Respuesta :
Answer:
velocity and pressure in a 2.6-cm:
P2 = 2.53x10^5Pa, v2 = 1.18m/s
Explanation:
Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A
Area = πr^2
From the question:
v1 = 0.5m/s
d1 = 4cm = 0.04m
r1 = d1/2 = 0.04/2 = 0.02m
Since water was pumped from basement, h1 = 0m
P1 = 3.03x10^5 Pa
A1 = π×0.02×0.02
A1 = 0.0004πm^2
v2 = unknown
d2 = 2.6cm = 0.026m
r2 = d2/2 = 0.026/2 = 0.013m
h2 = 5m
P2 = unknown
A2 = π×0.013×0.013
A2 = 0.000169πm^2
Using continuity equation:
A1v1 = A2v2
0.0004π * 0.5 = 0.000169π * v2
v2 = (0.0004π * 0.5)/(0.000169π)
v2 = 1.18m/s
Applying a Bernoulli principle
P + 1/2*density*v^2 + density*g*h =C
C = constant
P1 + 1/2*density*v1^2 + density*g*h1
= P2 + 1/2*density*v2^2 + density*g*h2
Let g = 9.81m/s
density of water = 1000kg/m^3
(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)
(P1-P2) = 1/2* density(v2^2 - v1^2) +
density* g(h2-h1)
(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))
(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050
3.03x10^5 - P2 = 571.2 + 49050
3.03x10^5 - P2 = 49621.2
3.03x10^5 - 49621.2 = P2
P2 = 253378.8
P2 = 2.53x10^5Pa
P2 = 2.53x10^5Pa, v2 = 1.18m/s
The velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above are 1.18m/s and 2.51 * 10^5 Pa
The given parameters are:
- Initial velocity: [tex]v_1 = 0.5ms^{-1}[/tex]
- Initial diameter [tex]d_1 = 4cm[/tex]
- Initial pressure: [tex]P_1 = 3.03 \times 10^5 Pa[/tex]
Start by calculating the radius using:
[tex]r = 0.5d[/tex]
So, we have:
[tex]r_1 = 0.5 \times 4cm[/tex]
[tex]r_1 = 2cm[/tex]
Express as meters
[tex]r_1 = 0.02m[/tex]
Next, calculate the area using:
[tex]A =\pi r^2[/tex]
So, we have:
[tex]A_1 = \pi \times 0.02^2[/tex]
[tex]A_1 = \pi \times 0.004[/tex]
[tex]A_1 = 0.004\pi[/tex]
Also from the question, we have:
[tex]d_2 = 2.6cm[/tex]
[tex]h = 5m[/tex]
Calculate the radius using:
[tex]r = 0.5d[/tex]
So, we have:
[tex]r_2 = 0.5 \times 2.6cm[/tex]
[tex]r_2 = 1.3cm[/tex]
Express as meters
[tex]r_2 = 0.013m[/tex]
The area is then calculated as:
[tex]A =\pi r^2[/tex]
So, we have:
[tex]A_2 = \pi \times 0.013^2[/tex]
[tex]A_2 = \pi \times 0.000169[/tex]
[tex]A_2 = 0.000169\pi[/tex]
The velocity is then calculated using:
[tex]A_1v_1 = A_2v_2[/tex]
Make v2 the subject
[tex]v_2 = \frac{A_1v_1}{A_2}[/tex]
So, we have:
[tex]v_2 = \frac{0.0004\pi \times 0.5}{0.000169\pi}[/tex]
[tex]v_2 = \frac{0.0004\times 0.5}{0.000169}[/tex]
[tex]v_2 = 1.18343195266[/tex]
Approximate
[tex]v_2 = 1.18[/tex]
The pressure is then calculated as follows:
[tex]P_1 + 0.5 \times density \times v_1^2 + density \times g \times h_1 = P_2 + 0.5 \times density \times v_2^2 + density \times g \times h_2[/tex]
Where:
[tex]g = 9.81ms^{-1}[/tex]
[tex]density = 1000kgm^{-3[/tex]
[tex]h_1 = 0[/tex]
So, we have:
[tex]3 \times 10^5 + 0.5 \times 1000 \times 0.5^2 + 1000 \times 9.8 \times 0 = P_2 + 0.5 \times 1000 \times 1.18^2 + 1000 \times 9.8 \times 5[/tex]
[tex]300000 + 125 + 0 = P_2 + 696.2 + 49000[/tex]
Collect like terms
[tex]300000 + 125 - 696.2 - 49000 = P_2[/tex]
[tex]250428.8 = P_2[/tex]
Rewrite as:
[tex]P_2 =250428.8[/tex]
Rewrite as:
[tex]P_2 = 2.51 \times 10^5\ Pa[/tex]
Hence, the velocity and pressure are 1.18m/s and 2.51 * 10^5 Pa
Read more about pressures and velocities at:
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