Example: Exercise 5.9.4 (Pump) Pump Oil (S=0.82) Flow Want: Rate at which energy is delivered to oil by pump Need to find h p associated with the pump:

Example: Exercise 5.9.4 (Pump)

Pump

D mm m r m1 1180 0 18 0 09 . .

p kN m1235 /

z 1 0

D mm m r m2 2120 0 12 0 06 . .

p kN m22120 /

z 2 0

Q Liters s m s 70 70 3/ /Oil (S=0.82)Flow

z

Want: Rate at which energy is delivered to oil by pump

VQ

A

m s

mm s1

1

3

2

0 07

0 092 7508

. /

( . ). /

V

Q

A

m s

mm s2

2

3

2

0 07

0 066 18936

. /

( . ). /

H H h h H Hp p2 1 2 1 Need to find hp associated with the pump:


h H Hp

zV

g

pz

V

gp

2 1

22

22

11

12

2 2

hkN m

kN m

m s

m s

kN m

kN m

m s

m sp

120

0 82 9 810

6 18936

2 9 81

35

0 82 9 810

2 7508

2 9 81

2

3

2

2

2

3

2

2

/

. ( . / )

( . / )

( . / )

/

. ( . / )

( . / )

( . / )

h mp 12 1334.

Rate of transfer of energy = pow er Q h p

( . . / ) ( . / )( . )0 82 9 81 0 07 12 13343 3kN m m s m

pow erm kN

skW

6 83 6 83. .


• Pumps (and also turbines) are characterized by their efficiency.

• Say, in exercise 5.9.4 the pump is 90% efficient and we require 6.83 kW of output, then

input = 6.83 kW / 0.9 = 7.59 kW

• Pumps (and also turbines) are characterized by their efficiency.

Efficiency = pow er ou tpu t

pow er inpu t

General Energy Equation for Steady Flow of Any Fluid

pz

V

gI h Q

pz

V

gIM H

1

11 1

12

12

22 2

22

22 2

First Law of Thermodynamics: For steady flow, external work done on any system plus the thermal energy transferred into or out of the system is equal to the change of energy of the system

(I) Using the first law of thermodynamics, (II) taking into account non-uniformvelocity at a cross-section of flow region, and (III) assuming flow goes from section 1 to section 2, we can derive the following:


pz

V

gI h Q

pz

V

gIM H

1

11 1

12

12

22 2

22

22 2

1

33

AVu dA V

Au dA

1

• is a correction factor accounting for non-uniform velocity in cross-section

• If velocity is uniform in cross-section, then

I

Q H

• This general equation also takes into account changes in density (via ) energy changes due to machines (via ) and due to heat transfer to or from outside the fluid (via )

h M

• It also accounts for the conversions of other forms of fluid energy into internal heat ( )

1


• On a unit weight basis, the change in internal energy is equal to the heat added to or removed from the fluid plus the heat generated by fluid friction:

I I I Q hH f ( )2 1

h I I Qf H( )2 1

• The head loss due to friction is equal to the internal heat gain minus any heat added from external sources, per unit weight of fluid

• Energy loss due to friction gets converted to internal energy (proportional to temperature)

Example: Exercise 5.3.5 (Friction Head Loss)

B

A

h m10 5.

p kN mA 170 2/ p kN mB 275 2/

V VA B ; Diameter at A = Diameter at B, thus by continuity V V VA B

Want: Pipe friction head loss and direction of flow

z

H H hB A f Assume flow goes fom A to B:

h H Hf A B

hp p

z zV V

gfA B

A BA B

( )( )

( )

2 2

2

S of liquid in pipe = 0.85

0 ; V VA B


B

A

h m10 5.

p kN mA 170 2/ p kN mB 275 2/

V VA B ; Diameter at A = Diameter at B, thus by continuity V V VA B z

hp p

z zkN m

kN mm mf

A BA B

( )

( )( ) /

. . /. .

170 275

0 85 9 8110 5 2 09

2

3


Thus flow goes from B to A and h mf 2 09.


B

A

h m10 5.

p kN mA 170 2/ p kN mB 275 2/

V VA B ; Diameter at A = Diameter at B, thus by continuity V V VA B z

hp p

z zkN m

kN mm mf

A BA B

( )

( )( ) /

. . /. .

170 275

0 85 9 8110 5 2 09

2

3


Thus flow goes from B to A and h mf 2 09.

Let . If flow goes from B to A, and h f 0

( )( )

( )( ) ( ) ( )

p pz z

p pz z p p z zA B

A BB A

A B B B A B

0

p pB A

Role of pressure difference (pressure gradient)

B

A

h z zA B ( )z

Thus flow will go from B (high pressure) to A (low pressure), only if

In general, the pressure force (resulting from a pressure difference) tends tomove a fluid from a high pressure region towards a low pressure region

( ) ( )p p z z hB B A B

p pB A

Otherwise flow will from A (low pressure) to B (high pressure)

For a flow to actually go from a high pressure region towards a low pressure region, the pressure force must be higher than other forces that could be trying to move fluid in opposite direction (e.g. gravitational force in exercise 5.3.5)

Energy Grade Line (EGL) and Hydraulic Grade Line (HGL)

Graphical interpretations of the energy along a pipeline may be obtained through the EGL and HGL:

EGLp V

gz

2

2

HGLp

z

EGL and HGL may be obtained via a pitot tube and a piezometer tube,respectively


EGLp V

gz

2

2HGL

pz

h hL f - head loss, say,

due to friction

piezometertube

EGL=HGL if V=0

V

g22

2

z 2z1

pitot tube

( )z0

EGL

HGL h L

p 2 /

Datum


Large V2/2g becausesmaller pipe here

Steeper EGL and HGLbecause greater hL per length of pipe

Head loss atsubmerged discharge

EGL and HGL EGL

HGL

EGL and HGLp /

z

z0

HGLp

z

EGLp V

gz

2

2h hL f

Positive


If then and cavitation may be possible P

0HGL z

EGL and HGL

HGL

EGLp /

Positive

V

g

2

2

p

Negative

p

z

z0

EGLp V

gz

2

2

HGLp

z

h hL f


Helpful hints when drawing HGL and EGL:

1. EGL = HGL + V2/2g, EGL = HGL for V=0

2. If p=0, then HGL=z

3. A change in pipe diameter leads to a change in V (V2/2g) due to continuity and thus a change in distance between HGL and EGL

4. A change in head loss (hL) leads to a change in slope of EGL and HGL

5. If then and cavitation may be possible HGL zP

0

6. A sudden head loss due to a turbine leads to a sudden drop in EGL and HGL

7. A sudden head gain due to a pump leads to a sudden rise in EGL and HGL

Helpful hints when drawing HGL and EGL (cont.):

8. A sudden head loss due to a submerged discharge leads to a sudden rise in EGL

Documents

Example: Exercise 5.9.4 (Pump) Pump Oil (S=0.82) Flow Want: Rate at which energy is delivered to oil by pump Need to find h p associated with the pump: