Solar power

Sep 15 21:21 UTC 1996

Solar power.  Everyone would love to have it, but it's too
expensive, intermittent, or otherwise difficult for many
applications.

What are some things that can be done with solar?  What
are the limits?

response 2 of 40:

Sep 15 21:57 UTC 1996

Solar power is the power source of choice for remote radio, telemetry,
data logging, even telephone, devices. 

response 4 of 40:

Sep 16 04:37 UTC 1996

My main calculator for the past decade or so is powered by "solar"
energy (is it still solar power you use electric lights to power 
it?)  I do sometimes have to turn on a light to use it, but
overall I prefer it to replacing/recharging batteries.

response 6 of 40:

Sep 16 14:11 UTC 1996

With only an initial cost, shouldn't the cost per kwh go down the more you
use it?  Or did that assume some finite usage period?  For regular batteries,
I could see coming up with a single figure like that...I bet that's high
compared to Detroit Edison, too.

response 8 of 40:

Sep 16 15:31 UTC 1996

Re #6:  I don't recall.  It was an interesting number.  Coin
cells are probably in that range too.

response 10 of 40:

Sep 16 16:32 UTC 1996

(I should note here that the roof of a 1000 ft^2 house in Arizona
receives almost 100 kilowatts of solar heat on a sunny summer day.
Capturing half of that and converting it to electricity at 20%
efficiency is 10 KW electrical, or about a buck an hour at retail
rates.  Such a rig could pay for itself at a rate of about $3000/year
over and above providing hot water for almost continuous showers.)

response 12 of 40:

Sep 16 21:44 UTC 1996

You can't do that with silicon solar cells, because the voltage
drops quite rapidly with increasing temperature.  And PV scales
*very* poorly because every square inch of cell costs about the
same.  Mirrors, pistons and pumps cost not very much more for
larger ones as smaller ones; cost depends largely on parts count.
This is why I think solar-steam is the future in suny areas.

response 14 of 40:

Sep 23 02:01 UTC 1996

This is an outgrowth of a discussion I was having elsewhere.  To wit,
could a solar-steam engine possibly break 15% thermal efficiency,
given some not-unreasonable assumptions?  (The caveat I entered was
that the output steam had to be at 450 F or greater temperature.
The other restriction is that the steam must not be allowed to have
excessive proportions of liquid water.)

Terms for the glossary:

Internal energy:  The energy of the molecular motion of the working fluid.

Enthalpy:  The internal energy of the working fluid, plus the
product of its pressure and volume (which has the units of energy).
This is used when the fluid crosses boundaries.

Entropy:  A measure of the disorder of the working fluid.  The units
of entropy are always (energy)/(absolute temperature)

All the above quantities are measured relative to some reference state,
which is arbitrarily assigned the value of zero.  In the case of the
steam tables, that is usually the triple point of water.  The quantities
are abbreviated:  h = enthalpy, s = entropy, u = internal energy (not
used below).  When mixtures of liquid and vapor are being handled, the
"quality" (mass-fraction which is vapor) is abbreviated x.

Here are 3 scenarios.  In each one, I assume that the expander
(piston or turbine) yields 60% of the available work as output,
the remaining 40% being lost to friction or heat transfer.  I
neglected heat loss (insulation is cheap and losses are too
dependent on physical dimensions).  I am assuming two expansions
with the steam being re-heated between expansions.

Units:  pressure in pounds per square inch absolute.
        Enthalpy in BTU per pound-mass.
        Entropy in BTU per pound-mass per degree Rankine
                (degrees Fahrenheit above absolute zero).

Unit conversions:  A BTU (British thermal unit) is the energy
        required to heat 1 lbm of water by one degree F.  It is
        equal to 1054.4 Joules (1 Joule = 1 Watt-second) or
        approximately 778 foot-pounds.

response 16 of 40:

Sep 23 02:01 UTC 1996

So it appears that 15% can be beaten.

But that's not the best that can be done.  A goodly part of the
heat added in the boiler is spent to bring the feedwater from
141 F up to the boiling point.  Adding heat at low temperatures
adds more entropy than adding it at high temperatures.  If some
of this heat could be taken from a source that is *already* at a
lower temperature, then this entropy isn't created, just moved.
This decreases the amount of entropy (and thus heat) that has to
be rejected at the condenser, increasing the amount that can be
output as work and thus the efficiency.

