Hydraulics 2 A4-1 David Apsley ANSWERS TO EXAMPLE SHEET ...

Hydraulics 2 A4-1 David Apsley

ANSWERS TO EXAMPLE SHEET FOR TOPIC 4 AUTUMN 2013

(Full worked answers follow on later pages)

Q1. (a) 208501.012 QH (H in m and Q in L s–1

)

(b) 10.7 L s–1

; 2.90 kW

(c) (i) 13.3 L s–1

; 5.97 kW; (ii) 14.4 L s–1

; 5.69 kW

Q2. (a) hf = 3.4710–3

Q2 (hf in m and Q in L s

–1)

(b) 28.1 L s–1

; 5.38 kW

(c) 1910 rpm

(d) (i) 4.80 kW; (ii) 1.15 kW; (iii) 0.47 kW; (iv) 0.47 kW; (v) 56%

Q3. (a) 77.1 L s–1

; 38.5 m; 57.7 kW

(b) 71.8 L s–1

(c) 36.9 kW

Q4. (a) 201251.080 QH (H in m and Q in L s–1

)

(b) 39.6 L s–1

; 99.6 m; 0.706; 54.8 kW

(c) 13.9 rpm; centrifugal pump

(d) 2510 rpm

Q5. (a) 2983206 QH (H in m and Q in m3 s

–1)

(b) 410 s; 109 kJ

(c) 154 s


–1)

(b) 8.93 L s–1

; 1.89 kW

(c) 1440 rpm


–1)

(b) (i) 0.106 m3 s

–1; (ii) 30.1 kW; (iii) –28.3 kPa gauge

(c) 0.158 m3 s

–1; 75.4 kW

Q8. (a) 2003470.02.3 QH (H in m, Q in L s–1

)

(b) pump B on efficiency grounds (67% for B as opposed to 49% for A at duty point)

(c) 3610 W

(d) 3.50 m; no cavitation

Q9. (a) 21.12915 QH (H in m, Q in m3 s

–1)

(b) 0.256 m3 s

–1; 34.4 MJ

(c) 0.16 m3 s

–1; 28.1 m; 21098QH (H in m, Q in m

3 s

–1)

(d) 1085 rpm; 21.0 MJ

Q10. 3170 rpm

Q11. (a) 16.4 m

(b) 62%


(c) 1.67 m s–1

(d) 11.0 m s–1

(e) 78%

Q12. (a) 20.5 MW

(b) 3

(c) 71.9 m s–1

; 33.1 m s–1

(d) 1.58 m

(e) 3.42 MW

(f) 1.55 m3 s

–1

(g) 0.166 m

(h) 85%


Q1.

(a)

System head = static lift + frictional head loss

Static lift:

m12214 sh

Frictional head loss:

g

V

D

Lh f

2λ

2

where 2π

4

D

Q

A

QV

With Q in m3 s

–1,

22

52

2

5285010

06.081.9π

4002.08

π

λ8QQQ

gD

Lh f

It is more convenient here to have Q in L s–1

. The system curve is then

208501.012 QH (H in m and Q in L s–1

)

(b) Plot Hpump and Hsystem as functions of Q. The relevant data is given in the table below.

Discharge (L s–1

): 0 3 6 9 12 15 18

Pump head (m) 30.0 29.5 27.6 24.4 19.7 13.5 5.9

System head (m) 12.00 12.77 15.06 18.89 24.24 31.13 39.54

0

10

20

30

40

50

60

0 5 10 15 20 25 30 35

H (m

)

Q (L s-1)

system characteristic

pumps in parallel

pumps in series

pump characteristic


0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20

eff

icie

nc

y (%

)

Q (L s-1)

Duty point:

Q = 10.7 L s–1

= 0.0107 m3 s

–1

H = 21.8 m

η = 0.79

Input power:

W2904788.0

8.210107.081.91000

η

ρ

gQHPin

Answer: Q = 10.7 L s–1

; Pin = 2.90 kW.


(c)

Pumps in parallel: same H, double Q.

Pumps in series: same Q, double H.

These are plotted in the H vs Q graph above. The duty points are as follows.

