Douglas: Fluid Mechanics, Solutions Manual, 5th edition
Chapter 2 : Pressure and Head 2.1
a . From eqn. 2.4 : p2 − p1
= −ρ g( z2 − z1) and taking sea level as datum point 1, then : −3 −2 p1 = 0 Nm , ρ = 1002 kgm , g = 9.81 ms , z 2 = −2000 m and and z1 = 0 m . −2 ⇒ p2 − 0 = −1002 × 9.81 × (−2000 − 0) , ∴ p2 = 19. 19.66 MNm −2
=ρ
dp
and from eqn. 2.17, p = ρ gz giving giving : dρ ρ2 g dz = K dρ . If it is assumed that z is measured from the surface and is negative
b . From eqn. 1.12, we know K
as depth increases, and that K is constant over small pressure ranges, we can then integrate z2
g
∫ K z1
ρ2
1 = dz = −∫ 2 dρ ⇒ ρ ρ K z ρ ρ ρ2
gz
1
z2
1
1
Since we know that z 1 Integrating, NB
=0
1
m th then,
1
gz K
=
1
−
1
ρ 2 ρ1
⇒ρ 2 =
K ρ1
ρ1gz + K
1
∫ 1 + ax dx = a ln(1 + ax )
ρ1 gz z ρ gz ⇒ p2 − p1 = −ρ1g ln 1 + ⇒ p2 − p1 = −K l n 1 + 1 2 ρ1 g K z =0 K 9 −2 −2 −3 We know know that p1 = 0 Nm , K = 2.05 × 10 Nm , ρ1 = 1002 kgm and z 2 = −2000 2000 m , hen hence ce : 1002 × 9.81 × (−2000) ∴ p = 19.7 9 −2 9.75 MNm p2 − 0 = − 2.05 × 10 ln 1 + 2 9 2.05 × 10 From eqn. 2.4 : p2 − p1 = −ρ g( z2 − z1) . Taking the free surface as datum point 1 −2 −3 −2 then, p1 = 0 Nm , ρ = 1000 kgm , g = 9.81 ms , z 2 = −12 m an and z1 = 0 m . −2 117.72 72 kNm ∴ p2 = 117. From eqn. 2.16 : p = ρ gh + patm , ie Absolute pressure = Gauge pressure + K
2
1
2.2
a.
b.
−2
Atmospheric pressure. Here, gauge gauge pressure pressure = 117720 Nm and atmospheric −2
−2 ∴ Absolute pressure = 218. 218.72 kNm kNm ρ ρ −3 Specific gravity of oil = oil ⇒ 0.8 = oil ⇒ ρ oil = 800 kgm ρH O 1000 3 −2 −3 From eqn. 2.17 : p = ρ gh , where p = 120 × 10 Nm , ρ = 800 kgm , 3 −2 g = 9.81 ms , ⇒ 120 × 10 = 800 × 9.81 × h , ∴ h = 15.3 m 3 −2 −2 −3 As before, p = 120 × 10 Nm andg = 9.81 ms , however, ρ = 1000 kgm . 3 From eqn. 2.17 : p = ρ gh ⇒ 120 × 10 = 1000 × 9.81 × h ∴ h = 1 2. 2 m 5 −2 We know that 1 bar = 1 × 10 Nm . From eqn. 2.4 : p2 − p1 = −ρ g( z2 − z1) . Taking the free surface as datum point 1 : −2 −2 −3 −2 5 p2 = 1 × 10 Nm , p1 = 0 Nm , ρ = 600 kgm , g = 9.81 ms and z1 = 0 m .
pressu pressure re = 101000 101000 Nm ,
2.3 a.
2
b.
2.4
4
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
2.5 a.
