RELATIVE EQUILIBRIUM OF FLUIDS
a) Moving vessel with constant acceleration (Horizontal Motion )
From figure A, the water surface lies horizontally when the vessel is not in
motion, but at the next figure B the water surface assumes an angle Ø with
the horizontal as it is being accelerated. To determine the value of Ø let us
consider a mass of liquid as shown
tan Ø =
tan Ø = a/g
b) Vertical Motion :
Ordinarily if the cointainer shown is at rest the pressure at point 1 is yh. But
if it will accelerate up, the pressure at the point changes. Let us assume a
vertical prism of liquid having a height “h” and an area of A:
(Motion going up )
c) Inclined Motion :
Problem 203
An open tank 1.82 m square weights 3425 N and contains 0.91 m of
water. It is acted by an unbalanced force of 10400 N parallel to a pair of
sides.
1) Find the acceleration of the tank.
2) What is the force acting on the side with the greatest depth?
3) What is the force acting on the side with the smallest depth?
Solution
1) Acceleration of the tank
F= ma
= w/g (a)
W= 3425 + 1.82(1.82)(0.91)(98100)
= 32995.13 N
10400= 32995.13 a / 9.81
a= 3.092 m/s
2) Force acting on side with the greatest depth
tan =
=
= 0.287
Greatest depth =0.91 + 0.287
Greatest depth = 1.197
P= wh A
= 0.5985
h=
= 0.5985
P= 9810(0.5989)(1.197)(1.87)
= 12791 N
Problem 204
An unbalanced vertical force of 270 N upward accelerates a volume of
0.044 m^3 of water. If the water is 0.9 m deep in a cylindrical tank.
1) What is the acceleration of the tank?
2) What is the pressure at the bottom of the tank in N/m^2?
3) What is the force acting on the bottom of the tank in N?
Solution:
1) Acceleration of the tank
Ww= 0.044(9810)= 431.64 N
F= ma
=a
270= a
a= 6.14 m/
2) Pressure at the bottom of the tank in N/
Pb= yw h (1 + )
= 9810(0.9)(1+ )
=14355 N/
3) Force acting on the bottom of the tank in N
V= Ah
0.044= A(0.90)
=0.0489
F= Pb A
=14355(0.0489)
=702 N
Problem 205
An open horizontal tank 2 m high, 2 m wide and 4 m long is full of water.
1) How much water is spilled out when the tank is accelerated
horizontally at 2.45 m/sec^2 in a direction parallel with its longest
side?
2) What is the force acting on the side with the greatest depth?
3) Compute the required accelerating force.
Solution:
1) Water spilled out
tan=
=
=
x= 1 m
Vol. of water spilled=
=4 cu. M
2) Force acting on the side with the greatest depth
= yw
=9.81(1)(2)(2)
= 39.21kN
3) Required accelerating force
= 9.81( )(1)(2)
= 9.81
Accelerating force= 39.24 – 9.81
Accelerating force= 29.43 kN
Problem 206
An open tank moves up an inclined plane as shown with constant
acceleration.
1) Compute the angle that the water surface in the tank makes with the
horizontal line.
2) Calculate the acceleration required for the water surface to move to the
position indicated.
3) Compute the vertical component of the acceleration.
Solution:
1) Angle that the water surface in the tank makes with the horizontal line.
tan=
ah= a cos
= 0.97 a
= a sin 14.04
= 0.2425 a
tan= =0.25
= 14.04
Tan( + ) =
= 0.5
= 0.50
tan + 0.25= 0.50-0.125 tan
= 12.53
2) Acceleration required for the water surface to move to the position
indicated
Tan 12.53
= 0.222(9.81 + 0.2425 a)= 0.97 a
2.18+ 0.054 a= 0.97a
a= 2.38 m/
3) Vertical component of the acceleration
= 2.38(0.2425)
= 0.577 m/
Problem 207
The tank shown in the figure contains oil with sp.gr of 0.88. If the tank is
10 m long the initial depth of oil is 2 m and the tank accelerates to the
right at 2.45 m/sec. Assume the tank walls are sufficiently high so that
there is no spillage.