(i) Pumps in parallel:

For both pumps:

Q = 13.3 L s–1

H = 27.0 m

For each individual pump:

Q = 6.65 L s–1

= 0.00665 m3 s

–1

H = 27.0 m

η = 0.590

W2985590.0

0.2700665.081.91000

η

ρ

gQHPin

The total power consumption (two pumps) is then

W5970298522 inP

Answer: total discharge = 13.3 L s–1

; total power consumption = 5.97 kW.

(ii) Pumps in series:

For both pumps:

Q = 14.4 L s–1

H = 29.7 m


Q = 14.4 L s–1

= 0.0144 m3 s

–1

H = 14.85 m

η = 0.737

W2846737.0

85.140144.081.91000

η

ρ

gQHPin

The total power consumption is then

W5692284622 inP

Answer: total discharge = 14.4 L s–1

; total power consumption = 5.69 kW.


Q2.

(a) Frictional head loss:

g

V

D

Lh f

2λ

2

where 2π

4

D

Q

A

QV

Hence

52

2

π

λ8

gD

LQh f

If hf is in m and Q in L s–1

:

2

2

52003470.0

10001.081.9π

2102.08Q

Qh f

Answer: head loss due to friction is 200347.0 Qh f (hf in m and Q in L s–1

).

(b) We require the system curve (static lift + losses).

The static lift is

m5.115.110 sH

Hence the system characteristic is

2003470.05.11 QH (hf in m and Q in L s–1

) (*)

The pump and system characteristics can be plotted graphically and the duty point

determined by their point of intersection.

0

5

10

15

20

0 10 20 30 40

H (m

)

Q (L s-1)


pump characteristic

duty point


0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 10 20 30 40

eff

icie

nc

y

Q (L s-1)

Q = 28.1 L s–1

= 0.0281 m3 s

–1

H = 14.2 m

728.0η

Input power:

W5380728.0

2.140281.081.91000

η

ρ

gQHPin

Answer: discharge = 28.1 L s–1

; power consumption = 5.38 kW.

(c) Let N1 = 1750 rpm and the new speed be N2.

Q2 = 34.0 L s–1

From the system characteristic (*),

m51.1500347.05.11 2

22 QH

On the scaling curve through this new duty point,

2

2

1

2

2

1

2

1

Q

Q

N

N

H

H

i.e.

2

11

0.3451.15

QH

or

2

11 01342.0 QH


0

5

10

15

20

0 10 20 30 40

H (m

)

Q (L s-1)


pump characteristic

new duty point

scaled duty point

This cuts the original pump characteristic at Q1 = 31.1 L s–1

. Then

093.11.31

0.34

1

2

1

2 Q

Q

N

N

Hence,

rpm19131750093.1093.1 12 NN

Answer: required pump speed = 1910 rpm.

(d) (i) When Q = 24 L s–1

the pump curves give H =15.5 m and η = 0.760.

The output power is

W36465.15)1000/24(81.91000ρ gQHPout

The input power is

W4797760.0

3646

η out

in

PP

Answer: power consumption of the pump = 4.80 kW.

(ii) Power dissipated in the pump:

W115136464797 outin PP

Answer: power dissipated in the pump = 1.15 kW.


(iii) Head lost due to friction, with Q in L s–1

:

m00.22400347.000347.0 22 Qh f

Power dissipated by friction:

W47100.2)1000/24(81.91000ρ ffriction gQhP

Answer: power dissipated by pipe friction = 0.47 kW.

(iv) Head lost at the control valve:

m00.200.25.115.15 systempumpvalve HHH

Power dissipated in the control valve:

W471ρ valvevalve gQHP

Answer: power dissipated in the control valve = 0.47 kW.

(v) Overall efficiency of the installation:

56.04802

47147111534802η

in

dissipatedin

overallPower

PowerPower

Answer: overall efficiency of the system = 56%


Q3.

The hydraulic scaling laws give

1

2

1

2

N

N

Q

Q ,

2

1

2

1

2

N

N

H

H

Here, N1 = 1000 rpm, N2 = 1400 rpm, so that

4.11

2 N

N

Hence, pump characteristics at 1400 rpm can be determined from those at 1000 rpm by

12 4.1 QQ

12 96.1 HH

This gives the following:

Pump characteristics at speed 1400 rpm:

Discharge (L s–1

) 0 28 56 70 84

Head (m) 98 88.2 64.68 49 27.44

Efficiency (%) – 60 69 60 40

Adding the static head (20 m) to the friction losses gives:

System characteristics:

