chap02

Chapter 2 : Pressure and Head

2.1 a. From eqn. 2.4 : p2 - p1 = -r g z2 - z1( ) and taking sea level as datum point 1, then :p1 = 0 Nm

- 2, r= 1002kgm- 3, g= 9.81ms- 2, z2 = -2000m and z1 = 0 m. p2 - 0= - 1002 9.81 - 2000- 0( )\ p2 =19.66MNm- 2

b. From eqn. 1.12, we know K = rdpdr

and from eqn. 2.17, p=r gz giving :

r 2gdz=K drIf it is assumed that z is measured from the surface and is negative as depthincreases, then :

- gdzK

=drr 2

If it is assumed that K is constant over small pressure ranges, we can then integrate gKz1

z2

dz= -1r 2r1

r 2

dr

gzK

z1

z2

=1r

r 1

r 2

Since we know that z1 = 0 m then, gzK

=1r 2

-

1r 1

r 2 =K r 1

r 1gz+ K

But we know that : dpdz

= -r g

dp= -Kr 1

K +r 1gz

g dz

dp= -r 1g1

1+r 1gzK

dz

Integrating, NB 1

1+axdx=

1a

ln1+ax( )

p2 - p1 = -r 1gK

r 1gln 1+

r 1gzK

z1=0

z2

p2 - p1 = - Kln 1+r 1gz2

K

We know that p1 = 0 Nm

- 2, K = 2.05 109 Nm-2 , r 1 =1002kgm- 3 and

z2 = -2000m, hence :

p2 - 0= - 2.05 109 ln 1+

1002 9.81 - 2000( )2.05 109

Elements of fluid mechanics 2

\ p2 = 19.75MNm- 2

2.2 a. From eqn. 2.4 : p2 - p1 = -r g z2 - z1( ) . Taking the free surface as datum point 1then, p1 = 0 Nm

- 2, r = 1000kgm- 3, g = 9.81ms- 2, z2 = - 12m and z1 = 0 m. p2 - 0 = - 1000 9.81 - 12- 0( )\ p2 = 117.72kNm

- 2

b. From eqn. 2.16 : p = r gh+ patmie Absolute pressure = Gauge pressure + Atmospheric pressureHere, gauge pressure = 117720Nm- 2 and atmospheric pressure = 101000Nm- 2.

\ Absolute pressure = 218.72kNm- 2

2.3 a. Specific gravity of oil = r oilr H2O

0.8=r oil

1000 r oil = 800kgm

- 3

From eqn. 2.17 : p = r gh

where p = 120 103 Nm- 2, r = 800kgm- 3, g = 9.81ms- 2

120 103 = 800 9.81 h\ h = 15.3m

b. As before, p = 120 103 Nm- 2 andg = 9.81ms- 2, however, r = 1000kgm- 3.

From eqn. 2.17 : p = r gh 120 103 = 1000 9.81 h\ h = 12.2m

2.4 We know that 1 bar = 1 105 Nm- 2.From eqn. 2.4 : p2 - p1 = -r g z2 - z1( ) . Taking the free surface as datum point 1 :p2 = 1 10

5 Nm- 2, p1 = 0 Nm- 2, r = 600kgm- 3, g = 9.81ms- 2 and z1 = 0 m.

1 105 - 0 = - 600 9.81 z2 - 0( ) z2 = - 17m\ depth = 17m

2.5 For each of the following : p = r gh (from eqn. 2.17) and h = 400 10- 3 m.

a. Specific gravity of mercury = r Hgr H2O

13.6=r Hg

1000 r Hg = 13600kgm

- 3

Hence, p = r gh p = 13600 9.81 400 10- 3

\ p = 53.4kNm- 2

b. Here, r = 1000kgm- 3, therefore p = r gh

p = 1000 9.81 400 10- 3

\ p = 3.92kNm- 2

c. Specific weight, w = r g = 7.9 103 Nm- 3. Therefore, p = r gh p = w h p = 7.9 103 400 10- 3

\ p = 3.16kNm- 2

Pressure and head 3

d. Here, r = 520kgm- 3, therefore p = r gh

p = 520 9.81 400 10- 3

\ p = 2.04kNm- 2

2.6 We know that : Force = mass x gravity. Here, mass = 50kg and g = 9.81ms- 2

F = 50 9.81\ F = 490.5N

Now, pressure = forcearea

where area = 0.01m2

p =490.50.01

Nm- 2

\ p = 4.905 104 Nm- 2

2.7 Referring to figure 2.7a, we can assume that manometers are connected at points 1and 2. Using eqn. 2.27 and equating pressures at XX, then :pgas + r gh( ) gas= patm= 2 + r gh( ) H2O (1) where r gas = 0.561kgm

- 3,hgas = 0.18m, r H2O = 1000kgm- 3 and h

H2O= 0.18m.