⇒ 1 × 10 5 − 0 = −600 × 9.81 × (z 2 − 0 ) ⇒ z2 = −17 m , ∴ depth = 17 m −3 For each of the following : p =ρ gh (from eqn. 2.17) and h = 400 × 10 m . ρHg ρ ⇒ 13.6 = Hg ⇒ ρHg = 13600 kgm −3 Specific gravity of mercury = ρH O 1000 −3 −2 Hence, p =ρ gh, ⇒ p = 13600 × 9.81 × 400 × 10 , ∴ p = 53.4 kNm −3 −3 Here, ρ= 1000 kgm , therefore p =ρ gh, ⇒ p = 1000 × 9.81 × 400 × 10 ∴ p = 3.92 kNm− 2 3 −3 Specific weight, w = ρg = 7.9 × 10 Nm . Therefore, p =ρ gh ⇒ p = w × h ⇒ p = 7.9 × 103 × 400 × 10 −3 , ∴ p = 3.16 kNm− 2 −3 Here, ρ= 520 kgm , therefore p =ρ gh −3 −2 p = 520 × 9.81 × 400 × 10 , ∴ p = 2.04 kNm −2 We know that : Force = mass x gravity. Here, mass = 50 kg and g = 9.81 ms 2
b. c. d. 2.6
⇒ F = 50 × 9.81, ∴ 2
0.01 m ,
2.7
⇒p=
490.5
force
F = 490.5 N , Now, pressure = −2
Nm ,
∴ p = 4.905 × 10 4
area
where area =
−2
Nm
0.01 Referring to figure 2.7a, we can assume that manometers are connected at points 1 and 2. Using eqn. 2.27 and equating pressures at XX, then : pgas + (ρgh )gas = patm =2 + (ρ gh)H O (1) 2
where
ρgas = 0.561 kgm
−3
, h gas
= 0.18
m , ρH2 O
= 1000
−3
kgm and h H O 2
= 0.18
m.
Equating pressures at YY :
[p
+ ( ∆p)gas (2 −1) ] + (ρ gh)gas = [p atm= 2 + ( ∆p)air ( 2−1 )] + (ρgh)H O where (∆ p)gas2( −1) = ρgas × g × 120 m , ⇒ ( ∆p)gas (2 −1) = 0.561 × 9.81 × 120 ⇒ ( ∆p)gas (2 −1) = 660.41 Nm − 2 and (∆ p)airgas( 2−1 ) = ρair × g × 120 m ⇒ ( ∆p)airgas (2 − 1) = 1.202 × 9.81 × 120 m , ⇒ ( ∆p)airgas (2 − 1) = 1415 Nm −2 ρgas and ρH O 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 ) −2 We know g = 9.81 ms , ∴ h = 103 mm gas
2
2
180mm 2 gas water
x
x
120m
gas 1 gas Y
h water Y
5
(2)
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
ρ= 13600 kgm− 3 , −3 −3 h =− 50 × 10 m , ⇒ p = 13600 × 9.81 × ( −50 × 10 ) ∴ p =− 6671 Nm −2 , ∴ Absolute pressure = 93.3 kNm−2
−2
2.8
From eqn. 2.17 : p =ρ gh , where
2.9
Pressure at base = pressure of air + pressure of oil + pressure of water (all gauge) Since the tank is open : pair
=0
g = 9.81 ms
and
−2
Nm (gauge)
ρoil ρ ⇒ 0.75 = oil ⇒ρ oil = 750 kgm−3 ρH O 1000 −2 Also, poil =ρ oil gh (from eqn. 2.17) where g = 9.81 ms and h = 2 m, hence −2 poil = 750 × 9.81 × 2 = 14715 Nm −3 −2 Similarly, for water, where ρ= 1000 kgm , g = 9.81 ms and h = 3 m : −2 pH O = 1000 × 9.81 × 3 = 29430 Nm ⇒ pressure at base = (0 + 14715 + 29430) Nm −2 ∴ pressure at base = 44.145 kNm−2 (gauge), so Absolute pressure = (44.145 × 10 3 ) + (1 × 10 5) Nm−2 = 144.145 kNm −2 −2 Gauge pressure at base = p Hg + pH O + p oil + p air . Using g = 9.81 ms , ρHg = 13600 kgm −3 , ρH O = 1000 kgm −3 , ρoil = 600 kgm −3 , and h Hg = 0.5 m , hH O = 2 m & h oil = 3 m , hence gauge pressure at base = (ρgh )Hg + (ρgh )H O + (ρgh) oil + p air , ∴ pair = 96 kNm −2 We know that specific gravity of oil =