One of the tricks in the steam trade is the "feedwater heater".
This taps off low-pressure steam and uses it to heat the water
going into the boiler.  The transfer of heat from steam to
water increases entropy, but not as much as adding the same
heat in the boiler because the temperature difference is smaller.

Re-running the previous analysis with a feedwater heater tapping
steam after the first expansion:

State 0:   Feedwater to the feedwater heater.  141 F, h = 109.39.
State 1:   Output of the feedwater heater.  The saturation
        temperature of steam at 60 psia is 293 F.  Assuming
        that the feedwater actually achieves this temperature
        the output water will have h = 262.25.  It picks up
        153.86 BTU/lbm.
State 2:  Output of the boiler, 250 psia, 450 F.  h = 1233.7, s = 1.6252
State 3:  first expander output @ 60 psia, saturated.  h = 1163.7,
        x = 0.984, s = 1.6252.  The work done is 70.0 BTU/lbm.

At this point, enough steam is tapped off to heat up the incoming
water to 293 F.  This requires 153.86 / (1163.7 - 109.39) = 0.146 
of the total flow, and is condensed in the process; it goes directly
to the feedwater pump.  The remainder (85.4%) goes to the reheater
and the second expander.

State 4:  output of the reheater, 60 psia, 450 F, h = 1258.3, s = 1.7413.
        The heat added is 94.6 BTU/lbm of steam, or (94.6*0.854)=80.8 BTU
        of feedwater.
State 5:  Output of the second expander, 3 psia, h = 1124.6.  The work
        output is 133.7 BTU/lbm of steam or (0.854*133.7) = 114.2 BTU/lbm
        of feedwater.

The second expander outputs to the condenser, where the remaining steam
is returned to the 141 F liquid state.  The heat rejected is 1015.2
BTU/lbm of steam, or 867.0 BTU/lbm of feedwater.

Total heat input:  (1233.7 - 262.25 + 80.8) = 1052.2 BTU/lbm
Total work output: 70 + 114.2 = 184.2 BTU/lbm = 17.5%
Total heat rejected: 867.0 BTU/lbm = 82.5% (more or less).

This picks up another 0.8 percent... not really impressive.  Something
else has to be done to get big improvements.  If the cost of the mirror
and boiler is the major expense of the system, it may pay to get fancier.

response 18 of 40:

Sep 23 14:11 UTC 1996

I've read, that in the hay-day of steam power, tripple expansion engines
were considered to be the most efficent.  It makes sense seeing how much
work the condenser sheds.  Are turbines more efficent? (Or is that what
the above discussion assumes.)

response 20 of 40:

Sep 23 22:06 UTC 1996

Well, I understand all that, but reviewing such proposals in detail is
something I get paid (well) to do, and I don't have time to pick through
this one in detail. However I think it misses the point, which is that the
sun is, for all practical thermodynamic purposes, at infinite temperature,
and its energy is nearly 100% convertible to other forms of free energy if
you can deal with the high temperatures. The closest that has been done so
far is a system built out west (naturally) where a large field of
steerable mirrors focus a bunch of sunlight on a "solar furnace" in a
tower, which runs at ca.  1000 C or so (and that's still peanuts compared
to the Stefan-Boltzman temperature of sunlight). Its been a while since I
read about this system - I'm not even sure that water is used - one of the
alkali metals would be a better working fluid. The material problems are
enormous, of course. 

response 22 of 40:

Sep 29 18:18 UTC 1996

I did not read or take the time to understand the thermodynamics
(Asuume I could remember enough of the basics to do so, which is iffy)
but in reading Rane's comments I found a thread of my own thinking.
Why is efficiency important? A big problem with Solar power is that it is
not particularly reliable (in most places where power is needed).

If this item was really not intended to address this type of issue, then
that's fine, but is thus reduced to a mere academic exercise.

response 24 of 40:

Sep 30 04:41 UTC 1996

More to the point, efficiency is important (relatively) because
mirror-area is one of the costlier parts of a system.

The point of the exercise was to try to get an idea of what is
possible, and it appears that the sunlight falling on the roof
of a home is more than capable of powering everything inside.