Is My Calculation for Bicycle Pump Pressure Incorrect?

[TSny]

So, is the following method correct?

Yes, that looks good. But, note that your expressions for $W_{3a}$ and $W_{3b}$ will be negative quantities since the arguments of the logarithms are less than one. This has to do with Chestermiller's expressions that represent the work done by the air instead of the work done on the air. I believe the problem is asking for work done on the air by the person operating the pumps.

 

[Hak]

Yes, that looks good. But, note that your expressions for $W_{3a}$ and $W_{3b}$ will be negative quantities since the arguments of the logarithms are less than one. This has to do with Chestermiller's expressions that represent the work done by the air instead of the work done on the air. I believe the problem is asking for work done on the air by the person operating the pumps.

Ok, thank you. Is the method for finding stroke 4 the same?

 

[TSny]

Ok, thank you. Is the method for finding stroke 4 the same?

Yes.

 

[Hak]

Yes.

So $W_1$ is the sum of all 4 contributions, right? As for $W_2$? How should it be?

 

[TSny]

So $W_1$ is the sum of all 4 contributions, right?

Yes

As for $W_2$? How should it be?

That should be easy. It's very similar to the first stroke of the small pump.

 

[Hak]

That should be easy. It's very similar to the first stroke of the small pump.

I'm getting:
$$W_2=P_{atm}(V_C+ 4V_p)\ln{\frac{V_C}{V_C+4V_p}}$$.

Is that correct?

 

[TSny]

I'm getting:
$$W_2=P_{atm}(V_C+ 4V_p)\ln{\frac{V_C}{V_C+4V_p}}$$.

Is that correct?

Yes

 

[Hak]

Yes

Maybe with a minus sign, right?

 

[TSny]

Maybe with a minus sign, right?

Yes. Or, you could just flip the fractions inside the logarithms.

 

[Hak]

Yes. Or, you could just flip the fractions inside the logarithms.

Ok, thanks. So, we have to demonstrate that $$W_1 = P_{atm}(V_C+ 4V_p)\ln{\frac{V_C}{V_C+4V_p}}$$, right?

 

[TSny]

Ok, thanks. So, we have to demonstrate that $$W_1 = P_{atm}(V_C+ 4V_p)\ln{\frac{V_C}{V_C+4V_p}}$$, right?

Right

 

[Hak]

Right

Ok. Unfortunately, I am still not seeing that. With the 3 results found (plus the fourth similar to that one), all with plus sign, I arrive at a different result. Should I consider all 4 results with equal sign or not? I will post the calculations later.

Edit: I arrive at the same result if and only if I consider the first stroke with opposite sign to the other three. Why?

 

[TSny]

Edit: I arrive at the same result if and only if I consider the first stroke with opposite sign to the other three. Why?

I'm not sure what particular expression you are using for the work done on the system during the first stroke. Are you using your expression from post #1 or @Chestermiller's expression in post #13? You know the work done on the system is positive, so the correct expression will be the one that gives positive work.

 

[Hak]

I'm not sure what particular expression you are using for the work done on the system during the first stroke. Are you using your expression from post #1 or @Chestermiller's expression in post #13? You know the work done on the system is positive, so the correct expression will be the one that gives positive work.

@Chestermiller's expression in post #13. Changing the sign of what @Chestermiller indicates with $W_1$, and keeping that of the other three work contributions, would lead to the correct result... Why don't they all have the same sign? Where am I going wrong?

Edit. I may have understood. The first contribution is the only positive one because the argument of the logarithm is greater than 1. For the result to adhere to the physical situation, the other 3 contributions must also be positive, so I must flip the fractions within the logarithms of these 3 contributions. In this case, the result matches. The work done on the air is thus obtained. Flipping only the fractions of the logarithm of the first contribution, this will be negative, so you get 4 contributions that are all negative, returning the work done by the air. Right?

 

[TSny]

Edit. I may have understood. The first contribution is the only positive one because the argument of the logarithm is greater than 1. For the result to adhere to the physical situation, the other 3 contributions must also be positive, so I must flip the fractions within the logarithms of these 3 contributions. In this case, the result matches. The work done on the air is thus obtained. Flipping only the fractions of the logarithm of the first contribution, this will be negative, so you get 4 contributions that are all negative, returning the work done by the air. Right?

Yes, I think you have it.

 

[Hak]

Yes, I think you have it.

Thanks. Have we overlooked something about this problem?

 

[TSny]

Thanks. Have we overlooked something about this problem?

I can’t think of anything that was overlooked.

 

[Chestermiller]

I think it was brilliant for @TSny to note that, since the initial and final states of the two processes are the same, the entropy change must be the same. And, since all the heat transfer takes place from a single constant temperature reservoir at T, the heat removed from the gas for each of the two reversible process must be the same. And, since the heat removed from the gas for each of the two reversible process must be the same, the total work done on the gas for each to the two processes must be the same. Bravo!!

 

[Hak]

I can’t think of anything that was overlooked.

OK, it was just to verify. My numerical result for item 1.) is:
$$p (n=4) = p_A \left(1+ \frac{4 V_p}{V_C} \right) = 101.3 kPa \left(1+ \frac{4 \times 0.3}{4.5} \right) \approx 128.3 kPa.$$

Is that correct? Why this formula is equal to the first I found out at first order (when applying the binomial formula)? Just for curiosity.

For item 2.), as we said, $$W_1-W_2 =0$$. If you want, I will post the long calculations.

In addition, I would like to make a graphic diagram of the physical situation and the procedure performed. Can anyone tell me how this can be done in this specific case? Thank you.

 

[TSny]

OK, it was just to verify. My numerical result for item 1.) is:
$$p (n=4) = p_A \left(1+ \frac{4 V_p}{V_C} \right) = 101.3 kPa \left(1+ \frac{4 \times 0.3}{4.5} \right) \approx 128.3 kPa.$$

Is that correct?

Looks good.

Why this formula is equal to the first I found out at first order (when applying the binomial formula)? Just for curiosity.

Yes, your original formula happens to agree with the correct formula to first order in the small quantity $V_p/V_C$.
From the binomial expansion we have $(1+x)^n \approx 1 + nx$ for small $x$.

 

[Hak]

Yes, your original formula happens to agree with the correct formula to first order in the small quantity $V_p/V_C$.
From the binomial expansion we have $(1+x)^n \approx 1 + nx$ for small $x$.

Thank you. I cannot, however, understand why. The most general possible formula should be @Chestermiller's, not mine.

 

[Hak]

I've also found the attached document. What do you think? Can it be useful to me in any way for some application of a similar physical situation? Do you recommend reading it carefully or is it a waste of time? Thanks.

 

[TSny]

Thank you. I cannot, however, understand why. The most general possible formula should be @Chestermiller's, not mine.

Let $x = V_p/V_C$. The correct formula for the final pressure when there are "n transfers" of the small pump is $$P = P_{atm}\left(1+\frac{nV_p}{V_C}\right) = P_{atm}(1+nx).$$
Your formula in post #1 is $$P= P_{atm}\left(\frac{V_C + V_p}{V_C}\right)^n = P_{atm}(1+x)^n. $$
Are you asking why these two formulas give approximately the same result when $x$ is much less than 1?

 

[Hak]

Let $x = V_p/V_C$. The correct formula for the final pressure when there are "n transfers" of the small pump is $$P = P_{atm}\left(1+\frac{nV_p}{V_C}\right) = P_{atm}(1+nx).$$
Your formula in post #1 is $$P= P_{atm}\left(\frac{V_C + V_p}{V_C}\right)^n = P_{atm}(1+x)^n. $$
Are you asking why these two formulas give approximately the same result when $x$ is much less than 1?

Thank you very much, but unfortunately that was not my question. What I meant was that, if it meant anything, the most general possible formula for $p$ would be the one I set out in post #1, i.e. the one with the power term. Doing the necessary approximations for small $x$, applying the binomial formula, you get @Chestermiller's formula (which is actually the correct one). So, if that were the case, the most correct expression possible would be mine, and that of @Chestermiller only a special approximation for small $x$. On the other hand, if the most correct possible formula is @Chestermiller's (which is indeed true), how to start from it to obtain my equation? One would be starting from a particular equation to obtain a general one, which is contradictory. This physical assumption (@Chestermiller's equation is an approximation of my equation for small $x$) would therefore not be correct. This is what I wanted to know, and why. Thank you.