Discharge (L s–1

) 0 20 40 60 80

Head loss (m) 20 21.0 24.0 30.0 40.0

0

20

40

60

80

100

120

0 10 20 30 40 50 60 70 80 90 100

H (m

)

Q (L s-1)

pump characteristic (1400 rpm)

pumps in parallel (1000 rpm)


duty point (a)

duty point (b)


0

10

20

30

40

50

60

70

80

0 10 20 30 40 50 60 70 80 90 100

Effi

cie

ncy

(%)

Q (L s-1)

single pump (1400 rpm)

single pump (1000 rpm)in parallel operation

Duty point (see graphs):

Q = 77.1 L s–1

= 0.0771 m3 s

–1

H = 38.5 m

η = 50.5 %

Then

W57660505.0

5.380771.081.91000

η

ρ

gQHPin

Answer: duty point is Q = 77.1 L s–1

, H = 38.5 m; power consumption, Pin = 57.7 kW.

(b) For pumps in parallel, double the flow and keep the head the same. This gives the

following characteristics.

Pumps in parallel at speed 1000 rpm:

Discharge (L s–1

) 0 40 80 100 120

Head (m) 50 45 33 25 14

This intersects the system curve at

Q = 71.8 L s–1

H = 35.7 m

Answer: maximum discharge rate for pumps in parallel, Q = 71.8 L s–1

.

(c) For both pumps:

Q = 71.8 L s–1

H = 35.7 m



Q = 35.9 L s–1

= 0.0359 m3 s

–1

H = 35.7 m

η = 0.681

W18460681.0

7.350359.081.91000

η

ρ

gQHPin

The total power consumption is then

W369201846022 inP

The discharge is similar to that of a single pump operating at a higher speed, but the power

consumption is substantially less (mainly because each pump operates closer to its maximum-

efficiency point). Assuming that the operating rather than capital and maintenance costs of

the system dominate, then this is the better option.

Answer: power consumption for pumps in parallel = 36.9 kW.


Q4.

(a)

System head = static lift + head losses

g

V

D

LHH s

2λ

2

where 2π

4

D

Q

A

QV

2

52π

λ8Q

gD

LHH s

If H is measured in m and Q in L s–1

then

2

52)

1000(

15.081.9π

57502.0880

QH

Hence, the system curve is

201251.080 QH

with H in m and Q in L s–1

.

(b) Plot the system and pump characteristics as a graph of H against Q. The corresponding

efficiency is obtained from a graph of η against Q.

0

20

40

60

80

100

120

140

0 10 20 30 40 50 60

H (m

)

Q (L s-1)

duty point


pump characteristic


0

10

20

30

40

50

60

70

80

90

100

0 10 20 30 40 50 60

eff

icie

nc

y (%

)

Q (L s-1)

Duty point:

Q = 39.6 L s–1

= 0.0396 m3 s

–1

H = 99.6 m

η = 0.706

Input power:

W54800706.0

6.990396.081.91000

η

ρ

gQHPin

Answer: Q = 39.6 L s–1

; H = 99.6 m; η = 0.706; Pin = 54.8 kW.

(c) Specific speed is

4/3

2/1

H

NQN s

calculated at the maximum-efficiency point (Q = 31 L s–1

= 0.031 m3 s

–1; H = 117 m). Thus

rpm9.13117

031.0rpm2800

4/3

2/1

sN

Answer: 13.9 rpm; this is a centrifugal pump.

(d) At some unknown new rotational speed N2 the new discharge is

Q2 = 30 L s–1

The corresponding head is (from the system characteristic)


H2 = 91.26 m

The scaling curve through this point is

2

2

2

22

Q

Q

N

N

H

H

or

22

2

2

2 1014.0 QQQ

HH

0

20

40

60

80

100

120

140

0 10 20 30 40 50 60

H (m

)

Q (L s-1)

scaled duty pointsystem characteristic

pump characteristic

new duty point

This scaling curve intersects the N1 = 2800 rpm curve at Q1 = 33.4 L s

–1. Hence,

1

2

1

2

Q

Q

N

N

rpm25104.33

302800

1

212

Q

QNN

Answer: N2 = 2510 rpm.


Q5.