Equating pressures at YY :

pgas + D p( ) gas2 - 1( )[ ] + r gh( ) gas = patm= 2 + D p( ) air 2- 1( )[ ] + r gh( ) H2O (2)where D p( ) gas2- 1( ) = r gas g 120m, D p( ) gas2 - 1( ) = 0.561 9.81 120 D p( ) gas2 - 1( ) = 660.41Nm

- 2 and D p( ) airgas2- 1( ) = r air g 120m D p( ) airgas2 - 1( ) = 1.202 9.81 120m, D p( ) airgas2 - 1( ) = 1415Nm

- 2

r gas and r H2O are unchanged.

Subtracting eqn. 2 from eqn. 1 gives :0.561 g 0.18( ) - 660.41- 0.561 g h( )= 1000 g 0.18( ) - 1415- 1000 g h( )We know g = 9.81ms- 2, \ h = 103mm

xgas

water

180mm

gas

2

x

120m

watergas

h1

Y Y

2.8 From eqn. 2.17 : p = r gh

where r = 13600kgm- 3, g = 9.81ms- 2 and h = - 50 10- 3 m


p = 13600 9.81 - 50 10- 3( )\ p = - 6671Nm- 2

Now, Absolute pressure = Gauge pressure + Atmospheric pressure

Absolute pressure = - 6671( ) + 1 105( ) Nm- 2

\ Absolute pressure = 93.3kNm- 2

2.9 Pressure at base = pressure of air + pressure of oil + pressure of water (all gauge) Since the tank is open :pair = 0 Nm

- 2 (gauge)

We know that specific gravity of oil = r oilr H2O

0.75=r oil

1000 r oil = 750kgm

- 3

Also, poil = r oilgh (from eqn. 2.17) where g = 9.81ms- 2 and h = 2 m, hence

poil = 750 9.81 2 = 14715Nm- 2

Similarly, for water, where r = 1000kgm- 3, g = 9.81ms- 2 and h = 3 m :

pH2O = 1000 9.81 3 = 29430Nm- 2

pressure at base = 0 + 14715+ 29430( ) Nm- 2

\ pressure at base = 44.145kNm- 2 (gauge)

Now, Absolute pressure = Gauge pressure + Atmospheric pressure

Absolute pressure = 44.145103( ) + 1 105( ) Nm- 2

\ Absolute pressure at base = 144.145kNm- 2

2.10 Gauge pressure at base = pHg + pH2O + poil + pairFrom eqn. 2.17 and using g = 9.81ms- 2, r Hg = 13600kgm

- 3, r H2O = 1000kgm- 3

and r oil = 600kgm- 3. We know hHg = 0.5m, hH2O = 2 m & hoil = 3 m, hence :

Gauge pressure at base = r gh( ) Hg + r gh( ) H2O + r gh( ) oil + pair 200 103 = 13600 9.81 0.5( ) + 1000 9.81 2( ) + 600 9.81 3( ) + pair\ pair = 96kNm

- 2

Pressure and head 5

2.11 Referring to figure 2.11a :

Original volume of water = 13p r1

2h1

If half this water is drained, then assume the remaining half fills a new cone withdimensions r2 and h2.

r2

r1

r3

R

12

13p r1

2h1

=

13p r2

2h2. But h1 = 0.5m 0.25r1r2

2

= h2

Due to similarity : r1h1=

r2h2

r1r2=

h1h2

hence :

0.25r1r2

2

= h2 0.25h1h2

2

= h2.

Substituting for h1 = 0.5m

h2 = hH2O = 0.397m.

