Kinetic energy of water pouring

[kox]

Hello. It's not a homework but it is kind of exercise. I was wondering what's the kinetic energy of water flow (for a certain amount) from a faucet. But I'm not sure if I got it right. So I counted how much volume of water is dropped in 30 seconds. It turned out it's about 5.6liters.
m=5.6kg // as 1 liter of water is almost 1kg
t=30s
l=1.256*10E-3m^2 // area of water dropping, cross-sectional area
Vol=5.6*10E-3m^3 // volume of water
Ek=mv^2/2
v=s/t
s=Vol/l
s=5.6*10E-3/1.256*10E-3 [m^3/m^2=m]
s=4.45m
v=4.45m/30s
v=0.148m/s
Ek=5.6kg*(0.148m/s)^2/2
Ek=5.6*0.0219kg*m^2/s^2
Ek=0.0613J
Ek=61.3mJ
So I got that pouring 5.6 liters of water realeses 61.3 milijoules. I thought it woul be a little more when I see water dropping at high pressure.
Did I get that right?
I don't know if I posted it in the right section. I'm ew here. It's not a homework but I'm afraid I would get my post deleted if I posted it in classic physics. Sorry I didn't use the template provided,but they didn't fit my post too good.

 

[mfb]

The kinetic energy of slowly moving masses is tiny. You get the same kinetic energy if you drop the water by a height of just ~1mm.

 

[kox]

I don't understand how is it equal to dropping water (how much?) from 1mm height? The water was dropping from more height and was at pressure. Could you explain?

 

[mfb]

In that case, some energy went to something else.

Dropping 5.6kg of water by a height of 1mm gives releases potential energy of mgh=5.6kg*10m/s²*0.001m = 0.056J = 56mJ.