(a)

g

V

D

LHH ssystem

2λ

2

where 2π

4

D

Q

A

QV

52

2

π

λ8

gD

LQHH ssystem

Substituting parameter values:

2983206 QH system (H in m, Q in m3 s

–1)

(b) Plot:

H vs Q (for pump and system) to find the duty point.

η vs Q and read off efficiency.

0

2

4

6

8

10

12

0 0.002 0.004 0.006 0.008 0.01

H (m

)

Q (m3 s-1)

Hpump

Hsystem


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 0.002 0.004 0.006 0.008 0.01

h

Q (m3 s-1)

From the graphs the duty point is:

Q = 0.00244 m3 s

–1

H = 6.583 m

η = 0.593

Time taken:

s8.40900244.0

10.1

QT

Power and energy:

W7.265593.0

583.600244.081.91000

η

ρ

gQHPin

J1089008.4097.265 TPE in

Answer: time taken = 410 s; energy used = 109 kJ.

(c) Scaling to from N1 = 1500 rpm to N2 = 2250 rpm:

11

1

22 5.1 QQ

N

NQ


11

2

1

22 25.2 HH

N

NH

The revised pump characteristics at 2250 rpm are given in the table below.

Discharge,

Q (L s–1

) 0 1.5 3 4.5 6 7.5 9 11.5

Head,

H (m) 20.07 18.00 15.80 13.48 11.05 8.46 5.76 2.95

Plot new H vs Q (for pump and system) to find the duty point.

0

5

10

15

20

25

0 0.005 0.01 0.015

H (m

)

Q (m3 s-1)

Hpump

Hsystem

At the duty point:

Q = 0.00651 m3 s

–1

The time taken is

s6.15300651.0

10.1

QT

Answer: time taken = 154 s.


Q6.

(a) The system head requirements are:

g

VK

D

LHH s

2)λ(

2

where 2π

4

D

Q

A

QV

42

2

π

8)λ(

gD

QK

D

LHH s

Substituting the given parameters (Hs = 12 m, λ = 0.04, L = 30 m, D = 0.08 m, K = 10) gives,

25043012 QH (H in m and Q in m3 s

–1)

or, if preferred,

2050430.012 QH (H in m and Q in L s–1

)

(b) Plot the graphs of H vs Q (pump and system characteristics) to determine the duty point.

Plot also η vs Q to find the efficiency. Either m3 s

–1 or L s

–1 may be used as units for Q.

0

5

10

15

20

25

30

0 0.005 0.01 0.015 0.02

Q (m3 s

-1)

H (

m)

Pump

SystemDuty point


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 0.005 0.01 0.015 0.02

Q (m3 s

-1)

h

The duty point is

Q = 0.00893 m3 s

–1

H = 16.0 m

η = 0.743

The power input is given by

inP

gQHρη

whence

W1886743.0

1600893.081.91000

η

ρ

gQHPin

Answer: discharge = 8.93 L s–1

; power consumption = 1.89 kW

(c) For the new arrangement, need to change Hs (to 20 m) and L (to 38 m). The new system

curve is

25850020 QH (H in m and Q in m3 s

–1)

or, if preferred,

205850.020 QH (H in m and Q in L s–1

)

It is not actually necessary to plot this curve because only the new duty point is required and

the discharge at the higher speed N2 is stated to be the same:

Q2 = 0.00893 m3 s

–1

whence

H2 = 24.7 m (from the new system curve).


The old and new duty points do not scale onto each other. To determine where this came

from on the original curve use the hydraulic scaling laws:

2

2222

,

N

N

H

H

N

N

Q

Q

or

2

22

Q

Q

H

H

With Q2 and H2 from above this gives (for H in m and Q in m3 s

–1):

2309700QH

Plot this to where it intersects with the original curve at speed N1.

0

5

10

15

20

25

30

0 0.005 0.01 0.015 0.02

Q (m3 s

-1)

H (

m)

Pump

System

Scaling curve

Pump (scaled)Old duty point

(Q2,H2)

New duty point

(Q1,H1)

This gives

Q1 = 0.00744 m3 s

–1

Then,

20.100744.0

00893.0

1

2

1

2 Q

Q

N

N

Hence, the new rotation rate of the pump is

rpm1440120020.1

Answer: new rotation rate = 1440 rpm.


Q7.