Before the pressure on the base of thecone can be calculated, the remainingheight of oil must be known.If the total volume before draining equals :

v =13p R2H

then the total volume of oil and water after draining equals :

voil&H2O =13p R2H -

12

13p r1

2h1

which fills volume, v =

13p r3

2h3

13p R2H -

12

13p r1

2h1 =13p r3

2h3

Substituting for h1 and H, then simplifying :


1- 0.25r1R

2

=r3R

2

h3

Again by similarity : r1h1=

RH

and r3h3=

RH

r1R=

h1H

and r3R=

h3H

giving :

1 - 0.25h1H

2

=h3H

2

h3

Substituting for h1 and H :

1- 0.250.51

2

=h31

2

h3

\ h3 = 0.979m.But this is the combined height ie hoil&H2O = hoil + hH2O 0.979= hoil + 0.397

\ hoil = 0.582m.

From eqn. 2.17, pressure at base = r gh( ) oil + r gh( ) H2OTaking r H2O = 1000kgm

- 3 and r oil = 900kgm- 3 ((since spec gravity of oil = 0.9)

pressure at base = 900 9.81 0.582( ) + 1000 9.81 0.397( )\ pressure at base = 9033Nm- 2

2.12a. Area of large piston = p r2 where r = 0.3 m A l = 0.283m2

Area of small piston = p r2 where r = 0.38= 0.0375m As = 4.418 10

- 3m2

m=3500kg

ratio 8:1

If the mass supported by the larger piston is 3500 kg, then the force on the largerarea is given by : force= mg= 3500 9.81= 34335N

Since : pressure=forcearea

pressure=343350.283

= 121.33kNm- 2

This is transmitted to the smaller piston with no difference in height, hence p is unchanged : force= p A s = 121.3310

3 4.418 10- 3

\ Force = 536 Nb. If the smaller piston is 2.6 m below the larger piston, then the additional pressure

on the smaller area is given by :

Pressure and head 7

p = r gh (from eqn. 2.17)

Here, we know spec. gravity = r fluid1000

= 0.8 r fluid = 800kgm- 3.

We also know : g = 9.81ms- 2 and h = 2.6 m,

p = 800 9.81 2.6= 20.4kNm- 2

Since this is in addition to the 121.33kNm- 2, the total pressure is :

pT = 121.33+ 20.4kNm- 2 = 141.73kNm- 2

This pressure is applied over the area As , hence the force can be found from :

force= pT As = 141.73103 4.418 10- 3

\ Force = 626.2 N

2.13a. We know that for a perfect gas, eqn. 1.13 applies ie : r =p

RTWe also know that if the atmosphere is isothermal, then temperature does not vary

with altitude. From eqn. 2.7 : dpdh= -r g

dpdh= -

pgRT

dpp= -

gRT

dh

Integrating from p = p1 when h = h1 to p = p2 when h = h2

logep2p1

= -

gRT

h2 - h1( )

p2p1= e

- gRT

h2 - h1( )

Also, since p2 = r 2RT and p1 = r 1RT then :p2p1=r 2RTr 1RT

=r 2r 1

\ p2p1=r 2r 1= e

- gRT

h2 - h1( )

b. In the stratosphere, the above equation applies and if the pressure is halved with

altitude, then : p2p1=

12

We know that : g = 9.81ms- 2, R = 287Jkg- 1K - 1 & T = -56.5+273 = 216.5 K :

p2p1= e

- gRT

h2 - h1( )

12= exp -

9.81287 216.5

h2 - h1( )

loge12= -

9.81287 216.5

h2 - h1( )\ h2 - h1 = 4390m


2.14a. For a uniformly decreasing temperature with increasing altitude, eqn. 2.14 applies

ie : p2p1= 1 -

d TT1

z2 - z1( )

gRd T

where we know : p2 = 45.5 103 Nm- 2, p1 = 101.5 10

3 Nm- 2 , g = 9.81ms- 2,

T1 = 273+ 15( ) K , T2 = 273+ - 25( )[ ] K and R = 287Jkg- 1K - 1.

We also know that : z2 - z1 = -T2 - T1d T

(based on eqn. 2.13).

