origin said:
You were certainly dishonest when you said:
Well, I guess this is goodbye then, since we'll obviously never see each other again, because oil prices are only going down
and when I called you on that prediction you said,
What I said to you in the context of that particular post was not a prediction, and you know it.
That sounds pretty dishonest to me!
Gosh, origin, when you put it that way, it sounds pretty damning. But when I go back to page 54 to have a look at the context of the interchange, what I actually find tells a completely different story than the one you keep trying to invent!:
Futilitist said:
"Values for $$E_{TP}$$ are derived from the solution of the Second Law statement, the
Entropy Rate Balance Equation for Control Volumes:
$$\frac{dS_{CV}}{dt}
=\sum_j\frac{\dot{Q}_{j}}{T_{j}}
+\sum_i\dot{m}_{i}s_{i}
-\sum_e\dot{m}_{e}s_{e}
+\dot{\sigma}_{cv}$$
"Where $$\frac{dS_{CV}}{dt}$$ represents the time rate of change of entropy within the control volume. The terms $$\dot{m}_{i}s_{i}$$ and $$\dot{m}_{e}s_{e}$$ account, respectively, for rates of entropy transfer into and out of the control volume accompanying mass flow. The term $$\dot{Q}_{j}$$ represents the time rate of heat transfer at the location on the boundary where the instantaneous temperature is $$T_{j}$$. The ratio $$\frac{\dot{Q}_j}{T_j}$$ accounts for the accompanying rate of entropy transfer. The term $$\dot{\sigma}_{cv}$$ denotes the time rate of entropy production due to irreversibilities within the control volume."
~
(Taken from Fundamentals of Engineering Thermodynamics by Moran and Shapiro)
There is only one temperature boundary, which is at the exit point of the reservoir, and there is no crude
entering the reservoir from the environment. So the equation reduces to:
$$\frac{dS_{CV}}{dt}=\frac{\dot{Q}_{j}}{T_{j}}-\dot{m}_{e}s_{e}+\dot{\sigma}_{cv}$$
giving: $$\frac{BTU}{sec*°R}$$
Since crude oil and water can be considered as incompressible substances for this application their specific entropy's ($$s_{c}$$ and $$s_{w}$$) are only affected by a change in temperature.
For specific heats: $$c_{v}=c_{p}=c$$, and $$s_{2}-s_{1}=c*\ln{\frac{T_{2}}{T_{1}}}$$ The reservoir temperature is constant so the entropy of the reservoir ($$S_{cv}/dt$$) must be decreasing (negative in sign) at the same rate that entropy is transferred by mass flow from the reservoir. The temperature of the mass transporting $$s_{e}$$ is the same within the reservoir as at the exit boundary. The exit boundary is where the well bore enters the reservoir. Therefore, as $$dS_{cv}/dt$$ and $$\dot{m}s_{e}$$ ($$dS_{cv}/dt$$ → 0 as $$\dot{m}s_{e}$$ → 0) must cancel, and the heat leaving the reservoir is negative in sign, the equation becomes:
$$\frac{\dot{Q}_{j}}{T_{j}}=\dot{\sigma}_{cv}$$
giving: $$\frac{BTU}{sec*°R}$$
The rate of entropy production in the petroleum production system is equal to the rate of heat extracted from the reservoir divided by the reservoir temperature.
The rate of irreversibility production in the
PPS therefore becomes:
$$\dot{I_{cv}}=T_{O}*\dot\sigma_{cv}$$
giving: $$\frac{BTU}{sec}$$
Where $$T_{O}$$ equals the standard reference temperature of the environment, 537 °R (77° F).
Therefore:
$$E_{TP}=\int_{t1}^{t2}\dot{I_{cv}}dt$$
giving: $$BTU$$
Because the mass removed from the reservoir is limited to crude oil and water, the increase in $$E_{TP}$$ per billion barrels (Gb) of crude extracted as $$ds=c\frac{dT}{T}$$ is:
(
Equation#7)
$$\frac{E_{TP/lb}}{Gb}
=\begin{bmatrix}\frac{(m_{c}*c_{c}
+m_{w}*c_{w})(T_{R}-T_{O})}{m_{c}} \end{bmatrix}/Gb$$
giving: BTU/lb/Gb
where: 0 ≤ ETp ≤ EG
$$m_{c}$$ = mass of crude, lbs.
$$c_{c}$$ = specific heat of crude, BTU/lb °R
$$m_{w}$$ = mass of water, lbs.
$$c_{w}$$ = specific heat of water, BTU/lb °R
$$T_{R}$$ = reserve temperature, °R
$$T_{O}$$ = standard reference temperature of the environment, 537 °R
$$s_{i}$$ = specific entropy into the control volume
$$s_{e}$$ = specific entropy exiting the control volume
BTU/gal/Gb for 35.7° API crude = BTU/lb/Gb * 7.0479 lb/gal
Evaluation of $$E_{TP}$$ from
Equation# 7 requires the determination of three variables: mass of the crude ($$m_{c}$$) mass of the water ($$m_{w}$$), and the temperature of the reservoir ($$T_{R}$$). These must be determined at time (t).
1) The mass of crude at time (t) is derived from the cumulative production function,
2) the mass of water is derived from the average % surface water cut (fw) of the reservoir,
3) temperature of the reserve is derived from the well depth. This assumes an earth temperature gradient of 1°F increase per 70 feet of depth.
So, Origin, since you are the math expert, what is wrong with all that?"
origin said:
"That is nice. Now please tell me why $$\dot{\sigma}$$ is not zero?
"
Futilitist said:
"Why should it be zero? Is the oil production process reversible?
In natural systems, the entropy production rate of every process is always positive (ΔS > 0) or zero (ΔS = 0). But only idealized adiabatic (perfectly insulated) and isentropic (frictionless, non-viscous, pressure-volume work only) processes actually have an entropy production rate of zero. Heat is produced, but not entropy. In nature, this ideal can only be an approximation, because it requires an infinite amount of time and no dissipation.
An irreversible process degrades the performance of a thermodynamic system, and results in entropy production. Thus, irreversible processes have an entropy production rate greater than zero (ΔS > 0), and that is really what the second law is all about (beyond the second law analysis of machines or devices). Every naturally occurring process, whether adiabatic or not, is irreversible (ΔS > 0), since friction and viscosity are always present.
"
origin said:
"I was going to discuss this but it is not really worth it.
I have decided I will just wait to see the price of oil increase and then come back to the thread and laugh at your hand waving explanation of how that was also predicted by the ept model.
"
Futilitist said:
"You sound like you almost believe that. Well, I guess this is goodbye then, since we'll obviously never see each other again, because oil prices are only going down.
"
So, origin, it is very clear that you are taking what I obviously intended as an off-handed joke and claiming that I was making a serious prediction instead. I have told you many times before that you are taking my statement *WAY* out of context, but you keep bringing it up over and over again. This is extremely childish. Please stop it. Thanks.
Ophiolite said:
I for one am prepared to accept this statement.
Great. Then we are moving on.
Does anyone have any serious comments?
---Futilitist
