SOLUTIONS TO THE TUTORIAL QUESTIONS FOR THE

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TEP 4195 TURBOMACHINERY

SOLUTIONS FOR TUTORIAL QUESTIONS

– COURSE TEP 4195

1)

Maximum pump displacement is

12cm

3

/rev

. Motor displacement is

120 cm

3

/rev

.

a)

At full pump displacement:

Ideal motor speed = 1450rev/min x 12/120 = 145 rev/min

The loss in motor speed due to the volumetric (slip) flow loss at 200 Nm motor torque is:

145 - 138 rev/min = 7 rev/min

At half pump displacement:

Ideal motor speed = 1450 x 6/120 = 72.5 rev/min

For a 100% increase in the motor load torque the pressure will have to increase by 100%.

The loss in motor speed due to the slip flow loss at 400 Nm is:

14 rev/min as the pressure has increased by 100%

The motor speed will be = 72.5 - 14 = 58.5 rev/min

b)

Load torque = 400 Nm

torque

With no losses, motor pressure =

displaceme nt ( m

3

/ rad )

2

120

400

10

6

210 bar

and

For 95%

 m

the motor pressure = 210/0.95 = 221 bar

Pump torque (pump displacement of

6cm

3

/rev

)

With no losses, the theoretical pump torque =

D

P

6

10

2

6

221

10

5

21 .

1 Nm

Pump mechanical efficiency,

 m

, is 95% at full pump displacement & 90% at half pump displacement.

The actual pump torque = 21.1/0.90 = 23.4Nm

Input power = pump torque x speed = 23.4 x 1450 x 2

/60 = 3.6 kW

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TEP 4195 TURBOMACHINERY

2) a) b) c)

Piston area =

A

( 50

10

3

)

2

4

1 .

96

10

3 m

2

2

Piston pressure =

1 .

96

10

4

10

3

102 bar

The piston flow, Q, is given by:

Q

u

A

0 .

98

0 .

5

10

3

1 .

96

10

3

60

10

3

L /

0 .

98 min

10

3

58 .

8 L / m

3 s

1 min

The pressure drop,

P

, across the restrictor valve will be 200 - 102 = 98bar. The required rated flow, Q

R

, at a pressure drop of 10bar across the restrictor valve is:

Q

R

58 .

5

10

98

19 L / min

The circuit in Figure 1 is a meter-in system, which can only be used for opposing load forces. For negative (pulling) forces the circuit can be modified by placing the restrictor valve in the outlet flow from the annulus side of the actuator which is referred to as meterout control.

A pressure compensated flow control valve is shown in Figure 1.

P

1

P

P

2

3

ISO Symbol

Figure 1 Pressure compensated flow control valve.

The force from the spring on the piston of the valve in Figure 1(s) causes the spool to be fully open. This force is opposed by the pressure drop across the adjustable restrictor downstream of the spool. Consequently if the force from the pressure drop is greater than that of the spring the valve spool will be moved to the left causing the inlet restriction created by the spool position to be increased.

The valve inlet pressure, P

1

, is kept constant by the relief valve and therefore the flow through the valve will be maintained at a constant pressure drop across the downstream restrictor. In this way the flow will be controlled at a constant value should there be changes in the actuator force and hence changes in the valve outlet pressure P

3

.

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TEP 4195 TURBOMACHINERY

3)

Valve flow

Q

2

p

V

;

A

dX

(X = valve displacement from null position)

 

P m

= P

1

-P

2

– load pressure drop = actuator force / actuator area

Total pressure in the valve is:

P

V

(

P

S

P

1

P

2

0)

P

S

(

P

1

P

2

)

P

S

P m

As

P

V

kQ

2

;

E = (

P s

- k

Q

2

)

Q dE

For max

E: dQ

P

S

3

kQ

2

and

as kQ

2

 

