Gauge pressure at different depths in a 56 m reservoir behind a dam

[DrMcDreamy]

Homework Statement

This is a two part problem:

a) A reservoir behind a dam is 56 m deep. What is the gauge pressure at the base of the dam? The acceleration of gravity is 9.8 m/s 2 and atmospheric pressure is 10100 Pa . Answer in units of kPa.

b) What is the gauge pressure 19 m from the top of the dam? Answer in units of kPa.

Homework Equations

I don't know what formula I am suppose to use. I know it has to do with gauge pressure, so a guess:

P₂=P₁ + $$\rho$$gh

OR

P₁+$$\rho$$gh₁+$$\rho$$$$\frac{v1^2}{2}$$=P₂+$$\rho$$gh₂+$$\rho$$$$\frac{v2^2}{2}$$

The Attempt at a Solution

But I don't know the exact pressure, I just know since its minimal, the top and bottom velocity (I believe) gets cancelled.

So are either of those right? If not, what do I use? TIA

 

[tiny-tim]

HI DrMcDreamy!

(have a rho: ρ )

Gauge https://www.physicsforums.com/library.php?do=view_item&itemid=80" is total pressure minus atmospheric pressure.

Either of those equations will do.

 

[DrMcDreamy]

^Are you assuming the reservoir is filled with water, so $$\rho$$ would be 1000 kg/m³?

 

[DrMcDreamy]

So would it be:

a) P₂=P₁+$$\rho$$gh

P₂-P₁=$$\rho$$gh

P₂-P₁=(1000 kg/m³)(9.80 kg m /s²)(56 m)

P₂-P₁=548,800 Pa

which would equal 548.8 kPa

b) P₂=P₁+$$\rho$$gh

P₂-P₁=$$\rho$$g

P₂-P₁=(1000 kg/m³)(9.80 kg m /s²)(37 m)

P₂-P₁= 362,600 Pa

which would equal 362.6 kPa

So is the work and answers right? If not, what did I do wrong? TIA

 

[tiny-tim]

Looks ok

 

[DrMcDreamy]

Thank you tiny-tim!

It seems the second part was wrong, because I figured it had wanted me to 56-19=37, so it would've been from the bottoms up, but it was supposed to be 19 so:

b) P₂=P₁+$$\rho$$gh

P₂-P₁=$$\rho$$g

P₂-P₁=(1000 kg/m³)(9.80 kg m /s²)(19 m)

P₂-P₁= 186,200 Pa

which would equal 186.2 kPa which is the correct answer to the 2nd part!