University Physics I · topic UPI-15.09
Fluids problem studio
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Fluids problem studio: predict a finite change
Hydrostatics, hydraulics, buoyancy, continuity, Bernoulli flow, and assumption checks. Use the supplied topic model Y_out = 0.5*(v1^2*(62/d2)^4 - v1^2), where Y_out is BERNOULLI DROP (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d2 is Throat bore d₂, measured in mm. Keep v1 (Inlet speed v₁) = 1.45 m s⁻¹; k (Loss coefficient K) = 0.26 fixed. For this exercise use only 27.4 mm ≤ d2 ≤ 52.6 mm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at d2 = 32.944 mm and 36.472 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
BERNOULLI DROP against d2
Predictions from the supplied Fluids problem studio model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BERNOULLI DROP / kPa
Horizontal axis (x): d2: Throat bore d₂ / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d2: Throat bore d₂ / mm | BERNOULLI DROP / kPa |
|---|---|
| 27.4 | 26.508198054898173 |
| 33.699999999999996 | 10.992262377610055 |
| 40 | 5.0165715703125 |
| 46.3 | 2.328999854175692 |
| 52.599999999999994 | 0.9779730656423262 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d2 = 32.944 mm, substitute into Y_out = 0.5*(v1^2*(62/d2)^4 - v1^2): Y₀ = 12.1364 kPa.
- At d2 = 36.472 mm, the same fixed-parameter model gives Y₁ = 7.72754 kPa.
- Signed change: ΔY_out = Y₁ − Y₀ = -4.40887 kPa; the input change is Δd2 = 3.528 mm.
- Average sensitivity is ΔY_out/Δd2 = -1.24968 kPa per mm. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 12.1364 kPa, Y₁ = 7.72754 kPa; ΔY_out = -4.40887 kPa; average sensitivity = -1.24968 kPa per mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Fluids problem studio: infer an input with uncertainty
Hydrostatics, hydraulics, buoyancy, continuity, Bernoulli flow, and assumption checks. Use the supplied topic model Y_out = 0.5*(v1^2*(62/d2)^4 - v1^2), where Y_out is BERNOULLI DROP (kPa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d2 is Throat bore d₂, measured in mm. Keep v1 (Inlet speed v₁) = 1.45 m s⁻¹; k (Loss coefficient K) = 0.26 fixed. For this exercise use only 27.4 mm ≤ d2 ≤ 52.6 mm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 3.5899954 kPa, with a bounded reading uncertainty ±0.3918 kPa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00252 mm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
BERNOULLI DROP against d2
Predictions from the supplied Fluids problem studio model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): BERNOULLI DROP / kPa
Horizontal axis (x): d2: Throat bore d₂ / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d2: Throat bore d₂ / mm | BERNOULLI DROP / kPa |
|---|---|
| 27.4 | 26.508198054898173 |
| 33.699999999999996 | 10.992262377610055 |
| 40 | 5.0165715703125 |
| 46.3 | 2.328999854175692 |
| 52.599999999999994 | 0.9779730656423262 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(d2) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(27.4) = 26.5082 kPa and Y_out(52.6) = 0.977973 kPa. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d2_line = 27.4 + (3.59 − 26.5082)(52.6 − 27.4)/(0.977973 − 26.5082) = 50.0218 mm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives d2 = 42.772 mm; substituting back gives Y_out = 3.5899954 kPa.
- The interpolation error in the input is |d2_line − d2| = 7.24976 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.1981954 kPa and 3.9817954 kPa. Sort the roots even if the response decreases: 41.9141 mm ≤ d2 ≤ 43.7255 mm.
- The lower and upper input deviations are 0.857868 mm and 0.953535 mm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer d2 ≈ 42.772 mm; compatible interval [41.9141, 43.7255] mm. Endpoint interpolation gives 50.0218 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.