Torricelli's Theorem and etc. help

[crhscoog]

Homework Statement

A large tank, 25m in height and open at the top, is completely filled with saltwater (density of 1025kg/m^3). A small drain plug witha cross- sectional area of 4.0 x 10^-5 m^2 is located 5 m from the bottom of the tank.

a. calculate the force exerted by the water on the plug before the plug breaks
b. calculate the speed of the water as it leaves the hole in the side of the tank.
c. calculate the volume flow rate of the water from the hole.

The Attempt at a Solution

for b i used bernoulli's equation to solve this.
P1 + pgy1 + (1/2)pv^2 = P2 + pgy2 + (1/2)pv^3
reduced down to: v = sqr(2gh)
v= sqr((2)(9.8)(20))
v is approx 19.8 m/s

for c, flow rate = av
(4.0 x 10^-5 m^2)(19.8)
7.9196 x 10^-4

for a, f= pa
f= 101300 x (4.0 x 10^-5 m^2)

this right?

 

[alphysicist]

Hi crhscoog,

Homework Statement

A large tank, 25m in height and open at the top, is completely filled with saltwater (density of 1025kg/m^3). A small drain plug witha cross- sectional area of 4.0 x 10^-5 m^2 is located 5 m from the bottom of the tank.

a. calculate the force exerted by the water on the plug before the plug breaks
b. calculate the speed of the water as it leaves the hole in the side of the tank.
c. calculate the volume flow rate of the water from the hole.

The Attempt at a Solution

for b i used bernoulli's equation to solve this.
P1 + pgy1 + (1/2)pv^2 = P2 + pgy2 + (1/2)pv^3
reduced down to: v = sqr(2gh)
v= sqr((2)(9.8)(20))
v is approx 19.8 m/s

for c, flow rate = av
(4.0 x 10^-5 m^2)(19.8)
7.9196 x 10^-4

for a, f= pa
f= 101300 x (4.0 x 10^-5 m^2)

I don't believe this is correct. This would give the force from the air on the plug, but you want to find the force from the water on the plug. Do you see how to find that?

 

[crhscoog]

Hi crhscoog,
I don't believe this is correct. This would give the force from the air on the plug, but you want to find the force from the water on the plug. Do you see how to find that?

hmm i see what you mean. ill look into my notes and see if i can find anything on it.

*from a glance at the notes, would it have to do something with absolute pressure?

 

[alphysicist]

hmm i see what you mean. ill look into my notes and see if i can find anything on it.

*from a glance at the notes, would it have to do something with absolute pressure?

You need to find the water pressure at that point, which is 20m below the surface. What would that be?

 

[crhscoog]

You need to find the water pressure at that point, which is 20m below the surface. What would that be?

pgh 1000 x 9.8 x 20 = 196000 Pa

 

[alphysicist]

pgh 1000 x 9.8 x 20 = 196000 Pa

That's the right idea; however they give a density (not 1000), and also what you have here is just the increase in pressure; remember that the full formula is:

$$P = P_0 + \rho g h$$

 

[crhscoog]

That's the right idea; however they give a density (not 1000), and also what you have here is just the increase in pressure; remember that the full formula is:

$$P = P_0 + \rho g h$$

ah right forgot about the given density

P= 101300 + (1025)(9.8)(20)
P= 200900 Pa

Using P=f/a, I multiply 200900 by (4.0 x 10^-5) to get:
f= 8.036 N

 

[alphysicist]

ah right forgot about the given density

P= 101300 + (1025)(9.8)(20)
P= 200900 Pa

I think you forgot to add the first term.

 

[crhscoog]

I think you forgot to add the first term.

ah snap. I am rushing this... sry =D

P(total)= 302200 Pa
F= 302200 x .00004
F= 12.088 N

 

[alphysicist]

ah snap. I am rushing this... sry =D

P(total)= 302200 Pa
F= 302200 x .00004
F= 12.088 N

That looks right to me for the water force on the plug.

 

[crhscoog]

That looks right to me for the water force on the plug.

thank you for your time, patience, and help. i really appreciate it!

 

[alphysicist]

You're welcome; I'm glad to help!