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Energy in compressed air 7

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freefallingbody

Electrical
Aug 18, 2003
55
Dear all,

While analysing the enegry efficiency performance of air compressor, I found difficult to understand the following:

When air is compressed, say from atmosphere to 8 bar(g), the air temperature rises. After the first stage of compression, some heat is removed in the intercooler before the air proceeds to the second stage. Some heat is again removed in the aftercooler after the second stage. The energy balance in this case looks like:

energy input to compressor = heat removed from the air + energy stored in compressed air in the receiver.

So far so good.

Literatures on compressed air systems published by US Dept. of Energy says that the compressed air is a very expensive utility BECAUSE A LOT OF HEAT IS REMOVED DURING AND AFTER COMPRESSION. ( 85% input energy is removed as heat)

Does this mean that efficiency is poor because I am removing lot of heat? What if I cool the compressed air to ambient temperature in the aftercooler? very low efficiency?

Or I am I just confusing myself unnecessarily?

 
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Questions like this come up a lot. They usually can be cleared up with a definition of "effeciency". You can define it in so many ways. My prefered way is the ratio of the potential energy of the compressed gas at pipeline (or receiver) temperature and pressure over the chemical (or electrical) energy used by the driver. For engine-driven compressors this number is typically under 30% since the engine output hp is generally under 40%. With this definition, the heat rejected in the cooler is a very small part of the total energy (and effeciency) losses.

To compare compression regardless of the driver technology, some people define effeciency as the shaft energy in to the gas energy out. In that case temperature lost in the cooler can be a big portion of the total losses. Unless your "amibient temperature" is approaching absolute zero, the added energy rejected cooling from something like 20F approach to ambient vs. cooling to ambient is generally insignificant. I tend to use a stabalized temperature in the reciever or pipeline to compute potential energy, so the impact is non-existant in my calculations.

David
 
As to your other question, if you are including heat loss as part of the efficiency, then the energy used for pre-cooling has to be included and the efficiency is essentially the same.

TTFN



 
Hi,

I will rephrase my query:

Consider the following sentences.

1. Compressed air is very expensive utility
2. Lot of heat is rejected in the process of compression

First sentence is true. But, is the second sentence the correct reason for that? If the compression is completed in the second stage, what does the heat removed in the aftercooler has to do with the efficiency, as the air has already obtained potential energy?

Rgds,

Dinesh
 
Two factors make compressed air an inefficient energy storage medium: one is the energy wasted as heat rejected to the atmosphere during/after compression, and the other is the inability to recover even part of this energy beneficially when you use the compressed air to do work.

You can only cool during inter-stage compression to a temperature as cold as your "cold reservoir", which in most cases is the atmosphere. In reality, you need a reasonable sized intercooler and some temperature difference to provide driving force for heat transfer, so you don't get to ambient temperature prior to your next compression step. Providing turbulence in this intercooler to promote heat exchange saps you of some of your stored energy too. Providing more compression steps and more intercoolers ultimately only gets you so far too. You're ultimately limited by Carnot efficiency.

Remember also that the compressed air will likewise cool when it expands adiabatically. If you do not provide adequate time and area for heat transfer to re-heat this air to ambient temperature (an "inter-heater" if you will), this will rob you of more of your theoretically stored energy. Considering the area required for efficient air-to-air heat transfer, you can see why this is not practical for most compressed air uses.
 
I think we're talking about degrees here. Think of AC power. I've never heard of a viable scheme for storing it in that form. You can convert it to DC and store it in batteries, but that is very ineffecient. The only way that I know of to store large scale quantities of energy that doesn't have a high, unrecoverable cost is pumping water into a water tower and thus avoiding the cost of pumping into a distributon system. Other than that, large-scale energy storage is expensive and inefficient.

I was recently doing an energy balance on a 2,300 hp gas compressor. The inter and after coolers required 90 hp of energy to reject the waste heat to the atmosphere. That is 4% of the namplate hp, and 1.8% of the energy contained in the fuel that was burned. That seems like a reasonable cost to me, but it may seem high to others. The heat energy removed from the product stream was 3 million BTU/hour. I didn't have a reason to calculate the total energy contained in the product stream, but it was a large number (if anyone wants to do it, it was pure methane flowing at 13.2 MMCF/d, interstage pressure was 140 psia, inlet temp to inter-cooler was 230F, temp out of the intercooler was 120F, after cooler is at 362 psia, inlet temp is 265F, and temp out is 120F). On the face of it, it seems like you're wasting a pretty small portion of the total energy.

David
 
This is a very interesting question and I never paid attention to the portion of compressed air energy, lost as heat. The basic idea of saying 'compressed air is costly' may be due to the fact that most people consider it as free as we are drawing it from atmosphere (free of cost).

I tried to do a simple calculation and got some weird(?) results.

