ROTATING VESSELS MEMBERS: AYON, LEO BENHUR RAMPING, WIMAR BLESS TANAGRAS, JAN TROY ALVIAR, BENJIE MOSQUERA, AILEEN JOY ARIBAS, HELEN GRACE GARCIA, ALPHSTER JANE RIVERA, CHESSA BARCELONA, KIMBERLY DE PERIO, CHINDAFYE CASILAGAN, RODEL
If an open vessel is partly filled with water or any liqu id. And is rotated at a certain velocity about its vertical axis, its free surface becomes concave in form. CF = man W = mg ; m=
an =
x
From the force polygon,
= x( ) Tan Ɵ = ( ) Since tan Ɵ = ) = dy = dx ∫ = ∫ y= Tan Ɵ =
DERIVATION OF THE VOLUME PARABOLOID
Volume of paraboloid
∫ y= K = V=
When y=h and x=r;
h= K
K=
∫ V=∫ V=
V=∫
If an open vessel is partly filled with water or any liqu id. And is rotated at a certain velocity about its vertical axis, its free surface becomes concave in form. CF = man W = mg ; m=
an =
x
From the force polygon,
= x( ) Tan Ɵ = ( ) Since tan Ɵ = ) = dy = dx ∫ = ∫ y= Tan Ɵ =
DERIVATION OF THE VOLUME PARABOLOID
Volume of paraboloid
∫ y= K = V=
When y=h and x=r;
h= K
K=
∫ V=∫ V=
V=∫
V=
[ ](from 0-h)
V=
There is no change in its volume as long as it will not overflow.
V1 = V2
h
π
1
=
h
h1 = h/2 SQUARED PROPERTY OF PARABOLA:
=
CASE 1: OPEN VESSEL •
LIQUID SURFACE CONDITIONS
- for open cylindrical containers partially filled and rotated about its vertical axis. a. ( no liquid spilled )
< D
b. Liquid surface just touching the top of the rim (no liquid spilled)
D
c. Some liquid spilled
D
d. Vortex at the bottom ( some liquid spilled)
y=H
e. Vortex ( imaginary) below the bottom (some liquid spilled) y>H
CASE 2: CLOSED CYLINDRICAL VESSEL a. partially filled
1. < D
2. ( liquid surface just touching the top rim)
=D
3. with imaginary paraboloid above
D
4. Vortex just touching the bottom
()() ()() ()() ()() π D = π By squared property of parabola:
= H =
By substitution,
π D = π H () y=
5. y > Vortex below the bottom
b. fully filled 1. without pressure at top
2. with pressure at top
c. pipes and tubes 1.without initial pressure inside
2. with initial pressure inside
3. U- tube revolved about its own axis: -the pressure head at any point in the tube is the vertical distance from the tube to the paraboloid. the pressure is positive if the paraboloid is above the point and negative if it is below the point. the limiting pressure is absolute zero.
SAMPLE PROBLEMS:
CASE 1: 1 .An open cylindrical vessel 0.60 m. in diameter and 0.90 m high is two – thirds full of liquid. If the vessel is rotated about its vertical axis: a. What is the greatest speed in revolutions per minute that it can have without causing any liquid to spill over the sides? b. What speed must it have in order that the depth at the center be zero? c. What speed must it have in order that there maybe no liquid within 150 mm from the vertical axis? SOLUTION : a. y=0.60 so that no liquid will spill out
() 0.60 = () y=
w = 11.44 rad/sec w=
()
w = 109. 244 rpm
b. When y = 0.90
() 0.90= () y=
w = 14.01 rad/sec w=
()
w= 133. 76 rpm
c.
y=
() (eq. 1)
() (eq. 2) () () + = 0.00344 = 0.90
= 16.17 rad/ sec w =
()
w = 154. 412 rpm
2. An open vessel in the form of a paraboloid having its vertex at the bottom is rotated at a speed of 85 rpm. At this speed the vertex of the paraboloid formed by the water surface was found to be 1m above the vertex of the containing vessel. Wh en at rest the container contains 254 liters of water when full. a.) Find the height of the open vessel. b.) Find the radius of the open vessel. c .) How many liters of water will remain in the vessel when rotated at the above mentioned speed? Solution:
a.Height of the open vessel:
Vol. of container = = 0.254 = y= =
()
8.9 rad/sec
y = H – 1 H – 1 =
() ()
H – 1=
H² - H – 0.1060 = 0 H = 1.097 m. y = H – 1 Y = 0.097 m. b.) Radius of the open vessel = 85 m.
() r²= =
r = 0.148 m. c.) Volume of water that remain in the vessel: Vol. of water when full = 254 liters Vol. of water spilled out = Vol. of small paraboloid
()() Vol. of water spilled out = Vol. of water spilled out = 0.0166 Vol. of water spilled out =
Vol. of remaining water = 254 – 16.6 Vol. of remaining water = 237.4 liters
3. A hemispherical bowl having a radius of 1m is full of water. If the hemispherical bowl is made to rotate uniformly about the vertical axis at the rate of 30 rpm. a . Determine the volume of water that is spilled out. b . Determine the remaining volume of water in the hemispherical bowl.
a.
Volume of water that is spilled out.
() = h=
= 3.14 ad/sec. h=
()() ()
Vol. of water spilled out.
v=
()()
v = 0.785
b.Remaining volume of water in the hemispherical bowl. volume of hemisphere:
dv = V = ∫ ( ) V = [∫ ∫ ] V=
* ,- [ ]+
,V = * [ ]+ V = [ ] v = () V(left) = V (hemisphere) – V(paraboloid)
() v = v=
4. An open vertical cylindrical vessel, 2m in diameter and 4m h igh is filled with water to the top. If rotated on its own vertical axis in order to discharge a quantity of water to uncover a circular area at the bottom of the vessel 1m in di ameter: a. Determine the angular speed in rpm. b .How much water is left in the cylinder after rotation .
SOLUTION:
a.) h=
Solve for h (by squared property)
() () h – 4 = 0.25 h 0.75 h = 4 h = 5.33 m 5.33 =
() ()
ω = 10.23
⁄ x
= 97.65 rpm
b.)
= (1)²(4) – (⁄ ()() ⁄ ()( ))
CASE 2: CLOSED CYLINDRICAL VESSEL 5. A closed cylindrical is rotated about its vertical axis. The diameter of the cylinder is 0.20 m and has a height of 0.12 m . It is filled with water at a depth of 0.10 m and the rest is filled up with air. a . What rotating speed is necessary for the water to just touch the bottom of the center of the tank in rpm? b . What is the pressure at the bottom of the periphery of the tank?
a. Rotational speed Volume of air does not change
() (0.02) = (0.12)
π
R= 0.0577 m y=
ω= 26. 59 rad/sec
()
ω=
= 253. 92 pm b. pressure at the bottom of the periphery of the tank. h=
( )() h= ()
h = .36 m
= = 9.81(0.36) = 3.53 kPa = 3530 Pa
6. 4. A closed cylindrical vessel 3 m. in diameter and 6 m high is filled with water to a height of 4.5 m. The rest is filled with air, the pressure of which is 105 kPa. If the vessel is rotated at 191 rpm about its axis, determine the maximum and minimum inside pressure at the base.
Speed of rotation
()
ω=
ω=20 rad/sec
TIP: Multiply rpm by π/30 for fast conversion to rad/sec. Notice that the above procedure is actually a multiplication of this amount. ω
h=
2
When x = r = 1.5 m, y = h 20 .5
h=
)
29.8
h= 45.871 m
Determine the position of the vortex. ( y(height of the paraboloid) = ) =y =y
()
y= 12m> 6m since >h
ω
=
2
=() ()
= (eq.1) ω
=
2
=() ()
= ; = + 6
() () h = ( ) ( ) ( )( ) ( ) = () (( )
π
=
( )
)
(
+ 2.7738 7. 8291 = 0.9246 = 8.468 m = 0.9246
The minimum pressure at the base occurs at all points within the circle of radius x 1 and is equal to the original air pressure. Pmin=105 kPa
The maximum pressure will occur anywhere along the circumference of the base. Pmax=105+ϒh−y=05+9.845.87−8.468 Pmax=471.92 kPa
7. A closed tank 2m in diameter and 3m in high is filled with water and rotates about its axis at 100 rpm. A pressure gage at the center of the top reads 30 kPa during rotation. a. Determine the equivalent height of fluid above the top along its periphery. b. Determine the pressure at the outer perimeter at the bottom of the tank. c. Determine the pressure at the bottom of the tank at the center.