2
2
2.10
2
2
2
2
2.11
Referring to figure 2.11a : Original volume of water =
1
πr1 2h1
3 If half this water is drained, then assume the remaining half fills a new cone with dimensions r2 and h 2 . R
r3
r1 r2
1 1 ⇒ × πr1 2h1 = πr2 2h 2 . But h 1 = 0.5 2 3 3 1
Due to similarity :
r1 h1
=
r2 h2
⇒
r1 r2
=
h1 h2
6
m
r1 2 ⇒ 0.25 = h2 r2
hence :
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
r1 2 h1 2 = h2 ⇒ 0.25 = h2 . 0.25 r2 h2 Substituting for h 1 = 0.5 m ⇒ h 2 = h H O = 0.397 m . 2
Before the pressure on the
base of the cone can be calculated, the remaining height of oil must be known. 1 2 If the total volume before draining equals : v = πR H 3 then the total volume of oil and water after draining equals : 1 1 1 2 1 2 2 v oil&H 2 O = πR H − × πr1 h 1 which fills volume, v = πr3 h3 3 2 3 3 1 1 1 1 ⇒ πR 2 H − × πr12 h1 = πr3 2h 3 3 2 3 3 r R r R Substituting for h 1 and H, and from similarity : 1 = and 3 = h1 H h3 H 2 2 h1 h3 ⇒ = and = giving : 1 − 0.25 = h 3 R H R H H H Substituting for h 1 and H gives h 3 = 0.979 m. But this is the combined height ie h oil&H O = h oil + h H O ⇒ 0.979 = h oil + 0.397 , ∴ h oil = 0.582 m. From eqn. 2.17, pressure at base = ( ρgh )oil + (ρgh )H O −3 −3 Taking ρH O = 1000 kgm and ρoil = 900 kgm ((since spec gravity of oil = 0.9) ⇒ pressure at base = (900 × 9.81 × 0.582) + (1000 × 9.81 × 0.397) ∴ pressure at base = 9033 Nm− 2 2 2 Area of large piston = πr where r = 0.3 m ⇒ A l = 0.283 m
r1
h1
r3
h3
2
2
2
2
2 . 1 2 a.
Area of small piston = πr where r = 2
0.3 8
= 0.0375
m
⇒ As = 4.418 × 10−3 m 2
m=3500kg
ratio 8:1 If the mass supported by the larger piston is 3500 kg, then the force on the larger area is given by : force = mg = 3500 × 9.81 = 34335 N force 34335 = 121.33 kNm − 2 Since : pressure = , ⇒ pressure = area 0.283 This is transmitted to the smaller piston with no difference in height, hence p is −3
unchanged : ⇒ force = p × A s = 121.33 × 10 × 4.418 × 10 ∴ Force = 536 N b . If the smaller piston is 2.6 m below the larger piston, then the additional pressure on the smaller area is given by : p =ρ gh (from eqn. 2.17) 3
7
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
Here, we know spec. gravity =
ρfluid
1000 Now: g = 9.81 ms and h = 2.6 m,, −2
= 0.8 ⇒ ρfluid = 800
−3
kgm .