Hence, substituting gives :

45.5 103

101.5 103= 1 -

d T288

-T2 - T1d T

9.81287d T

0.4483= 1+248- 288

288

9.81287d T

0.4483= 0.86111( )9.81

287d T

Taking the natural log of each side :

loge0.4483=9.81

287 d Tloge 0.86111

d T = 6.37 10- 3K m- 1

\ d T = 6.37oC per 1000 m.

b. Now, at z2 - z1( ) = 3000 m, p2 is unknown. But we know from eqn. 2.14 that :

p2p1= 1 -

d TT1

z2 - z1( )

gRd T

where p1, T1, g and R remain unchanged.

Also, d T = 6.37 10- 3K m- 1 giving :

p2101.5 103

= 1-6.37 10- 3

288 3000

9.812876.37 10- 3

\ p2 = 70.22kNm- 2

c. To find r 2, we refer to the equation of state : r 2 =p2RT2

Substituting T2 = T1 - d T z2 - z1( )[ ] gives :r 2 =

p2R T1 - d T z2 - z1( )[ ]

r 2 =70.22 103

287288- 6.37 10- 3 3000( )[ ]\ r 2 = 0.91kgm

- 3

Pressure and head 9

2.15a. Letting p and T = pressure and temperature at level z and p0 and T0 = pressure and

temperature at level z0, we know from eqn. 2.13 : T = T0 - d T z - z0( ) . Rearranging gives :

T0 - Td T

= z - z0( ) and substituting into eqn. 2.14 :

pp0= 1 -

d TT0

z - z0( )

gRd T

pp0= 1 -

d TT0

T0 - Td T

gRd T

pp0=

TT0

gRd T

But we know that g

Rd T is a constant, say n,

\ pp0

=

TT0

n

b. We know that :

pp0= 1 -

d TT0

z - z0( )

gRd T

where d T = 6.5 10- 3K m- 1, z - z0( ) = 10700m, T0 = 15+ 273( ) K ,g = 9.81ms- 2 and R = 287Jkg- 1K - 1, giving :

pp0= 1 -

6.5 10- 3

288

10700( )

9.81287 6.5 10- 3

\ pp0= 0.2337

c. From eqn. 2.15 : rr 0= 1 -

d TT0

z - z0( )

gRd T

- 1

rr 0= 1 -

6.5 10- 3

288

10700( )

9.81287 6.5 10- 3

- 1

\ rr 0= 0.3082

2.16a. Before the pressure at 14 500 m can be calculated, the pressure at the start of thestratosphere ie that at the end of the troposphere must be known. Since thetroposphere experiences a uniform temperature decrease, eqn. 2.14 is applicable up

to a height of 11 000 m ie : p2p1= 1 -

d TT1

z2 - z1( )

gRd T

where from eqn. 2.17, p1 = r Hggh= 13600 9.81 0.76= 101396Nm- 2.


Also, d T = 6.5 10- 3K m- 1, T1 = 288K , z2 - z1( ) = 11000m, g = 9.81ms- 2

and R = 287Jkg- 1K - 1, giving :

p2101396

= 1-6.5 10- 3

28811000( )

9.81287 6.5 10- 3

p2 = 22610Nm- 2

This pressure now becomes p1 in the stratosphere in which the following equation

applies : p2p1= exp -

gRT z2 - z1( )

where z2 - z1( ) = 14500- 11000= 3500m and T = 216.5K , giving :p2

22610= exp -

9.81287 216.5

3500( )

p2 = 13011Nm- 2

From eqn. 2.17 : p = r Hggh

13011= 13600 9.81 h\ h = 97.52mm

b. From the equation of state : r 2 =p2RT

r 2 =13011

287 216.5\ r 2 = 0.2094kgm

- 3

2.17 Taking the fluid/air level in the right hand arm of the U-tube as the level XX, then atXX : Sum of pressures in left arm = Sum of pressures in right arm pA + r gh1( ) H2O + r gh2( ) Hg = patm where g = 9.81ms

- 2, r H2O = 1000kgm- 3,

h1 = 0.15m, h2 = 0.3m and patm = 101300Nm- 2.

Also, since spec. weight of Hg = 13.6 spec. weight of water r g( ) Hg = 13.6 r g( ) H2O r Hg = 13.6 1000= 13600kgm

- 3. Hence, summing pressures :pA + 1000 9.81 0.15( ) + 13600 9.81 0.3( ) = 101300\ pA = 59.8kNm

- 2

2.18 Taking the interface on the left arm of the U-tube as the level XX, then at XX : Sum of pressures in left arm = Sum of pressures in right arm pA + r 1ga( ) = pB + r 1g b - h( )[ ] + r 2gh( ) (1)We are told that : a = 1.5 m, b = 0.75 m, h = 0.5 m and that r 2 = 13.6r 1.

