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Sagot :
Solution :
The volume rate of flow is given by : R = 5.0 L/s
[tex]$ = 5.0 \times 10^{-3} \ m^3/s$[/tex]
The radius of the pipe, [tex]$r_1= 5 \times 10^{-2} \ m$[/tex]
∴ [tex]$ 5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_1 = \frac{5.0 \times 10^{-3}}{(3.14)(5 \times 10^{-2})^2}$[/tex]
= 0.637 meter per second
Then the speed of the water at wider section,
[tex]$R=A_1v_1$[/tex]
Similarly, the speed of water at narrow pipe.
The radius of the [tex]$r_2 = 2.5 \times 10^{-2}$[/tex] m
[tex]$5.0 \times 10^{-3} = \pi (2.5 \times 10^{-2})^2 v_1$[/tex]
then, [tex]$v_2 = \frac{5.0 \times 10^{-3}}{(3.14)(2.5 \times 10^{-2})^2}$[/tex]
= 2.55 meter per sec
Now from Bernoulli's theorem,
[tex]$P_1 + \frac{1}{2} \rho v_1^2 =P_2 + \frac{1}{2} \rho v_2^2 $[/tex]
[tex]$P_1 = P_2 + \frac{1}{2} \rho (v_2^2 - v_1^2)$[/tex]
[tex]$= 50 \kPa + (0.5)(10^3)[(2.55)^2-(0.637)^2]$[/tex]
= 50 kPa + 3.05 kPa
= 53.05 kPa
or 53000 Pa
This question involves the concepts of Bernoulli's Theorem and Volumetric Flowrate.
The pressure reading in the wide section is "53.05 KPa".
First, we will use the volumetric flow rate to find the velocities of the water at wide and narrow sections.
[tex]V = A_1v_1[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₁ = radius of narrow section = 5 cm/2 = 2.5 cm = 0.025 m
A₁ = Area of narrow section = πr₁² = π(0.025 m)²
v₁ = velocity at narrow section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.025\ m)^2](v_1)\\\\v_1=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.025\ m)^2}\\\\v_1=2.55\ m/s\\[/tex]
Similarly,
[tex]V = A_2v_2[/tex]
where,
V = Volumetric Flow Rate = 5 L/s = 5 x 10⁻³ m³/s
r₂ = radius of wide section = 10 cm/2 = 5 cm = 0.05 m
A₂ = Area of wide section = πr₁² = π(0.05 m)²
v₂ = velocity at wide section = ?
Therefore,
[tex]5\ x\ 10^{-3}\ m^3=[\pi(0.05\ m)^2](v_2)\\\\v_2=\frac{5\ x\ 10^{-3}\ m^3}{\pi (0.05\ m)^2}\\\\v_2=0.64\ m/s\\[/tex]
Now, we will use Bernoulli's Theorem to find out the pressure wide section.
[tex]P_1 + \frac{1}{2}\rho v_1^2=P_2 + \frac{1}{2}\rho v_2^2[/tex]
where,
[tex]\rho[/tex] = density of water = 1000 kg/m³
P₁ = pressure in narrow section = 50 KPa = 50000 Pa
P₂ = pressure in wide section = ?
Therefore,
[tex]50000\ Pa + \frac{1}{2}(1000\ kg/m^3)(2.55\ m/s)^2=P_2 + \frac{1}{2}(1000\ kg/m^3)(0.64\ m/s)^2[/tex]
P₂ = 50000 Pa + 3251.25 Pa - 204.8 Pa
P₂ = 53046.45 Pa = 53.05 KPa
Learn more about Bernoulli's Theorem here:
https://brainly.com/question/13098748?referrer=searchResults
The attached picture shows Bernoulli's Theorem.
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