Ideal vs Real Fluid: The Difference

An ideal fluid is incompressible and has no viscosity. A real fluid changes density under pressure and loses energy to internal friction. AP Physics 1 states in a boundary statement and again on the equation sheet that fluids are assumed to be ideal unless a question says otherwise.

AP Physics: Unit 8 (topics 8.1 Internal Structure and Density, 8.4 Fluids and Conservation Laws). The definition on this page is AP Physics 1 essential knowledge 8.1.A.4, which states that an ideal fluid is incompressible and has no viscosity, and it is the only place the word viscosity appears in that course framework. Essential knowledge 8.2.A.3 supplies the consequence of incompressibility, stating that the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid. The assumption itself is stated in two places. Unit 8's only boundary statement, printed under Topic 8.4, reads that all fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated; Topics 8.1, 8.2 and 8.3 carry no boundary statement. The box of exam conventions on the AP Physics 1 equation sheet holds four bullets, of which the fourth reads that fluids are assumed to be ideal, and pipes are assumed to be completely filled by fluid, unless otherwise stated. A third list exists and does not include fluids: the CED appendix's discussion of models summarizes the idealizations in five bullets covering inertial frames, negligible air resistance, negligible frictional and drag forces, negligible edge effects of charged plates, and ideal strings, springs and pulleys. That appendix also states that the models are elaborated on within the boundary statements provided in the course frameworks, as well as in the conventions for the AP Exams listed on the equation sheets, and it names Bernoulli's equation as an example of a mathematical model while calling the model of a fluid flow as a steady flow of particles a conceptual model. The equations the assumption supports are 8.4.A.2's continuity equation for incompressible fluids, 8.4.B.2's Bernoulli equation described as conservation of mechanical energy in fluid flow, and 8.4.B.3's derived Torricelli result. Unit 8 is weighted at 10 to 15 percent of the multiple-choice section across a suggested 12 to 17 class periods. The AP Physics 2 framework spans Units 9 to 15 and contains no fluid topics, though its equation sheet prints the same Mechanics and Fluids panel and its own conventions box of nine bullets mentions no fluid assumption.

The distinction, stated once

The CED defines the ideal fluid in nine words. Essential knowledge 8.1.A.4 reads: an ideal fluid is incompressible and has no viscosity.

That is the whole definition, and it names exactly two properties. A real fluid has both of the things the definition removes.

Compressibility. Squeeze a real fluid hard enough and its density rises. Squeeze an ideal one and nothing happens: 8.2.A.3 states that the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid.
Viscosity. A real fluid resists being sheared. Layers moving past each other rub, energy leaves the flow as thermal energy, and pressure falls along a pipe even where nothing about the pipe changes. An ideal fluid has none of that.

So this is not a comparison of two substances. It is a comparison of a model with the thing it models, and the useful question is not which is correct but which terms the model deletes and what you are allowed to conclude once they are gone.

The CED is explicit that this is the right way to read it. Its appendix, discussing models, says that the models chosen to simplify the universe have been done so with alignment to their respective AP Physics courses, and that these models are elaborated on within the boundary statements provided in the course frameworks, as well as in the conventions for the AP Exams, listed on the equation sheets. It also names fluid flow as a modeling example: Bernoulli's equation is an example of a mathematical model, while the model of a fluid flow as a steady flow of particles is a conceptual model.

The practical upshot for an exam is short. Of the seven fluid lines on your equation sheet, two are definitions that hold for any fluid at all, $\rho = m/V$ and $P = F_{\perp}/A$ . The other five are written for the ideal one: $P = P_0 + \rho g h$ , $P_{\text{gauge}} = \rho g h$ and $F_b = \rho V g$ all need a density that does not vary, $A_1 v_1 = A_2 v_2$ needs incompressibility, and Bernoulli's equation needs incompressibility and zero viscosity together. None of the five carries a correction term you could apply to a real fluid.

