In a section of horizontal pipe with a diameter of 3.0 cm,the pressure is 100 kPa and water is flowing with a speed of 1.5 m/s.The pipe narrows to 2.0 cm.What is the pressure in the narrower region? Treat the water as an ideal incompressible fluid.
A)95 kPa
B)48 kPa
C)44 kPa
D)230 kPa
E)67 kPa

Question

Quizplus · Accepted Answer

To find the pressure in the narrower region, we can use the Bernoulli's equation, which in one of its forms can be written as $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P_1$ and $P_2$ are the pressures at points 1 and 2, $v_1$ and $v_2$ are the velocities at points 1 and 2, and $\rho$ is the density of the fluid (water in this case). The equation assumes the fluid is incompressible and flow is steady.Given:- $P_1 = 100$ kPa- $v_1 = 1.5$ m/s- Diameter at point 1, $D_1 = 3.0$ cm or 0.03 m- Diameter at point 2, $D_2 = 2.0$ cm or 0.02 mFirst, we find the velocities using the continuity equation $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas at points 1 and 2, respectively. The area of a circle is given by $A = \pi r^2$, so we can find $v_2$ as follows:$$A_1 = \pi (0.015)^2$$$$A_2 = \pi (0.01)^2$$$$v_2 = v_1 \times \frac{A_1}{A_2} = 1.5 \times \frac{\pi (0.015)^2}{\pi (0.01)^2} = 1.5 \times \frac{(0.015)^2}{(0.01)^2} = 1.5 \times 2.25 = 3.375 \, \text{m/s}$$Now, applying Bernoulli's equation to solve for $P_2$:$$100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 = P_2 + \frac{1}{2} \times 1000 \times (3.375)^2$$Solving for $P_2$:$$P_2 = 100 \times 10^3 + \frac{1}{2} \times 1000 \times (1.5)^2 - \frac{1}{2} \times 1000 \times (3.375)^2$$$$P_2 = 100000 + 1125 - 5681.25 = 100000 - 4556.25 = 95443.75 \, \text{Pa} = 95.44 \, \text{kPa}$$Rounding to the nearest kPa gives us 95 kPa, which corresponds to choice A.

In a Section of Horizontal Pipe with a Diameter of 3.0