(a) System head requirement is

g

V

D

Lh

lossesliftstaticH

s2

λ2

With 2π

4

D

Q

A

QV this is

2

52π

λ8Q

gD

LhH s

Substituting numerical values (hs = 12 m, λ = 0.025, L = 35 m, D = 0.15 m) gives

21.95212 QH

when H is in m and Q is in m3 s

–1. Note that L includes both length of pipe; alternatively add

the losses from separate parts – you will get the same answer.

(b) The system-head values may be calculated for the Q values in the table:

Q (m3 s

–1) 0.00 0.03 0.06 0.09 0.12 0.15 0.18

Hsystem (m) 12.00 12.86 15.43 19.71 25.71 33.42 42.85

The H vs Q and η vs Q graphs can then be plotted as below.

0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 0.05 0.1 0.15 0.2 0.25 0.3

Q (m3 s

-1)

H (

m)

pump

system


0.0

0.2

0.4

0.6

0.8

1.0

0 0.05 0.1 0.15 0.2 0.25 0.3

Q (m3 s

-1)

h

At the duty point,

Q = 0.106 m3 s

–1

H = 22.70 m (from the system characteristic in part (a))

η = 0.785

(i) Answer: Q = 0.106 m3 s

–1.

(ii) Since

P

gQHρη

one has

W30070785.0

70.22106.081.91000

η

ρ

gQHpowerin

Answer: powerin = 30.1 kW.

(iii) At pump inlet,

moverlossesheadHH sumpinlet 10

Hence, in terms of gauge pressures

g

V

D

L

g

Vz

g

p inlet

inlet

inlet

2λ0

2ρ

22

Hence,


2

21 ρ)λ1(ρ V

D

Lgzp inlet

inletinlet

Now,

m0.2inletz

m10inletL

1

22sm998.5

15.0π

106.04

π

4

D

QV (quite quick!)

Hence,

Pa28350

998.51000)15.0

10025.01()2(81.91000 2

21

inletp

Answer: –28.3 kPa gauge.

(c) If all else (especially speed) is constant then the hydraulic similarity laws imply:

3DQ

2DH η remains constant (because both ρgQH and the input power scale as D

5)

For each value of Q in the original table there will be new Q values multiplied by

1.23 (= 1.728) and new H values multiplied by 1.2

2 (= 1.44). The new pump characteristics

are given below.

Q (m3 s

–1) 0.00 0.052 0.104 0.156 0.207 0.259 0.311

H (m) 43.2 42.9 40.5 36.0 29.4 20.3 8.9

η (%) – 29 54 73 80 70 38

The new characteristics may be plotted (or added to the original graphs). The system curve

might also need be extended in length (but is still the same function of H vs Q). Note that you

must plot a new efficiency graph.


0.0

5.0

10.0

15.0

20.0

25.0

30.0

35.0

40.0

45.0

50.0

0 0.05 0.1 0.15 0.2 0.25 0.3

Q (m3 s

-1)

H (

m)

pump (original)

system

pump (new)

0.0

0.2

0.4

0.6

0.8

1.0

0.00 0.05 0.10 0.15 0.20 0.25 0.30

Q (m3 s

-1)

h

The new duty point is:

Q = 0.158 m3 s

–1

H = 35.77 m

η = 0.735


The new input power is

W75430735.0

77.35158.081.91000

η

ρ

gQHpowerin

Answer: discharge = 0.158 m3 s

–1; input power = 75.4 kW.


Q8.


g

V

D

LHH s

2λ

2

where 2π

4

D

Q

A

QV

52

2

π

λ8

gD

LQHH s

Substituting the given parameters (Hs = 3.2 m, L = 21 m, D = 0.1 m, λ = 0.02) gives, with

heads in m and discharges in L s–1

(for convenience, to be consistent with the data in the

tables):

2

52 10001.081.9π

2102.082.3

QH

2003470.02.3 QH (H in m, Q in L s–1

)

(b) Plot graphs of H vs Q and η vs Q (both pumps and, for head, the system requirements on

the same graphs) to determine the duty points.

0

5

10

15

20

25

0 5 10 15 20 25 30 35 40

H (

m)

Q (L s-1)

Pump A

Pump B

System


0

10

20

30

40

50

60

70

80

90

100

0 5 10 15 20 25 30 35 40

h (

%)

Q (L s-1)

Pump A

Pump B

Duty points:

Pump A:

Q = 32.9 L s–1

; H = 6.96 m; η = 49%

Pump B:

Q = 34.1 L s–1

; H = 7.23 m; η = 67%

The pumps provide almost exactly the same discharge (and the question tells you that both

are adequate), but pump B is considerably more efficient than pump A and should be

selected.