Ideal against real, property by property

Property	Ideal fluid	Real fluid
Definition source	8.1.A.4, incompressible and no viscosity	anything actually in a pipe
Density under pressure	fixed, per 8.2.A.3	rises, by a little for liquids and a lot for gases
Internal friction	none	present, and it converts kinetic energy to thermal energy
Pressure along a straight pipe of constant area	unchanged	falls in the direction of flow
Continuity, $A_1 v_1 = A_2 v_2$	exact, per 8.4.A.2	needs the mass form, with the densities kept
Bernoulli's equation	an exact conservation statement, per 8.4.B.2	an approximation with a missing loss term
Pressure against depth	a straight line of slope $\rho g$	curves, because $\rho$ grows with depth
Speed out of a hole	$v = \sqrt{2g\Delta y}$ , per 8.4.B.3	slower, and it depends on the liquid
Does the answer depend on which liquid	only through $\rho$ , and often not even that	yes, strongly
On the AP sheet	five of the seven fluid lines are written for it	no equation on the sheet describes one

The last row is the one to keep. There is no viscosity symbol anywhere in the Mechanics and Fluids panel, in either algebra-based course, and none in that panel's symbol list either. The $\mu$ that does appear on the sheet is the coefficient of friction, in the left-hand column, and it has nothing to do with fluids. If a question wanted you to account for viscous loss it would have to hand you the machinery, and the sheet does not have any.

Row seven and row four are the two that surprise people, because both describe things a real fluid does that feel like they should be in the physics. Water at the bottom of a deep ocean genuinely is denser than water at the top, and water genuinely does lose pressure running along a hose. The model says neither happens, and the exam is written to the model.

Where the assumption is written down, and where it is not

AP Physics 1 states this assumption twice, in two different places, and the two do not say quite the same thing. Both are worth having in front of you.

In the course framework, as a boundary statement. Unit 8 carries exactly one boundary statement in the whole unit, and it is printed under Topic 8.4:

> All fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated.

Topics 8.1, 8.2 and 8.3 print none. That single sentence covers the unit.

On the equation sheet, in the box of exam conventions below the trigonometric table. The AP Physics 1 box holds four bullets in total:

The frame of reference of any problem is assumed to be inertial unless otherwise stated.
Air resistance is assumed to be negligible unless otherwise stated.
Springs and strings are assumed to be ideal unless otherwise stated.
Fluids are assumed to be ideal, and pipes are assumed to be completely filled by fluid, unless otherwise stated.

Four, and the fluids one is the fourth. So the assumption is in your hand during the exam, not only in a document you read in September.

There is a third list, and it is the reason to check rather than to recall. The appendix's discussion of models gives its own summary of what students may assume unless otherwise stated, and that list has five bullets: frames of reference are inertial; air resistance is negligible; frictional and drag forces are negligible; edge effects of charged plates are negligible; strings, springs, and pulleys are ideal. Fluids are not on it. The appendix is summarizing across the AP Physics courses rather than reproducing the Physics 1 conventions box, so the omission is not a contradiction, but anyone who learned the idealizations from that page alone would not know the fluid assumption exists.

Two further notes on the wording of the boundary statement, because both matter.

First, it has a second clause. All pipes are assumed to be completely filled by the fluid. That is a separate assumption from ideality, and it is what lets you use the pipe's cross-sectional area as the flow area in $A_1 v_1 = A_2 v_2$ without asking how full the pipe is. It also underwrites 8.4.A.1.i, which says the rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.

Second, it ends unless otherwise stated. A question is free to switch the assumption off, and if it does, the sentence that does so is the point of the question.

What each assumption buys you

Three assumptions, and each one is load-bearing for specific equations on your sheet. Knowing which is which tells you what a question is testing when it withdraws one.

Incompressible buys you a constant density. Once $\rho$ cannot change, three things follow.

$A_1 v_1 = A_2 v_2$ becomes exact. Conservation of mass says $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$ ; the densities cancel only because they are equal. 8.4.A.2 says as much in words, calling the continuity equation a statement of conservation of mass flow rate in incompressible fluids.
$P = P_0 + \rho g h$ is linear in depth, so a pressure against depth graph is a straight line with slope $\rho g$ and intercept $P_0$ . That is why 8.2's suggested skill 1.C, sketching qualitative graphs, is a reasonable thing to ask.
$F_b = \rho V g$ has one value of $\rho$ rather than a value that depends where in the fluid the object is sitting.

Zero viscosity buys you an exact energy conservation law. Bernoulli's equation, per 8.4.B.2, describes the conservation of mechanical energy in fluid flow. Mechanical energy is only conserved if nothing is draining it, and viscosity is the drain. With it gone:

The pressure at two points of equal area and equal height in a horizontal pipe must be equal, so a straight uniform pipe has no pressure drop along it at all.
The efflux speed from a hole, 8.4.B.3's derived $v = \sqrt{2g\Delta y}$ , contains neither $\rho$ nor any material property, so water, oil and mercury all leave the same tank at the same speed.
No question can ask you how much energy the flow lost, because in this model the answer is zero.