SOLUTION: a. Equivalent height of fluid above the top along its periphery
h h
h= 3.06 m
() () ()
y= 5.59 m
height = h + y
height = 3.06 + 5.59 height = 8.65 m
b. Pressure at the outer perimeter at the bottom of the tank H = 5.59 + 3.06 + 3 H = 11.65 m P = H P = 9.81(11.65) P= 114. 29 kPa
c. Pressure at the bottom of the tank at the center P =
P = 9.81(3 + 3.06) P=59.45 kPa
CASE 3: PIPES AND TUBES
8. A 75mm diameter pipe , 2 m long is filled with water and capped at both ends . It is held on a plane inclined 60 ° with the horizontal and rotated about a vertical axis through its lower end with constant angular speed of 5 rad/ sec . Compute the pressure at the upper end of the pipe and determine the minimum pressure and its location in the pipe.
SOLUTION:
Since there is no initial pressure in the pipe the pressure head at the lower end of the pipe will remain equal to the static pressure head of 1.73 m , and therefore the vortex of the paraboloid will be 1.73m above the lower end.
() h= 1.274
a. pressure at the upper end
= (9.81)1.274 = 12. 497 kPa
b. minimum pressure
solve for h’ h’ = y+z z= 1.73 – x tan 60° ==
+ (1.73 – x tan 60°) eq. 1
y= 1. 274 h’ = 1.274
minimize h’ :
x= 0.68 m a = xsec 60 a= 0.68 sec 60 = 1.36 m
from eq 1 h’ = .274
() + 1.73 – (0.68) tan 60
h’ = .4 m
= 9.81(1.41) = 11. 196 kPa located 1.36 m from the lower end (along the pipe)
9. 1.) At what angular velocity must a U-tube 0.30 m. in width and filled with water to a depth of 0.30 m. be rotated about its axis to cause the water at a to vaporize?
2.) Find the absolute pressure at B in kPa. 3.) Find the absolute pressure at B in bars.
Solution:
To cause the water at A to vaporize, we assume the pressure of water at A to be absolute zero or the required pressure head to be 760mm of Hg. = 0.76 (13.6) = 10.34
h = 10.34 + 0.30 = 10.64 h=
ωr
10.64 =
() ()
ω= 96.32 rad/sec ω=
()
= 919.81 rpm
b.) Absolute pressure at B in kPa
h = 9.81 (10.64) = 104.4 kPa c.) Absolute pressure at B in bars = 1.044 bars w = 154. 412 Note : 1 bar = 100 kPa
10. a.) How fast must the U-tube shown be rotated about y-axis so that the difference of water levels in the tubes is 30 cm. b.) What is the original height of the mercury when at rest? c.) what is the maximum pressure at the bottom of the tube on the right side when it was rotated about the y-axis? Solution:
a.Speed must the U-tube shown to be rotated about y-axis so that the difference of waterlevels in the tubes is 30 cm since:
= y = 4
2 () () =
= 7 rps ( ) = 66. 88 rpm b.) Original height of the mercury when at rest: Original height of water
y =
() 2
y = 9.99 + 30. y = 39.99 cm c.) Maximum pressure at the bottom of the tube on the right side when it is rotated about the y-axis.
y 9.81 (39.99) 3.92 kPa
DOCUMENTATION:
This group who was assigned to report and explain the topic: ROTATING VESSEL. The sub-topic/ cases was expounded by each member and some problems was solved to enlighten the listeners.
LEO BENHUR AYON. He introduced his groupmates to the class. He also made an introduction to the report and derived the first equation.
JAN TROY TANAGRAS. Tubes when rotated was assigned to him. He explained it in class.
WIMAR BLESS RAMPING. V(air initial) = V (air final) : the magic formula in case two. He explained its conditions and its relationship.Moreover, he derived important equations to remember.
RODEL CASILAGAN. He solved the 2 nd eample’s second problem iven.
BENJIE ALVIAR. He answered the 2 nd eample’s first problem iven.
KIMBERLY BARCELONA. She explained closed cylindrical vessel with pressure and without pressure.
CHINDAFYE DE PERIO. She derived the second formula which is the height of the paraboloid.
CHESSA RIVERA. She derived the volume of the paraboloid.
ALPHSTER JANE GARCIA. She was assigned to answer the first problem given.
HELEN GRACE ARIBAS. She explained the first case in the topic. Then derived the equation relating to the case.
AILEEN JOY MOSQUERA. She solved rd
the second eample’s 3 sub question.
