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In this Physics tutorial, you will learn:
Suppose you want to wet the garden but the hose is not long enough to reach all parts of the garden. What do you do in such a case?
Do you think water velocity is the same if you open two holes in the lateral sides of a container filled with water if holes are at different heights? Why?
These questions belong to Hydrodynamics - a part of fluid dynamics that studies liquids in motion. In this tutorial, we will explain the basics of this theory accompanying it with examples.
Since the mathematical apparatus used to study the behaviour of real fluid is too complicated, we use a simplified concept that helps us study fluids with a satisfactory approximation. This concept is known as "Ideal Fluid" and obviously, it contains some restrictions compared to real fluids. These restrictions are:
It is a kind of flow, in which the velocity of the fluid at a particular fixed point does not change with time. Thus, the flow of ideal fluids is steady.
For example, all liquid particles passing through the point A will have the same speed vA if the flow of liquid is steady.
Real fluids are slightly compressible. However, since this compression is very small, we neglect it when dealing with ideal gases and thus, we consider them as incompressible. This helps us a lot because we can take the volume of ideal fluids as constant when the number of molecules doesn't change.
In simple words, viscosity represents the resistance of a fluid to the flow. Thus, ideal fluids are considered as non-viscous as they do not make any resistance to the flow.
This means fluid particles do not rotate during the flow but they move only forward. In simple words, this means the fluid streams do not interfere with each other even when they have to pass through an obstacle as shown in the figure.
Let us consider again the first scenario discussed in the "Introduction" paragraph. Thus, from experience we know that if we want to increase the flowing speed of water in order to reach the farthest parts of the garden, we must partially close the hose opening using the thumb. In this way, we decrease the flowing area in the output. This means the cross sectional flowing area and the flowing speed are somehow inversely proportional to each other (smaller cross-sectional area → higher flowing speed and larger cross sectional area → lower flowing speed.
But, how can we prove mathematically the above assumption? For this, let's consider a kind of bottle opened in both ends as shown below.
Let's consider a liquid sample flowing in the two ends of the tube (the coloured sections); one at the beginning and the other at the end of the tube. Since there is the same amount of liquid in both sections (m1 = m2 = m and ρ1 = ρ2 = ρ), we have
This means that
Since the volume of a cylinder is calculated by the formula V = A × Δx (where Δx here stand for height of cylinder), we obtain for volume in both sections
Here the cylinder height Δx acts as a kind of particles displacement. From uniform motion we know that
where v here represents the flowing speed and Δt the time interval during which this flow occurs.
Since there is the same fluid in both coloured part and given that ideal liquids are incompressible, we have for the flowing rate in both sections
Therefore, Δt1 = Δt2 = Δt. Hence, substituting ∆x = v × ∆t in the equation
for both sections, we obtain
Simplifying Δt from both sides, we obtain
Thus, we proved the above assumption, i.e. the cross sectional area of a flowing fluid and its flowing speed are inversely proportional to each other. The above equation is known as the "equation of continuity for ideal fluids".
A 4 cm thick hose is connected to a water tap from which water comes out at 0.8 m/s. The hose is 12 m long if stretched. How much area in the output part of the hose must be block with thumb so that water reaches a point, which is 15 m away from the tap? The hose is taken in the horizontal position at 1 m above the ground. Take g = 10 m/s2 if needed.
First, we have to determine by what speed the water must come out from the hose, i.e. to determine v2 in the equation of continuity. This speed is calculated by kinematic methods. Thus, since water comes out horizontally from the hose (v0y = 0) and the height from the ground is h = 1 m, we obtain for the falling time t:
Hence, given that the horizontal distance of water is Δs = 15 m - 12 m = 3 m, we obtain for the output speed v2 (which corresponds to v0x in kinematics):
Now, let's calculate the flowing area A2 allowed at the output by blocking the rest with thumb. Given that A1 = 4 cm2, v1 = 0.8 m/s and v2 = 6.67 m/s, we obtain from the equation of continuity
Thus, the area we must block by thumb is
Consider the situation described in the figure below, in which a liquid is flowing from left to right through a hose with different heights (h1 and h2) and different thicknesses (A1 and A2).
Obviously, we don't expect the liquid have the same flowing speed in both sides, so we write these speeds as v1 and v2 respectively.
Let's consider a water sample of volume V (V1 = V2 = V). It is clear that if we want to send this sample from the left to the right end of the hose, we must use an input force F1 which when multiplied by the input area A1 gives the input pressure P1 (F1 = P1 × A1), otherwise, the liquid will not raise on the right part, because of the gravity. However, since the right end of the hose is open and therefore it is in contact with air, we have an opposing force F2 caused by the atmospheric pressure P2 on the output area A2 (F2 = P2 × A2).
Both the above forces do some work on the liquid. Thus, the source does the work W1 = F1 × Δx1 on the liquid, while atmosphere does the work W2 = F2 × Δx2 on the liquid. Since the input work is positive and the output work is negative, we obtain for the resultant work done on the system
This change in work contributes in the change of the mechanical energy ME of the system. Thus, since mechanical energy is the sum of kinetic and gravitational potential energy, we can write
Given that KE = m × v2 / 2 and GPE = m × g × h, we obtain
Substituting forces F1 and F2 by P1 × A1 and P2 × A2 respectively, and also expressing mass as a product of density and volume (m = ρ × V), we obtain
Also, since there is the same liquid in both sections considered, we have the same volume as stated earlier, i.e. V1 = V2 = V = A1 × Δx1 = A2 × Δx2.
