Question

Water flows through a 4.0-cm-diameter horizontal pipe at a speed of 1.3 m/s. The pipe then narrows down to a diameter of 2.0 cm. Ignoring viscosity, what is the pressure difference between the wide and narrow sections of the pipe?

Answer 1

Answer:

$P_{1}-P_{2}=12675Pa$

Explanation:

From the equation of continuity we know that:

$v_{1}A_{1}=v_{2}A_{2}\\Given \\r_{1}=2.0cm\\r_{2}=1.0cm\\v_{1}=1.3m/s\\Density of water p=1000kg/m^{3}\\Now\\A_{1}=\pi r_{1}^{2} \\A_{2}=\pi r_{2}^{2}\\ Therefore\\v_{2}=v_{1}\frac{A_{1}}{A_{2}}\\v_{2}=v_{1}\frac{\pi r_{1}^{2} }{\pi r_{2}^{2}}\\v_{2}=v_{1}\frac{r_{1}^{2} }{r_{2}^{2}}\\v_{2}=1.3*\frac{2.0^{2} }{1.0^{2} } \\v_{2}=5.2m/s$

From Bernoulli equation we know that:

$P_{1}+1/2pv_{1}^{2}+pgy_{1}=P_{2}+1/2pv_{2}^{2}+pgy_{2}$

Now assuming $y_{1}=y_{2}$

$P_{1}+(1/2)pv_{1}^{2}=P_{2}+(1/2)pv_{2}^{2}\\ P_{1}-P_{2}=1/2pv_{2}^{2} -1/2pv_{1}^{2}\\ P_{1}-P_{2}=1/2p(v_{2}^{2}-v_{1}^{2} )\\ P_{1}-P_{2}=1/2*1000(5.2^{2}-1.3^{2} )\\ P_{1}-P_{2}=12675Pa$

Answer 2

This question involves the concepts of Bernoulli's Theorem and continuity equation. Substituting the respective values the pressure difference can be found.

The difference in the water pressure between the wide and the narrow of the pipe is "12.78 KPa".

Applying the continuity equation to find the speed of the water at the narrow section:

$A_1v_1=A_2v_2$

where,

A₁ = Area of wider section = $\pi \frac{d_1^2}{4} = \pi \frac{(0.04\ m)^2}{4} = 1.26\ x\ 10^{-3}\ m^2$

A₂ = Area of narrow section = $\pi \frac{d_2^2}{4} = \pi \frac{(0.02\ m)^2}{4} = 3.14\ x\ 10^{-4}\ m^2$

v₁ = speed of flow at wider section = 1.3 m/s

v₂ = speed of flow at narrow section = ?

Therefore,

$(1.26\ x\ 10^{-3}\ m^2)(1.3\ m/s)=(3.14\ x\ 10^{-4}\ m^2)v_2\\\\v_2=\frac{(1.26\ x\ 10^{-3}\ m^2)(1.3\ m/s)}{3.14\ x\ 10^{-4}\ m^2}\\\\v_2=5.22\ m/s$

Applying Bernoulli's Theorem to this situation:

$P_1+\rho g h_1+\frac{1}{2}\rho v_1^2=P_2+\rho g h_2+\frac{1}{2}\rho v_2^2\\\\P_1-P_2=\Delta P=-\rho g h_1-\frac{1}{2}\rho v_1^2+\rho g h_2+\frac{1}{2}\rho v_2^2$

where,

ΔP = Pressure Difference between top and bottom = ?

$\rho$ = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.81 m/s²

h₁ = h₂ = h = heights of both the ends  

 Therefore,

$\Delta P=-\rho g h_-\frac{1}{2}(1000\ kg/m^3)(1.3\ m/s)^2+\rho g h_+\frac{1}{2}(1000\ kg/m^3)(5.22\ m/s)^2$

ΔP = 12779.2 Pa = 12.78 KPa

 

Learn more about Bernoulli's Theorem here:

brainly.com/question/13098748?referrer=searchResults

The attached picture shows Bernoulli's Theorem.