Question

Q-1 For the two compartment flow system as shown below, the water tank (open and immersed) is draining water out in the left compartment. The left and right compartments are filled with fluids of spor of 0.75 and 0.65, respectively. Both the compartments are exposed to the atmosphere. The two compartments are separated by 2 inch circular opening. Please answer the following questions

Answer 1

Kindly refer to the following image::

Open to atmosphere Open to atmosphere Fluid A 0.75 sp gr 20 ft Fluid A 0.65 sp gr 26 ft 10 ft water 1 2 6 ft 20 ft 25 ft

E)

The flow will move from left to right.

A fluid flows from a certain place (A) to another place (B) only when energy difference exists, energy of (A) being higher than energy of (B).

Coming to fluid mechanics, the "energy head" consists of primarily 3 sub-categories:

• Pressure head
• Elevation head
• Velocity Head

Open to atmosphere Open to atmosphere Fluid A 0.75 sp gr 20 ft Fluid A 0.65 sp gr 26 ft 10 ft water А B 6 ft 20 ft 25 ft

In the above case, A and B lie in the same height, therefore elevation head difference is ZERO (0). The magnitude of height is irrelevant when they both lie at the same height.

Initially, if the circular gate is not open, then the fluids are not moving, thus velocities are ZERO (0) and therefore, Velocity heads of both A and B are ZERO (0) then.

Now, we shall calculate Pressure head (P.H)

PH A =SXH

where

• s = Specific gravity of fluid in the left tank = 0.75
• H = 26 ft = 7.9248 m (given)

Thus,

P.HA = S XH = 0.75 x 7.9248 = 5.9436 m

[Note that we did not find the Pressure at A, but instead we found the pressure head at A. "Head" is an ideal way of expressing energy in Fluid Mechanics while Pressure isn't always an ideal manner..]

P.HB = 51 x H1 + S2 x H-

where

• s1 = Specific gravity of upper fluid in the right tank = 0.65
• H1 = 20 ft = 6.096 m
• s2 = Specific gravity of lower fluid in the right tank = 0.75
• H2 = 6 ft = 1.8288 m

Thus,

P.HB= si X H1 + S2 X H2 0.65 x 6.096 + 0.75 x 1.8288 = 5.334 m

We can see that the energy head is greater in the left side (5.9436 m) of the tank as compared to that of the right side (5.334 m) of the tank, therefore, the flow will be from left to right.

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F)

Open to atmosphere Open to atmosphere Fluid A 0.75 sp gr 20 ft Fluid A 0.65 sp gr 26 ft 10 ft water 1 2 6 ft 20 ft 25 ft

We are asked to calculate velocities at points 1 and 2.

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NOTE:

A certain part of the data is missing in the question. For instance, the 10 ft water tank is told to be draining out water. And if we consider the points 1 and 2 to be on the floor of the both sides of the tank, then water is going to fill up the bottom of the left side tank because water has a greater specific gravity (S = 1) as compared to the fluid in the left side of the tank (S = 0.75 ). Denser fluid falls at the bottom. This will cause the pressure head at the bottom of the tank to be different from the calculated 5.9436 m (because even water's pressure will be acting at the bottom). This will result in an increase in the velocity. But since no such information is relayed in the question, we will assume that such a phenomena doesn't occur.

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Point 1:

We will assume that the water flows out into the open (atmosphere) from the left side tank.

Open to atmosphere Fluid A 0.75 sp gr 26 ft 10 ft water 20 ft

Velocity at point 1 =

V1= V29H

where

• g = 9.81 m/s2
• H = 26 ft = 7.9248 m

V1= 29H = V2 x 9.81 x 7.9248 = 12.469 m/s

Point 2:

We will again assume that the water flows out into the open from the right side tank.

Open to atmosphere 20 ft Fluid A 0.65 sp gr 3 2 6 ft 25 ft

We will apply Bernoulli's Theorem bewteen 3 and 2.