Answer: select pump B on efficiency grounds.

(c)

inin

out

power

gQH

power

power ρη

W361067.0

23.70341.081.91000

η

ρ

gQHpowerin

Answer: input power = 3610 W.

(d) Considering the total head between the sump and the pump inlet:


head at pump = head at start – losses along half the pipe

g

V

D

L

g

p

g

Vz

g

p atm

2λ)00

ρ(

2ρ

2

212

The velocity in the pipeline is

1

22sm342.4

1.0π

0341.04

π

4

D

QV

Hence,

m179.681.92

342.4)

1.0

5.1002.01(2.30

2)λ1(

ρρ

22

21

g

V

D

Lz

g

p

g

p atm

The NPSH is then

m505.381.91000

95000179.6

ρ

g

pp cav

This is well above zero and unlikely to cause cavitation.

Answer: net positive suction head = 3.50 m; no cavitation.


Q9.


g

V

D

LHH s

2λ

2

where 2π

4

D

Q

A

QV

52

2

π

λ8

gD

LQHH s

Substituting the given parameters (Hs = 15 m, L = 800 m, D = 0.4 m, λ = 0.02) gives, with

heads in m and discharges in m3 s

–1:

2

52 4.081.9π

80002.0815 QH

21.12915 QH (H in m, Q in m3 s

–1)

(b) Plot graphs of H vs Q and η vs Q to determine the duty point.

0

5

10

15

20

25

30

35

40

45

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

H (m

)

Q (m3 s-1)

H (pump)

H (system)


0

10

20

30

40

50

60

70

80

90

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

h (

%)

Q (m3 s-1)

Efficiency

Efficiency

Duty point:

Q = 0.256 m3 s

–1; H = 23.5 m; η = 67%

Input power:

W8809067.0

5.23256.081.91000

η

ρ

gQHpowerin

Time taken to pump 100 m3 is

s6.390256.0

100

The energy used is then

J1044.36.39088090 7

Answer: discharge = 0.256 m3 s

–1; energy consumed = 34.4 MJ.

(c) From the efficiency graph the most efficient operation (η = 79.5%) is for Q = 0.16 m3 s

–1,

H = 28.1 m.

Since the hydraulic scaling laws give

2

14001.28

NH and

140016.0

NQ

we have, eliminating the speed ratio,

2

16.01.28

QH


or

21098QH (H in m, Q in m3 s

–1)

Answer: Q = 0.16 m3 s

–1; H = 28.1 m; 21098QH (H in m, Q in m

3 s

–1).

(d) Since efficiency is unchanged by hydraulic scaling, the scaling line from part (c) will

always indicate the maximum-efficiency conditions as the rotational speed changes. This line

is plotted below. It will form the duty point when it meets the operating requirements (i.e. the

system curve) at a rotational speed such that

Q = 0.124 m3 s

–1

H = 17.0 m

0

5

10

15

20

25

30

35

40

45

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

H (m

)

Q (m3 s-1)

H (pump)

H (system)

Scaling line

*** Alternative method

In this instance the intersection point can also be found analytically as the intersection of two

quadratics:

21.12915 QH (system curve)

21098QH (scaling line)

Equating gives (in metre-second units throughout):

22 1.129151098 QQ

159.968 2 Q

124.0Q (as above)

*** End of alternative method

The speed ratio is


775.016.0

124.0

1400

N

rpm10851400775.0 N

The power consumption is (noting that efficiency is unchanged by scaling):

W26010795.0

0.17124.081.91000

η

ρ

gQHpowerin

Time taken to pump 100 m3 is

s5.806124.0

100

The energy used is then

J1010.25.80626010 7

Answer: operating speed = 1085 rpm; energy consumed = 21.0 MJ.


Q10.

The pump and system characteristics are initially given in different units. Convert them both

to metres of water.