⇒ p = 800 × 9.81 × 2.6 = 20.4
kNm
−2
−2
Since this is in addition to the 121.33 kNm , the total pressure is : pT
= 121.33 + 20.4
kNm
−2
= 141.73 kNm−2
This pressure is applied over the area A s , hence the force can be found from : force = pT × As = 141.73 × 10
3
× 4.418 × 10−3 , ∴ Force
2 . 1 3 a. We know that for a perfect gas, eqn. 1.13 applies ie :
ρ=
= 626.2 N p
RT We also know that if the atmosphere is isothermal, then temperature does not vary dp dp pg dp g = −ρ g , ⇒ = − ⇒ =− with altitude. Since dh dh dh RT p RT Integrating from p = p1 when h = h1 to p = p 2 when h = h2 g − (h − h ) p2 g p2 RT ⇒ log e =− (h − h ) , ⇒ = e . Also, since p1 RT 2 1 p1 g − RT (h − h ) p2 ρ 2 p2 ρ 2RT ρ2 = = , ∴ = =e and p1 = ρ1 RT then : ρ1RT ρ1 p1 ρ 1 p1 2
1
2
p2
= ρ2 RT
1
b . In the stratosphere, the above equation applies and if the pressure is halved with p 1 −2 −1 −1 altitude, then : 2 = . We know that : g = 9.81 ms , R = 287 Jkg K & T = p1 2
56.5+273 = 216.5 K : 1
⇒ log e = −
p2 p1
=e
g − RT (h 2 − h1)
,
1 9.81 ⇒ = exp − (h 2 − h1) 287 × 216.5 2
9.81
(h 2 − h 1) , ∴ h2 − h1 = 4390 m 2 287 × 216.5 2 . 1 4 a. For a uniformly decreasing temperature with increasing altitude, eqn. 2.14 applies g
δT = 1 − ( z2 − z1) . p1 T1 3 −2 −2 p1 = 101.5 × 10 Nm , g = 9.81 ms ,
ie
:
p2
We
know
:
p2
= 45.5 × 103
−2
Nm ,
T1 = ( 273 + 15) K , T2 = [ 273 + ( −25)] K
− T1 . Substituting gives : δ T 9.81 9.81 3 287 T ×δ 287 45.5 × 10 1 − δT − T2 − T1 1 + 248 − 288 ×δ T , ⇒ = ⇒ = , 0.4483 3 288 δT 101.5 × 10 288 9.81 9.81 ⇒ 0.4483 = ( 0.86111)287 ×δ T . Taking ln: ⇒ log e 0.4483 = log e 0.86111, 287 × δT ⇒δ T = 6.37 × 10−3 K m −1 , ∴ δT = 6.37 o C per 1000 m. Now, at (z 2 − z1 ) = 3000 m, p2 is unknown. But we know from eqn. 2.14 that : and R
b.
RδT
= 287
−1
−1
Jkg K , and : z 2 − z1
=−
8
T2
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
g
RδT p2 δT = 1− ( z2 − z1) where p1 , p1 T1 −3 −1 Also, δT = 6.37 × 10 K m giving :
T1 , g and R remain unchanged.
9.81
6.37 × 10−3 287×6.37 ×10− −2 × 3000 , ∴ p2 = 70.22 kNm 3 = 1 − 101.5 × 10 288 p To find ρ2 , we refer to the equation of state : ρ2 = 2 3
p2
c.
Substituting T2 = [ T1 −δ T( z2
− z1 )] ⇒ρ 2 =
RT2 70.22 × 103
−3
287[288 − 6.37 × 10 (3000)]
∴ ρ2 = 0.91 kgm −3 2 . 1 5 a. Letting p and T = pressure and temperature at level z and p 0 and T0 = pressure and
temperature at level z 0 , we know from eqn. 2.13 : T = T0 − δT(z − z 0 ) . T −T = (z − z0 ) and substituting into eqn. 2.14 : Rearranging gives : 0 δT g
g
δT RδT δT T − T RδT p = 1 − (z − z0 ) , ⇒ = 1 − 0 p0 T0 p0 T0 δT g R T δ T p T n p = ⇒ = ,∴ p 0 T0 p0 T0 p
g
b.