Data

Capacity - 100 cfm
Suction Pressure - 14.7 psia
Discharge Pressure - 80 psig or 94.7 psia
Inlet air temperature - 80F

Procedure

Considering single stage compression, the discharge temperautre should be 540 x (94.7/14.7)(1.4-1)/1.4= 919.48 R or 459.48F

If this air is to be cooled down to 80F then the amount of heat to be removed is 100cfm x 0.075 lb/cu.ft x 0.254btu/lb.R x (459.48-80)R = 722.9 btu/min = 17.28 HP

The theoretical adiabatic power consumption can be calculated by

HP = (4.36/1000)[k/(k-1)](cfm)(p1)[(p2/p1)(k-1)/k-1]

Therefore, HP = 15.76 (???)

Can anybody point me where I went wrong (I know and tested the compressor power equation many times)?

 
In response to the many posts in this thread, my response is as follows:

The rigorous computation of a centrifugal compressor requires invoking the fundamental requirements from thermodynamics:

Work = integral of v dP, for pressure (P) going from the suction to the discharge value.

This integration must be performed along the polytropic path, defined by P*(v^n) = constant, where

n is the polytropic exponent, and
v is the molar volume

At each point along the path, the following things must be done, using a proper equation of state (EOS) that is suitable for describing the behavior of the mixture at hand:

(1) Solve for Zv, the vapor compressibility root of the EOS, and then the molar volume (v), density (rho), and finally the actual suction volumetric flow

(2) Using the Zv from step 1, solve for both Cp and Cv for the given EOS. This requires applying the rigorous relationships, derived from thermodynamics, to the EOS.

(3) Calculate k=Cp/Cv

(4) Using the compressor efficiency curves, at the current suction volumetric flow and RPM, find the polytropic efficiency, eta

(5) Use the relationship (n-1)/n = (k-1)/k/eta to find the current value of the polytropic exponent, n

(6) Integrate a small step along the polytropic path. Calculate the incremental work and the new temperature by heat balance. If you have not reached the discharge pressure, return to step 1. Otherwise, you are done.

CAUTION: Use small enough integration step sizes to ensure that the final error is within some desired limit. Typically, 15-20 steps are enough.

Obviously, each of the quantities mentioned in the steps above varies from point to point along the path. The simplified methods presented in undergraduate textbooks assume, invariably, that the polytropic exponent, n, remains constant from suction to discharge. This enables providing a closed form solution for the required integration. Unfortunately, especially for highly non-ideal gases, this simplified equation can be significantly in error.

Some textbooks do not even do an adequate job explaining why one must use a polytropic path to perform this computation. By the way, the compressor curves for polytropic efficiency can be applied to any gas, within a reasonable molecular weight range of the design case, provided the gas velocity is well within the sonic limit.

Therefore, the proper answer to quark’s question is that this problem really cannot be solved properly except by use of a computer to do the tedious integration work, with iterative solution of the EOS to get the vapor root and all derived thermodynamic quantities at each step.
 
The method of process calculation UmeshMathur pointed to, is pretty common. However, why cannot you integrate the equation of polytropic process as a e= v* dP/dH without all that simplifying assumptions which force you to use k and n? I agree, that usually equations of state use (P, T) as a argument, not (P, H), so small additional function should be written to integrate the process equation. But doing that you can avoid using of k what is highly desirable from my point of view for real gas calculations. k IS NOT EQUAL TO Cp/Cv for real gas! EOS should provide you true value of k and then, yes, the result of using that k will be the same as if you solve the process equation directly. But very often EOS providers calculate k as a Cp/Cv which is wrong and you even don't know about it.

 
MichaelPn:

Re: your statement that "k IS NOT EQUAL TO Cp/Cv for real gas!"

With respect, I couldn't disagree more. We have a serious issue with what the thermodynamic definition of k is.

In ALL the thermodynamic textbooks that I possess (over 30), k = Cp/Cv. For a real gas, k is NOT equal to Cp/(Cp - R), which follows from assuming that Cp - Cv = R (valid only for ideal gases).

Both Cp and Cv must be corrected for deviations from the ideal gas to get the correct Cp/Cv ratio. This requires choosing an appropriate equation of state that is valid for the system at hand, not simply a "small additional function". All thermodynamic quantities MUST be calculated consistently from the same equation of state. If the EOS that you have chosen does not represent your data accurately, choose a different one until you have satisfaction. In my experience, the BWR has been a good choice for EVERY compressor calculation I have ever done for over 30 years.

Your last sentence is, therefore, COMPLETELY ERRONEOUS. I notice that your affiliation is with mechanical engineering and it is understandable that you are unaware of the many types of equations of state and their properties. As I understand it, this subject is not normally covered extensively in mechanical engineering thermodynamics courses, but it is the lifeblood for all chemical engineering undergraduates.