Completely filled pipes buy you a well-defined area. The $A$ in the continuity equation is the pipe's cross-section. If a pipe were half full the flow area would be some fraction of it that no printed equation gives you, and the free surface inside would need its own treatment.

Run the logic backwards and you have a fast way to read fluid questions. A question about flow rates is leaning on incompressibility. A question that gives you a pressure at one point and asks for it at another is leaning on zero viscosity. A question involving a tube or a hose is leaning on the pipe being full.

The case that separates them: the pipe with no pressure drop

Here is the single prediction that most cleanly divides the model from the substance, and you can check it against a garden hose.

Take a long horizontal pipe of constant cross-section carrying water steadily. Apply Bernoulli's equation between a point near the inlet and a point near the outlet. Same height, so the $\rho g y$ terms cancel. Same area, so continuity forces the same speed, and the $\frac{1}{2}\rho v^2$ terms cancel too. What is left is

P_1 = P_2

The ideal fluid says the pressure at the far end of a straight uniform pipe equals the pressure at the near end, however long the pipe is. No pump would be needed to keep water moving along a level pipe once it was already moving, because nothing is taking energy out of it.

Every real plumbing system is built around the fact that this is false. Pressure falls along a hose, the fall grows with length and with flow speed, and the loss is exactly the viscous dissipation the model deleted. Attach a longer hose to the same tap and the flow at the end weakens, which the ideal model forbids.

AP Physics 1 has no equation for that loss and does not ask you to compute it. What it does ask, and this is the exam-relevant half, is that you apply Bernoulli between two points and get a definite answer, which is only possible because the loss term is absent. When a Bernoulli question gives you $P_1$ at a wide section and asks for $P_2$ at a narrow one, it is not being generous with information. It is telling you that the model's bookkeeping is complete, so the geometry alone determines the answer.

The same structure appears in the efflux problem. Torricelli's result, $v = \sqrt{2g\Delta y}$ , is derived in 8.4.B.3 from conservation of energy, and it has no material property in it. Fill a tank with water and fill an identical tank with heavy oil, punch the same hole at the same depth in each, and the model says both jets leave at the same speed. Anyone who has poured honey knows that is not what happens. The prediction is not a small error, it is the model announcing which physics it left out.

When it costs a mark

Adding a viscous loss term to Bernoulli. There is none on the sheet, there is none in the CED, and a question that expects one would have to supply it. Writing an energy-loss term into a Bernoulli line loses marks and time.
Keeping the densities in the continuity equation. For an incompressible fluid they are equal, so $A_1 v_1 = A_2 v_2$ is the printed form. Carrying $\rho_1$ and $\rho_2$ separately is not wrong, but it invites you to imagine they might differ.
Missing the phrase that switches the assumption off. The boundary statement and the conventions box both end with unless otherwise stated. When a question does state otherwise, that clause is the question. Read it as a given, not as scene setting.
Assuming the pipe might be partly full. The second half of the boundary statement rules it out. The flow area is the pipe's whole cross-section.
Treating an ideal fluid as frictionless in every sense. Zero viscosity removes the fluid's internal friction. It says nothing about friction between an object and a surface, and the conventions box's separate bullet about air resistance is a separate assumption about a different situation.
Assuming ideal means constant speed. An ideal fluid speeds up in a narrower pipe and slows in a wider one. What is constant is the density, and the volume flow rate.
Concluding that a real fluid result contradicts the physics. It contradicts the model, which is the point of a model. The CED's own appendix frames the idealizations as simplifications that later work refines, describing the addition of detail to a simplified model as a normal scientific process.

When the model is close enough, and why that hides it

The reason the ideal fluid survives an entire unit without seeming like an approximation is that for the situations AP Physics 1 draws, it very nearly is not one.

Liquids really are almost incompressible. Water's density barely moves over the depths a textbook problem uses, so $P = P_0 + \rho g h$ with a fixed $\rho$ is not a sacrifice, it is accurate. This is the assumption you would notice least if it were withdrawn, and it is also the one the CED states twice, once in the definition at 8.1.A.4 and again in 8.2.A.3.