δT RδT p = 1 − (z − z0 ) We know that : p0 T0 −3 −1 where δT = 6.5 × 10 K m , (z − z 0 ) = 10700 −2 −1 −1 g = 9.81 ms and R = 287 Jkg K , giving : p p0
6.5 × 10 = 1 − (10700) 288 −3
c . From eqn. 2.15 :
⇒
9.81 287 × 6.5×10−3
,
∴
p p0
m, T0 = (15 + 273) K ,
= 0.2337
g −1 RδT
ρ δT = 1 − (z − z 0 ) ρ 0 T0
9.81 −1 287 ×6.5 ×10−3
ρ 6.5 × 10 = 1 − (10700 ) ρ0 288 −3
,
∴
ρ = 0.3082 ρ0
2 . 1 6 a. Before the pressure at 14 500 m can be calculated, the pressure at the start of the stratosphere ie that at the end of the troposphere must be known. Since the troposphere experiences a uniform temperature decrease, eqn. 2.14 is applicable up g
δT RδT = 1 − ( z2 − z1) p1 T1 −2 p1 = ρHg gh = 13600 × 9.81 × 0.76 = 101396 Nm .
to a height of 11 000 m ie : where from eqn. 2.17,
p2
9
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
δT = 6.5 × 10 −3 K m −1 , T1 = 288 K , (z 2 − z1 ) = 11000 −1 −1 and R = 287 Jkg K , giving : Also,
p2
9.81 287 × 6.5×10−3
−3
6.5 × 10 = 1 − (11000 ) 288
−2
m , g = 9.81 ms
⇒ p2 = 22610 Nm −2
101396 This pressure now becomes p1 in the stratosphere in which the following equation p g ( z − z ) where (z − z ) = 14500 − 11000 = 3500 m applies : 2 = exp − 2 1 RT 2 1 p1 p2 9.81 = exp − and T = 216.5 K , giving : (3500) 287 × 216.5 22610 ⇒ p2 = 13011 Nm−2 . From eqn. 2.17 : p =ρ Hggh, ∴ h = 97.52 mm p 13011 b . From the equation of state : ρ2 = 2 ⇒ρ 2 = RT 287 × 216.5 −3 ∴ ρ2 = 0.2094 kgm 2.17
Taking the fluid/air level in the right hand 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 + ( ρgh1)H O + (ρgh 2 )Hg = patm where g = 9.81 ms− 2 , ρH O = 1000 kgm −3 , −2 h 1 = 0.15 m, h 2 = 0.3 m and patm = 101300 Nm . Also, since spec. weight of Hg = 13.6 × spec. weight of water ⇒ (ρ g)Hg = 13.6 × (ρg)H O ⇒ρ Hg = 13.6 × 1000 = 13600 kgm−3 . Hence, summing pressures : −2 pA + (1000 × 9.81 × 0.15) + (13600 × 9.81 × 0.3) = 101300, ∴ pA = 59.8 kNm 2
2
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 + ( ρ1ga) = p B + [ρ1g ( b − h)] + (ρ2 gh )
(1)
We are told that : a = 1.5 m, b = 0.75 m, h = 0.5 m and that ρ2 = 13.6ρ1 . ρ 13600 −3 = 1000 kgm −3 . As fluid 2 is mercury (ρ2 = 13600 kgm ) ⇒ ρ1 = 2 = 13.6 13.6 Rearranging equation 1 gives : pA − p B = ρ1g ( b − h − a ) +ρ 2 gh −2
We know that g = 9.81 ms , 2.19
∴ pA − pB = 54.4
−2
kNm
Firstly, let us calculate the density of both fluids : Specific gravity of oil = 0.98 =
ρoil ρH O@4 C
=
o
2
Also, specific gravity of water = 1.01 =
ρoil 1000
ρH O 2
ρH O@4 C o
2
Referring to figure 2.19a, we see that at level XX:
10
⇒ ρoil = 980 =
ρH O 2
1000
−3
kgm .
⇒ ρoil = 1010 kgm −3 .
