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
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 and
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 1

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
.
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 2

TEP 4195 TURBOMACHINERY
3)
Valve flow
Q

C A
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
2

P s
6
Cq

0

62 and Q = A U
Actuator velocity U

C q

X max d P
A 3 s

=
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
 uA
For zero load force
1
; Q
2
 uA
2
P A

P A
2
For the extend stroke:
Q
1

K
1
P s

P
1
; Q
2

K
2
P
2
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 3

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 2
1
K
2

P s

2 3
2
K A
1
 u
E

K
A
1
1
P s

A
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

P ) so:
Q
2
1
K
2

P
1

P
2
A
2 
A
2
A
1
A
1
(P
S

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

A
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

2 78 10

7
Q
1

A
1 u 25 10

4 
49

A
1 Q
2
A
2
 
Q
2

12

2

10

4 m
3 s

2 Q
2

P
V 1

Q
( )
K
1 2  
 
4

7


2
 

P v 1

192 bar

P v 2



P v
2
1 
192
P
1

P
S
 
P v 1

4

48 bar

P
A
 
58 bar
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 4

TEP 4195 TURBOMACHINERY
Force

P A

P A
     
10

. KN
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
(2)
For adiabatic expansion of the gas from P
2
to P
1
the gas law is PV
  cons tan t so we get:
V
1



P
2
P
1


1

V
2
Substituting equations (2) and (1) into (3) gives:
V V
0



P
P
0
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 .
5
1
1 .
6 

1
0 .
77 L

5 .
58 L
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 5

TEP 4195 TURBOMACHINERY
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 m / rad

80 85 cm / rev
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

N D m
 v

  
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.
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 6

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 P D

(1)
We have
2 P
 r
D m

  5 
.
D m
 6
D m
So (1) gives:
Equations (3) and (4) give.

U
  6
D m
For the flow we have:
D m
 m
 r
 
MV
Q m
 

MV
Q r
D m
For Q
M
= 25L/min, r = 0.2m,

V
= 0.95 we get:

6
U

.
 
.


4 D m
D m or
  
6
6
D m
 
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 m / rad

67 8
(2)
(3)
(4)
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 7

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)
2


6

Motor inlet pressure P

T

MM
D
M


Pump outlet flow Q
P

N
P

D
P
 
PV


6 m / rad

175
.
 
6

146 bar
  
3 
.

.
L / min
3)
4)
Motor speed N
M

Q
P
D
M
 
MV


3

0 93

294 rev / min
Cooler inlet flow (point A in Figure 2) Q
C

Q
P

Q
DP
where Q
DP
is the drain flow from the pump, which is 50% of the total pump leakage flow.
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 8

TEP 4195 TURBOMACHINERY
Thus:
And
Q
DP
Q
C

  
PV
)N D
P

.

.
.

.

.
L / min
 
.
L / min b)
1) The total heat generated by the losses is given by W ( 1

T
)W
I
where W
I
is the power input to the pump.
The system efficiency

T
     
.

.

.

.

.
The input power W
I

N D
P

10
60

PM

6
P
Total heat generated = W

.
(





.
)

.
kW

6
  5 
.
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

1 9
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
 
20 3
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
P Chapple aPRIL 2006 Solutions to Tutorial Questions TEP 4195 9