Thus, we can write
Simplifying volume V from both sides, we obtain
Rearranging the terms of the above equation, we obtain
The above formula gives the Bernoulli Equation for a fluid in two states, 1 and 2. It is named after the Swiss scientists Daniel Bernoulli, who analysed the flowing properties in fluids during the 18th century. Generalizing this equation for all possible states, we obtain
A cylinder completely filled with water is 1.1 m high. If we open a small hole in the lateral side of cylinder (at point A), 40 cm below the water surface, calculate:
Take the upper surface of water as still. Also, take g = 10 m/s2 and ρwater = 1000 kg/m3.
a Let's try to cancel out some of the terms from the Bernoulli equation
This can be achieved by trying to choose an appropriate point of reference for the initial state. Thus, if we choose the point B at the water surface as a reference point, we write the index B instead of 1 and A instead of 2. In this case, we can cancel PB and PA from both sides of Bernoulli equation as both point are under the influence of atmospheric pressure Patm (both of them are in contact with the atmosphere). Thus, the Bernoulli equation becomes
Since we have chosen the point B as a reference, we have hB = 0 and hA = - 40 cm = - 0.40 m. Also, we can cancel out the water density ρ from both sides of equation. Finally, we have vB = 0 as water in the upper surface is still. Hence, we obtain
b The same reasoning (and procedure) can be used to find the emerging speed of water from the point C. Then, since water trajectory is parabolic, we can use the equations of projectile motion to determine the horizontal displacement Δx.
This time, the Bernoulli equation is
Thus, choosing now the position A as reference point, we obtain hA = 0 and hC = - 20 cm = - 0.20 m. Also, we cancel out again the density of water ρ and pressures PA and PC as both points are in contact with the atmosphere, i.e. PA = PC = Patm. Thus, the Bernoulli equations for this situation becomes
Multiplying both sides by 2, we obtain
Given that the height of the point C from the ground is hC = 1.1 m - 0.4 m - 0.2 m = 0.5 m, we can determine the falling time t using the kinematic equation
as in the vertical direction, we consider water as falling freely. Thus,
Therefore, the horizontal distance from the cylinder lower base in which the water coming out from the point C falls on the ground is
Thus, water coming out from the point C falls 1.11 m on the right of the cylinder.
In situations involving the Bernoulli Equation
the terms ρ × g × h1 and ρ × g × h2 are often cancelled out because the changes in flowing height are small, i.e. h1 ≈ h2. Therefore, we obtain a simplified version of Bernoulli equation:
It is obvious that the terms ρ × v21/2 and ρ × v22/2 represent pressure as we cannot add two quantities that are not of the same type. More precisely, they represent the dynamic pressure as they involve the flowing speed of liquid v. On the other hand, the terms P1 and P2 represent the static pressure we have discussed earlier. Therefore, we can obtain a simplified version of Bernoulli equation, which states that:
This statement is similar to the definition of mechanical energy we have discussed in Section 5 (ME = KE + PE = constant).
Not all terms of the Bernoulli Equation
are always present in a given situation. Thus, for example, in static liquids (when liquids are not flowing), the terms related to flowing velocities are not considered. Therefore, the Bernoulli Equation becomes
Furthermore, if we take the reference point at one of the given positions (for example at the position 1, the term related to the height h1 is cancelled out as h1 = 0. Therefore, the Bernoulli Equation is further simplified and becomes
On the other hand, when height difference is negligible as discussed in the previous section, the height-related terms cancels out from Bernoulli Equation, so it becomes
Finally, if there is the same pressure in both parts examined (for example, if both positions are in contact with the atmosphere as in the previous example, the static pressure-related terms P1 and P2 cancel out from the Bernoulli equation, and we therefore obtain
Since in most cases there is the same liquid flowing throughout the tube, the density cancels out. Also, if we take one of the positions as a reference point (for example the position 1, so h1 = 0), the Bernoulli Equation becomes
If the pushing force of the tap in the figure below is 12 N and the hose is 4 cm2 thick, what is the water speed at the output if we block three quarters of the hose's end by placing the thumb on it? Water comes out from the tap at 2 m/s and the entire hose is in the horizontal position. Take g = 10 m/s2 and ρwater = 1000 kg/m3. Also take the atmospheric pressure Patm = 100 000 Pa.
Let's convert the input and output areas into m2 first. Thus, A1 = 4 cm2 = 0.0004 m2 and A2 = A1 - 3/4 A1 = 1/4 A1 = 1/4 × 4 cm2 = 1 cm2 = 0.0001 m2.
Since there is the same height in both ends of the hose, the height-related term in the Bernoulli Equation
cancel out. Thus, we have
The input pressure is obtained by the equation
(Here we must consider the total pressure as input, not only water pressure, because water is in contact with the atmosphere at the source. If it were not so, water would turn back to the source when we close the tap due to the pushing force exerted by the atmosphere at the output.)
Now let's substitute the values in the (reduced) Bernoulli Equation. Thus,
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