 

[TSny]

What I meant was that, if it meant anything, the most general possible formula for $p$ would be the one I set out in post #1, i.e. the one with the power term. Doing the necessary approximations for small $x$, applying the binomial formula, you get @Chestermiller's formula (which is actually the correct one). So, if that were the case, the most correct expression possible would be mine, and that of @Chestermiller only a special approximation for small $x$.

I don't understand your argument here or what you mean when you say that your formula is "the most general possible formula".

The correct formula for $P$ is $P = P_A(1+nx)$. There are an infinite number of incorrect formulas that would approximate the correct formula for small $x$. For example, $P = P_A\left(1+\frac{1}{2} nx \right)^2$, $P = P_Ae^{nx}$, and $P = P_A\left[1 + n\tan(x)\right]$. They are certainly incorrect formulas for the final pressure. But they do give approximately the right answer for small $x$. How do you decide which of these incorrect formulas is "the most general"? Even if you did have a criterion for deciding "most general", what does being "the most general expression" have to do with being "the most correct expression"?

On the other hand, if the most correct possible formula is @Chestermiller's (which is indeed true), how to start from it to obtain my equation?

I don't understand the motivation for trying to obtain an incorrect formula starting from the correct formula.

 

[Hak]

I don't understand your argument here or what you mean when you say that your formula is "the most general possible formula".

The correct formula for $P$ is $P = P_A(1+nx)$. There are an infinite number of incorrect formulas that would approximate the correct formula for small $x$. For example, $P = P_A\left(1+\frac{1}{2} nx \right)^2$, $P = P_Ae^{nx}$, and $P = P_A\left[1 + n\tan(x)\right]$. They are certainly incorrect formulas for the final pressure. But they do give approximately the right answer for small $x$. How do you decide which of these incorrect formulas is "the most general"? Even if you did have a criterion for deciding "most general", what does being "the most general expression" have to do with being "the most correct expression"?I don't understand the motivation for trying to obtain an incorrect formula starting from the correct formula.

Thank you. I wanted to say that if my expression was correct, it would be as general as possible, since from it, via the binomial formula, we get @Chestermiller's formula. If my expression is wrong (which it unquestionably is), that makes no sense. When I had asked "why, via the binomial formula, do I get @Chestermiller's correct one from my result?", you understood my question in a purely mathematical sense, telling me what binomial formula had to be used for this to make sense. Instead, I wanted to know if my solution could somehow be correct, since there was this particularity. I now had the answer to my doubt: my solution is incorrect, full stop. Binomial calculus and Taylor series have no relevance here from a physical point of view. Thank you for clarifying my doubt.
You and @Chestermiller have been very helpful.

Also, I wanted to add that the author of the exercise said that he had not yet found an intuitive solution for point 2.) I think the intuitive method may be your brilliant solution, @TSny, or am I wrong? I will refer it to him. As you can see, everyone was convinced to do calculations, no one but you would ever think there was a faster solution. Thanks again.

 

[TSny]

Also, I wanted to add that the author of the exercise said that he had not yet found an intuitive solution for point 2.) I think the intuitive method may be your brilliant solution, @TSny, or am I wrong? I will refer it to him. As you can see, everyone was convinced to do calculations, no one but you would ever think there was a faster solution. Thanks again.

I appreciate the compliments on my solution. But, I did not see that solution at first! Initially, I went through essentially the same calculations as outlined by @Chestermiller. When I saw that $W_1$ and $W_2$ came out the same, I stepped back to see if there was a different way. It took me a while.

Here's another way to see the answer. The volume of the large pump is four times the volume of the small pump. So, we can think of the initial air in the large pump as consisting of four "parcels" of air each of which has an initial volume equal to that of the small pump.

First step: Push the knob of the large pump until the first parcel of air is transferred to the container. The three parcels of air remaining in the large pump are now at the new pressure, $P_1$, of the container. So, each remaining parcel has been compressed from atmospheric pressure to $P_1$. The total work done so far is the work done to transfer the first parcel plus the work done to compress the other 3 parcels. Note that the work done on just the first parcel equals the work that is done by the small pump during its first stroke. So, the total work done by the large pump at this point is greater than the work done by the first stroke of the small pump.