P3 P2 V3? 29 + Z3 = + + Z + 29

where

• P3 = Pressure at point 3 = Specific weight of above (lighter) fluid x Height = 650 x 9.81 x H(lighter) = 650 x 9.81 x 6.096 = 38871.144 Pa
  • H(lighter) = 20 ft = 6.096 m
• V3 = 0 (not moving)
• Z3 = 6 ft = 1.8288 m
• P2 = 0 (assuming all the fluid drains in the atmosphere)
• V2 = ?
• Z2 = 0 (datum)
• $\omega$ = Specific weight of heavier fluid = 750 x 9.81 = 7357.5 N/m3 (The fluid on which the pressure is acting)

P3 P2 V3? 29 + Z3 = + + Z + 29

38871.144 V2 +0+1.8288 = 0+ +0=> V2 = 11.8125 m/s 7357.5 29

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G)

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CONCEPT:

This is a very tricky part. What we will essentially do is that on the left hand side, of the tank, we have the space completely filled with the fluid of SG = 0.75. But on the right hand side, we do not have this homogeneity. We have water and this fluid. Now the problem is, we can find the pressure on the right hand side too by conventional means. But that will lead to a lot of confusion. Therefore, we will convert the entire right hand side liquids into one liquid (that is water). Now if we convert all the right hand side liquid to water, it will naturally result in a lower height [represented by "h"] as water is denser and thus, will take lesser volume for the same pressure force. This lower height is represented in the second image (below). So now, the first image (below) helps us calculate the force on the left hand side while the second image (below that) helps us calculate the force on the right hand side of the tank. Since these two forces act horizontally, we can then find the resultant force acting on the tank if we happen to subtract the lesser force from the greater force.

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For this, we are not given any "width" of the tank in the question. However, we will assume that it is a square sided tank which has a width of 10 ft (perpendicular to the plane of paper).

Also, we will find the pressure acting ON ONE SIDE of the water tank.

Open to atmosphere х Fluid A 0.75 sp gr Right Left 7.925 m 3.048 m water 1 20 ft

In the above picture, we have converted all of the required data from feet to m.

The pressure acting on the left side of the tank is straight forward. We will find the hydrostatic force on the left side in the following manner.

Fleft =W.AT

where

• $\omega$ = Specific weight of the fluid =  750 x 9.81 = 7357.5 N/m3
• Area = height x width = 10 x 10 = 100 ft2 = 9.29 m2
• 
  T
  = depth of centre of gravity =
  3.048 +7 m 2
  

Thus,

Fleft = WAT = 7357.5 x 9.29 x (1.524 + 2) = 104167.19 +68351.175.0

Now, for right hand side, we will convert the entire column into a water column.

Thus,

(S.G(fluid) x 2 + S.G(water) x 3.048)

• SG(fluid) = 0.75
• SG(water) = 1

=> $(0.75 \times x + 1 \times 3.048)$ = 3.048 + 0.75x (represented by "h" in the image below - also, free surface level changes)

Open to atmosphere X Fluid A 0.75 sp gr Right Left New water level h 7.925 m 3.048 m water 1 20 ft

Now, again,

Fright = WAT

• $\omega$ = Specific weight of the water=  9810 N/m3
• Area = height x width = 10 x 10 = 100 ft2 = 9.29 m2
• 
  T
  = depth of centre of gravity =
  3,048 3.048 + (h-3.048) + (3.048 +0.75.2 -3.048) 2 3
  
• 
  T
  =
  3,048 +(3.048 +0.75.7 -3.048) = 1.524+0.75.7 2
  

Thus,

Fright = WAT = 9810 x 9.29 x (1.524+0.75.7) = 138889.587+68351.175.0

The final force = F(right) - F(left) = 138889.587 + 68351.175x - 104167.19 - 68351.175x = 34722.397 N

This force acts on the side (which is of area 9.29 m2). Thus

Pressure Force Area 34722.397 9.29 3737.61 Pa

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This question has taken me a really long time to complete. Kindly upvote if you are satisfied with my efforts. :)