5 – strainer; 6 – flow control valve; 7 – cylinder

Determine:

1. The pump pressure and its power consumption, if the pump efficiency is η.

2. Resistance coefficient ζ₄ throttle 4.

3. Piston velocity and pump pressure in the case the bypass pipe is closed.

Note: Take d₂ = d₃ during the calculation.

Solution technique:

1. Find the liquid discharge into the cylinder and necessary pressure to move piston against the load.

2. Estimate the pressure at point A.

3. Calculate the pump pressure and the power consumption of the driver.

4. Estimate liquid flow rate through the bypass pipeline and resistance coefficient of the throttle.

Data	Variant
Q, m /s3	0,0020	0,0019	0,0018	0,0011	0,0016	0,0015	0,0014	0,0013	0,0012	0,0011
l0, m	4,0	2,5	3,0	3,5	4,0	4,5	5,0	5,5	6,0	6,5
D0, m	0,060	0,055	0,055	0,048	0,045	0,042	0,040	0,038	0,035	0,032
l1, m	3,0	3,5	4,0	4,5	5,0	5,5	6,0	6,5	7,0	7,5
d1, m	0,038	0,035	0,032	0,030	0,028	0,025	0,024	0,022	0,020	0,018
l2, m
d2, m	0,040	0,038	0,035	0,032	0,030	0,028	0,025	0,024	0,022	0,020
l3, m
ζ2
V, m/s	0,065	0,060	0,055	0,050	0,045	00,40	0,035	0,030	0,025	0,020
F, m2	0,019	0,018	0,017	0,016	0,015	0,014	0,013	0,012	0,011	0,010
η	0,95	0,94	0,93	0,92	0,91	0,90	0,89	0,88	0,87	0,86
R, kN	1,1	1,2	1,4	1,6	1,8	2,0	2,2	2,4	2,6	2,8

Table 12

Data	Variant
Q, m /s3	0,0020	0,0019	0,0018	0,0011	0,0016	0,0015	0,0014	0,0013	0,0012	0,0011
l0, m	4,0	2,5	3,0	3,5	4,0	4,5	5,0	5,5	6,0	6,5
D0, m	0,060	0,055	0,055	0,048	0,045	0,042	0,040	0,038	0,035	0,032
l1, m	3,0	3,5	4,0	4,5	5,0	5,5	6,0	6,5	7,0	7,5
d1, m	0,038	0,035	0,032	0,030	0,028	0,025	0,024	0,022	0,020	0,018
l2, m
d2, m	0,040	0,038	0,035	0,032	0,030	0,028	0,025	0,024	0,022	0,020
l3, m
ζ2
V, m/s	0,065	0,060	0,055	0,050	0,045	00,40	0,035	0,030	0,025	0,020
F, m2	0,019	0,018	0,017	0,016	0,015	0,014	0,013	0,012	0,011	0,010
Η	0,95	0,94	0,93	0,92	0,91	0,90	0,89	0,88	0,87	0,86
R, kN	1,1	1,2	1,4	1,6	1,8	2,0	2,2	2,4	2,6	2,8

ASSIGNMENT 10

Pump 8 (Fig.10.) supplies liquid to the cylinders, loaded by forces R₁ and R₂, through the non-return valve 7 and filter 6.

Calculation data are given in the table 13.

Determine:

1. Pump entrance (point A) and pump exit (point B) pressure.

2. Local resistance coefficients for throttles ζ₂ and ζ₃, if moving velocities of both pistons are equal.

3. Draw hydraulic gradient for CK and EK sectors.

Note: Local resistance coefficients indexes correspond with local resistance number in Fig.10.

Liquid head in the tanks may be ignored.

Fig.10. Schematic diagram for hydraulic drive: 1 – service tank;

2, 5 – throttles; 3, 4 – cylinders; 6 – filter; 7 – non-return valve; 8 – pump

Solution technique:

1. Obtain piston velocity and liquid in the drain pipeline.

2. Calculate the pressure at points A, B, D (it is necessary to consider roughness Δ₆ pipeline 6).

3. Estimate cylinders drain pressure p₃ and p₄.

4. Find local resistance coefficients ζ₂ and ζ₅.

5. Plot hydraulic gradients.

Table 13

Data	Variant
P0, kPa
P1=P2, MPa					16,5	14,5	12,5	12,5		13,5
Q,m3/s	8,0*10-4	8,2*10-4	8,4*10-4	8,5*10-4	8,0*10-4	7,8*10-4	8,2*10-4	8,3*10-4	8,4*10-4	8,5*10-4
R1,N	2,0*104	2,2*104	2,3*104	2,0*104	2,1*104	2,2*104	2,0*104	2,5*104	2,2*104	2,3*104
R2,N	2,5*104	2,1*104	2,8*104	2,6*104	2,7*104	2,8*104	2,3*104	2,2*104	2,2*104	2,1*104
l1,m	3,0	4,0	5,5	5,5	4,5	4,0	3,5	3,0	3,5	4,0
l1,m	3,0	4,5	6,5	6,3	5,2	5,0	4,0	4,0	4,0	4,6
l 2, m	4,0	5,0	6,0	6,0	8,0	7,5	6,5	5,5	6,0	7,0
l 3, m	2,0	2,5	3,0	3,5	4,0	4,5	4,0	5,0	6,0	6,5
l4, m	5,0	6,0	7,0	8,0	9,0	9,5	10,0	10,5	10,0	8,5
l5, m	4,0	4,5	4,2	4,4	5,0	6,0	4,5	7,0	7,0	6,0
l6,m
d1,m	0,040	0,040	0,040	0,042	0,044	0,044	0,042	0,042	0,040	0,040
d2,m	0,012	0,012	0,012	0,014	0,014	0,014	0,014	0,016	0,016	0,016
d3,m	0,008	0,008	0,008	0,008	0,010	0,010	0,010	0,010	0,012	0,012
d4,m	0,012	0,012	0,012	0,010	0,008	0,008	0,008	0,010	0,010	0,010
d5,m	0,010	0,010	0,010	0,008	0,008	0,008	0,010	0,008	0,008	0,010
d6,m	0,018	0,018	0,014	0,014	0,012	0,012	0,016	0,012	0,016	0,018
Δ6,mm	0,14	0,14	0,14	0,14	0,12	0,12	0,13	0,13	0,12	0,12
D, m	0,060	0,062	0,058	0,060	0,060	0,058	0,058	0,062	0,062	0,062
dr, m	0,038	0,038	0,038	0,040	0,040	0,040	0,042	0,042	0,042	0,042
ν•10-6 m2/s	1,2*10-5	1,2*10-5	1,2*10-5	1,4*10-5	1,4*10-5	1,1*10-5	1,1*10-5	1,3*10-5	1,0*10-5	1,0*10-5
ρ,kg/ m3