FFR - More on Fluids
Original textbook reading.
Enrichment beyond the AP course.
Gas equations, non-ideal gases, head loss and fluid momentum are extra exploration. AP Physics 1 uses ideal incompressible-fluid models; the buoyancy recap remains a useful reference.
In this section, we will learn about other relevant topics that are related to our chapter of Fluids. In this section we will see how the Ideal Gas Laws are actually not that “Ideal” and also that fluids flowing through a pipe actually has some resistance to it.
In section of this chapter, we covered the concepts behind ideal gas laws. Well, those laws only work in “ideal” situations. What do we mean by ideal? The ideal gas laws work under the assumptions that:
- The collisions between the gas molecules and the wall are completely elastic, therefore there is no loss in kinetic energy.
- That gas molecules take up no volume
- That there are no attractive or repulsive forces-also known as Van der Waals forces: felt between molecules.
At higher temperature, thermal kinetic energy can make intermolecular attractions less important relative to molecular motion. Attractions do not cease to exist.
Ideal-gas behavior is often a good approximation at low density and temperatures well above condensation. Molecular volume then occupies a small fraction of the container, and intermolecular potential energies are small compared with thermal kinetic energy.
These non ideal situations of gases occur in two scenarios: one at low temperature and high pressure.
At low temperature, intermolecular attraction can be more significant relative to thermal kinetic energy, and condensation may occur. The collision rate also depends on density and molecular speed.
At high pressures, the volume is really small, so you would be very concerned about the volume of your gas molecules.
We already know that the ideal gas law utilizes Equation 8.1.1 We know what you’re thinking, if there is an ideal gas law equation, what is the non-ideal gas equation. Well, such an equation exists to help characterize how real gas molecules act, and it is known as the van der Waals equation:
In the van der Waals model, a(n/V)² corrects for attractions that reduce the measured pressure, and nb is an excluded-volume correction. The constants a and b depend on the gas; b is not simply one molecule’s literal volume per mole.
Here water is flowing through a pipe with no head loss, so it is flowing naturally
Head loss is a reduction in mechanical energy per unit weight of flowing fluid, often expressed as a length. Friction in straight pipe and local disturbances at fittings, valves, entrances, and bends can dissipate energy.
- Frictional head loss is mechanical energy dissipated by viscous flow through pipe. In a uniform pipe carrying steady incompressible flow, continuity keeps mean speed constant; a pressure gradient supplies the dissipated energy.
- Minor head losses are local energy losses at entrances, exits, bends, valves, and other fittings. They depend on flow speed and the geometry of the disturbance; the word minor does not mean they are always smaller than straight-pipe losses.
Figure 8.4.1 Fluid Momentum
Fluid momentum is exactly what it sounds like. It means that fluids also have an associated momentum, just like objects with a mass do. Water coming out of a hose has a a stream of water coming out with some velocity. At the same time, the water has an associated mass per unit time exiting the hose. If you were to put your hand to partially cover the opening on the hose, the water velocity would increase. Imagine that water hitting your bare skin. It would hurt right? That’s because that fluid has an associated momentum with it.
A fluid parcel has momentum mv. In steady flow, mass flow rate is ṁ = ρQ, and momentum transported per unit time in a uniform jet is ṁv. The net force depends on the change in momentum flux. A smaller outlet raises speed if Q is held fixed, but partly closing a real hose can also reduce Q.
Pop Quiz 8.4.1 Gas Laws, Head Loss and Fluid Momentum
Question 1 of 5 · 0 correct
Extra exploration · Read the worked solution
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Question 1
Question 2
Question 3
Which pictured route contains the most approximately right-angle bends? Use this as a qualitative illustration of local-loss sources; the images alone do not determine numerical head loss.
Question 4
1234Match these labels to the numbered figure targets. Write your matches before opening the separate worked answer.
Question 5
Key Note: Now, fluid momentum is a bit out of scope for most physics 1 classes as you usually go more in depth with it in a dedicated fluids class that you’ll take in your college curriculum, therefore we won’t cover how to solve it.
F_B = ρ_water V_object,submerged g (Equation 8.3.1)
These are magnitudes. Notice that buoyancy points upward, opposite to the weight of the displaced water: F⃗_B = −ρ_water V_object,submerged g⃗.
Original problem 1
The pressure is unusually low but is retained. The gas law requires absolute pressure; a gauge reading would require atmospheric pressure as another input.
Problem text
A 3 L tank contains an ideal gas at 35 °C and an absolute pressure of 30 Pa. Find the amount of gas in moles.
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Original problem 2
This is the ideal introductory model. Real acceleration in water also depends on hydrodynamic added mass and later on drag.
