DOC: add lab demos

docs/lab_demos24/IssaMaoPathak/Hydraulic_hysteresis/index.md

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+## Theory
+
+In this experiment, we explore the dynamics of channel flow featuring an obstacle positioned midway along the channel. The experimental setup involves maintaining a constant inflow rate, represented by $ q $, throughout the channel. Upon reaching equilibrium, it is observed that the water level at the obstacle is $ d $ above the height of the obstacle.
+
+When the inflow rate is altered, conservation laws dictate that $ d $ will also change. However, the system requires a certain duration to adjust $ d $ and achieve a new equilibrium configuration. This phenomenon of time delay in adjusting $ d $ is termed hydraulic hysteresis.
+
+
+
+## Methods
+
+1. **Experimental Setup:**
+   - A channel flow apparatus was assembled with an obstacle positioned halfway down the channel.
+   - The downstream reservoir was intentionally kept at a low level to ensure the channel started dry.
+
+2. **Flow Speed Setting:**
+   - A moderately slow flow speed was set to initiate the experiment.
+   - The evolution of the flow was observed and recorded.
+
+3. **Observations Recorded:**
+   - The following data points were noted:
+     - The time taken for water to start spilling over the obstacle.
+     - The time taken to reach steady state flow conditions.
+
+4. **Inflow Rate Variation:**
+   - The inflow rate was systematically varied.
+   - Changes in the flow behavior were observed and recorded.
+
+
+
+
+## Observations 
+1. - Water level when water just starts to spill = 11.8 cm
+   - First inflow rate = 1000 mL/(1:36 - 1:30) = 1L/6 sec = 166.67 $cm^3 \, s^{-1}$
+   - Maximum inflow rate
+   - Water height = 13.7 cm
+   - d = 13.7 cm - 11.8 cm = 1.9 cm
+   - Time it takes to attain equilibrium = (0:45 - 0:15) = 30 sec
+
+![Equilibrium configuration with maximum inflow rate](1st.png)
+
+2. - Medium Inflow rate: Decreased the inflow rate
+   - Second inflow rate = 1L/(2:45 - 2:31) = 1L/14 sec = 71.43 $cm^3 \, s^{-1}$
+   - Max height = 13.2 cm
+   - d = 13.2 cm - 11.8 cm = 1.4 cm
+   - Time it takes to attain equilibrium = (2:11 - 1:52) = 19 sec
+
+![Equilibrium configuration with intermediate inflow rate](2nd.png)
+
+3. - Lowest inflow rate: Started from zero flow
+   - Third inflow rate = 1L / (6:29 - 5:56) = 1L/33 sec = 30.3 $cm^3 \, s^{-1}$
+   - Water height = 12.1 cm
+   - d = 12.1 cm - 11.8 cm = 0.3 cm
+   - Time it takes to attain equilibrium = (5:43 - 5:08) = 35 sec
+
+![Equilibrium configuration with minimum inflow rate](3rd.png)
+
+
+
+
+
+4.  - Measuring total time : Started from zero flow and new inflow rate
+    - First video
+    - Inflow rate = 1L / (0:38 - 0:21) = 1L/17 sec = 58.82 $cm^3\, s^{-1}$
+    - Total time to fill the reservoir = (4:54 - 0:21) = 1L/473 sec
+
+ ## Discussion
+   - Here is the graph of d vs q:
+   ![d vs q](d_vs_q.bmp)
+
+   - Our data points are almost in a good agreement with the theoretical prediction of $d \propto q^{2/3}$. Discrepancies in the plot are due to error in our measurements. 
+   - It takes 473 sec for the water to start spilling over the obstacle. We can use the dimensions of the tank and inflow rate to estimate this time. Theoretical calculation to calculate time taken to fill the tank is 
+   - Time = Volume ($cm^3$)/ q ($cm^3 s^{-1}$) = l*b*h/q = 70 * 26.5 * 11.8 / 58.82 $\approx$ 372 s. There is a discrepancy in the theoretical tie and the calculated time because we have not precisely measured the dimensions and the inflow rate.
+
+   - When we decrease the inflow rate, it takes some time for the system to attain the equilibrium as d will decrease as a result of decreasing q as a result of mass conservation.
+
+   - When we used a shallower object for the same system, we did not see much change probably because the change in the ddepth of the object was quite small to make a noticeable change.
+
+   - Also when we filled the downstream reservoir so the water level is higher there, again we couldn't see any noticeable change. 