As a standard reference for this matter, I recommend the following for an explanation of real gas Cp and Cv calculations:

B. E. Poling, J.M. Prausnitz, and J.P. O'Connell: "The Properties of Gases and Liquids", especially equation 6-5.1 and pages 6.16-6.17, 5th Edition, (McGraw-Hill, 2001). These three gentlemen are among the premier thermodynamic authorities of the world. The required derivatives in equation 6-5.1 are found by applying partial differentiation to the chosen EOS.

Ironically, the compressor calculation method I have described was published originally in a slightly simplified form in an ASME journal by Huntington, if I remember correctly, over 40 years ago. I have always called it "Huntington's method".

Therefore, my position for accurate compressor work remains unchanged:
(a) k = Cp/Cv, with both Cp and Cv being found from a proper EOS.
(b) The required integration cannot be done analytically and requires use of numerical methods, a computer being necessary to preserve sanity owing to the extremely elaborate trial and error involved.
 
UmeshMathur (Chemical)States the process follows a path along pv^n with n varying along the path.
Relating to steps 4 and 5, isn't eta related to the overall process? And obtained from the beginning and endpoints of the process? Therefore the integration should include allowance for a eta that varies along the flow path.
Additionally, for the polytropic process that you are describing,is there heat transfer?

With regard to your comments for VMichaelPn:

"With respect, I couldn't disagree more. We have a serious issue with what the thermodynamic definition of k is"


Many MEs use gamma as Cp/Cv and k as the gamma for the ideal gas.

Regards
 
Hi, sailoday28:

A polytropic process, not being adiabatic, does have heat transfer associated with it. This heat transfer is less than the theoretical limit for an isothermal compression process.

The polytropic exponent, n, is known only by back-calculation from the real gas Cp/Cv ratio. The Cp/Cv ratio is labeled as k or gamma, depending on your preference. My point to MichaelPn was simply that Cp/Cv for a real gas can be found ONLY using an appropriate equation of state. His post stated categorically that (a) "k IS NOT EQUAL TO Cp/Cv for real gas!", and (b) "But very often EOS providers calculate k as a Cp/Cv which is wrong and you even don't know about it."

In my view, this represents a fundamental misunderstanding of the role of Cp/Cv (whether you call it k or gamma) in this problem. The basic point is that, for a real adiabatic process, the gas mixture follows the law:

P*(v^k) = constant, where

k is the adiabatic exponent (IDENTICAL to real gas Cp/Cv), and
v is the molar volume.

For a polytropic process, however:
(a) k is replaced by n, and
(b) n is related to k via the polytropic efficiency, as explained in my post of 6 Apr 06 17:13 in step 5. You may rename k as gamma, but it is still ALWAYS equal to the real gas Cp/Cv.

Doubtless, there are many theoretically equivalent ways to formulate the integration problem. However, any abandonment of the EOS approach to derive the underlying thermodynamic departure functions (from ideality) will prove erroneous, if not disastrous, especially if the gas mixture is very far from ideal.
 
UmeshMathur (Chemical)I quote from Thermodynamics" by Jos. Keenan

A polytropic process is one for which the pressure-volume relation is given by pv^n= constant......
........"
He then shows that for a perfect gas with a reversible polytropic process
Q/(T1-T2)=(Cp-nCv)/(n-1)

Clearly, I'm looking at the case of a perfect gas and the exponent n. For the adiabatic case, Q=0 and n=gamma.
Perhaps I may be missing something relating to use of n in the polytropic process (or reversible polytroic process) as related to compressors.

I do note that in the I-R handbook of compressed air and gas data 2nd ed --- That the polytropic process is irreversible. The hand book further states-"Although n is actually a changing value during compression, an average or effective value, as calculated from experimental information is used."

At any rate, I am stilll curious as to how in your step 6, eta is obtained.

Regards


Regards
 
sailoday28:

There are two entirely distinct issues here: (a) non-ideality of gas, and (b) thermodynamic irreversibility.

I believe that, in the present discussion, we should focus on (a) alone, as the specification of the path (reversible v/s irreversible) is not being disputed.

The vendor's curves typically show that stage polytropic efficiency varies as a function of actual suction volume and RPM. Once the suction compressibility factor is known, the suction volumetric flow is fixed. Once the RPM is also fixed, the polytropic efficiency does not change. However, the Cp/Cv, and consequently, the polytropic exponent (n) do change along the path, by virtue of the following relationship:

(n-1)/n = (k-1)/k/eta

This is why a numerical integration along the path is necessary.
 
How would one calculate the elastic potential energy content of 1 standard cubic foot of compressed air at 25 degrees C (~room temperature) at, for example, 5 psig?

Is there a simple formula for this?
 