Buoyancy and hydrostatics never touch viscosity at all. Nothing in a static fluid is shearing, so a floating block or a pressure at depth is described identically by the ideal model and the real one. Topics 8.1, 8.2 and most of 8.3 are therefore exact, and the idealization is doing no work in them.

Viscosity is where the model actually bites, and it bites only when the fluid is moving: Bernoulli, Torricelli, and flow through pipes, which is Topic 8.4. That is also where the unit's only boundary statement is printed, under Topic 8.4 rather than under 8.1 where the definition sits. That placement is not an accident.

So the honest summary is that the assumption is nearly free in three quarters of the unit and expensive in the last quarter, and it is stated in the quarter where it costs something.

One last thing that lulls. Because the model is good for water in short wide pipes at low speed, and because those are the pipes in the problems, a student can finish Unit 8 believing that fluid physics simply is this. It is worth knowing what sits outside: blood in a capillary, honey through a funnel, air compressed in a cylinder, and lubricating oil in a bearing are all situations where one of the deleted terms dominates the answer. None of them is on the AP Physics 1 exam, and knowing why is a better grasp of the unit than knowing the equations.

What the CED asks, and what it keeps out of scope

The ideal fluid is introduced in Topic 8.1, Internal Structure and Density, the first topic of Unit 8. Unit 8 is weighted at $10$ to $15\ \%$ of the multiple-choice section across a suggested $12$ to $17$ class periods.

Learning objective 8.1.A is to describe the properties of a fluid, and its four pieces of essential knowledge are:

8.1.A.1, distinguishing properties of solids, liquids, and gases stem from the varying interactions between atoms and molecules.
8.1.A.2, a fluid is a substance that has no fixed shape.
8.1.A.3, fluids can be characterized by their density, defined as a ratio of mass to volume, with $\rho = m/V$ .
8.1.A.4, an ideal fluid is incompressible and has no viscosity.

The word viscosity appears in that one sentence and nowhere else in the AP Physics 1 CED. There is no learning objective about viscosity, no equation for it, and no boundary statement limiting how it is treated, because it is not treated at all. The same is true of turbulence and of laminar flow, neither of which the framework names.

What the unit does with the assumption is spend it. Topic 8.4, Fluids and Conservation Laws, is where every equation that needs it lives: 8.4.A.2's continuity equation for incompressible fluids, 8.4.B.2's Bernoulli equation, and 8.4.B.3's Torricelli result. The unit's only boundary statement sits at the end of that topic. Topic 8.4 has the full framing, and Topic 8.1 carries the definition.

A note on the other algebra-based course, because the equation sheet invites the question. The AP Physics 2 framework runs from Unit 9 to Unit 15 and contains no fluid topics: the words buoyant, Bernoulli and viscosity appear nowhere in it. Its equation sheet nonetheless prints the same seven fluid lines in its Mechanics and Fluids panel, because that panel carries the prior mechanics a Physics 2 student is assumed to bring. The AP Physics 2 conventions box has nine bullets and none of them is about fluids, which fits: it does not need the assumption because it does not set the questions.

Gauge vs absolute pressure and buoyant force vs weight both sit inside this assumption, and the fluids guide works the standard procedures it makes possible.

A narrowing pipe, with each assumption flagged as it is spent

Water of density $1000\ \text{kg/m}^3$ flows steadily through a horizontal pipe. The wide section has area $1.2 \times 10^{-3}\ \text{m}^2$ and the water there moves at $1.5\ \text{m/s}$ with a pressure of $1.80 \times 10^5\ \text{Pa}$ . The pipe narrows to $4.0 \times 10^{-4}\ \text{m}^2$ . (a) Find the volume flow rate. (b) Find the speed in the narrow section. (c) Find the pressure there. (d) Find what the speed in (b) would be if the water were compressible and its density in the narrow section were $20\ \%$ higher.