docs/lab_demos24/IssaMaoPathak/HydrostaticPressure/index.md

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+# Lab Demo: Hydrostatic Pressure and Nozzle Velocity
+
+## Theory
+
+In this experiment, we are considering the scenario where water is filled in a rectangular tank up to a height H. There is a small circular hole at the bottom of the tank through which water is exiting. We assume that the water is incompressible. The flow of water out of the tank can be analyzed using principles of fluid dynamics.
+
+### Bernoulli’s Equation Applied to the System
+
+For an incompressible fluid, Bernoulli’s equation along a streamline from the water surface (point 1) to the hole at the bottom (point 2) is:
+
+$$ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 $$
+
+Given:
+- $P_1 = P_2 = P_{\text{atm}}$, the atmospheric pressure at both the water surface and the exit of the hole.
+- $h_1 = H$, the height of the water column.
+- $h_2 = 0$, considering the reference level at the hole.
+- $v_1$ and $v_2$ are the velocities of water at the surface and at the hole, respectively.
+
+Simplifying the equation by canceling out $P_{\text{atm}}$ and rearranging terms:
+
+$$ \frac{1}{2}\rho v_1^2 + \rho gH = \frac{1}{2}\rho v_2^2 $$
+
+### Continuity Equation
+
+The continuity equation for an incompressible fluid states that the volume flow rate must be constant throughout the flow:
+
+$$ A_1v_1 = A_2v_2 $$
+
+Where:
+- $A_1$ is the cross-sectional area of the tank.
+- $A_2$ is the cross-sectional area of the hole.
+- $v_1$ and $v_2$ are the velocities of water at the surface and at the hole, respectively.
+
+### Solving the Equations
+
+From the continuity equation, we can express $v_1$ in terms of $v_2$:
+
+$$ v_1 = \frac{A_2}{A_1}v_2 $$
+
+Substituting $v_1$ in Bernoulli’s equation:
+
+$$ \frac{1}{2}\rho \left(\frac{A_2}{A_1}v_2\right)^2 + \rho gH = \frac{1}{2}\rho v_2^2 $$
+
+Solving for $v_2$, we get:
+
+$$ \frac{1}{2}\rho \frac{A_2^2}{A_1^2}v_2^2 + \rho gH = \frac{1}{2}\rho v_2^2 $$
+
+$$ \left(\frac{1}{2}\frac{A_2^2}{A_1^2} + \frac{1}{2}\right)\rho v_2^2 = \rho gH $$
+
+$$ \left(\frac{A_2^2 + A_1^2}{A_1^2}\right)\frac{1}{2}\rho v_2^2 = \rho gH $$
+
+$$ v_2^2 = \frac{2gHA_1^2}{A_2^2 + A_1^2} $$
+
+$$ v_2 = \sqrt{\frac{2gHA_1^2}{A_2^2 + A_1^2}} $$
+
+
+
+### Taking the Small Area Limit
+
+Now, if we consider the limit where $A_1 >> A_2$, the term $A_2^2$ becomes negligible compared to $A_1^2$, and the equation simplifies to:
+
+$$ v_2 \approx \sqrt{2gH} $$
+
+This result aligns with the simplified case where $v_1 \approx 0$ (assuming the area of the tank is much larger than the area of the hole). It shows how the velocity of water exiting the hole is primarily determined by the height of the water column, assuming the hole's area is much smaller than the tank's cross-sectional area.
+
+To find the distance the water travels after exiting the small hole at the bottom of the tank, we can use the equation for projectile motion. When the water exits the hole, it behaves like a projectile under the influence of gravity. Given that the tank is above the height $H_{\text{tank}}$ above the ground, the total distance to the ground from the hole is $H_{\text{tank}}$.
+
+### Projectile Motion
+
+When the water exits the hole, it has a horizontal velocity component $v_{2}$ (as derived earlier) and a vertical component that is initially 0 since it exits horizontally. The distance $R$ (range) that the water travels can be found using the formula for the range of a projectile launched from a height:
+
+$$ R = v_{2} \times t $$
+
+Where $t$ is the time it takes for the water to hit the ground. To find $t$, we use the equation for the vertical motion:
+
+$$ H_{\text{tank}} = \frac{1}{2}gt^2 $$
+
+Solving for $t$, we get:
+
+$$ t = \sqrt{\frac{2H_{\text{tank}}}{g}} $$
+
+Substituting the value of $v_{2}$ from the previous derivation:
+
+$$ v_{2} = \sqrt{\frac{2gH}{A_2^2 + A_1^2/A_1^2}} $$
+
+And considering the limit where $A_1 >> A_2$, simplifying $v_{2}$ to:
+
+$$ v_{2} \approx \sqrt{2gH} $$
+
+The range $R$ can now be calculated as:
+
+$$ R = \sqrt{2gH} \times \sqrt{\frac{2H_{\text{tank}}}{g}} $$
+
+$$ R = \sqrt{4H \cdot H_{\text{tank}}} $$
+## Procedure:
+
+1. **Filling the Reservoir:**
