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I understand that work = force $\cdot$ distance, but I'm not sure of the units and conversions to use.

What I know is that I have a piston head of area $\pu{28cm^2}$ (~$\pu{4.4in^2}$) traveling a distance of $\pu{6cm}$ (2.4") perpendicular to an opposing force of $40,000$ psi . How much work is being done? I'm guessing I am looking for an answer in joules?

My thinking so far is that the answer to this question is as follows: work in $\pu{Joules}$ = $\pu{Pa * m^3}$ displacement.

Displacement = area of piston head $\cdot$ distance travelled $= \pu{28cm^2} \cdot \pu{6cm} = \pu{168cm^3} = \pu{1.68\cdot 10^{-3} m^3}$. Force in $\pu{Pa}$ is $\pu{40000psi} \cdot 6895 = \pu{275.8MPa}$.

So force $\cdot$ displacement = $\pu{46,334 Joules}$.

I'm not sure if this is correct and am confused by some of the conversions.

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    $\begingroup$ A question being put on hold does not stop people from commenting. Therefore, I don’t agree with the claim that “had [we] put the question on hold before he had the opportunity to answer [you] would still be in the dark”. Based on your edit I have reopened the question (that is the point of closing things; it is to encourage people to edit their questions). I removed your meta commentary; if you want to discuss site policies, please consider posting on Chemistry Meta. Lastly, if you don’t want your question to be misconstrued as schoolwork, then try not to make it look like schoolwork - eg context... $\endgroup$ – orthocresol Feb 28 at 10:33
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Loong Feb 28 at 12:46
  • $\begingroup$ I know: (The Displacement = The distance traveled), but : (area of piston head $\cdot$ distance traveled = $\Delta{V}$) $\endgroup$ – Adnan AL-Amleh Feb 28 at 17:07
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    $\begingroup$ Adnan, according to Merriam webster at least, displacement is a volume, (area x distance), not merely the distance travelled, which is what is calculated above. $\endgroup$ – user1023110 Feb 28 at 18:38
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    $\begingroup$ @user1023110 People here volunteer their time to answer questions. What you wrote initially was a problem that very easily could have been pulled directly from an introductory physics assignment. There are dozens of people per day on this site who simply copy and paste an assignment and hope someone will do it for them. Its much easier to have the default position be to close these questions and reopen them when the user clarifies the post then to leave them open to clutter the site. $\endgroup$ – Tyberius Feb 28 at 19:59
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You might want to familiarize yourself first with the correct use of quantities, and then with the correct use of the corresponding units. In particular, $40\,000\ \mathrm{psi}$ is the value of a pressure and not of a force. Furthermore, you should avoid confusing quantities, names of quantities, and units in equations as in “work in joules = Pa * m^3 displacement”. You might want to use proper quantity calculus instead. You should also avoid the wrong use of SI prefixes as in “40000 × 6895 = 275.8M”.

Your initial assumption is correct. Work is force times distance, assuming the movement is in the direction of the force. In quantity calculus, this is expressed as the following quantity equation (for simplicity’s sake ignoring the vectorial character of the quantities):

$$W=F\cdot x\tag1$$ where
$W$ is work,
$F$ is force, and
$x$ is distance.

You do not know the value of the force $F$; however, you know the value of the pressure $p$ and the value of the surface area $A$. And you know that pressure is defined as

$$p=\frac FA\tag2$$

Thus, the force $F$ acting on the piston head is

$$F=p\cdot A\tag3$$

You could use this equation to calculate an intermediate value for the force $F$; however, it is generally preferable to insert such intermediate equations into the total equation for the final result; i.e. in this case, insert Equation (3) into Equation (1), which yields:

$$W=p\cdot A\cdot x\tag4$$

Now you have only a single equation for the work $W$ that only depends on known independent quantities (the given pressure $p$, area $A$, and distance $x$).

It is only now that you should insert actual numerical values for the quantities. In order to avoid the need for conversion factors, you might want to convert the given values to a coherent system of units first, preferably SI base units. Thus

$p=40\,000\ \mathrm{psi}=2.7579\times10^8\ \mathrm{Pa}$,
$A=28\ \mathrm{cm^2}=0.0028\ \mathrm{m^2}$, and
$x=6\ \mathrm{cm}=0.06\ \mathrm m$.

Inserting these values into Equation (4) yields

$$\begin{align}W&=p\cdot A\cdot x\\&=2.7579\times10^8\ \mathrm{Pa}\times0.0028\ \mathrm{m^2}\times0.06\ \mathrm m\\&=46333\ \mathrm J\\&\approx5\times10^4\ \mathrm J\end{align}$$

(Note that $1\ \mathrm{Pa}=1\ \mathrm{kg\ m^{-1}\ s^{-2}}$ and that $1\ \mathrm J=1\ \mathrm{kg\ m^2\ s^{-2}}$.)

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  • $\begingroup$ Thank you. , conceptually I think I understand all of this, suggested by the fact that we come up with the same answer, although not sure I understand what is wrong with saying “work in joules = Pa * m^3 displacement”, though perhaps better if I said "work in joules = pressure in Pa * m^3 volume displaced"? From what I can see this appears to be a true statement. $\endgroup$ – user1023110 Feb 28 at 18:32
  • $\begingroup$ $\mathrm{Pa}\cdot\mathrm{m^3} =\frac{\text{newton}}{\mathrm{m^2}}\cdot\mathrm{m^3}=\text{newton}\cdot\mathrm{m}=\text{Joule}$ $\endgroup$ – Adnan AL-Amleh Feb 28 at 20:40

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