Second step: Push the knob of the large pump farther until the second parcel of air has been transferred. The pressure of the system increases from $P_1$ to $P_2$. The total work done in this step is the work done on parcel number 2 as it was transferred to the container plus the work done on the two remaining parcels to compress them so that their pressure increases to $P_2$.

Note that the total work that was done on parcel number 2 is the work that was done on it in the first step to raise its pressure from atmospheric pressure to $P_1$ plus the work that was done on it as it was transferred to the container in the second step. Thus, this total work on parcel 2 matches the work done by the small pump during its second stroke.

Third step: Parcel number 3 will be transferred and the total work done on parcel 3 will be the work done on it during steps 1 and 2 to compress it from atmospheric pressure to $P_2$ plus the work done in step 3 to transfer it to the container. So, the total work done on parcel 3 matches the work done by stroke 3 of the small pump.

Fourth step: (Left for you).

 

[Hak]

I appreciate the compliments on my solution. But, I did not see that solution at first! Initially, I went through essentially the same calculations as outlined by @Chestermiller. When I saw that $W_1$ and $W_2$ came out the same, I stepped back to see if there was a different way. It took me a while.

Here's another way to see the answer. The volume of the large pump is four times the volume of the small pump. So, we can think of the initial air in the large pump as consisting of four "parcels" of air each of which has an initial volume equal to that of the small pump.

First step: Push the knob of the large pump until the first parcel of air is transferred to the container. The three parcels of air remaining in the large pump are now at the new pressure, $P_1$, of the container. So, each remaining parcel has been compressed from atmospheric pressure to $P_1$. The total work done so far is the work done to transfer the first parcel plus the work done to compress the other 3 parcels. Note that the work done on just the first parcel equals the work that is done by the small pump during its first stroke. So, the total work done by the large pump at this point is greater than the work done by the first stroke of the small pump.

Second step: Push the knob of the large pump farther until the second parcel of air has been transferred. The pressure of the system increases from $P_1$ to $P_2$. The total work done in this step is the work done on parcel number 2 as it was transferred to the container plus the work done on the two remaining parcels to compress them so that their pressure increases to $P_2$.

Note that the total work that was done on parcel number 2 is the work that was done on it in the first step to raise its pressure from atmospheric pressure to $P_1$ plus the work that was done on it as it was transferred to the container in the second step. Thus, this total work on parcel 2 matches the work done by the small pump during its second stroke.

Third step: Parcel number 3 will be transferred and the total work done on parcel 3 will be the work done on it during steps 1 and 2 to compress it from atmospheric pressure to $P_2$ plus the work done in step 3 to transfer it to the container. So, the total work done on parcel 3 matches the work done by stroke 3 of the small pump.

Fourth step: (Left for you).

Thank you. For the fourth step, I would follow the same path as the third step. I would say that parcel number 4 will be transferred and the total work done on parcel 4 will be the work done on it during steps 1, 2, and 3 to compress it from atmospheric pressure to $\displaystyle P_3$ plus the work done in step 4 to transfer it to the container. So, the total work done on parcel 4 matches the work done by stroke 4 of the small pump.

Therefore, we can see that the total work done by one stroke of the large pump is equal to the sum of the work done by each parcel of air, which is equal to the sum of the work done by each stroke of the small pump. This would confirm our previous result that $W_{four} = W_{one}.$

If I have made mistakes, please point them out to me. I would appreciate any subsequent response from you that better formulates the answer to Fourth step. Please do not hesitate to let me know. Thanks.

 

[TSny]

Thank you. For the fourth step, I would follow the same path as the third step. I would say that parcel number 4 will be transferred and the total work done on parcel 4 will be the work done on it during steps 1, 2, and 3 to compress it from atmospheric pressure to $\displaystyle P_3$ plus the work done in step 4 to transfer it to the container. So, the total work done on parcel 4 matches the work done by stroke 4 of the small pump.

Therefore, we can see that the total work done by one stroke of the large pump is equal to the sum of the work done by each parcel of air, which is equal to the sum of the work done by each stroke of the small pump. This would confirm our previous result that $W_{four} = W_{one}.$

Yes, that looks good.

 

[Hak]

Yes, that looks good.

Thank you so much!