+   - We filled the large orange-colored reservoir with water.
+   - The reservoir dimensions were: 
+     - Length: $70\,cm$
+     - Breadth: $26.5\,cm$
+   - Diameter of the smaller opening at the bottom of the tank $\approx 0.6\,cm$
+
+2. **Initiating Water Flow:**
+   - We opened the bottommost screw, positioned approximately 1 cm above the tank's bottom, to allow water to flow out.
+
+3. **Recording Measurements:**
+   - As the tank emptied, we recorded the following measurements multiple times:
+     - Height of the water relative to the hole.
+     - Distance from the hole to where the water squirts before striking the ground.
+   - We also measured $H_{\text{tank}} = 25 cm$.
+## Results
+
+- The experimental setup is illustrated in the figure below, depicting the process where water ejects through a smaller aperture and undergoes projectile motion until impacting a metallic scale positioned beneath to ascertain the projectile's range. Concurrently, another transparent scale is utilized for gauging the water level within the tank.
+![Experimental Setup](range.png)
+
+- The entire sequence of our experiment is documented in a video available at the following link: [Experiment Video](https://drive.google.com/file/d/1liYaju8dAr2Ju6-Raz8xzO9SSOhJ7lc2/view?usp=drive_link).
+
+- The subsequent figure shows the relationship between the distance from the aperture to the point of impact and the water level at various instances.
+![Distance vs. Water Level Plot](plot.bmp)
+
+## Discussion
+
+The experiment conducted to examine hydrostatic pressure and nozzle velocity delved into the fluid dynamics governing the egress of water through a diminutive orifice situated at the tank's base. Predicated on Bernoulli's equation and the continuity equation, the endeavor aimed to elucidate the exit velocity of water and its ensuing projectile motion.
+
+Contrary to theoretical predictions suggesting a correlation of the form $y = A\sqrt{x}$ between water level and projectile distance, empirical findings diverged from this model. Potential sources of discrepancy include:
+
+- The dispersion of water upon impact, forming a circular pattern on the ground, was not meticulously accounted for when measuring projectile ranges, possibly skewing data.
+
+- Height measurements of the water within the tank were conducted using a transparent scale, where refraction effects complicated the accurate determination of water surface height from the tank's bottom.
+
+- A methodological enhancement to reduce measurement error could have been the averaging of multiple projectile range readings at consistent water levels.
+
+- The range of water level heights ($H$) explored was confined to 4-9 cm. A broader spectrum of measurements, particularly at elevated levels, might have yielded a trend more congruent with theoretical projections of $y = A \sqrt{x}$. The experimental design precluded higher water level assessments due to the configuration of an obstruction at the tank's terminus.
+
+## Conclusion
+
+The laboratory demonstration focusing on hydrostatic pressure and nozzle velocity aimed to investigate the dynamics of water flowing through a small aperture at the bottom of a tank, utilizing principles of fluid dynamics and the application of Bernoulli's and continuity equations. 
+
+Experimental findings, however, revealed discrepancies between theoretical predictions and observed data. These discrepancies could be attributed to several factors, including measurement inaccuracies due to water dispersion upon exit, the effects of refraction on height measurements, and the limited range of water levels tested. Despite these challenges, the experiment offered valuable insights into the complexities of fluid dynamics in practical applications.
+
+Recommendations for future experiments include adopting more precise measurement techniques to account for dispersion and refraction effects, and exploring a broader range of water heights to fully capture the relationship between water column height and exit velocity. Additionally, further investigation into the effects of varying aperture sizes on exit velocity and projectile motion could provide deeper understanding of fluid dynamics principles in real-world scenarios.
+
+
+