Problem text
A 10 kg sphere of radius 15 cm is fully submerged in water and held at rest by a string attached below it, as shown. Find the string tension and the initial acceleration just after it is cut. Use water density 1000 kg/m³ and g = 9.81 m/s²; neglect drag and added-mass effects.
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Original problem 4
Problem text
A block has weight 10.4 N in air and apparent weights 5.6 N in water and 6.5 N in gasoline. It is fully immersed in each liquid. Find the density of the block and gasoline. Neglect air buoyancy; use water density 1000 kg/m³.
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Original problem 3
The reviewed wording preserves 15 °C and both measured volumes, but treats the second bath as unnamed. If the initial temperature were explicitly changed to 0 °C, the same volume ratio would instead imply 163.89 °C, still not boiling fluorine. No actual bath temperature is inferred.
Problem text
A fixed amount of ideal xenon gas occupies 150 cm³ in a bath at 15 °C. At the same pressure in a second bath, it occupies 240 cm³. Calculate the second temperature. Check the source descriptions of an “ice-water bath at 15 °C” and “boiling liquid fluorine” against this result.
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Original problem 5
The result assumes the vertical equilibrium shown, not a claim that every possible orientation is stable.
Problem text
A cylindrical rod floats vertically in water. Its lower 5 cm has density 1500 kg/m³ and its upper 7 cm has density 400 kg/m³. Both sections have the same radius R. Find the depth of the rod’s bottom below the water surface. Use water density 1000 kg/m³.
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Original problem 6
“Top floor” suggests a possible height typo, but the image clearly says 40 cm. The main answer preserves it. The 40 m result is explicitly an alternative interpretation. If 4.6 atm is gauge pressure, the reported upper pressure is also gauge; if it is absolute, the result is absolute.
Problem text
Water at street level has pressure 4.6 atm and speed 0.09 m/s in an 8.0 cm diameter pipe. It rises 40 cm to a 3.4 cm diameter pipe. Find the upper speed and pressure, assuming steady incompressible flow, no head loss, water density 1000 kg/m³, and g = 9.81 m/s². Keep the pressure convention the same at both points.
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Original problem 7
The van der Waals result is a model estimate, not an exact experimental “non-ideal” pressure. The constants and their litre/atmosphere units are made explicit.
Problem text
Find the pressure of 0.7000 mol of helium in 0.4000 L at −30 °C, first as an ideal gas and then using the source table’s van der Waals constants a = 0.034 atm·L²/mol² and b = 0.024 L/mol. Use R = 0.0820574 L·atm/(mol·K).
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Original problem 8
The source does not specify a sprinkler design. Fixed spray heads need not rotate; the reaction-torque explanation applies to angled rotating arms.
Problem text
Explain how a water sprinkler produces jets and, for a rotating sprinkler, how its arms turn.
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Original problem 9
The very high stream speeds are retained as source data. This is a mass-conservation exercise, not a realistic hydraulic river model.
Problem text
Two streams merge without water storage or loss. Stream 1 has width 5.6 m, mean depth 4.5 m, and mean speed 24 m/s. Stream 2 has width 8.9 m, mean depth 7.8 m, and mean speed 15 m/s. Approximate each cross-section as rectangular. The downstream river has area 65 m². Find its mean speed.
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Original problem 10
The tank diameter, discharge coefficient, and landing height are omitted. Therefore the source does not determine an exact time-varying drain rate or a general maximum range. The numerical result is conditional on the stated large-tank idealization and initial 10 m head.
Problem text
An open cylindrical tank initially has a water surface 10 m above a side opening of diameter 0.9 m, as shown. Estimate the initial ideal volume flow rate and the horizontal distance of an initially horizontal jet during a given 3 s flight. Assume the tank area is much larger than the opening area, negligible losses, and g = 9.81 m/s².
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Reviewed physics notes (14)
These corrections keep the original lesson’s purpose while fixing errors in the source edition.
