On Dec 14, 2009, at 8:56 AM, Horace Heffner wrote:
The following is in regards to the bubble lift idea at:
http://mtaonline.net/~hheffner/SLVN.pdf
The density of water is 1000 kg/m^3. The pressure of 1 atmosphere is
about 10,000 kgf/m^2.
At a depth of x meters, the weight of water above a square meter is
equal to a column of water x meters high, or x * 1000 kgf/m^3. The
pressure of water at depth x is given by:
P(x) = (1 atm) + (1000 kgf/m^2) *x = (1 atm) * (1000 kgf/m^2)*
[(1 atm)/(10,000 kgf/m^2)] * x
P(x) = (1 atm) + (0.1 atm)/m * x
So, the volume V(x) of a bubble at depth x, having risen from being
at volume V0 at depth D is:
V(x,D) = V0 * P(x)/P(D) = (V0/P(D)) * P(x)
and the work done per cubic meter of water an increment dx is g*
((1000 kg)/m^3)*dx so the work dW(x,D) by V(x,D) gas moving up an
increment dx is:
dW(x,D) = V(x,D) * [g*(1000 kg/m^3) * dx]
dW(x,D) = [(V0/P(x)) * P(D)] * [g*(1000 kg/m^3) * dx]
but with D constant:
dW(x) = (V0 * P(D) * g * (1000 kg/m^3)) * 1/((1 (atm) + (0.1 atm/
m) x) dx
Let k = (V0 P(d) * g * (1000 kg/m^3) / (1 atm)) so:
dW(x) = k * ((1 + (0.1/m) x) dx
To obtain total work W(x) integrating dW(x) from x=-D to 0 we have:
W(x) = [integral x=D to 0] k * 1/(1 + (0.1/m) * x) dx
W(x) = [eval x=D to 0 m] k * ln(|(0.1/m) * x + 1|)* (10 m)
W(x) = k [ln(|(0.1/m) * D + 1|)* (10 m)] - k [ln(|(0.1/m) * 0 +
1|)* (10 m)]
W(x) = k * (10m) * ln(|(0.1/m) * D + 1|)
Example: for v0 = 1 m^3 and D = 2000 m we have:
P(D) = (1 atm) + 0.1 atm/m * (2000 m) = 201 atm
k * (10 m) = (V0 * g * (1000 kg/m^3) * (P(D) / (1 atm)))* (10 m)
k = ((1 m^3) * g *(1000 kg/m^3) * ((201 atm) / (1 atm))) * (10 m)
k = 1.9711x10^7 J
W(2000 m) = (1.9711x10^7 J) * ln(201) = 1,045x10^8 J = 29 kWh
A commercial electrolyzer produces hydrogen using 4.8 kWh per cubic
meter (Peavy, Fuel from Water, p. 22). However, we can include an
equal volume of O2, so that is 2.3 kWh per cubic meter at the
surface. However, at depth, we need P(D)/(1 atm) as much gas, and
this takes V0*(2.3 kWh/m^3)*P(D)/(1 atm) energy, so at 2000 m depth,
that is 201 * 2.3 kWh = 462 kWh, even if the H2 expansion energy is
free. However, this can be offset by recombining the H2 and O2 at the
surface to drive the electrolysis.
If we don't want to recover the chemical energy and use:
COP = [V0*(2.3 kWh/m^3)*P(D)/(1 atm)] /
[V0 * g * (1000 kg/m^3) * (P(D) / (1 atm)))* (10 m) * ln
(|(0.1/m) * D + 1|)
Then COP = (2.3 kWh)/[g*(1000 kg)*10m*ln(|(0.1/m) * D + 1|)]
and if we want COP of 1 then:
(1000 kg/m^3)*10m*ln(|(0.1/m) * D + 1|) = (2.3 kWh)
ln(|(0.1/m) * D + 1|) = (2.3 kWh) / [g*(1000 kg)*10m] = (2.3
kWh)/0.0272 kWh) = 84.4
(0.1/m D * + 1 = exp(84.1) = 4.5x10^26
thus D = 4.5 x 10^27 m. Pretty deep ocean!
This is not a very practical idea!
Best regards,
Horace Heffner
http://www.mtaonline.net/~hheffner/
