Pressures at ends of a horizontal pipe are given for water. Find speed v at end 2 if speed at end 1 is 10 m/s. (density of water = 1000 kg/m³). Find v (in m/s).

The correct answer is 11.

Step-by-step Explanation:

From the provided image of the horizontal pipe, we identify the following given data points:

At end 1 (left side):
Pressure, $P_1=13\text{kPa}=13000\text{Pa}$
Speed of water, $v_1=10\text{m/s}$

At end 2 (right side):
Pressure, $P_2=2.5\text{kPa}=2500\text{Pa}$
Speed of water, $v_2=v$

Density of water, $\rho =1000\text{kg/m}{3}^{}$

Since the pipe is horizontal, we can apply Bernoulli's Equation for a horizontal flow where height differences are negligible:

$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

Substitute the given values into the equation:

$13000+\frac{1}{2}\left(1000\right){\left(10\right)}^2=2500+\frac{1}{2}\left(1000\right)v^2$

Simplify both sides of the equation:

$13000+500\left(100\right)=2500+500v^2$

$13000+50000=2500+500v^2$

$63000=2500+500v^2$

Isolate the term containing $v^2$ by subtracting 2500 from both sides:

$63000-2500=500v^2$

$60500=500v^2$

Divide both sides by 500:

$v^2=\frac{60500}{500}$

$v^2=121$

Take the square root of both sides to find the speed $v$ at end 2:

$v=\sqrt{121}=11\text{m/s}$