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
oil
h1 X
X
h2 h
0.075m
water
A
B
Sum of pressures in left arm = Sum of pressures in right arm
⇒ pA − ρH Og( h − h1) = pB −ρ H Og( h − h2 ) − ρoil g(h 2 − h1 ) This reduces to : pA − p B = g(ρH O h 2 − ρoil h2 +ρ oil h1 − ρH Oh1 ) But we know that : h 2 − h1 = 0.075 m ⇒ h 2 = 0.075 + h1 . 2
2
2
2
Hence substitution gives : pA − p B = ρH 2O ( 0.075 + h1 ) − ρoil ( 0.075 + h1 ) +ρ oil h1 −ρ H2 O h1 g p − pB = 0.075 (ρ H2O − ρoil ) which then reduces to : A g −2 −2 giving for g = 9.81 ms : pA − pB = 22 Nm
[
2.20
]
If the accuracy required is ±3% of 3mm, then the manometer must be able to measure a pressure of : ±3% of 3 mm = 0.09 mm of water, hence : Pressure to be measured, using eqn 2.17 and
ρH O = 1000 2
−3
−2
kgm & g = 9.81 ms
⇒ p =ρ gh = 1000 × 9.81 × 0.09 × 10− 3 = 0.8829 Nm−2 .
For
an
inclined
manometer, we know that :
d 2 p1 − p 2 = ρgz 1 + where z = x sin θ D −2 −3 Here, p1 − p 2 = 0.8829 Nm , ρ= 740 kgm , d = 0.008 x = 0.0005 m giving : 0.008 2 0.8829 = 740 × 9.81 × 0.0005 × sin θ 1 + 0.024 ∴ θ = 12 o 39' 2.21
m , D = 0.024 m and
Volume of oil transferred from cylinder to sloping manometer tube Vol = 50 *
11
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
(22/7) * 52/4 = 982.143 mm 3 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 tube h = 50 * sin 15 = 12.941 mm Pressure in duct relative to atmosphere = ρ g ∆ h where ∆ h = 12.941 + 1.02 the vertical separation of the fluid levels pressure = 0.785 * 9.81 * (12.941 + 1.02) = 107.5 N/m 2. 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/m 2
2.22
If the specific gravity of oil = 0.83 =
ρoil ρ = oil ⇒ ρoil = 830 ρH O 1000
kgm
−3
2
If the movement in the 7 mm tube is 1 cm, then the volume displaced is : v = π r1 h1 where r1
= 0.0035 m and h 1 = 0.01 m ⇒ v = π(0.0035)2 × 0.01 = 3.85 × 10 −7 m 3 44mm ∅ 2
A
B
water
oil h3
h2
h4
X X1 7mm ∅
h1 =1cm Y1 Y
This causes a height change in the large end of h 2 ie : vol
= πr 22 h 2 where r2 = 0.022
m giving :
−7
= π (0.022)2 h 2 ⇒ h 2 = 0.253 × 10 −3 m 3.85 × 10
12
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
= py ⇒ pA + ρH Ogh 3 = p B +ρoil gh 4 If (p A − p B ) = ∆p1 then : ∆ p1 = ρoil gh 4 − ρH O gh3 At the interface XY : px 2
2
'
'
After movement to the position X Y then : px'
= py
'
⇒ pA + ρH Og[h 3 + (0.01) + (0.253 × 10 −3 )] = p B +ρoil g[h 4 + (0.01) − (0.253 × 10−3 )] Similarly, if we let (p A − p B ) = ∆p 2 then : ∆ p2 = ρoil g[ h 4 + (0.01) − (0.253 × 10 − 3)] −ρ H O g[h3 + (0.01) + (0.253 × 10 −3 )] −2 Hence, the overall difference in pressure to cause movement of 1 × 10 m is : ∆ pT = ∆p1 −∆ p 2 which after substitution and reduction leaves : ∆ pT = −ρoilg (0.01 − 0.253 × 10−3 ) +ρ H Og( 0.01 + 0.253 × 10 −3 ) ⇒ ∆p T = −830 × 9.81(0.01 − 0.253 × 10 −3) + 1000 × 9.81(0.01 + 0.253 × 10 −3 ) ∴ ∆ pT = 21 Nm−2 a Pressure on the bottom of the vessel = p = ρgh 1 + from eqn. 2.23, where : g ρ= 840 kgm −3 , g = 9.81 ms− 2 , h = 0.8 m and a = 4 ms −2 giving : 4 −2 p = 840 × 9.81 × 0.8 × 1 + 9.81 = 9280.32 Nm Now, force = p × area = 9280.32 × (1.4 × 2) ∴ Force = 25985 N 2
2
2
2 . 2 3 a.