(a) The volume flow rate is the area times the speed, which 8.4.A.1.ii gives as the derived $V/t = Av$ : $Q = (1.2 \times 10^{-3})(1.5) = 1.8 \times 10^{-3}\ \text{m}^3\text{/s}$ . This step uses the completely-filled assumption, because it takes the flow area to be the whole cross-section.
(b) The continuity equation $A_1 v_1 = A_2 v_2$ applies because the fluid is incompressible: $v_2 = \dfrac{A_1 v_1}{A_2} = \dfrac{1.8 \times 10^{-3}}{4.0 \times 10^{-4}} = 4.5\ \text{m/s}$ .
Sanity check: the area fell by a factor of $1.2 \times 10^{-3} / 4.0 \times 10^{-4} = 3.0$ , so the speed should rise by the same factor. $1.5 \times 3.0 = 4.5\ \text{m/s}$ . It does.
(c) Bernoulli's equation between the two points, with $y_1 = y_2$ so the height terms cancel: $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ . This step uses the zero-viscosity assumption, because Bernoulli conserves mechanical energy exactly.
$v_1^2 = 2.25\ \text{m}^2\text{/s}^2$ and $v_2^2 = 20.25\ \text{m}^2\text{/s}^2$ , so $\frac{1}{2}\rho(v_1^2 - v_2^2) = (500)(2.25 - 20.25) = (500)(-18.0) = -9000\ \text{Pa}$ .
$P_2 = 1.80 \times 10^5 - 9000 = 1.71 \times 10^5\ \text{Pa}$ . Faster water, lower pressure, as the model requires.
(d) Now suspend incompressibility. Conservation of mass in general is $\rho_1 A_1 v_1 = \rho_2 A_2 v_2$ , and with $\rho_2 = 1.2\rho_1$ this gives $v_2 = \dfrac{\rho_1 A_1 v_1}{1.2 \rho_1 A_2} = \dfrac{4.5}{1.2} = 3.75\ \text{m/s}$ .
That is lower than the ideal answer by $(4.5 - 3.75)/4.5 = 16.7\ \%$ . Denser fluid packs more mass into each cubic meter, so less volume has to pass per second to carry the same mass.
Note what part (d) is not. It is not a correction you could apply on the exam, because nothing on the sheet tells you how much a real fluid's density rises. It is a demonstration that the printed $A_1 v_1 = A_2 v_2$ is a consequence of 8.1.A.4 rather than a law of nature, and that the densities cancelled only because the model made them equal.

(a) $Q = 1.8 \times 10^{-3}\ \text{m}^3\text{/s}$ . (b) $v_2 = 4.5\ \text{m/s}$ . (c) $P_2 = 1.71 \times 10^5\ \text{Pa}$ . (d) With a density $20\ \%$ higher in the narrow section the speed would be $3.75\ \text{m/s}$ , $16.7\ \%$ below the ideal prediction.

Torricelli, and the material property that is not there

A large open tank of liquid has a small hole in its side, $1.8\ \text{m}$ below the liquid surface. The hole has area $3.0 \times 10^{-4}\ \text{m}^2$ . Take $g = 9.8\ \text{m/s}^2$ . (a) Derive the efflux speed from Bernoulli's equation, naming the assumption each cancellation uses. (b) Compute it. (c) Compute the volume flow rate out of the hole. (d) State how the answer changes if the liquid is heavy oil instead of water, and say why.

(a) Apply Bernoulli's equation between the liquid surface, point 1, and the hole, point 2: $P_1 + \rho g y_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho g y_2 + \frac{1}{2}\rho v_2^2$ . Using it at all assumes zero viscosity, since 8.4.B.2 calls it a statement of the conservation of mechanical energy in fluid flow.
Both the surface and the hole are open to the air, so $P_1 = P_2$ and the pressure terms cancel. Note that this cancellation does not care whether those are gauge or absolute pressures, only that they are the same kind on both sides.
The tank is large, so the surface drops slowly and $v_1 \approx 0$ . That leaves $\rho g y_1 = \rho g y_2 + \frac{1}{2}\rho v_2^2$ , or $\rho g \Delta y = \frac{1}{2}\rho v_2^2$ with $\Delta y = y_1 - y_2$ .
Now $\rho$ appears in every remaining term, so it cancels. That cancellation is the incompressibility assumption doing its work: one single density for the whole fluid. The result is 8.4.B.3's derived equation, $v = \sqrt{2g\Delta y}$ .
(b) $2g\Delta y = 2(9.8)(1.8) = 35.28\ \text{m}^2\text{/s}^2$ , so $v = \sqrt{35.28} = 5.9397\ \text{m/s}$ , which is $5.9\ \text{m/s}$ to two significant figures.
Check by an energy argument on a single particle: a mass falling freely from rest through $1.8\ \text{m}$ reaches $v = \sqrt{2g h} = 5.94\ \text{m/s}$ . Torricelli's result is exactly free fall from the surface, which is what mechanical energy conservation with no losses has to give.
(c) The volume flow rate is $Q = A v = (3.0 \times 10^{-4})(5.9397) = 1.782 \times 10^{-3}\ \text{m}^3\text{/s}$ , or $1.8 \times 10^{-3}\ \text{m}^3\text{/s}$ to two significant figures.
(d) In the model, nothing changes. The density cancelled in step four, and viscosity never entered, so heavy oil, water and mercury all leave a $1.8\ \text{m}$ head at $5.9\ \text{m/s}$ . The efflux speed contains no property of the liquid at all.
That is a strong prediction and a real fluid does not obey it, as anyone who has poured honey can confirm. The AP Physics 1 exam nonetheless expects the model's answer, because the unit's boundary statement assumes ideal fluids unless a question states otherwise, and there is no printed equation that would let you produce any other number.