P

V

P m

2

3

P

S

i) Equal Area Actuator

Load Pressure

P m

2

3

P s

for maximum power

Valve Pressure Drop =

1

3

P s

Actuator force =

2

3

P s

A

=

2

3

250

10

5

12

5

10

4

= 20

8

10

3

N

Pressure Drop across each port =

1

6

P s

Flow =

C q

X max d

Actuator velocity

U

2

P s

6

Cq

0

62 and Q = A U

C q

X max d

A

P s

3

=

0

62

12

6

 

5

5

10

4

10

6

250

3

10

870

5

= 0.46m/s

And the power = 0

46

20

8

10

3

= 9

6

k W

ii)

Flow

Q

1

For zero load force

uA

1

;

Q

2

P A

2

uA

2

For the extend stroke:

Q

1

K

1

P s

P

1

; Q

2

K

2

P

2

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TEP 4195 TURBOMACHINERY

K

1

K

2

K

for a valve having symmetric metering =

C

Q

dX

2

.

Hence

Q

2

1

K

2

P s

P

1

P s

P

2

A

2

A

1

P s

K

2

2

2

A

1

2

K

2

2

1

P s

2 3

2

K A

1

u

E

K

A

1

1

P s

2

3

A

1

extending actuator

Using a similar analytical approach the retract velocity

u = u

R

can be obtained:

Q

1

K P ; Q

2

K (P

S

 so:

Q

2

1

K

2

P

1

P

2

A

2

A

1

A

2

(P

S

A

1

Q

2

2

K

2

)

and:

u

K

2

R

2

( A

2

1

A

3

2

A

1

)

P

S

A

2

A

1

u

R

K

A

2

1

P s

1

3

A

2

These velocities can be rearranged to show that: i.e. for 2:1 area ratio

u

E u

E u

R

A

1

A

2

1 .

41

u

R

iii) Unequal Area Actuator

Extend Velocity of 0.49

m/s

K

max

2

7

Q

1

A

1

12

2

10

4

m

3

4



49

A

1

Q

2

A

2

Q

2

s

2

Q

2

P

V

1

(

Q

1

K

)

2

 

 

4

7

2



P v

1

192

bar

P v

2

P

1

P

S

P v

2

1

192

 

P v

1

4

48

bar

P

A

58

bar

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TEP 4195 TURBOMACHINERY

Force

  

 

10

4) a)

The oil volume available for discharge is

V

V

1

V

2

(1)

For isothermal compression of the gas from P

0

to P

2

we get

P

0

V

1

P

2

V

2

Thus:

V

2



P

0

P

2



V

0

For adiabatic expansion of the gas from P

2

to P

1

the gas law is

PV

(2)

cons tan t

so we get:

V

1



P

2

P

1



1

V

2

Substituting equations (2) and (1) into (3) gives:

V

0

P

0

P

2

 

 

 

P

2

P

1

1

1

b)

The data for the accumulator application are:

P

0

P

2

0 .

9

75

P

1

75 bar

100 bar

67 .

5 bar

Operating time/cycle = 8s

Volume required for the actuator = 0.9L

Adiabatic index,

for the gas = 1.6

(3)

(4)

Pump flow = 1 L/min

Pump volume delivered in 8s = 0.13L and therefore the volume required from the accumulator is:

 

L

Applying the values to equation 4 gives:

V

0

100

100

75

67

1

1 .

6

.

5

1

0 .

77 L

5 .

58 L

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5) a)

Vertical weight component of the excavator =

W sin 20

0

5400

0 .

34

1847 N

For starting the excavator the track force required

1847

Torque required at the track wheels,

T

T

2297

0 .

2

450

459 Nm

2297 N

Taking a maximum pressure of

210bar

, the torque from each motor is given by:

T

M

P max

D m

ms

=

T

T

2

ii) iii)

D m

459

5

 

6

For a speed of

6m/s

the speed of the track wheels

6

0 .

4

4 .

8 rev / s

288 rev / min

The motor type

B

in the table will provide the required performance. The motor displacement

D m

is

83cm

3

/rev = 13.2 x 10

-6

m

3

/rad

and has a maximum speed of

300rev/min

.

b)

i) The pump flow required for each motor

Q

P

v m

3

25

L / min

The displacement of each pump

D m

Q

P

vp

N

P

3

Force required to drive the excavator at

6ms

-1

F

250

250

6

1750 N

Pressure required creating this force

Fr

2

D

 

6

 

5

144

bar

The input power to each pump

pm

4 

5

.

kW

c)

The total volumetric efficiency for the transmission

0 .

