In this problem we are asked to do an energy
balance on a heat exchanger.
We're given the feed which is superheated
steam and were're told how much energy is
removed from that steam to heat some reactor
feed.
We want to know what's the outlet temperature
of that steam, what phases are present and
then what's the enthalpy.
So first thing we want to do is draw a diagram
so we can input the information we already
know about the system.
So here we are showing the mas flow rate in,
m dot, that's a constant so the same mass
is leaving.
We don't know the outlet temperature and the
heat removed is - if we take as my system
is the steam so I'm removing 450 kJ per second
which corresponds to 450 kW.
So what i want to do is the energy balance.
It's a steady state system and so the energy
balance is going to be steady state, we have
energy flowing into the system, the inlet
stream.
Same mass flowrate out, enthalpy is going
to be different and then we have a rate of
heat removal.
Of course in general we would have work but
there is no work in this system.
First thing we are going to do is look up
inlet enthalpy using steam tables.
So 200 C, 10 bar, go to superheated steam
tables.
And we know it's superheated, one because
this pressure is less than saturated pressure
and the problem statement states it out.
And so the enthalpy I can just read directly.
The enthalpy coming in, 2828 kJ per kg.
I can solve this equation for the enthalpy
out.
So let me first rearrange the equation and
the energy balance.
So I rearranged the energy balance by taking
these two terms to the other side of the equation
and then I substituted in the values for Q
dot and M dot.
So H out is just H in minus 225.
H in, something we just looked up in the steam
tables, 2828 kJ per kg minus the 225 kJ per
kg.
And so the enthalpy out, in order to determine
the temperature out we go back to the steam
tables and we look at enthalpy at 10 bar for
saturated vapor.
And the enthalpy: 2777 kJ per kg.
Which means, since we we're at the value less
than that but we're at 10 bar, we must be
in the two phase region.
so that means we're at saturation conditions,
we can calculate the quality, because the
enthalpy is a mixture of so much vapor.
So saturated vapor, so x is the fraction of
vapor, 1-x is the fraction that's liquid.
Enthalpy liquid saturation conditions.
We know the total enthalpy, that's 2603 kJ/kg
so I'm going to write down the values for
saturated liquid, saturated vapor at 10 bar.
So we can solve for x the quality.
And is 0.91 so that means we know the temperature
leaving, it's one of the things we wanted
to determine and that's the saturation temperature,
10 bar 179.9, which, significant figures,
180.
We now know the quality so it's wet steam
that's leaving. and it's enthalpy 2603 kJ/
kg.
That's the information we are asked to determine.