b . At constant velocity, acceleration = 0 and hence eqn. 2.23 reduces to : p =ρ gh
⇒ p = 840 × 9.81 × 0.8 = 6592.32 Nm −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,
13
ω , given by :
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
ω=
2 πN 60
=
2π (600) 60
= 62.83
rad s
−1
∅=0.025 0.6m
∅=0.5 The force exerted on the top of the drum is caused by two components ie a force due to rotation and a force caused by hydrostatic pressure. The force due to rotation is calculated as follows: Force =
r2
∫ ρgz dA r
r1
+ const (from eqn. 2.32) 2g The free surface is open to atmosphere, hence at r = 0 , p = 0 and z r ω 2 r2 Substituting into eqn. 2.32 gives const = 0, hence : z r = 2g We also know that dA = 2πr dr . But we know that z r
=
ω 2 r2
= 0.
r ω2 r 2 2 3 ⇒ Force = ∫ r ρg 2π r dr = ρω π∫ r dr r 2g 4 r 2 r Integrating gives : Force =ρω π 4 r 0.025 = 0.0125 m (upper pipe radius) Here, r1 = 2 0.5 & r2 = 2 = 0.25 m (radius of the drum). 4 0.25 2 r ⇒ Force =ρω π 4 0.0125 r2
2
1
1
2
1
Substituting for the limits of integration, for
ω and putting ρ= 1000
Force
= 1000 × 62.83
2
−3
kgm gives :
0.25 4 0.01254 ×π × − 4 4
⇒ Force = 12111 N To calculate the force caused by the hydrostatic pressure : p −3
−2
ρ= 1000 kgm , g = 9.81 ms and h = 0.6 m. ⇒ p = 1000 × 9.81 × 0.6 = 5886 Nm −2
where
14
=ρ gh
from eqn. 2.17
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
The force due to this pressure is found from : force = pressure
π ( r2 − r ) and r1 & r 2 = 0.0125 m & 0.25 m 2 2 Force = 5886 ×π ( 0.25 − 0.0125 ) = 1153 N where area =
2
2 1
× area
respectively, giving :
This means that the total force ie the sum of the rotational and hydrostatic forces is given by : FT
= 12111 + 1153 N
∴ Force 2.25
= 13.26 kNm
−2
Since the point C is on the centre line of rotation, then the pressure at C is purely hydrostatic and can be calculated using eqn. 2.17 : −3
p =ρ gh where ρ= 1000 kgm , g = 9.81 ms
−2
and h = (50
+ 250) mm
⇒ pC = 1000 × 9.81 × 300 × 10 −3 = 2943 Nm − 2 250mm A 50mm D
B
∆z r 250mm
C We know that the pressure at point D is generated by both hydrostatic and rotational forces. Referring to figure 2.25a, we know that the hydrostatic pressure at point D can also be calculated from eqn. 2.17 : −3
−2
p =ρ gh where ρ= 1000 kgm , g = 9.81 ms and h = 50 mm.