As fluid 2 is mercury r 2 = 13600kgm- 3( ) r 1 = r 213.6=

1360013.6

= 1000kgm- 3.

Rearranging equation 1 gives :pA - pB = r 1g b - h - a( ) + r 2ghWe know that g = 9.81ms- 2

Pressure and head 11

pA - pB = 1000 9.81 0.75- 0.5- 1.5( )[ ] + 13600 9.81 0.5( )\ pA - pB = 54.4kNm

- 2

2.19 Firstly, let us calculate the density of both fluids :

Specific gravity of oil = 0.98 = r oil

r H2O@4oC = r oil

1000 r oil = 980kgm

- 3.

Also, specific gravity of water = 1.01 = r H2Or H2O@4oC

= r H2O1000

r oil = 1010kgm- 3.

Referring to figure 2.19a, we see that at level XX:

X

h1oil

water

h2

h

X

BA

0.075m

Sum of pressures in left arm = Sum of pressures in right arm pA - r H2Og h - h1( ) = pB - r H2Og h - h2( ) - r oilg h2 - h1( )This reduces to :

pA - pB = g r H2Oh2 - r oilh2 + r oilh1 - r H2Oh1( )But we know that : h2 - h1 = 0.075m h2 = 0.075+ h1.Hence substitution gives :pA - pB

g= r H2O 0.075+ h1( ) - r oil 0.075+ h1( ) + r oilh1 - r H2Oh1

which then reduces to :pA - pB

g= 0.075r H2O - r oil( )[ ]

giving for g = 9.81ms- 2 :

pA - pB = 0.0751010- 980( )[ ] 9.81\ pA - pB = 22Nm

- 2


2.20 If the accuracy required is 3% of 3mm, then the manometer must be able tomeasure a pressure of : 3% of 3 mm = 0.09 mm of water, hence :Pressure to be measured, using eqn 2.17 and r H2O = 1000kgm

- 3 & g = 9.81ms- 2

p = r gh = 1000 9.81 0.09 10- 3 = 0.8829Nm- 2

For an inclined manometer, we know that :

p1 - p2 = r gz 1 +dD

2

where z = x sinq

Here, p1 - p2 = 0.8829Nm- 2 , r = 740kgm- 3, d = 0.008m, D = 0.024m and

x = 0.0005m giving :

0.8829= 740 9.81 0.0005 sinq 1+0.0080.024

2

\ q = 12o 39'

2.21 Volume of oil transferred from cylinder to sloping manometer tube Vol = 50 *(22/7) * 52/4 = 982.143 mm3

Drop in oil level in cylinder = Vol/((22/7)*352/4)

= 982.143/962.50 = 1.02 mm

Vertical height of oil in sloping manometer tubeh = 50 * sin 15 = 12.941 mm

Pressure in duct relative to atmosphere = r g D h

where D h = 12.941 + 1.02 the vertical separation of the fluid levels

pressure = 0.785 * 9.81 * (12.941 + 1.02) = 107.5 N/m2.

If the cylinder surface movement is ignored the error is due to the 1.02 mm head =1.02 * 9.81 * 0.785 = 7.85 N/m2

2.22 If the specific gravity of oil = 0.83 = r oilr H2O

= r oil

1000 r oil = 830kgm

- 3

If the movement in the 7 mm tube is 1 cm, then the volume displaced is :v = p r1

2h1 where r1 = 0.0035m and h1 = 0.01m

v = p 0.0035( ) 2 0.01= 3.85 10- 7 m3


h1=1cm

h3 h4

h2

XX1

YY1

oilwater

A B44mm

7mm

This causes a height change in the large end of h2 ie : vol = p r2

2h2 where r2 = 0.022m giving :

3.85 10- 7 = p 0.022( ) 2h2 h2 = 0.253 10

- 3 m

At the interface XY : px = py pA + r H2Ogh3 = pB + r oilgh4

If pA - pB( ) = D p1 then :D p1 = r oilgh4 - r H2Ogh3

After movement to the position X'Y' then :

px' = py'

pA + r H2Og h3 + 0.01( ) + 0.253 10- 3( )[ ] = pB + r oilg h4 + 0.01( ) - 0.253 10- 3( )[ ]

