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Chapter 3: Problem 60

An inclined manometer is a useful device for measuring small pressure differences. The formula given in Section 3.4 for the pressure difference in terms of the liquid-level difference \(h\) remains valid, but while \(h\) would be small and difficult to read for a small pressure drop if the manometer were vertical, \(L\) can be made quite large for the same pressure drop by making the angle of the inclination, \(\theta,\) small. (a) Derive a formula for \(h\) in terms of \(L\) and \(\theta\) (b) Suppose the manometer fluid is water, the process fluid is a gas, the inclination of the manometer is \(\theta=15^{\circ},\) and a reading \(L=8.7 \mathrm{cm}\) is obtained. What is the pressure difference between points? and?? (c) The formula you derived in Part (a) would not work if the process fluid were a liquid instead of a gas. Give one definite reason and another possible reason.

Short Answer

Expert verified
The pressure difference can be calculated using the formula \( \Delta p = pgL\sin(\theta)\). It results in \(226.8\:\)Pa for the given thickness, inclination and fluid. The formula wouldn't work if the fluid was a liquid because it is nearly incompressible.

Step by step solution

01

Derive a formula for h in terms of L and 胃

Starting with the relationship between \(h\) and \(L\) in a right-angled triangle, which is: \(h = L \sin(\theta)\). This formula was derived using the fact that sin(胃) is equal to the opposite side (which is h here) over the hypotenuse (which is L here) of a right-angled triangle.
02

Calculate the pressure difference

By observing that the pressure at the bottom of the liquid column must be the same, the pressure difference \( \Delta p \) between two points at the same level in connected tubes is given by \( \Delta p = pgL\sin(\theta)\), where \(p\) is the density of the liquid, \(g\) is the acceleration due to gravity and \(L\) and \(\theta\) are as defined previously. For water \(\rho_w = 1000 \: kg/m^3\), \(g = 9.81 m/s^2\), \(L= 8.7cm = 0.087m\) and \(\theta = 15^{\circ}\) we just need to plug the numbers into the formula to find \(\Delta p = pgL\sin(\theta) = 1000 \cdot 9.81 \cdot 0.087 \cdot \sin(15) = 226.8 \: Pa\).
03

Explain why the formula won't work for a liquid

One definite reason that the formula would not work if the process fluid were a liquid instead of a gas is that gases are very much compressible i.e., their volume can be changed with the application of pressure but liquids are nearly incompressible. Hence, in the formula \(h = L \sin(\theta)\), \(h\) tends to be very small in case of a liquid as \(L\) and \(\sin(\theta)\) changes with change in volume are not significant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Measurement
Pressure measurement is an essential aspect of fluid mechanics, used to determine the force exerted by a fluid per unit area. Accurate pressure readings are crucial in many applications, including engineering, weather forecasting, and even medical tools like blood pressure monitors. Pressure can be measured in different units, such as Pascals, atmospheres, or psi (pounds per square inch), depending on the application and region. The most common instruments for measuring pressure include manometers, barometers, and pressure gauges. Each device uses different principles to assess fluid pressure accurately. Manometers are particularly used when precision is vital, such as in laboratory settings or detailed engineering assessments. They measure pressure by balancing a column of liquid against the pressure to be measured. This measurement reflects the physical principles of pressure equilibrium and fluid statics.
Understanding how to measure pressure precisely allows engineers and scientists to facilitate better designs and predictions in various fields.
Inclined Manometer
The inclined manometer is a specialized tool designed to measure small pressure differences with high sensitivity. Unlike traditional vertical manometers, inclined manometers are tilted at an angle, \(\theta\), allowing for more precise readings.The key advantage of an inclined manometer is its ability to amplify small changes in pressure. By inclining the measuring tube, the length change \(L\) becomes longer for the same pressure change, making it easier to read and more accurate overall. This inclination improves the readability especially when dealing with minimal pressure differences, such as those found in gases.In a typical inclined manometer setup, the liquid inside the tube doesn't require a large height change to indicate a pressure difference. This is because the length of the liquid column along the tube (which increases as the inclination decreases) provides a bigger scale for the interpretation of pressure differences.
  • Inclination angle \(\theta\) allows for better resolution.
  • Suitable for measuring minute pressure differences in gaseous samples.
  • Precision makes it ideal in lab environments where accuracy is crucial.
Pressure Difference Calculation
Calculating pressure differences accurately is crucial for many scientific and engineering tasks. An inclined manometer provides a straightforward method for determining these differences by utilizing the formula derived from basic trigonometry.The relationship, derived from the geometry of a right-angled triangle, gives us the formula: \(h = L \sin(\theta)\). Here, \(h\) represents the vertical height difference of the liquid column in the manometer, \(L\) is the length of the liquid along the inclined tube, and \(\theta\) is the inclination angle.
Additionally, the pressure difference, \(\Delta p\), is given by the equation: \[\Delta p = \rho g L \sin(\theta)\]where \(\rho\) is the density of the manometer liquid, and \(g\) is the acceleration due to gravity. By using this formula, specific pressure differences can be calculated accurately by substituting the appropriate values. It's crucial to note, however, that this formula is primarily useful when the process fluid is a gas. The compressibility of gases allows for the significant movement of the liquid in the manometer. In contrast, the near-incompressibility of a liquid would require different considerations in calculating pressure differences.
  • Precise \(\theta\) measurement enhances accuracy.
  • The formula applies effectively with gaseous fluids.
  • Different parameters necessary for liquid-based measurements.

