Homebrewed Combined Heat & Power

[Peter Mounsey]

Our off-grid home has a significant winter heating requirement. We have a
woodburner with backboiler feeding radiators but want to also fit an outdoor
hot-tub.
My idea is to fit a water cooled generator, I'm thinking of a marine type
with water cooled exhaust, and making a hot water main fed by the generator
and / or back boiler and feeding domestic hot water, radiators and then the
hot tub in that priority. I reckon that I can manage the switching and
plumbing OK.
Can anyone tell me how much heat the generator should produce while it is
turning out its maximum 10KW? This would be 'worst case' when we wanted the
hot tub rapidly warmed by pushing 6KW heating load plus pumps in
electrically and then the generator hot water too (via heat exchanger of
course). I'm hoping that it will be around 30KW total heat input which would
warm it up nicely!
Has anyone tried this arrangement please?
Pete.

 

[[email protected]]

Your generator will turn out more than 100Kw of heat for it's max load
of 10Kw electrical...

No. Maybe 30-40 kW.

Nick

 

[[email protected]]

'Fraid you're wrong. Internal combustion engine converts less than 10%
of the fuel to mechanical energy.

Wrong.

Nick

 

[Me]

'Fraid you're wrong. Internal combustion engine converts less than 10%
of the fuel to mechanical energy. Small generator converts less than
50% of the mechanical energy to electricity. Oveall efficiency less
than 5%.
ie 95% of the fuel is wasted as heat (of which only some could be
recovered)
*********

I don't know where you learned your Diesel Engineering but the "Big Boys"
who design CoGeneration Plants, use this "Rule of Thumb" when they talk
to "Bean Counters".

1/3 of the fuel BTU's go out the Crankshaft
1/3 of the fuel BTU's go out the Radiator
1/3 of the fuel BTU's go out the Exhaust Stack.

So using that "Rule of Thumb" if the OP was using a 10Kw Diesel Genset
at 95% Rated Load, he could recover about 80% of the Radiator BTU's,
and maybe 35% of the Exhaust BTU's in a very good Marine Style
Water Jacket Cooled Exhaust Manifold, IF he pays good attention to
the recovery system. That would be in the neighborhood of of:

33% Crankshaft + Friction loss = ~29% of the fuel BTU's
33% Radiator *.80 = ~26% of the fuel BTU's
33% Exhaust *.35 = ~12% of the fuel BTU's
_________________ _______________________
Total BTU's used = ~67% of the fuel BTU's@ 95% Load

That matches just about the limit of recoverable energy
in a coGeneration situation, plus or minus a few percent.


Me

 

[AJH]

My idea is to fit a water cooled generator, I'm thinking of a marine type
with water cooled exhaust

I did much the same on a project 10 years ago, we had a lister 3cyl
running an capacitively excited induction generator. The voltage
control left a lot to be desired. The marinised exhaust is really to
prevent the engine space overheating rather than an attempt to get as
much heat as possible from the exhaust.

Diesel exhaust temperature varies with load, at high power expect 66%
of the waste heat to be in exhaust 33% in coolant. At low power the
proportions will differ but the problem is the temperature is lower
for about the same massflow ( unless the generator is asynchronous) so
the heat exchange efficiency is lower.

As I have said before our big mistake was sizing for a peak load with
a low average load. This meant our conversion efficiency was around
20%.

 

[[email protected]]

... the "Big Boys" who design CoGeneration Plants, use this "Rule of Thumb"

1/3 of the fuel BTU's go out the Crankshaft
1/3 of the fuel BTU's go out the Radiator
1/3 of the fuel BTU's go out the Exhaust Stack.

Intelligen's 11 HP Lister diesel system had a measured 93% CHP efficiency.

Nick

 

[Me]

The frictional loss shows up as heat, which must be disipated by the
radiator. Call the work out at the crank 29% if you like, but add the
4% to the radiator if you want to keep track of where the energy went.

Friction Losses, here, are external to the engine, so just how do they
show up anywhere in the radiator???

Why is only 80% recovered from the radiator? It may not be 100%
efficient, but the heat not recovered in a pass through is just fed
back into the engine to go through the cycle again. Where else do you
think the lost 20% is going?

The 20% losses are due to heating loss after the heat exchanger. You
can't get 100% use of those BTU's in ANY system, and 80% actuall recovery
for usefull output is actually a bit high, in most cases.