- High temperature does not eliminate collisions or finite molecular volume. — Reference 1
- Slow motion does not create attraction by providing “time” for it, and temperature alone does not determine collision count. — Reference 1
- Clarifies the pressure sign and meaning of the excluded-volume parameter. — Reference 1
- Head loss is energy dissipation, not a force or necessarily a progressive decrease of speed in a uniform steady pipe. — Reference 1
- The source says wall friction must progressively slow the steady flow in the pipe. — Reference 1
- A bend does not consume energy merely because a change in direction occurs. Loss is due to viscous dissipation and flow disturbances. — Reference 1
- The source conflates momentum with momentum rate and assumes a real nozzle restriction cannot change flow rate. — Reference 1
- Corrects the claim that hot molecules have no time to attract or repel. — Reference 1
- This frame repeats the no-head-loss caption despite depicting the minor-loss topic. — Reference 1
- Corrects “headless” and avoids suggesting speed must decline along a uniform steady pipe. — Reference 1
- Qualifies the fixed mass-flow assumption. — Reference 1
- Numerical minor head loss cannot be ranked without speeds, dimensions, and fitting loss coefficients. Image 2 remains the answer to the stated geometric comparison. — Reference 1
- Provides the source table constants and explicitly identifies a model estimate and absolute pressure. The marked 640 °C is numerically correct in this model. — Reference 1
- The fluid recap previously equated the buoyancy vector to the displaced-water weight vector. Their magnitudes are equal but their directions are opposite. The recap now uses scalar magnitudes and states F⃗_B = −ρ V g⃗, consistent with the retained buoyancy lesson. — Reference 1, Reference 2
Glossary and deeper reading · 17 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Fluid
A fluid is a material that flows and continuously deforms under a sustained shear stress. Liquids and gases are both fluids, although their compressibility and other properties can differ greatly. For example, both water in a pipe and air flowing through a duct are fluids.
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Ideal gas
An ideal gas model treats particles as having negligible volume and no sustained intermolecular forces between brief elastic collisions. The equilibrium state obeys the ideal gas law using absolute pressure and kelvin temperature. For example, a dilute gas far from condensation often approximates this model better than a dense gas near a phase change.
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Mechanical work
Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.
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Kinetic energy
Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.
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Mass density
Mass density is mass per unit volume of a material or sample. It is commonly measured in kilograms per cubic meter and can vary with temperature or pressure. For example, equal-volume samples of water and oil generally have different masses because their densities differ.
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Temperature
Temperature is a state quantity that determines the direction of net energy transfer through thermal contact. For a classical ideal gas it is proportional to the average translational kinetic energy per particle, rather than to the total energy of the sample. For example, a small hot metal sample can have a higher temperature but less total internal energy than a large cooler sample.
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Pressure
Pressure is the normal force per unit area exerted on a surface. It is a scalar measured in pascals, with one pascal equal to one newton per square meter. For example, a gas exerts pressure throughout its volume through many microscopic collisions, even when its bulk motion is zero.
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Ideal gas law
The ideal gas law relates absolute pressure, occupied volume, amount of gas, and kelvin temperature. The forms using moles and using particle count express the same state relationship with different constants. For example, heating a fixed amount of ideal gas in a rigid sealed container raises its pressure.
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Mole
A mole is an amount of substance containing Avogadro’s number of specified entities. Those entities must be named, such as atoms, molecules, or ions, and a mole is not itself a mass unit. For example, one mole of helium atoms and one mole of oxygen molecules contain the same number of specified entities but have different masses.
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Head loss
Head loss describes a decrease in a flowing fluid’s mechanical energy per unit weight. Friction and fittings can produce losses that the ideal Bernoulli model omits. For example, a long rough pipe can require a greater pressure difference than a short smooth one to maintain the same flow.
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Mechanical energy
Mechanical energy is the sum of the kinetic and potential energies included in a model. It stays constant when transfers to or from other energy stores and across the chosen boundary are absent. For example, a pendulum’s mechanical energy can remain approximately constant while energy shifts between motion and gravitational configuration.
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Viscosity
Viscosity measures a fluid’s resistance to relative motion between neighboring layers. Viscous effects can dissipate mechanical energy and alter the velocity profile of flowing fluid. For example, honey flows more reluctantly than water under comparable conditions because it is more viscous.
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Incompressible fluid
An incompressible-fluid model treats density as constant despite pressure changes in the modeled situation. This approximation often works for liquids over modest pressure ranges but is not a universal property of all fluids. For example, water flowing through a narrowing pipe can often be treated as incompressible.
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Linear momentum
Linear momentum is a vector describing an object’s motion and equals mass times velocity in the nonrelativistic model. The vector sum of momenta is conserved when the net external impulse on the system is zero. For example, two identical carts moving equally fast in opposite directions have zero total linear momentum but nonzero total kinetic energy.
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Velocity
Velocity describes how position changes with time and includes direction. Instantaneous velocity describes one moment, while average velocity is displacement divided by elapsed time. For example, a car turning a corner can change velocity while maintaining the same speed.
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Net force
Net force is the vector sum of all forces acting on the object or system under consideration. For constant mass in an inertial frame, it equals mass times acceleration. For example, equal upward and downward forces give zero net force even though neither force is zero.
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Buoyancy
Buoyancy is the net upward force arising from pressure differences over an object in a fluid under gravity. Its magnitude equals the weight of the displaced fluid in the usual hydrostatic model. For example, a fully submerged object can rise, sink, or remain suspended depending on how buoyancy compares with its weight.