--- title: "Streaklines compared with streamlines" subject: "Fluid Mechanics" chapter: 4 key: "semester_3/fluid_mechanics:streaklines-vs-streamlines-bx" source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/streaklines-vs-streamlines-bx site: BCT Engineering --- # Streaklines compared with streamlines Fluid Mechanics · Chapter 4: Hydrokinematics - Asked in: 2078 Kartik, 4+4 marks - Video: [What is the difference between Pathline, Streakline and Streamline](https://www.youtube.com/watch?v=grMmkSP637w) — Cowan Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/streaklines-vs-streamlines-bx - Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-4 ## Question Consider the flow described by the velocity field $\vec V = Bx(1+At)\hat i + Cy\hat j$, with A = 0.5 s⁻¹, and B = C = 1 s⁻¹. Coordinates are measured in meter. Plot the streak lines traced out by the particle that passes through the point (1, 1) during the interval from t = 0 to t = 3 s. Compare with streamlines plotted through the same point at the instants t = 0, 1 and 2 s. (no need of graph paper, plot in answer copy in precision as far as possible) ## Short answer Streamlines through (1,1): y=x (t=0), y=x^{2/3} (t=1 s), y=x^{1/2} (t=2 s). Streakline at t=3 s: from (1,1) to (190.6,20.1) m, equation x=y^{2.5}e^{-0.25(ln y)²}. ## Answer **Velocity field:** $u = Bx(1+At) = x(1+0.5t)$, $v = Cy = y$. #### Streamlines through (1, 1) at fixed t $$ \frac{dy}{dx} = \frac{y}{x(1+At)}\;\Rightarrow\; \ln y = \frac{1}{1+At}\ln x\;\Rightarrow\; y = x^{1/(1+0.5t)} $$ | Time | Streamline | |---|---| | t = 0 | $y = x$ | | t = 1 s | $y = x^{2/3}$ | | t = 2 s | $y = x^{1/2}$ | Points: for $x = 1, 2, 3, 4$: $t=0$: $y = 1, 2, 3, 4$; $t=1$: $y = 1, 1.587, 2.080, 2.520$; $t=2$: $y = 1, 1.414, 1.732, 2.0$. #### Streakline Take the streakline observed at $t = 3$ s, formed by all the particles that passed through (1, 1) at release times $\tau$ from 0 to 3 s. Integrate the path of a particle released at $\tau$: $$ \frac{dy}{dt} = y \Rightarrow y = e^{(t-\tau)},\qquad \frac{dx}{dt} = x(1+0.5t)\Rightarrow x = \exp\left[(t-\tau) + 0.25(t^2-\tau^2)\right] $$ At $t = 3$: | $\tau$ (s) | 0 | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 | |---|---|---|---|---|---|---|---| | x (m) | 190.57 | 108.58 | 54.60 | 24.23 | 9.49 | 3.28 | 1.00 | | y (m) | 20.09 | 12.18 | 7.39 | 4.48 | 2.72 | 1.65 | 1.00 | Eliminating $\tau$ (with $s = \ln y = t-\tau$): $x = y^{1+0.5t}e^{-0.25(\ln y)^2}$, and at $t=3$, $x = y^{2.5}e^{-0.25(\ln y)^2}$. ```text y (m) 20 | * (190.6, 20.1) tau = 0 | 12 | * (108.6, 12.2) tau = 0.5 | 7 | * (54.6, 7.4) tau = 1 4 | * (24.2, 4.5) tau = 1.5 1 |*(1,1) tau = 3 [(3.3, 1.65) and (9.5, 2.7) lie close to the start] +------------------------------------------------- x (m) 0 50 100 150 200 ``` **Comparison:** a streamline is the instantaneous curve tangent to the velocity. Through (1, 1) it is the straight line $y = x$ at $t = 0$, and it becomes flatter with time ($y = x^{1/(1+0.5t)}$). The streakline at $t = 3$ s is much longer and lies well below the line $y = x$ (at $y = 20$ m, $x$ is already 190 m), because particles released earlier have been carried far in the $x$-direction by the growing $u$. It does not lie on any one streamline. The flow is unsteady, so streaklines, pathlines and streamlines are different curves; they would coincide only in steady flow. **Answer: streamlines $y = x,\ x^{2/3},\ x^{1/2}$ at $t = 0, 1, 2$ s; the streakline at $t = 3$ s runs from $(1,1)$ to $(190.6, 20.1)$, as given by the table above.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are streamlines of the unsteady field obtained at a given instant? A. Integrate dy/dx = v/u with t held constant B. Integrate dx/dt = u and dy/dt = v with time varying C. Join all particles that passed one point earlier D. Set the acceleration to zero Answer: A. A streamline is an instantaneous curve tangent to velocity, so t is treated as a constant. (Key point: Streamline definition) 2. What is the streamline through (1,1) at t = 1 s? A. y = x^(3/2) B. y = x^(2/3) C. y = x D. y = x^(1/2) Answer: B. Exponent is 1/(1+0.5t) = 1/1.5 = 2/3 at t=1. (Key point: Streamline at t=1) 3. How is the streakline at t = 3 s plotted? A. Plot the path of one particle for 3 s B. Plot velocity vectors at t = 3 s C. Follow particles released from (1,1) at each time tau from 0 to 3 and plot where they all are at t = 3 D. Use y = x^(1/2.5) only Answer: C. A streakline is the locus at one instant of all particles that passed through the injection point earlier. (Key point: Streakline construction) 4. Why do the streakline and streamlines differ here? A. The flow is incompressible B. The velocity is zero at (1,1) C. Streamlines always bend D. The flow is unsteady (u depends on t) Answer: D. For steady flow they coincide; here u = x(1+0.5t) varies with time, so they are different curves. 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