(a) The pressure terms cancel because both surfaces are open to the air, the surface speed is negligible for a large tank, and the density cancels because the fluid is incompressible, leaving $v = \sqrt{2g\Delta y}$ . (b) $v = 5.9\ \text{m/s}$ . (c) $Q = 1.8 \times 10^{-3}\ \text{m}^3\text{/s}$ . (d) Unchanged. The result carries no material property, which is the clearest sign of what the ideal model deleted.

Why the pressure-depth graph is a straight line

A tank holds fresh water of density $1000\ \text{kg/m}^3$ , open to the atmosphere at $P_0 = 1.0 \times 10^5\ \text{Pa}$ . Take $g = 9.8\ \text{m/s}^2$ . (a) Find the gauge and absolute pressures at depths of $10.0\ \text{m}$ , $20.0\ \text{m}$ and $30.0\ \text{m}$ . (b) State the slope and intercept of a graph of absolute pressure against depth. (c) Identify which assumption makes that graph straight, and say what shape a compressible fluid would give.

(a) 8.2.B.3 gives the gauge pressure of a vertical column as $P_{\text{gauge}} = \rho g h$ , and 8.2.B.2 gives the absolute pressure as $P = P_0 + \rho g h$ . The product $\rho g = (1000)(9.8) = 9800\ \text{Pa}$ per meter of depth.
At $10.0\ \text{m}$ : $P_{\text{gauge}} = (9800)(10.0) = 98{,}000\ \text{Pa}$ , and $P = 1.0 \times 10^5 + 9.8 \times 10^4 = 1.98 \times 10^5\ \text{Pa}$ .
At $20.0\ \text{m}$ : $P_{\text{gauge}} = 196{,}000\ \text{Pa}$ , and $P = 2.96 \times 10^5\ \text{Pa}$ .
At $30.0\ \text{m}$ : $P_{\text{gauge}} = 294{,}000\ \text{Pa}$ , and $P = 3.94 \times 10^5\ \text{Pa}$ .
Look at the increments. From $10$ to $20\ \text{m}$ the absolute pressure rises by $2.96 \times 10^5 - 1.98 \times 10^5 = 0.98 \times 10^5\ \text{Pa}$ , and from $20$ to $30\ \text{m}$ it rises by $3.94 \times 10^5 - 2.96 \times 10^5 = 0.98 \times 10^5\ \text{Pa}$ . Equal depth steps give equal pressure steps.
(b) Equal increments mean a straight line. Written as $P = \rho g h + P_0$ against $h$ , the slope is $\rho g = 9800\ \text{Pa/m}$ and the vertical intercept is $P_0 = 1.0 \times 10^5\ \text{Pa}$ .
The same graph for gauge pressure has the identical slope and passes through the origin. Slope identifies the fluid; intercept identifies which pressure you are plotting.
(c) The line is straight only because $\rho$ is the same at every depth, which is 8.2.A.3: the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid.
In a compressible fluid the density at the bottom would exceed the density at the top, so each successive meter of depth would add more than the one above it. The graph would curve upward, and no printed equation on the sheet describes that curve.
This is the assumption paying for a skill rather than for an equation. Topic 8.2 lists suggested skill 1.C, creating qualitative sketches of graphs, and a question can only ask you to sketch pressure against depth because the model has already decided the shape.

(a) Gauge: $9.8 \times 10^4$ , $1.96 \times 10^5$ and $2.94 \times 10^5\ \text{Pa}$ . Absolute: $1.98 \times 10^5$ , $2.96 \times 10^5$ and $3.94 \times 10^5\ \text{Pa}$ . (b) Slope $9800\ \text{Pa/m}$ , intercept $1.0 \times 10^5\ \text{Pa}$ . (c) Incompressibility, per 8.2.A.3. A compressible fluid would give an upward-curving graph.