UmeshMathur:

About using k in the real-gas calculations. Let's leave aside theoretical discussion for a while and consider the isentropic process as a example. I have calculated P V^k for three cases for two points on the isentropic path each (please see the table below). All of them for pure CH4; BWR was used as a equation of state. As you can see the futher the gas from the ideal the futher from the constant the P V^k value. If it doesn't work for the isentropic process it cannot work for the polytropic one also. That's why I try to avoid using of k in the compressor's calculations for non-ideal gase cases.

P, bar T, K z S, J/kg/K V, m3/kg k P V^k
Case 1 0.5 288.15 0.99901 282.4033 2.983824 1.3072 208734.163
1 337.764 0.99888 282.4033 1.748563 1.286 205157.846
Case 2 5 288.15 0.99013 -922.0423 0.2957308 1.3077 101639.246
10 338.29 0.98905 -922.042 0.1734044 1.2887 104562.695
Case 3 50 288.15 0.90453 -2234.609 0.0270163 1.3448 38888.494
100 342.177 0.91916 -2234.606 0.0163003 1.4111 30007.536


 
MichaelPn:

Your numerical example shows clearly that, even for a "permanent" gas like methane, the variation in k is highly significant over modest changes in pressure and temperature. For other gases such as heavier hydrocarbons and ammonia, these errors can easily get much worse.

Therefore, since k is related to the polytropic exponent n, it stands to reason that n will also vary significantly over the polytropic path. This is the reason why using ideal values of k [=Cp/(Cp-R)] leaves you susceptible to huge errors in compression power. Some even use an ideal gas value for Cp, i.e., fail to correct Cp for deviations from ideality.

Unfortunately, this error is ubiquitous, even in otherwise highly regarded sources of technical information. For example, the 1998 edition of the GPSA Engineering Data Book, widely used in the oil and gas industry, contains a numerical example (Fig. 13-7, page 13-5, “Calculation of k”) where they recommend use of the average temperature to (a) get the mixture ideal gas Cp, (b) find Cv from the ideal gas assumption, and thus (c) calculate an ideal gas k. This value is then recommended (page 13-22 under “Calculating Performance”) for use in real gas compression calculations. Regrettably, the fact that increasing pressure can increase deviations from ideality is completely ignored, and no mention is made anywhere of the use of equations of state for such work. Ironically, the section labeled “Calculating Performance” begins with the phrase “When more accurate information is required …”.

In my opinion, all serious work should always use theoretically sound procedures so that compressor rating and re-rating is done in a professional and defensible manner. With the widespread availability of computers and desktop software, there really is no justification for avoiding use of the proper tools. It is conceivable that contractual disputes can arise with respect to performance guarantees if the evaluation is done with improper or insufficiently rigorous methods.
 
UmeshMathur:
You wrote: "Your numerical example shows clearly that, even for a "permanent" gas like methane, the variation in k is highly significant over modest changes in pressure and temperature. For other gases such as heavier hydrocarbons and ammonia, these errors can easily get much worse." I 100% agree with that. I entirely agree also that ideal gas calculations for k are unacceptable. I'd like to underscore, however, that BWR EOS I used was bought by our company from well known provider and the cost was tens thousands dollars! We can expect a real accurate calculations for such money, can we? You can see how accurate they are even for very moderate case.
Please, consider my point of view: we can solve the polytropic process equation as a polytropic efficiency = v* dP/dH directly, without using polytropic exponent. Of course H and v should be re-calculated on each step (I use 4th order Runge-Kutta method which is good enough). I use S1=S2 condition for isentropic calculations. Such approach provides at least repeatability if not an accuracy.

One of the best sources I know (and PTC-10 references also) is: Journal of Engineering for Power; Vol 84; January 1962; J.M. Schultz; "The Polytropic Analysis of Centrifugal Compressors".


 
MichaelPn:

If you disregard the polytropic efficiency and simply assume S1=S2, you are treating your compressor as an isentropic process. Therefore, you will always over-estimate the required horsepower for a given pressure ratio. Recall the theoretical limits: (a) the isentropic process requires the most horsepower, while (b) the isothermal process requires the least. The polytropic process approximates the real world most closely, and lies somewhere in between, depending on the polytropic efficiency.

The numerical technique you use to solve equations of state (EOS) is not the issue in the present discussion. Rather, it is the inaccuracy of the "traditional" but highly simplistic methods that seem to be used most often, and this is the basis of my objections.

The BWR is a very well established EOS, but is not the only one. Personally, I think your company paid a lot of money if all they got was a compressor program with BWR imbedded. For example, there is a completely rigorous process simulator called PD-Plus, which can simulate not only polytropic compressors but also entire chemical processes, available for a lot less money at
At this time, PD Plus does not have the BWR EOS option, but it does have the Soave-Redlich-Kwong cubic EOS which should be close enough for most real-world applications with light hydrocarbon systems and permanent gases.
 
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