Similarly, if we let pA - pB( ) = D p2 then :D p2 = r oilg h4 + 0.01( ) - 0.253 10

- 3( )[ ] - r H2Og h3 + 0.01( ) + 0.253 10- 3( )[ ]Hence, the overall difference in pressure to cause movement of 1 10- 2 m is :D pT = D p1 - D p2 which after substitution and reduction leaves :

D pT = -r oilg 0.01- 0.253 10- 3( ) + r H2Og 0.01+ 0.253 10- 3( )

D pT = - 830 9.810.01- 0.253 10- 3( ) + 1000 9.810.01+ 0.253 10- 3( )

\ D pT = 21Nm- 2

2.23a. Pressure on the bottom of the vessel = p = r gh 1+ag

from eqn. 2.23, where :

r = 840kgm- 3, g = 9.81ms- 2, h = 0.8 m and a = 4 ms- 2 giving :

p = 840 9.81 0.8 1+4

9.81

= 9280.32Nm

- 2

Now, force = p area = 9280.32 (1.4 2)\ Force = 25985 N

b. At constant velocity, acceleration = 0 and hence eqn. 2.23 reduces to :


p = r gh p = 840 9.81 0.8= 6592.32Nm- 2

Similarly, force = p area = 6592.32 (1.4 2)\ Force = 18458 N

2.24 If the speed, N is 600 rpm, then the angular velocity, w , given by :

w =2p N60=

2p 600( )60

= 62.83rads- 1

0.6m =0.025

=0.5

The force exerted on the top of the drum is caused by two components ie a forcedue to rotation and a force caused by hydrostatic pressure. The force due to rotationis calculated as follows:

Force = r gzrr1

r2

dA

But we know that zr =w 2r2

2g+ const (from eqn. 2.32)

The free surface is open to atmosphere, hence at r = 0, p = 0 and zr = 0.

Substituting into eqn. 2.32 gives const = 0, hence : zr =w 2r2

2gWe also know that dA = 2p r dr.

Force= r gw 2r2

2g

r1

r2

2p r dr = rw 2 p r3r1r2

dr

Integrating gives : Force= r w 2 pr4

4

r1

r2

Here, r1 =0.025

2

= 0.0125m (upper pipe radius)

& r2 =0.52 = 0.25m (radius of the drum).

Force= r w 2 pr4

4

0.0125

0.25

Substituting for the limits of integration,for w and putting r = 1000kgm- 3 gives :


Force= 1000 62.832 p 0.254

4-

0.01254

4

Force= 12111NTo calculate the force caused by the hydrostatic pressure : p = r gh from eqn. 2.17where r = 1000kgm- 3, g = 9.81ms- 2 and h = 0.6 m.

p = 1000 9.81 0.6= 5886Nm- 2

The force due to this pressure is found from : force = pressure areawhere area = p r2

2 - r12( ) and r1 & r2 = 0.0125 m & 0.25 m respectively, giving :

Force = 5886p 0.252 - 0.01252( ) = 1153NThis means that the total force ie the sum of the rotational and hydrostatic forces isgiven by :FT = 12111+ 1153N

\ Force = 13.26kNm- 2

2.25 Since the point C is on the centre line of rotation, then the pressure at C is purelyhydrostatic and can be calculated using eqn. 2.17 :p = r gh where r = 1000kgm- 3, g = 9.81ms- 2 and h = 50+ 250( ) mm pC = 1000 9.81 300 10

- 3 = 2943Nm- 2

A

D z

2 50mm

r

C

D B

50mm

250mm

We know that the pressure at point D is generated by both hydrostatic and rotationalforces. Referring to figure 2.25a, we know that the hydrostatic pressure at point Dcan also be calculated from eqn. 2.17 :p = r gh where r = 1000kgm- 3, g = 9.81ms- 2 and h = 50 mm.

