Working out work with calculus
You’ve landed on this page because you’ve either accidentally clicked on the button on the Water Rocket Physics – Work page, or you’ve accidentally ended up here. Fear not; it’s only calculus!
Boyle’s Law Again
Here’s a recap of Boyle’s Law if you’ve forgotten it since the last page. In a container with no holes, pressure is inversely proportional to volume. We write this as:
PV = constant
So if we think about it in terms of initial pressure and volume and then we go changing stuff, we end up with a final pressure and volume for the same amount of air so that we can write that as:
PiVi = PfVf
Of course, the world is not simple, and we must consider how heat moves around. In the water rocket, we can pretend that no heat is lost to the outside world as things happen quickly. For air, we have to add another bit to the formula so it approximates the real world adequately, so we raise the Volume to the power of the ratio of specific heats – this turns out to be 1.4 for normal air and we use the symbol ϒ. The new formula is therefore:
PiViϒ = PfVfϒ
We have to now write that in terms of Pf because we calculate that based on the volume. We do that by dividing both sides by Vfϒ to get:
Pf = PiViϒ/Vfϒ
This can be simplified to:
Pf = Pi(Vi/Vf)ϒ
Making a Function
We can put some numbers to these terms such as the starting volume of air (Vi) is 0.00085 m3, remember there’s 150ml of water in our 1 litre bottle leaving 850 ml for the air. We also know the initial pressure (Pi), in this case we have pumped the bottle up to 25 psi, that’s 172369 pascals. We also know that γ is 1.4. We can rewrite our formula as a function, P(V), and include the known numbers:
P(V) = 172369 x (0.00085/V)1.4
Now that looks a bit simpler. This formula is the function for the curve on the graph below.
The Integral
To find the area underneath the curve of:
P(V) = 172369 x (0.00085/V)1.4
We have to integrate the function and then solve the integral between 850 ml and 1000 ml. This is the point where most run and hide, so if you’re still here then keep reading as we are nearly there.
The area of the underneath the curve is Work so we can depict that as:
W = ∫ P.dV
All that means is that we need to sum all of the tiny rectangles of width dV, the ∫ symbol is kind of like the ∑ symbol. We already know that P is the function P(V) that we worked out before so we can rewrite the formula as:
W = ∫ Pi(Vi/Vf)ϒ.dV
constants don’t have to be integrated and we have some of those so lets put them in:
W = ∫ 172369 x (0.00085/V)1.4.dV
Now we’ll move the constants out of the way:
W = 172369 x 0.000851.4 ∫ 1/V1.4.dV
and simplify:
W = 8.66 ∫ V-1.4.dV
To find the integral of V-1.4 we need to use:
xn+1/n+1
so all together that’s:
W = 8.66 x ((V-1.4 + 1/1+-1.4) + C)
W = 8.66 x ((V-0.4/-0.4) + C)
We are solving the integral between 1 litre and 850ml, we have to convert those to m3 and put them in the formula, because we are solving the integral we don’t have to worry about the C.
W = (8.6/-0.4) x (Vbottle-0.4 – Vair-0.4)
Then we plug the number in:
W = (8.6/-0.4) x (0.001-0.4 – 0.00085-0.4)
W = (-21.66) x (-1.063)
W = 23.02 Joules
