Solution #3
posted on Aug 01, 2007
Powe33 - usenet poster
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Peter,
start with the first law of thermodynamics for a control volume (the tank)
dE/dt = (mh)in - (mh)out
where dE/dt is the rate of increase of energy (heating rate) in the tank, m
is a mass flow rate, h is an enthalpy, in is at the tank inlet, out is at the
tank outlet, inlet and exit potential and kinetic energy effects are
neglected, and any heat transfer through the tank wall is neglected.
The inlet and outlet enthalpies can be determined by going into a steam table
with the inlet and outlet temperatures and pressures.
dE/dt = dU/dt, with negligible potential and kinetic energies within the tank,
where U is the instantaneous internal energy of the batch.
dU/dt can be approximated as (m2u2 - m1u1)/(t2 - t1), where m2 is the final
mass in the batch (i.e., at time t2), u2 is the internal energy per unit mass
of the batch at t2, m1 is the initial mass in the batch (i.e., at time t1 =
0), u1 is the internal energy per unit mass of the batch at t1. u1 and u2
can be looked up in a steam table , or inversely, temperature data can be
determined from a computed u2
Also, (m2 - m1)/(t2 - t1) can be approximated by min - mout. Note that m1 and
m2 are masses, whereas min and mout are mass flow rates.
If there is no outlet, the problem is not too bad.
Tod
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