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Created: 2024/12/20 22:20
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# Experiment No.: 02
# Experiment Name: Flow Through a Venturimeter
# Objective
The objective of this experiment is to,
1. Determine the coefficient of discharge $C_d$.
2. Plot $Q_a$ against $H$ on log-log graph paper to determine the value of $n$ and $C$.
3. Calibrate the meter by determining actual coefficient of discharge $C_d$.
# Scope
The scope of venturimeter and this experiment includes:
1. Demonstrate the relation between flow rate ($Q_a$) and pressure head loss ($H$).
2. Measure flow in pipelines using the same meter for future uses of both ideal and real fluids.
3. Application in industries such as water supply systems, oil and gas processing.
4. Limitation includes high installation cost and permanent frictional loss in the pipeline.
# Datasheet and graph
# Calculation
From the two points ($P_1$ and $P_2$) on log-log graph, we get:
1. $H_1 = 2.0\times 10^{-1}$, $Q_1 = 3.1\times 10^{-4}$ and
2. $H_2 = 4.4\times 10^{-1}$, $Q_2 = 4.6\times 10^{-4}$
The exponent of $H$ can be calculated using the formula,
$$
n = \dfrac{\log{\dfrac{Q_2}{Q_1}}}{\log{\dfrac{H_2}{H_1}}}
$$
Applying the value of $Q_1$, $Q_2$, $H_1$, and $H_2$, we can get,
$$
\space n = \dfrac{\log{\dfrac{4.6\times 10^{-4}}{3.1\times 10^{-4}}}}{\log{\dfrac{4.4\times 10^{-1}}{2.0\times 10^{-1}}}} = 0.501
$$
The calibration parameter $C$ can be calculated using the formula,
$$
C = 10^{\log{Q} - n\log{H}}
$$
Applying the value of $n$, $Q = Q_2$ and $H = H_2$, we can get,
$$
C = 10^{\log{4.6\times 10^{-4}} - 0.501\times \log{4.4\times 10^{-1}}} = 6.940\times 10^{-4}
$$
Now, the cross-sectional area of the intel,
$$
A_1 = \pi(\dfrac{0.0284}{2})^2 = 6.335\times 10^{-4}\space m^2
$$
Comparing with the calibration equation $Q_a = CH^n$ and $Q_a = C_dKH^n$, the average coefficient of discharge can be calculated using,
$$
C_d = \dfrac{C}{K} = \dfrac{C}{A_1K_1}
$$
Applying the value of $C$, $A_1$ and $K_1$, we can get,
$$
C_d = \dfrac{6.940\times 10^{-4}}{6.335\times 10^{-4}\times 1.109} = 0.988
$$
# Discussion
In this experiment, the venturimeter was calibrated by determining the average coefficient of discharge $C_d$ and the exponent $n$. The values of average $C_d$ and $n$ were determined to be $0.988$ and $0.501$, respectively. These constants are unique to the venturimeter used. Now, the actual discharge can be calculated by only measuring pressure head loss, $H$ using this formula:
$$
Q_a = C_dKH^n = CH^n = 6.940\times 10^{-4}H^{0.501}
$$
## Probable errors
There are several potential sources of error when performing this experiment:
1. The experiment involves the use of fluid friction apparatus, venturimeter, stopwatch — damaged or poorly calibrated equipment can lead to inaccuracies.
2. There can be inaccuracies when taking manometer readings due to parallax errors.
3. Human errors include delays in starting or stopping the stopwatch.
4. Presence of air bubbles in the manometer can affect pressure head readings.
## Formation of cavitations
Cavitation is the formation of liquid bubbles when pressure in a liquid drops too low. While it was not observed, it is highly possible that cavitation formed near the throat region when performing this experiment.
It typically occurs in region of high velocity (and thus low pressure). As pressure drops below the liquid's vapor pressure, vapor bubbles form in the fluid. And when these bubbles move to region with higher pressure, they implode violently.
## Practical vs experimental results
The experiment yielded values of discharge coefficient, $C_d$, ranging from $0.916$ to $1.034$. The theoretical expectation is that $C_d$ should be below and close to $1$ due to frictional losses:
$$
H = H_{actual} + H_L
$$
Here, $H_L$ is the head loss due to frictions. Hence, the $H$ we considered during calculation of theoretical discharge ($Q_t$) was slightly higher than actual value causing some $C_d$ values to drop below $1$.
On the other hand, the $C_d$ had exceed $1$ possibly due to:
1. Misreading of pressure heads.
2. Poorly calibrated or damaged apparatus.
3. Errors in measurement of $Q_a$ such as timing errors.
However, the average discharge coefficient was determined to be $0.988$. It aligns with the concept of frictional losses and can be used to produce near accurate flow rate.
# Assignments
## Diverging angle smaller than the converging angle
The diverging angle is kept smaller to prevent flow separation and turbulence. A gradual divergence minimizes energy loss by ensuring smooth conversion of velocity head back into pressure head.
## Increase accuracy of venturimeter
The accuracy of venturimeter can be increased by:
1. Properly calibrate and maintaine the apparatuses.
2. Take extra caution when taking manometer readings to avoid parallax errors.
3. Similarly take extra caution when stopping or starting stopwatch.
4. Avoid air bubbles.
## Factors that meter coefficient depend on
The meter coefficient ($C_d$) depends on:
1. Shape of venturimeter — throat size and angle. Large meter may produce around 0.99 while smaller ones produce 0.97 to 0.98.
2. Fluid properties — viscosity and density.
3. Surface roughness causing frictional losses.
## Cavitation and its effect
Cavitation occurs when the fluid pressure drops below its vapor pressure, forming vapor bubbles. It typically occurs near the throat of a venturimeter, where velocity is high (and therefore, pressure is low). These bubbles collapse when they move to regions of higher pressure, causing damage and disrupting flow measurements.
Cavitation can be avoided by ensuring throat pressure remains above the fluid’s vapor pressure by controlling flow velocity and avoiding sharp reductions in cross-sectional area.
## Mathematical problem
Given,
- $d_1 = 10\space in = 0.2540\space m$
- $d_2 = 3\space in = 0.0766\space m$
- $H = 7\space in = 0.1778\space m$
Therefore area,
$$
A_1 = \pi(\frac{d_1}{2})^2 = \pi(\frac{0.2540}{2})^2 = 0.0507\space m^2
$$
Now,
$$
Q_t = A_1\sqrt{\frac{2g}{(\frac{d_1}{d_2})^4-1}}\times H^\frac{1}{2} = 0.0507\sqrt{\frac{2\times 9.8}{(\frac{0.2540}{0.0766})^4 - 1}}\times 0.1778^{0.5} = 8.644\times 10^{-3}\space m^3/s
$$
Hence,
$$
C_d = \frac{Q_a}{Q_t} = \frac{1\times 10^{-3}}{8.644\times 10^{-3}} = 0.1157
$$