The pump characteristic at N1 = 2950 rpm is then

)0.1(24.0 11 QH (H in m of water, Q in m3 s

–1) (*)

The system characteristic in the same units is

75.175.1 896.181.91000

18600QQH system

Pump characteristics at other speeds can be determined using the hydraulic scaling laws:

2

1

2

2

1

2

1

2

Q

Q

N

N

H

H (**)

From the system characteristic, the duty point at the higher speed is

m2043.028.0896.1sm28.0 75.1

2

13

2 HQ

Substituting in (**),

2

11

28.02043.0

QH

or

2

11 606.2 QH (***)

The scaled duty point at the lower speed may be found by solving (*) and (**)

simultaneously – either graphically or, here, by solving a quadratic:

)0.1(24.0606.2 1

2

1 QQ

whence

13

1 sm2609.0 Q

Then

073.12609.0

28.0

1

2

1

2 Q

Q

N

N

Hence,

rpm31702950073.1

073.1 12

NN

Answer: estimated fan speed = 3170 rpm.

0.00

0.05

0.10

0.15

0.20

0.25

0.30

0.0 0.1 0.2 0.3 0.4 0.5

H (

m)

Q (m3 s-1)

pump characteristicat N2

systemcharacteristic

duty point

scaledduty point


Q11.

(a)

Total head = piezometric head + velocity head

Need velocities at suction (inlet) and discharge (outlet):

1

22sm415.1

15.0π

025.04

π

4

inin

inD

Q

A

QV

1

22sm183.3

1.0π

025.04

π

4

outout

outD

Q

A

QV

Hence,

m898.381.92

415.14

2ρ

* 22

g

V

g

pH inin

in

m52.1281.92

183.312

2ρ

* 22

g

V

g

pH outout

out

Hence the difference in total head across the pump is

m42.16 inout HH

Answer: head difference = 16.4 m.

(b) Overall efficiency,

6195.06500

)898.352.12(025.081.91000)(ρη

in

inout

P

HHgQ

Answer: η = 62%.

(c)

1sm667.1015.0

025.0 A

QVr

Answer: Vr = 1.67 m s–1

.

(d)

Ideal whirl velocity (i.e. assuming leaves parallel to blades) is comprised of forward motion

Rω and backward component Vr / tan β. The actual whirl velocity is 80% of this. Hence,

)βtan/ω(8.0 rt VRV

But,

m125.02/25.02/ DR

1srad6.14660/π2140060/π2ω N

Hence,

1sm00.11)20tan/667.16.146125.0(8.0 tV

Answer: Vt = 11.0 m s–1

.


(e)

7787.06.146125.000.11

))4(12(81.9

ω

)**(

RV

ghh

headEuler

headcpiezometriefficiencyManometric

t

inout

Answer: manometric efficiency = 78%.


Q12.

(a) Total available head is

m28020300 lostgross HHH

Specific speed:

4/5

2/1

H

NPN s

Inverting for the power (per wheel):

MW50.20kW20500280400

50 2/5

2

2/5

2

H

N

NP s

Answer: output power per wheel = 20.5 MW.

(b) Number of wheels required:

9.250.20

60

wheelperPower

powerTotal

Answer: 3 wheels required.

(c) Jet velocity:

1sm90.7128081.9297.02 gHcv v

Bucket velocity:

1sm07.3390.7146.046.0 vu

Answer: jet velocity = 71.9 m s–1

; bucket velocity = 33.1 m s–1

.

(d) Use u = Rω.

1srad89.4160

π2ω

N

m7894.089.41

07.33

ω

uR

m579.17894.022 RD

Answer: wheel diameter = 1.58 m.

(e)

W10417.36

105.20 66

jetsofnumber

wheelperpowerjetperPower

Answer: power per jet = 3.42 MW.


(f) Inverting P = ηρgQH:

136

sm555.128081.910008.0

10417.3

ηρ

gH

PQ

Answer: quantity of flow per jet = 1.55 m3 s

–1.

(g) Since

4

π 2DvvAQ

the jet diameter is

m1659.090.71π

555.14

π

4

v

QD

Answer: jet diameter = 0.166 m.

(h) Hydraulic efficiency is

851.097.024523.0

)165cos85.01(46.0)46.01(ρ

)θcos1()(ρ

2

2

gH

v

gQH

kuuvQ

Answer: hydraulic efficiency = 85%.

Hydraulics 2 A4-1 David Apsley ANSWERS TO EXAMPLE SHEET ...

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Transcript of Hydraulics 2 A4-1 David Apsley ANSWERS TO EXAMPLE SHEET ...