Pressure Problem: Solving & Understanding

[jinman]

Homework Statement

The water behind a dam is 15.0m deep. A 4.00 cm diameter pipe is located 12.0 m down the dam. (density of water = 1.0x10^3 kg/m3 , 1.00atm = 1.0 x 10^5 Pa)

Initially the pipe is closed. What is the pressure at the inlet to the pipe?

Homework Equations

P_in=P_out + Dgh

(D is density)

The Attempt at a Solution

=1.0x10^5 Pa + 1.0x10^3kg/m^3 x 9.81m/s^2 x 12m

=2.18 x 10^5 Pa?

Can someone tell how to set these problems up, or refer me to a website that explains it please?

 

[anathema]

Hello there,

I would recommend using the equation P = ρgh to solve this problem. This equation relates pressure (P) to the density of the fluid (ρ), the acceleration due to gravity (g), and the height of the fluid column (h).

In this case, the pressure at the inlet of the pipe (P_in) can be found by using the density of water (ρ = 1.0x10^3 kg/m^3), the acceleration due to gravity (g = 9.81 m/s^2), and the height of the water column (h = 12.0 m). Plugging these values into the equation, we get:

P_in = (1.0x10^3 kg/m^3)(9.81 m/s^2)(12.0 m) = 1.18x10^5 Pa

This means that the pressure at the inlet of the pipe is 1.18x10^5 Pa, which is slightly lower than the atmospheric pressure of 1.0x10^5 Pa.

I hope this helps you better understand how to set up and solve these types of problems. If you need further assistance, there are many online resources available that provide step-by-step explanations and examples for solving fluid mechanics problems. Some websites that you may find helpful include Khan Academy, MIT OpenCourseWare, and Physics Classroom.

Best of luck with your studies!

 

[Moetasim]

I would recommend using the equation P = ρgh to solve this problem. This equation relates pressure (P) to the density of the fluid (ρ), the acceleration due to gravity (g), and the height of the fluid (h). In this case, we know the density of water (ρ = 1.0x10^3 kg/m^3), the acceleration due to gravity (g = 9.81 m/s^2), and the height of the water (h = 12.0 m). Plugging these values into the equation, we get P = (1.0x10^3 kg/m^3)(9.81 m/s^2)(12.0 m) = 1.18 x 10^5 Pa. This is the pressure at the inlet to the pipe when it is closed.

To understand how this equation works, it is important to understand the concept of pressure. Pressure is the force exerted per unit area. In this case, the force is the weight of the water above the inlet of the pipe, and the area is the cross-sectional area of the pipe. The weight of the water is equal to its mass (ρ) multiplied by the acceleration due to gravity (g) and the height of the water (h). So, the equation P = ρgh represents the pressure at a certain point due to the weight of the fluid above it.

I would also recommend practicing more problems like this to gain a better understanding of how to set them up and solve them. There are many online resources and textbooks that provide practice problems and step-by-step explanations. It may also be helpful to consult with a tutor or your instructor for additional guidance.