Respuesta :
Answer:
Mean velocity = 0.292 m/s
Maximum velocity = 0.584 m/s
The flow is laminar as Re = 229.2
Explanation:
D = 1.5 inches = 0.0381 m
Q = volumetric flow rate = 20 L/min = 0.000333 m³/s
Q = A × v
A = Cross sectional Area = πD²/4 = π(0.0381)²/4 = 0.00114 m²
v = average velocity
v = Q/A = 0.000333/0.00114 = 0.292 m/s
For flow in circular pipes, maximum velocity = 2 × average velocity = 2 × 0.292 = 0.584 m/s
To check if the flow is laminar or turbulent, we need its Reynolds number
Re = (ρvD)/μ
ρ = 1030 kg/m
v = 0.292 m/s
D = 1.5 inches = 0.0381 m
μ = 50 cP = 0.5 poise = 0.05 Pa.s
Re = (1030 × 0.292 × 0.0381)/0.05 = 229.2
For laminar flow, Re < 2100
For turbulent flow, Re > 4000
And 229.2 < 2100, hence, this flow is laminar.
The average velocity of the liquid in the pipe is 0.292 m/s.
The flow of the liquid in the pipe is laminar.
The given parameters;
- flow rate of the liquid, Q = 20 L/min = 0.000333 m³/s
- diameter of the pipe, d = 1.5 in = 0.0381 m
The average velocity of the liquid in the pipe is calculated as follows;
Q = Av
- where;
- A is the area of the pipe
[tex]A = \frac{\pi d^2}{4}\\\\A = \frac{\pi \times 0.0381^2}{4} \\\\A = 0.00114 \ m^2[/tex]
[tex]v = \frac{Q}{A} \\\\v = \frac{0.000333}{0.00114} \\\\v = 0.292 \ m/s[/tex]
The Reynolds's number for the given
[tex]R_e = \frac{\rho vD}{\mu} \\\\R_e = \frac{1030 \times 0.292 \times 0.0381}{0.05} \\\\R_e = 229.2 \[/tex]
For turbulent flow:
[tex]R_e \ > 4000[/tex]
For laminar flow;
[tex]R_e \ < 2,100[/tex]
Thus, the flow of the liquid in the pipe is laminar.
Learn more here:https://brainly.com/question/13221363