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Most popular questions from this chapter

The chemical reactor shown below has a cover that is held in place by a series of bolts. The cover is made of stainless steel ( \(\mathrm{SG}=8.0\) ), is 3 inches thick, has a diameter of 24 inches, and covers and seals an opening 20 inches in diameter. During turnaround, when the reactor is taken out of service for cleaning and repair, the cover was removed by an operator who thought the reactor had been depressurized using a standard venting procedure. However, the pressure gauge had been damaged in an earlier process upset (the reactor pressure had exceeded the upper limit of the gauge), and instead of being depressurized completely, the vessel was under a gauge pressure of 30 psi. (a) What force ( \(\left(\mathrm{b}_{\mathrm{f}}\right)\) were the bolts exerting on the cover before they were removed? (Hint: Don't forget that a pressure is exerted on the top of the cover by the atmosphere.) What happened when the last bolt was removed by the operator? Justify your prediction by estimating the initial acceleration of the cover upon removal of the last bolt. (b) Propose an alteration in the turnaround procedure to prevent recurrence of an incident of this kind.

A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5\% and 15\%. A mixture containing 9.0 mole\% methane in air flowing at a rate of 7.00 \(\times 10^{2} \mathrm{kg} / \mathrm{h}\) is to be diluted with pure air to reduce the methane concentration to the lower flammability limit. Calculate the required flow rate of air in mol/h and the percent by mass of oxygen in the product gas. (Note: Air may be taken to consist of \(\left.21 \text { mole } \% \mathrm{O}_{2} \text { and } 79 \% \mathrm{N}_{2} \text { and to have an average molecular weight of } 29.0 .\right)\)

Coal being used in a power plant at a rate of \(8000 \mathrm{lb}_{\mathrm{m}} / \mathrm{min}\) has the following composition: $$\begin{array}{lc}\hline \text { Component } & \text { Weight } \% \text { (dry basis) } \\ \hline \text { Ash } & 7.2 \\\\\text { Sulfur } & 3.5 \\\\\text { Hydrogen } & 5.0 \\\\\text { Carbon } & 75.2 \\ \text { Nitrogen } & 1.6 \\\\\text { Oxygen } & 7.5 \\\\\hline\end{array}$$ In addition, there are \(4.58 \mathrm{lb}_{\mathrm{m}} \mathrm{H}_{2} \mathrm{O}\) per \(\mathrm{lb}_{\mathrm{m}}\) of coal. Determine the molar flow rate of each element in the coal (including water) other than ash.

An open-end mercury manometer is connected to a low-pressure pipeline that supplies a gas to a laboratory. Because paint was spilled on the arm connected to the line during a laboratory renovation, it is impossible to see the level of the manometer fluid in this arm. During a period when the gas supply is connected to the line but there is no gas flow, a Bourdon gauge connected to the line downstream from the manometer gives a reading of 7.5 psig. The level of mercury in the open arm is \(900 \mathrm{mm}\) above the lowest part of the manometer. (a) When the gas is not flowing, the pressure is the same everywhere in the pipe. How high above the bottom of the manometer would the mercury be in the arm connected to the pipe? (b) When gas is flowing, the mercury level in the visible arm drops by \(25 \mathrm{mm}\). What is the gas pressure (psig) at this moment?

A storage tank containing oil ( \(\mathrm{SG}=0.92\) ) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of oil it contains can be determined from the gauge pressure at the bottom. (a) A pressure gauge connected to the bottom of the tank was calibrated with the top of the tank open to the atmosphere. The calibration curve is a plot of height of oil, \(h(\mathrm{m}),\) versus \(P_{\text {geuge }}(\mathrm{kPa}) .\) Sketch the expected shape of this plot. What height of oil would lead to a gauge reading of \(68 \mathrm{kPa} ?\) What would be the mass (kg) of oil in the tank corresponding to this height? (b) An operator observes that the pressure gauge reading is 68 kPa and notes the corresponding liquid height from the calibration curve. What he did not know was that the absolute pressure above the liquid surface in the tank was \(115 \mathrm{kPa}\) when he read the gauge. What is the actual height of the oil? (Assume atmospheric pressure is \(101 \mathrm{kPa}\).)
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