⇒ p = 1000 × 9.81 × 50 × 10 −3 = 490.5
Nm
−2
Furthermore, the pressure generated by the rotational force is given by : ρω 2 r 2 + const (from eqn. 2.33) where r = 0.25 p= 2 −3 But at r = 0, p = 0, hence const = 0. Thus for ρ= 1000 kgm :
1000 × ω 2 × 0.252 +0 p= 2 ⇒ p = 31.25ω2 Since the pressure at point C equals the pressure at point D ie pC = pD
⇒ 2943 = 490.5 + 31.25ω 2 ⇒ω = 8.86 rad s− 1 15
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
ω=
But we know that :
2 πN 60
2πN
⇒ 8.86 = 60 ∴ N = 84.6 rpm b . We have established that for any radius, r, at height ∆ z from point D, the pressure can be calculated from : pT = hydrostatic press. at B + hydrostatic press. due to ∆ z+ rotational press. at r ρω 2r 2 ⇒ pT = 490.5 +ρ g∆z + 2 −3 −2 −1 Knowing ρ= 1000 kgm , g = 9.81 ms and ω = 8.86 rad s
⇒ pT = 490.5 + 9810 ×∆ z + 39249.8 × r 2 Hence, for varying heights from point D at the corresponding radius, r, the total pressure head can be calculated. Knowing that the equation of this quadrant can be 2
expressed as : ( ∆ z) + r = 0.25 (from the equation of a circle ie x graph shown below was plotted. 2
2
2
+ y 2 = r 2 ), the
4000 3500 3000 a P , e r u s s e r P
Total hydrostatic pressure
2500 2000
Rotational pressure
1500
Total pressure
1000 50 0 0 0
0.05
0.1
0.15
0.2
0.25
0.3
Height down from D, m
This shows the line representing the total hydrostatic head (which obviously increases with distance from point D), as well as that for rotational pressure. The curve representing the sum of the two is also shown and from this the value and the position of the maximum pressure head is easily obtained.
⇒ Maximum pressure head = 3556 Since, from eqn. 2.17 : p =ρ gh p 3556 ⇒h= = ρg 1000 × 9.81 ∴ h = 0.362 m of water.
Nm
−2
From the graph we see that this occurs at 0.12 m below point D on the curved portion CD.
16
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
2.26
We know that if the tank is rotating at 180 rpm, then angular velocity can be 2 πN 2π × 180 = = 18.85 rad s−1 calculated from : ω = 60 60 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=40kNm-2
πr 2 × depth = π( 0.5) 2 × 3.3 = 2.592 m 3
L0
4m 3.3m L1
1m
Since i) no water is lost ii) volume of a paraboloid = half volume of the circumscribing cylinder iii) whilst rotating, the water wets the top of the tank : Volume of water = Volume of tank - Volume of paraboloid where : 1 1 2 2 3 Volume of paraboloid = × ( πr × Lo ) = × π (0.5) × Lo = 0.3927 × Lo m 2 2 2 ⇒ 2.592 = π(0.5) × 4 − [0.3927 × Lo ]
[
[
⇒ L o = 1.4
]
]
m
Hence, the height of water left in the tank is : L 1 = 4 − L o = 4 − 1.4 = 2.6 m At 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 , ρH O = 1000 2
kgm
−3
−2
and g = 9.81 ms , the head due to the air pressure can be calculated as :
ρg
=
40 × 103
= 4.08 m of water : 1000 × 9.81 ⇒ Pressure head at the centre of the tank = 2.6 m + 4.08 m (gauge) h
=
p
−2
We know that atmospheric pressure = 101325 Nm which again using eqn. 2.17 p 101325 = = 10.33 m . equates to a head of h = ρ g 1000 × 9.81 Thus, absolute pressure at the centre of the base of the tank = 2.6 + 4.08 + 10.33 m ∴ Absolute pressure = 17.01 m To calculate the pressure at the circumference, we need to know the pressure due to rotation, which from eqn. 2.33 is given by : ρω 2 r 2 + const p= 2
17
Douglas: Fluid Mechanics, Solutions Manual, 5th edition
To find the constant, we know that at r = 0, p
= 40 × 103 Nm −2
⇒ const = 40 × 103 Nm −2 ρω 2r 2 + 40 × 103 Substituting into eqn. 2.33 : ⇒ p = ⇒ 40 × 10 3 = 0 + const
2
At the circumference, r = 0.5, hence : 1000 × 18.852 × 0.52 ⇒p= + 40 × 103 = 84415 Nm−2 2 Again using eqn. 2.17, this equates to : p 84415 = = 8.6 m head, h = ρ g 1000 × 9.81 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
18