Frequently asked questions

What is an ideal fluid in AP Physics 1?

Essential knowledge 8.1.A.4 defines it in a single sentence: an ideal fluid is incompressible and has no viscosity. Incompressible means its density does not change no matter what pressure is applied, which essential knowledge 8.2.A.3 restates by saying the volume and density of a given amount of an incompressible fluid are constant regardless of the pressure exerted on that fluid. No viscosity means the fluid has no internal friction, so no mechanical energy is lost as it flows. Those two properties are the whole definition. Of the seven fluid lines printed on the AP Physics 1 equation sheet, two are definitions that hold for any fluid, density as mass over volume and pressure as perpendicular force over area. The other five depend on the assumption: the two pressure-at-depth equations and the buoyant force all need a density that does not vary with depth, the continuity equation needs incompressibility, and Bernoulli's equation needs incompressibility and zero viscosity together. None of them carries a correction term for a fluid that lacks either property.

Does the AP Physics 1 exam ever use real fluids?

Only if a question says so. Unit 8 carries exactly one boundary statement, printed under Topic 8.4, and it reads that all fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated. The same assumption appears again in the box of exam conventions on the equation sheet itself, as the fourth of its four bullets. So the default is the ideal model, in the framework and in your hand during the exam. The trailing clause, unless otherwise stated, does leave the door open, and if a question walks through it that sentence is the point of the question rather than background detail.

Why does the continuity equation not have density in it?

Because the densities cancel for an incompressible fluid. Conservation of mass in general says rho one times A one times v one equals rho two times A two times v two. When the fluid is incompressible those two densities are the same number, so they divide out and leave the printed A one v one equals A two v two. Essential knowledge 8.4.A.2 says exactly this, describing the continuity equation as conservation of mass flow rate in incompressible fluids. If the density really did change between the two sections, the volume form would fail: a fluid whose density rose by 20 percent in a narrow section would move 16.7 percent slower there than the volume form predicts.

Is there a viscosity equation on the AP equation sheet?

No. There is no viscosity term anywhere in the Mechanics and Fluids panel of either algebra-based sheet, and no symbol for viscosity in that panel's symbol list. The Greek letter mu that does appear on the sheet is the coefficient of friction, in the mechanics column, and it is unrelated. The word viscosity appears exactly once in the whole AP Physics 1 course framework, inside the definition at 8.1.A.4 that removes it. There is no learning objective about it, no boundary statement limiting it, and no exam question that could require it, because you would have no equation to use.

What does it mean that pipes are assumed to be completely filled?

It is the second half of Unit 8's boundary statement, and it is a separate assumption from ideality. It means the flow area in the continuity equation is the pipe's entire cross-section, so you can use the pipe's area without asking how much of it is occupied. It also underwrites essential knowledge 8.4.A.1.i, which says the rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube. A half-full pipe would have a free surface inside it and a flow area no printed equation gives you, which is why the assumption is stated rather than left implicit.

Where does the ideal fluid model actually break down?

Wherever the fluid is moving and something is shearing. The clean test case is a long horizontal pipe of constant cross-section. Bernoulli's equation between the two ends has the height terms cancel and, by continuity, the speed terms cancel too, leaving the prediction that the pressure at the far end equals the pressure at the near end however long the pipe is. Real plumbing is designed around the fact that this is false: pressure falls along a hose, and the loss is exactly the viscous dissipation the model deleted. Hydrostatics is different. A floating block or a pressure at depth involves no shearing at all, so Topics 8.1, 8.2 and most of 8.3 are described identically by the model and by reality.

Why does the AP Physics 2 equation sheet print fluid equations if fluids is a Physics 1 unit?

Because that panel carries the mechanics a Physics 2 student is assumed to bring with them. The AP Physics 2 course framework runs from Unit 9 to Unit 15 and contains no fluid topics: the words buoyant, Bernoulli and viscosity appear nowhere in it. Its Table of Information nonetheless includes the same Mechanics and Fluids panel as the Physics 1 sheet, ending with the same seven fluid lines, from density and pressure through the buoyant force to the continuity and Bernoulli equations. Consistently with that, the Physics 2 exam conventions box has nine bullets and none of them mentions fluids, because that course does not set the questions the assumption would govern.