Torricelli's law problem about a sprinkler system

[spark706]

Homework Statement

Sprinkler in a hotel is supplied by gravity from a cylindrical tank on the roof. Suppose the tank has a radius of 25' and the diameter of the outlet is 4". The system must provide 2270 psf for at least 5 minutes. What height should the engineer specify for the tank to meet these requirements.

Homework Equations

So I'm using Torricelli's law dV/dt=-a√(2gh) and pressure=density x depth
a=area of the outlet=.087266 ft^2
A=area of tank=1963.495 ft^2
g=gravity=32.2
h=height

The Attempt at a Solution

dh/dt*A = -a√(2g)
dh/dt = (-a√(2g))/A * √h
∫-.00035663dt = ∫dh/√h
-.000356663t + C = 2√h
(-.0001783315t+C)^2 =h

pressure / density = depth 2270/62.5 = 36.32' after five minutes

And from here I'm lost. MAybe I'm on the wrong track? Can anybody please help?

 

[mighty2000]

Hello there,

First of all, good job on starting with the correct equations and attempting to solve the problem! It seems like you're on the right track, but there are a few things to consider.

First, let's clarify the units for the given values. The radius of the tank is given in feet, so we'll stick with feet for all other dimensions. The diameter of the outlet is given in inches, so we'll need to convert that to feet before using it in our calculations. This means that the area of the outlet is not 0.087266 ft^2, but rather 0.000603 ft^2. Also, the given pressure of 2270 psf should be converted to feet of water, since we'll be using the depth in our calculations. This gives us a pressure of 37.12 ft of water.

Next, let's look at the equation you're using for Torricelli's law. It should be dV/dt = -a√(2gh), where V is the volume of water leaving the tank, a is the area of the outlet, g is the acceleration due to gravity, and h is the depth of the water in the tank. We can rearrange this equation to solve for h:

h = (V^2)/(2ga^2)

Now, we have two variables - V and h - and we need to find the relationship between them. This is where the given pressure and time come into play. We know that the pressure is equal to the depth, and we want the depth to be 37.12 ft after 5 minutes. So, we can set up the following equation:

37.12 ft = (V^2)/(2ga^2)

We also know that the volume of water leaving the tank is equal to the volume of the tank itself, since the tank is being emptied completely. The volume of a cylinder is given by V = πr^2h, where r is the radius and h is the height. Substituting in the given values, we get:

37.12 ft = (π(25 ft)^2h)/(2(32.2 ft/s^2)(0.000603 ft^2)^2)

Solving for h, we get h = 72.68 ft. This is the height that the engineer should specify for the tank in order to meet the given requirements.

I hope this helps! Let me