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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 little-known rare earth element nauseum (atomic weight \(=172\) ) has the interesting property of being completely insoluble in everything but 25 -year- old single-malt Scotch. This curious fact was discovered in the laboratory of Professor Ludwig von Schlimazel, the eminent German chemist whose invention of the bathtub ring won him the Nobel Prize. Having unsuccessfully tried to dissolve nauseum in 7642 different solvents over a 10 -year period, Schlimazel finally came to the \(30 \mathrm{mL}\) of The Macsporran that was the only remaining liquid in his laboratory. Always willing to suffer personal loss in the name of science, Schlimazel calculated the amount of nauseum needed to make up a 0.03 molar solution, put the Macsporran bottle on the desk of his faithful technician Edgar P. Settera, weighed out the calculated amount of nauseum and put it next to the bottle, and then wrote the message that has become part of history: "Ed Settera. Add nauseum/" How many grams of nauseum did he weigh out? (Neglect the change in liquid volume resulting from the nauseum addition.)

A mixture of methanol (methyl alcohol) and water contains \(60.0 \%\) water by mass. (a) Assuming volume additivity of the components, estimate the specific gravity of the mixture at \(20^{\circ} \mathrm{C} .\) What volume (in liters) of this mixture is required to provide 150 mol of methanol? (b) Repeat Part (a) with the additional information that the specific gravity of the mixture at \(20^{\circ} \mathrm{C}\) is 0.9345 (making it unnecessary to assume volume additivity). What percentage error results from the volume- additivity assumption?

An object of density \(\rho_{\mathrm{a}},\) volume \(V_{\mathrm{a}},\) and weight \(W_{\mathrm{a}}\) is thrown from a rowboat floating on the surface of a small pond and sinks to the bottom. The weight of the rowboat without the jettisoned object is \(W_{\mathrm{b}}\). Beforethe object was thrown out, the depth of the pond was \(h_{\mathrm{pl}}\), and the bottom of the boat was a distance \(h_{\mathrm{b} 1}\) above the pond bottom. After the object sinks, the values of these quantities are \(h_{\mathrm{p} 2}\) and \(h_{\mathrm{b} 2}\). The area of the pond is \(A_{\mathrm{p}}\); that of the boat is \(A_{b} . A_{b}\) may be assumed constant, so that the volume of water displaced by the boat is \(A_{\mathrm{b}}\left(h_{\mathrm{p}}-h_{\mathrm{b}}\right)\). (a) Derive an expression for the change in the pond depth \(\left(h_{\mathrm{p} 2}-h_{\mathrm{p} 1}\right) .\) Does the liquid level of the pond rise or fall, or is it indeterminate? (b) Derive an expression for the change in the height of the bottom of the boat above the bottom of the pond \(\left(h_{b 2}-h_{b 1}\right) .\) Does the boat rise or fall relative to the pond bottom, or is it indeterminate?

The feed to an ammonia synthesis reactor contains 25 mole \(\%\) nitrogen and the balance hydrogen. The flow rate of the stream is \(3000 \mathrm{kg} / \mathrm{h}\). Calculate the rate of flow of nitrogen into the reactor in \(\mathrm{kg} / \mathrm{h}\). (Suggestion: First calculate the average molecular weight of the mixture.)

Limestone (calcium carbonate) particles are stored in \(50-\mathrm{L}\) bags. The void fraction of the particulate matter is 0.30 (liter of void space per liter of total volume) and the specific gravity of solid calcium carbonate is 2.93. (a) Estimate the bulk density of the bag contents ( \(\mathbf{k g}\) CaCO \(_{3}\) /iter of total volume). (b) Estimate the weight ( \(W\) ) of the filled bags. State what you are neglecting in your estimate. (c) The contents of three bags are fed to a ball mill, a device something like a rotating clothes dryer containing steel balls. The tumbling action of the balls crushes the limestone particles and turns them into a powder. (See pp. \(21-64\) of Perry's Chemical Engineers' Handbook, 8 th ed.) The limestone coming out of the mill is put back into \(50-\mathrm{L}\) bags. Would the limestone (i) just fill three bags, (ii) fall short of filling three bags, or (iii) fill more than three bags? Briefly explain your answer.
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