1) Determine the slope of the water surface.
2) Determine the max. pressure at the bottom of the tank.
3) Determine the min. pressure at the bottom of the tank.
Solution:
1) Slope of the water surface
Tan =
=
= 14.02
2) Max. pressure at the bottom of the tank
Tan
x= 1.25
= 1.25 + 2
= 3.25 m
= yw
=9.81(3.25)(0.88)
= 20 kN/
3) Min. pressure at the bottom of the tank
= 2-1.25
= 0.75 m
P2= 9.81(0.88)(0.75)
= 6.47 kN/
Problem 208
A rectangular car tank 20 m long 4 m wide and 3 m deep is completely
open at the top. If it is initially filled to the top.
1) How much liquid will be spilled if its given a horizontal acceleration of
0.3 g in the direction of its length.
2) What is the max force acting on the side opposite to the direction of
the acceleration.
3) Determine the volume of the remaining water tank.
Solution:
1) vol. of liquid spilled out
Tan =
=
Tan = 0.3
= tan
= 0.30
y= 6 m
Vol. spilled out = (3)(4)
= 180
2) Max force acting on the side opposite to the direction of the acceleration
P= Ywh A
=9.81 (1.5)(3)(4)
=176.58 kN
3) Volume of the remaining water of the tank
Remaining volume = 20(4)(3)-180
Problem 209
An open vessel of water accelerates up to 30 degree plane at 2 m/s^2.
1) What is the horizontal component of the acceleration?
2) What is the vertical component of the acceleration?
3) What is the angle that the water surface makes with the horizontal?
Solution:
1) Horizontal component of the acceleration
= 2 cos 30
= 1.73 m/
2) Vertical component of the acceleration
= 2 sin 30
= 1 m/
3) Angle that the water surface maes with the horizontal
tan=
=
=9.09
Problem 210
A 1x1x1 m tank 50% filled with oil (sp. gr 0.82) and the remaining is
water. When it is translated vertically upward at 5.10 m/s^2?
1) Determine the pressure at the bottom.
2) Determine the velocity at the orifice 50 mm diam. located at the
bottom of the tank.
3) Determine the discharge if the coefficient of discharge C is equal to
0.6.
Solution:
1) Pressure at the bottom
P= p static (1 + )
P static = w h
h= 0.5 (0.82)+ 0.5(1)
= 0.91 m
P= 9.81(0.91)(1 + )
=13.57 kPa
2) Velocity at the orifice 50 mm diam. Located at the bottom of the tank
h= 0.91( 1+ )
= 1.38 m
V=
= 5.20 m/s
3) Discharge if the coefficient of discharge C is equal to 0.6
Q= CA
=
Q= 0.0061 /s
Problem 210 -A
An open tank 1.8 m x 1.8 m and 1.30 m deep contains water at a depth
of 0.90 m. The weight of the tank is 3500 N. The tank is acted by an
unbalanced force of 12000 N parallel to a pair of sides.
1) Determine the acceleration of tank.
2) Determine the max. pressure at the bottom of the tank.
3) Determine the force acting on the side with greatest depth.
Solution:
1) Acceleration of tank
F= m a
=a
W= 3500 +(1.8)(1.8)(0.9)(9.81)
= 32106
12000 = a
a= 3.67 m/
2) Max pressure at the bottom of the tank
tan=
y= 0.34 m
h= 0.90 + 0.34
=1 .24
P= yw h
= 9.81 (1.24)
P= 12.16 kPa
3) Force acting on the side with greatest depth
P= (1.24)(1.8)
P= 13.57 kN
Problem 210-B
An open tank 3 m by 3 m in horizontal section weights 3.6 kN and
contains water to a depth of 1 m. If is acted by an unbalanced force of 16
kN parallel to a pair of side.
1) Determine the acceleration of the tank
2) What must be the height of the tank so that no water will spill out.
3) If the acceleration is increased by 4 m/s^2, how much water will be
spilled out.
