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) 44 kPa
B) 67 kPa
C) 230 kPa
D) 95 kPa
E) 48 kPa

Question

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)  44 kPa
B)  67 kPa
C)  230 kPa
D)  95 kPa
E)  48 kPa

Quizplus · Accepted Answer

To find the pressure in the narrower region, we can use the Bernoulli's equation, which for a horizontal flow can be simplified to $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$, where $P_1$ and $P_2$ are the pressures, $v_1$ and $v_2$ are the velocities in the wider and narrower sections of the pipe, respectively, and $\rho$ is the density of water (approximately $1000 \, \text{kg/m}^3$). The continuity equation gives us $A_1v_1 = A_2v_2$, where $A_1$ and $A_2$ are the cross-sectional areas of the wider and narrower sections, respectively. From this, we can solve for $v_2$, and then use Bernoulli's equation to find $P_2$.Given:- $D_1 = 3.0 \, \text{cm} = 0.03 \, \text{m}$- $D_2 = 2.0 \, \text{cm} = 0.02 \, \text{m}$- $P_1 = 100 \, \text{kPa}$- $v_1 = 1.5 \, \text{m/s}$First, find the areas:- $A_1 = \pi r_1^2 = \pi (0.015)^2 \, \text{m}^2$- $A_2 = \pi r_2^2 = \pi (0.01)^2 \, \text{m}^2$Then, use the continuity equation to find $v_2$:- $A_1v_1 = A_2v_2$- $v_2 = \frac{A_1}{A_2}v_1 = \frac{\pi (0.015)^2}{\pi (0.01)^2} \cdot 1.5 \, \text{m/s}$- $v_2 = \frac{(0.015)^2}{(0.01)^2} \cdot 1.5 = \frac{0.225}{0.1} \cdot 1.5 = 2.25 \cdot 1.5 = 3.375 \, \text{m/s}$Finally, use Bernoulli's equation to find $P_2$:- $100000 + \frac{1}{2}(1000)(1.5)^2 = P_2 + \frac{1}{2}(1000)(3.375)^2$- Solving for $P_2$, we get $P_2 = 100000 + \frac{1}{2}(1000)(1.5)^2 - \frac{1}{2}(1000)(3.375)^2$- $P_2 = 100000 + 1125 - 5681.25 = 100000 - 4556.25 = 95443.75 \, \text{Pa} = 95 \, \text{kPa}$Therefore, the pressure in the narrower region is approximately 95 kPa.

In a Section of Horizontal Pipe with a Diameter of