Respuesta :
Answer:
(a) 21.44 ft/s
(b) 0 ft/s
(c) 19.51 ft/s
Explanation:
2 in = 2/12 ft = 0.167 ft
For steady laminar flow, the function of the fluid velocity in term of distance from center is modeled as the following equation:
[tex]v(r) = v_c\left[1 - \frac{r^2}{R^2}\right][/tex]
where R = 0.167 ft is the pipe radius and [tex]v_c[/tex] is the constant fluid velocity at the center of the pipe.
We can integrate this over the cross-section area of the in order to find the volume flow
[tex]\dot{V} = \int\limits {v(r)} \, dA \\= \int\limits^R_0 {v_c\left[1 - \frac{r^2}{R^2}\right]2\pi r} \, dr\\ = 2\pi v_c\int\limits^R_0 {r - \frac{r^3}{R^2}} \, dr\\ = 2\pi v_c \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]^R_0\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^4}{4R^2}\right)\\= 2\pi v_c \left(\frac{R^2}{2} - \frac{R^2}{4}\right)\\= 2\pi v_c R^2/4\\=\pi v_c R^2/2\\A = \pi R^2\\\dot{V} = Av_c/2\\[/tex]
So the average velocity
[tex]v = \dot{V} / A = v_c/2 = 10.72[/tex]
[tex]v_c = 10.72*2 = 21.44 ft/s[/tex]
b) At the wall of the pipe, r = R so [tex]v(R) = v_c(1 - 1) = 0 ft/s[/tex]
c) At a distance of 0.6 in = 0.6/12 = 0.05 ft
[tex]v(0.05) = v_c(1 - 0.05^2/0.167^2) = 0.91v_c = 0.91*21.44 = 19.51 ft/s[/tex]
Answer:
The answers to the questions are;
(a) The velocity at the center of the pipe is 21.44 ft/s
(b) The velocity at the wall of the pipe is 0 ft/s
(c) The velocity at a distance of 0.6 inches from the center-line is 19.63 ft/s.
Explanation:
To solve the question, we note that
The velocity profile in the cross section of a circular pipe with laminar flow is given by
U = 2×v×[1 - (r/r₀)²]
Where
U = The sought velocity at a point
r = Pipe radius where velocity is sought
r₀ = Internal radius of pipe = for 2-in Schedule 40 steel pipe = 2.067 in 52.6 mm
v = Average velocity of flow = 10.72 ft/s = 3.2675 m/s
Therefore we have
(a) The velocity at the center of the pipe
At the center r = 0 so we have
U = 2×v×[1 - (r/r₀)²]
At center U = 2×10.72 ft/s×[1 - (0/2.067 in)²] = 2×10.72 ft/s = 21.44 ft/s
(b) The velocity at the wall of the pipe is given by
r = r₀ ⇒ U = 2×v×[1 - (r/r₀)²] ⇒ U = 2×v×[1 - (r₀/r₀)²]
= U = 2×v×[1 - (1)²] = 2×v×0 = 0
The velocity at the wall of the pipe is 0 ft/s
(c) The velocity at a distance of 0.6 inches from the center-line is given by
U = 2×v×[1 - (r/r₀)²] = 2×10.72 ft/s×[1 - (0.6/2.067)²] = 19.63 ft/s.
