Respuesta :
A) Speed in the lower section: 0.638 m/s
B) Speed in the higher section: 2.55 m/s
C) Volume flow rate: [tex]1.8\cdot 10^{-3} m^3/s[/tex]
Explanation:
A)
To solve the problem, we can use Bernoulli's equation, which states that
[tex]p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2[/tex]
where
[tex]p_1=1.75\cdot 10^4 Pa[/tex] is the pressure in the lower section of the tube
[tex]h_1 = 0[/tex] is the heigth of the lower section
[tex]\rho=1000 kg/m^3[/tex] is the density of water
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]v_1[/tex] is the speed of the water in the lower pipe
[tex]p_2[/tex] is the pressure in the higher section
[tex]h_2 = 0.250 m[/tex] is the height in the higher pipe
[tex]v_2[/tex] is hte speed in the higher section
We can re-write the equation as
[tex]v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}[/tex] (1)
Also we can use the continuity equation, which state that the volume flow rate is constant:
[tex]A_1 v_1 = A_2 v_2[/tex]
where
[tex]A_1 = \pi r_1^2[/tex] is the cross-section of the lower pipe, with
[tex]r_1 = 3.00 cm =0.03 m[/tex] is the radius of the lower pipe (half the diameter)
[tex]A_2 = \pi r_2^2[/tex] is the cross-section of the higher pipe, with
[tex]r_2 = 1.50 cm = 0.015 m[/tex] (radius of the higher pipe)
So we get
[tex]r_1^2 v_1 = r_2^2 v_2[/tex]
And so
[tex]v_2 = \frac{r_1^2}{r_2^2}v_1[/tex] (2)
Substituting into (1), we find the speed in the lower section:
[tex]v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s[/tex]
B)
Now we can use equation (2) to find the speed in the lower section:
[tex]v_2 = \frac{r_1^2}{r_2^2}v_1[/tex]
Substituting
v1 = 0.775 m/s
And the values of the radii, we find:
[tex]v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s[/tex]
C)
The volume flow rate of the water passing through the pipe is given by
[tex]V=Av[/tex]
where
A is the cross-sectional area
v is the speed of the water
We can take any point along the pipe since the volume flow rate is constant, so
[tex]r_1=0.03 cm[/tex]
[tex]v_1=0.638 m/s[/tex]
Therefore, the volume flow rate is
[tex]V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s[/tex]
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Answer:
(a)Speed of flow in the lower section is [tex]v_1=0.638\rm m/s[/tex]
(b)Speed of flow in the upper section is [tex]v_2=\frac{0.03^2}{0.15^2}\times0.638=2.55\rm m/s[/tex]
(c)Volume flow rate through the pipe is [tex]Q_1=A_1v_1=1.8\times\ 10^{-3} \rm m^3/s[/tex]
Explanation:
Given information:
[tex]\rho =1000\rm kg/m^3[/tex]
[tex]g=9.8\rm m/s^2[/tex]
At lower point,
pressure is
[tex]p_1=1.75\times 10^4 \rm kPa[/tex]
pipe diameter is
[tex]d_1=6\rm cm =0.06m[/tex]
At higher point,
pressure is
[tex]p_2=1.20\times 10^4 \rm kPa[/tex]
pipe diameter is
[tex]d_2=3\rm cm=0.03m[/tex]
speed of the pipe at lower point=
[tex]v_1[/tex]
area of the pipe at lower side =
[tex]A_1=\frac{\pi \times d_1^2}{4}[/tex]
speed of the pipe at higher point=
[tex]v_2[/tex]
area of the pipe at higher side =
[tex]A_2=\frac{\pi \times d_2^2}{4}[/tex]
Flow is irrotational so by use of Bernoulli's equation,
[tex]p_1+\rho g h_1+\frac{1}{2}\rho v_1^2=p_2+\rho g h_2+\frac{1}{2}\rho v_2^2[/tex]
We can write it as,
[tex]v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}[/tex]
By continuity equation,
[tex]A_1v_1=A_2v_2\\\\\pi \frac{d_1^2}{4}\times v_1=\pi \frac{d_2^2}{4}\times v_2\\\\\frac{v_2}{v_1}=\frac{d_2^2}{d_1^2}=\frac{0.03^2}{0.06^2}=0.25[/tex]
On substitution,
[tex]{v_2}=0.25{v_1}[/tex],
[tex]v_1^2-0.25^2=\frac{2(1.75\times10^4-1.20\times10^4)+1000\times 9.8\times 0.250}{1000}[/tex]
[tex]v_1=0.638\rm m/s[/tex]
(b) By this [tex]A_1v_1=A_2v_2[/tex]
[tex]v_2=\frac{0.03^2}{0.15^2}\times0.638=2.55\rm m/s[/tex]
(c)Volume flow rate is,
[tex]Q_1=A_1v_1=\pi r_1^2\times v_1=\pi 0.03^2\times 0.638=1.8\times\ 10^{-3} \rm m^3/s[/tex]
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