docs/lab_demos24/IssaMaoPathak/index.md

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+---
+layout: default
+title: Issa, Mao, Pathak
+---
+
+
+- [Stability](./stability/)
+- [Hydrostatic Pressure](./HydrostaticPressure/)
+- [Hydraulic hysteresis](./Hydraulic_hysteresis/)
+- [Ventuir](./venturi/)

docs/lab_demos24/IssaMaoPathak/siphon.md

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+# Siphon
+Repeat the experiment but with a hose that goes higher than the surface of the reservoir, but outlets below the surface of the reservoir.
+## Experiment Setup
+We established an upstream reservoir (the orange tank) to estimate the velocity of the siphon by measuring the projectile motion distance of the water flow from the siphon's outlet. As a control group in the experiment, a nozzle is positioned at the bottom of the upstream reservoir. The relationship between the nozzle's outlet water flow velocity and the water level was previously discussed.
+
+## Theory
+
+There are two main theories regarding how a siphon operates: the atmospheric pressure theory and the cohesion tension theory. The atmospheric pressure theory suggests that atmospheric pressure drives the water up the siphon, a concept that fails to explain siphon functionality in a vacuum. The cohesion tension theory proposes that water is pulled up the siphon like a chain, although it also faces challenges.
+
+Bernoulli's equation provides a close approximation of flow in an idealized, friction-free siphon operation. Considering point 1 at the water surface and point 2 at the siphon's outlet, we have:
+
+$$\frac{1}{2} v_1^2 + gz_1 + \frac{P_1}{\rho} = \frac{1}{2} v_2^2 + gz_2 + \frac{P_2}{\rho}$$
+
+where $v_1$ is the velocity at the surface, $z_1$ is the height of the water surface, $P_1$ is the pressure at the surface, $v_2$ is the velocity at the outlet, $z_2$ is the height of the outlet, and $P_2$ is the pressure at the outlet. Given that the siphon opens to the atmosphere, we have $P_1 = P_2 = 0$, and $v_1 = 0$. Thus:
+
+$$\frac{1}{2}  v_2^2 = g (z_1 - z_2)$$
+
+Solving for $v_2$, we find:
+
+$$v_2 = \sqrt{2 g (z_1 - z_2)}$$
+
+## Experiment Goals
+
+Our objective is to set up a siphon alongside the nozzle, aligning the outflows at the same height, to compare their velocities by measuring the distance water travels through the air. We aim to test the hypothesis that:
+1. The outflow velocity of the siphon matches that of the nozzle.
+2. The outflow velocity of the siphon is independent of the inlet height.
+
+## Experiment Results
+
+In the experiment, we positioned the siphon and the nozzle at the same height and observed that the siphon's outflow velocity was less than that of the nozzle, as indicated by the shorter air travel distance of the water. We also encountered difficulties in simultaneously measuring the outflow velocities of both the siphon and the nozzle due to alignment and measurement tool limitations.

docs/lab_demos24/IssaMaoPathak/stability/index.md

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+# Pressure and Buoyancy
+
+For this part of the lab, we tested the stability to tipping for multiple objects.
+1. Styrofoam cup (with and without an added weight)
+2. A 3-D printed object with a heavy metal carriage acting as a ballast
+3. A ball with an elongated helmet
+4. A cylinder
+
+In the following video, the stability of each of the objects is showcased.
+
+<video src="stability.mov" width="320" height="240" controls></video>
+
+As shown in the video, the styrofoam cup has three points where it will float:
+i) the larger opening ii) the smaller bottom iii) the side of the cup.
+The cup can be tipped at a larger angle and still return to the same point for the larger opening than the smaller bottom.
+This makes sense because there is a larger area for the cup to find equilibrium. 
+When the weight is added to the cup, it falls to a new stability point which is at an angle, rather than flatly on the side, because the extra weight is not fixed and shifts to push the cup under the water rather than float on top.
+
+Next, the object with the ballast has its stability point when the ballast is in the water. 
+Even if the object was completely flipped on its back, it rotates around so that the ballast will be underneath again.
+The ballast keeps the object from tipping over, although not displayed in the video, such that even when pushing down on the side of the boat it will return back to its stability point. 
+The reason that the ballast works is because it is heavy enough to shift the centre of gravity lower than the object, meaning that any torques that object + ballast experience are around the ballast, which wants to sit flatly in the water.
+
+The ball with the helmet has a stability point where the ball is in the water and the helmet is sitting flatly.
+The object will even undergo multiple rotations to find this position, as shown in the video. 
+No other point could be found where the object was stable, it always wanted to return to this stability point no matter its orientation.
+This is due to the geometry of the helmet and that the ball is acting similarly to a ballast.
+
+Finally, we have the cylinder. The major stability point for this object was on its side, but for very small angles we got it to float on its surface once (although not shown in the video). 
+Another thing we tested, although forgot to film, was adding a ballast to one side of the cylinder.
+When placed on its side or the opposite face, the object would rotate so that the ballast would be floating under the water and allow for the cylinder to remain upright.