# Homework Help: The power measured by a hydrophone in Watts or Watts/cm^2

1. Aug 18, 2016

### rwooduk

1. The problem statement, all variables and given/known data

2. Relevant equations
RL = 20 log10(Vrms) - OCV
For this hydrophone OCV = -211dB

3. The attempt at a solution
1. RL = 20 log10(Vrms) - OCV
RL = 20 log10(3.5mV) - (-211)
RL = 161dB re uPa

2. 20 log10 (uPa) = dB
Pa = 10^(dB/20) X 10^(-6) = 10^(161/20) X 10^(-6) = 112 kPa

3. ?

4. ?

5. Impedance

So I'm basically stuck in how to convert pascals to watts, and watts per cm^2. Would I use the volume of the water? but wouldn't that be cm^3? why is power measured in W/cm^2? and how do I convert the original 20W into this unit?

Thanks for any help with this!

p.s. If I use the online converter here:

http://www.sengpielaudio.com/calculator-soundlevel.htm

I get 112 kPa = 194 db(SPL) = 25118864 W/m^2 = 2511 W/cm^2 which is HUGE, the value should be ~10W/cm^2 maximum

2. Aug 19, 2016

### rwooduk

err why has this been moved to Introductory? am I being that dumb? if there is a simple answer to this problem i'd love to hear it. aahhhhh yeah im being dumb ... W = Pa m^3 /s

hmmm nope that doesnt work, if I use Pa = 112kPa the volume of the liquid = 400mL and divide by 1us, the value is huge.

3. Aug 26, 2016

### rwooduk

yeah it was me being dumb, i'll just leave this here for anyone else who has a similar moment of dementure