Solution:
1) Acceleration of the tank
F= m a
=a
= 3.6 + 3(3)(1)(9.81)
= 91.89 kN
16= a
a= 1.71 m/sec
2) Height of the tank
tan= a/g
= 0.26
h= 0.26+ 1
= 1.26 m
3) Volume of the water spilled out if
a= 1.71 + 4
=5.1 m/s
Problem 211
The tank shown in the figure is acceleated to the right. Width of the tank is
1 m.
1) Determine the acceleration needed to cause the free surface to touch
point A.
2) Determine the pressure at B.
3) Determine the total force acting on the bottom of the tank.
Solution:
1) Acceleration needed and after moving is the same
2(0.2)(1)=
= 0.667 m
tan =
= 1.2/ 0.667
= 17.66 m/
2) Pressure at B
Tan =
=
y+ 1.2 =
=y+1.2 = 3.6 m
= Yw(y + 1.2)
= 9.81(3.6)
= 35.32 kPa
3) Total force acting on the bottom of the tank
F= area of bottom
= 35.32 kN
Problem 212
The tank shown is accelerated to the right at 10 m/s^2.
1) Find the pressure at A
2) Find the pressure at B
3) Find the pressure at C
Problem 213
An open tank 10 m long is supported on a car moving on a level track
and uniformly accelerated from rest to 30 kph. The tank was filled with
water to within 15 cm of its top, when it was accelerated.
1) Compute the acceleration of the car.
2) Find the shortest time in which the acceleration maybe
accomplished without liquid spilling over the edge of the tank.
3) The height of water is 1 m when the tank is at rest. Determine the
total force acting on the bottom of the tank when the tank was
accelerated if it has a width of 2 m.
Solution:
1) Acceleration of the car
tan=
=
a= 0.29 m/s^2
2) Shortest time in which the acceleration maybe accomplished without
liquid spilling over the edge.
=0
=
= 8.33 m/s
= + at
8.33 = 0 + 0.29 t
t= 28.74 sec
3) = yw
= 9.81(1.15)
=11.28 kPa
= yw h2
= 9.81(0.85)
=8.34 kPa
F= A
= (10)(2)
=196.2 kN
Problem 214
The container shown has a width of 500 mm and contains water at a
depth of 300 mm.
1) At what angle will the water surface lakes with the horizontal when
the tank surface makes with the horizontal when the tank is
accelerated to the right so that water will just touch point A.
2) Determine the acceleration of the tank to right at this instant.
3) What is the maximum pressure at point B of the tank at this instant.
Solution:
1) Angle the water surface maes with the horizontal
AB=
=424.26 mm
BC= 700-424.26
= 275.26 mm
= 1112.20 mm
Sin law:
= 10
2) Acceleration of the tan to the right so that no water will be spilled out
Tan =
tan10=
a= 1.74 m/s (acceleration)
3) Maximum pressure at the bottom of the tank at this instant
tan10=
= 213.39 mm
= yw h
= 9.81(0.21339)
=2.09 kPa
Problem 215
A cubical box 1 m on an edge, open at the top and half filled with water
is placed on an inclined plane making a 30’ angle with the horizontal.
The box alone has a gravity force of 500 N and has a coefficient of
friction with the plane of 0.30.
1) Determine the acceleration a x of the box.
2) Determine the acceleration a v of the box.
3) Determine the angle the free water surface makes with the horizontal.
Solution:
1) Acceleration of the box
tan=
= a cos 30
=0.866 a
= a sin 30
tan=
Total weight of water + tank
= 0.5(1.0)(1)(9810) +500
=5405 N
N= 5405 Cos 30
= 4680.87 N
= N
= 0.30(4680.87)
= 1404.26 N
F= 5405 sin 30 -1404.26
= 1298.24 N
F= ma
1298.24 = a
= 2.36 m/s
= 0.866 a
=0.866(2.36)
= 2.04 m/
2) Acceleration of the box
= a sin 30
=-2.3(0.5)
=-1.18 m/
= -1.18 m/
3) Angle the free water surface maes with the horizontal
tan=
=
=
=13.3