95

0 .

95

0 .

9

. Thus the leakage loss in the pumps motors and to the external drain is

10%

of the total flow at an oil viscosity of

32cSt

. For an oil viscosity of

20cSt

the leakage loss will increase to

32

20

10

16 %

. This will give a volumetric efficiency of

84%

.

d)

The hydraulic circuit for each pump and motor that provides boost flow input, relief valves and a brake control valve are shown in the Figure. A single pump could supply the boost flow to both circuits.

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TEP 4195 TURBOMACHINERY

e)

The value of reduced motor displacement that can be used for operation with maximum flow and pressure on level ground is determined by using the torque and flow equations.

The torque equation gives:

(

250

250

U )r

2 (1)

We have

2

P

r

D m

5

D m

6

D m

So (1) gives:

U

 6

D m

For the flow we have:

D m

m

U

4

MV

Q m r

For

Q

M

=

25L/min, r = 0.2m,

V

= 0.95

we get:

.

D m

MV

D m

D m

6

Equations (3) and (4) give.

or

6

D m

6

250

0

D m

3

D

2

m

D m

79 10

6

0

The solution for this quadratic is (taking only the positive root):

D m

1

3

 

6

67 8

(2)

(3)

(4)

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TEP 4195 TURBOMACHINERY

6)

A

Figure 2 Pump and motor system

Data

Torque required at maximum motor speed

Motor displacement

Motor mechanical efficiency

Motor volumetric efficiency

Pump mechanical efficiency

Pump volumetric efficiency

175Nm

82cm

3

/rev (D

M

)

92%

(

MM

).

93% (

mV

)

95%

(

PM

)

96% (

PV

)

Pump speed

Pump displacement

1800rev/min (N

P

)

15cm

3

/rev (D

P

)

2100J/kg/

0

C (C

P

)

Fluid specific heat

Heat dissipated in the cooler for a water inlet temperature of

20 0 C

3

T

OW

40

kW

(

T

OW

is the difference between the cooler oil inlet and water inlet temperatures).

=

50%

of the total pump leakage Pump external drain leakage flow

a)

Motor displacement

D

M

1)

2

6

Motor inlet pressure

P

T

MM

D

M

2)

Pump outlet flow

Q

P

N

P

D

P

PV

6

175

.

6

146

bar

3

 

L / min

3)

4)

Motor speed

N

M

Q

P

D

M

 

MV

3

Cooler inlet flow (point

A

in Figure 2)

Q

C

Q

P

294

rev / min

Q

DP

where

Q

DP

is the drain flow from the pump, which is

50%

of the total pump leakage flow.

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TEP 4195 TURBOMACHINERY

Thus:

And

b)

1)

Q

DP

Q

C

 

PV

.

 input to the pump.

The system efficiency

T

The input power

W

I

Total heat generated =

W

P

10

60

PM

6

P

(

.

P

)

.

L / min

.

kW

6

The total heat generated by the losses is given by

W

.

L / min

.

5

(

1

T

)W

I

where

W

I

is the power

.

kW

2) The temperature increase in the fluid between the pump inlet and the cooler inlet is given by:

T

S

W

Q C

P

3

60

3

2100

0

C

This assumes that there is perfect mixing of the flows at point

A

.

3) The temperature that is required at point

A

(cooler inlet) to dissipate the heat generated in 1) is obtained from the cooler performance parameter:

Thus for the cooler to dissipate the heat generated of

1.52kW

the difference between the fluid temperature at the cooler inlet and the water inlet temperature must be:

T

OW

 

0

C

3000

The fluid temperature at the cooler inlet will be

T

C

   0

C

4) For the value of

T

OW

obtained in 3) the reduction in the fluid temperature through the cooler will be the same as the temperature increase in the system. Therefore the reservoir temperature will be:

T

T

 

.

0

C

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