# Thermodynamics- determining mechanical power

1. Sep 8, 2013

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

Hello all,

I've worked through the problem below, but I feel my answer is incorrect. I would appreciate it if someone could check my work.

Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 20kW of shaft power. The free surface of the upper reservoir is 45m higher than that of the lower reservoir. If the flow rate of water is measured to be 0.03 m^3/s, determine mechanical power that is converted to thermal energy during this process due to frictional effects.

2. Relevant equations

$e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)$

$\dot{E}=\dot{m}e_m$

$\dot{m}=\rho\dot{V}$

$\eta=\frac{energy-out}{energy-in}$

3. The attempt at a solution

$e_m = ke =(\frac{P}{\rho})(\frac{1}{2}\vec{V}^2)(gz)=gz=(9.81\frac{m}{s^2})(45m)=441.45\frac{m^2}{s^2}=441.45\frac{J}{kg}$

$\dot{m}=\rho\dot{V}=(1 \frac{kg}{m^3})(0.03\frac{m^3}{s})=0.03\frac{kg}{s}$

$\dot{E}=\dot{m}e_m=(0.03\frac{kg}{s})(441.45\frac{J}{kg})=13.24\frac{J}{s}=13.24W = 13.24*10^{-3}kW$

$\eta=\frac{13.24*10^{-3}kW}{20kW}=.66*10^{-3}=0.066$%

$20kW-20\eta= 19.99 kW$

So 19.99 kW out of 20 are wasted to heat in this problem? doesn't seem correct.

2. Sep 8, 2013

### SteamKing

Staff Emeritus
You have the density of water as 1 kg / m^3. This is some really, really light water.

3. Sep 8, 2013

Ha, how did I miss that? Thanks.

How's the rest of my method look?

4. Sep 8, 2013

### Ygggdrasil

You have the wrong value for the density of water. [edit: nevermind, Steam King beat me to it]

5. Sep 8, 2013