The marine style exhaust may only recover 35% of the BTU's in the
exhaust, but it wouldn't be hard to build a heat exchanger that could
do much better. How hot is the hot tub? Say the engine air intake was
60 deg. F, the exhaust temp at the valve is 1400 deg. F and the exhaust
temp leaving the heat exchanger is 100 deg. F. The losses would be in
the 40 deg. difference between inlet and exhaust. That could be
further reduced by passing the 100 deg air through a heat exchanger to
heat the house or the inlet air for the engine. If you really want to
go nuts you could use some of the electricity produced to run a heat
pump, drawing the heat out of the exhaust. With the exhaust discharged
at the same temp as the inlet 100% of the exhaust energy is being
recovered.

Again, the "USEFULL" recovered BTU's for this runs about 35% from
Exhaust Heat thru a water cooled Marine Exhaust Manifold.

As for the previous poster's claim, "Small generator converts less than
50% of the mechanical energy to electricity", where is the other 50%
going? If it is going to heat it would soon melt the generator. I
suspect 90% is a lot closer to reality. Even if the generator was only
50% efficient, the waste heat could be used to heat the house.

When you say "Small Generator" what are you talking about? Define what a
"Small Generator" is. Are you talking the Genend itself or the whole
system, engine, Genend, ect? Mechanical to electrical efficency in all
the 20Kw Genends that I run is in the 85 to 95% range, and just how are
you planning to recover any waste heat from the genend itself, build a
pile of duct work and blow the hot air somewhere? Not really a very
efficent way to move BTU's around, and very lossy as well to the outside
enviorment.

I would say the 30 to 40 KW was a good guess. Might want to consider
using electric hot water heaters and electric space heaters to provide
a load for the generator to work against.

Bruce Richmond

Me

 

[daestrom]

The frictional loss shows up as heat, which must be disipated by the
radiator. Call the work out at the crank 29% if you like, but add the
4% to the radiator if you want to keep track of where the energy went.

Why is only 80% recovered from the radiator? It may not be 100%
efficient, but the heat not recovered in a pass through is just fed
back into the engine to go through the cycle again. Where else do you
think the lost 20% is going?

The marine style exhaust may only recover 35% of the BTU's in the
exhaust, but it wouldn't be hard to build a heat exchanger that could
do much better. How hot is the hot tub? Say the engine air intake was
60 deg. F, the exhaust temp at the valve is 1400 deg. F and the exhaust
temp leaving the heat exchanger is 100 deg. F. The losses would be in
the 40 deg. difference between inlet and exhaust. That could be
further reduced by passing the 100 deg air through a heat exchanger to
heat the house or the inlet air for the engine. If you really want to
go nuts you could use some of the electricity produced to run a heat
pump, drawing the heat out of the exhaust. With the exhaust discharged
at the same temp as the inlet 100% of the exhaust energy is being
recovered.

Actually, even that isn't true. If you measure the volumetric flow of air
going into the engine, and the volumetric flow of exhaust leaving the
engine, you'll find the volume of the exhaust gasses (after being cooled to
exactly the same temperature as the inlet) is *larger*. This is because of
the chemical changes in the gasses caused by burning fuel. This larger
volume means a loss as the pressureXvolume product of the gasses is now
larger.

But I would guesstimate the effort to recover this energy is not worth it.

daestrom

 

[daestrom]

Surely some substanial portion of the exhaust heat will be in the
form of heat of vaporization of the water formed during combustion,
are you taking that into account?

That makes the efficiency even lower. By not condensing the vapor, one can
only extract what is known as the Lower Heating Value (LHV). By condensing
all the water vapor, one can extract the HHV (higher heating value). But
condensing *all* the vapor is a challenge. Even if cooled to exactly the
same temperature as the inlet, unless the inlet air was already at 100% RH,
the exhaust will have a higher humidity than inlet, and that means the heat
of vaporization for the added moisture is lost.

A condensing exhaust system would recover a lot of that energy. But now you
have to watch out for some folks playing a 'numbers' game. If they claim
the chemical energy of the fuel is the LHV, and they have a good condensing
exhaust as well as other heat recovery systems, they could conceivably
extract more than the LHV from the fuel and claim over-unity. But they
could never reach 100% of the HHV.

Some home furnaces play this game a bit. They will claim the efficiency of
the furnace is 95%. That is, they recover 95% of the LHV from the natural
gas. They do this with a condensing system and recover some of the HHV
energy. So it seems like some ultra-efficient heating system when in fact
they are getting energy from two components (the sensible heat and latent
heat in the combustion products) and only comparing it with the sensible
heat in the combustion products.

daestrom