# Thermodynamics: Control Volume analysis using energy

## Homework Statement

Steam flows through an uninsulated pipe at 0.5 kg/s, entering at 5 bar, 440 C and exiting at 3 bar, 320 C. How much energy is lost from the steam per hour?

## Homework Equations

Which terms in the energy eqn can drop out? I'm having trouble with that.

Is there only one W?

## The Attempt at a Solution

Starting w/ the full energy eqn:

de/dt = Q - W_ + (m_i)[u_i + (p_i)(v_i) + (v_i)^2 + g*z_i] - (m_e)[u_v + (p_e)(v_) + (v_e)^2 + g*z_e]

where Q = vol. flow rate
W = work rate
m = mass flow rate
subscript e = out
subscript i = in
v = velocity
u = specific internal energy
p = pressure

So z_i = z_e = 0

Q = mv = (0.5)(0.6548) = 0.3274?

I would treat this as a steady state problem. Meaning your mass flow rate in is going to equal your mass flow rate out. and de/dt is zero.

Since we're not given any sorts of elevation, the potential energy will be drop out. No work is happening, so W will also drop out.

You will be left with
Q/m=hb-ha+(V^2/2)b-(V^2/2)a

h is of course equal to u+pv

a=in
b=out
m=mass flow rate
Q=energy rate
v=specific volume
h=enthalpy
V=velocity

I can't check any of your values because I don't have a thermo book with me.

ok that makes sense, but how do i find velocity?

You know what, based on the information given, I would let velocity drop out as well. In a basic thermo class most prevalent place you're going to see velocity -not- being negligible is in nozzle and diffuser problems, or where it is expressively given to you in the problem statement.