p = 1000 9.81 50 10- 3 = 490.5Nm- 2

Furthermore, the pressure generated by the rotational force is given by :

p =rw 2r2

2+ const (from eqn. 2.33) where r = 0.25

But at r = 0, p = 0, hence const = 0. Thus for r = 1000kgm- 3 :


p =1000 w 2 0.252

2+ 0

p = 31.25w 2

Since the pressure at point C equals the pressure at point D ie pC = pD 2943= 490.5+ 31.25w 2

w = 8.86rads- 1

But we know that : w =2p N60

8.86=2p N60

\ N = 84.6 rpm

b. We have established that for any radius, r, at height D z from point D, the pressurecan be calculated from :pT = hydrostatic press. at B + hydrostatic press. due to D z+ rotational press. at r

pT = 490.5+ r gD z +rw 2r2

2Knowing r = 1000kgm- 3, g = 9.81ms- 2 and w = 8.86rads- 1

pT = 490.5+ 9810D z + 39249.8 r2

Hence, for varying heights from point D at the corresponding radius, r, the totalpressure head can be calculated. Knowing that the equation of this quadrant can beexpressed as : D z( ) 2 + r2 = 0.252 (from the equation of a circle ie x2 + y2 = r2), thegraph shown below was plotted.

0

500

1000

1500

2000

2500

3000

3500

4000

Pre

ssu

re,

Pa

0 0.05 0.1 0.15 0.2 0.25 0.3

Height down from D, m

Total pressure

Rotational pressure

Total hydrostatic pressure

This shows the line representing the total hydrostatic head (which obviouslyincreases with distance from point D), as well as that for rotational pressure. Thecurve representing the sum of the two is also shown and from this the value and theposition of the maximum pressure head is easily obtained.


Maximum pressure head = 3556Nm- 2

Since, from eqn. 2.17 : p = r gh

h =pr g=

35561000 9.81

\ h = 0.362 m of water.From the graph we see that this occurs at 0.12 m below point D on the curvedportion CD.

2.26 We know that if the tank is rotating at 180 rpm, then angular velocity can be

calculated from : w =2p N60=

2p 18060

= 18.85rads- 1

We are told that the tank (diameter 1 m) originally contains water to a depth of 3.3 m, hence the volume of water = p r2 depth= p 0.5( ) 2 3.3= 2.592m3

p=40kNm-2

1m

4m

L1

3.3m

L0

Sincei) no water is lostii) volume of a paraboloid = half volume of the circumscribing cylinderiii) whilst rotating, the water wets the top of the tank :

Volume of water = Volume of tank - Volume of paraboloid where :

Volume of paraboloid = 12 p r2 Lo( ) =

12 p 0.5( ) 2 Lo[ ] = 0.3927 Lo m3

2.592= p 0.5( ) 2 4[ ] - 0.3927 Lo[ ] L o = 1.4m

Hence, the height of water left in the tank is : L 1 = 4 - L o = 4 - 1.4= 2.6mAt the centre of the tank, the pressure is due only to the hydrostatic head and the air

pressure. Using eqn. 2.17 and knowing p = 40 103 Nm- 2, r H2O = 1000kgm- 3

and g = 9.81ms- 2, the head due to the air pressure can be calculated as :

h =pr g=

40 103

1000 9.81= 4.08m of water :

Pressure head at the centre of the tank = 2.6 m + 4.08 m (gauge)


We know that atmospheric pressure = 101325Nm- 2 which again using eqn. 2.17

equates to a head of h =pr g=

1013251000 9.81

= 10.33m.

Thus, absolute pressure at the centre of the base of the tank = 2.6 + 4.08 + 10.33 m\ Absolute pressure = 17.01 mTo calculate the pressure at the circumference, we need to know the pressure due torotation, which from eqn. 2.33 is given by :

p =rw 2r2

2+ const

To find the constant, we know that at r = 0, p = 40 103Nm- 2

40 103 = 0 + const const= 40 103Nm- 2

Substituting into eqn. 2.33 : p =rw 2r2

2+ 40 103

At the circumference, r = 0.5, hence :

p =1000 18.852 0.52

2+ 40 103 = 84415Nm- 2

Again using eqn. 2.17, this equates to :

head, h =pr g=

844151000 9.81

= 8.6m

Since the air pressure has been included in the calculation here, only the hydrostatic pressure need be added to get :Gauge pressure = 8.6 + 2.6 = 11.2 Absolute pressure = 11.2 + 10.33 m \ Absolute pressure = 21.53 m

chap02

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