Thread: vacuum pump poser

hello again.

i have a question i wonder whether some one will give a hint as to how best solve it ?

A vacuum pump removes 15% of the air that is in a container at the

beginning of a stroke. If

mr is the mass of air in the container after the r th

stroke:
(i) express

m2 and m3 in terms of the original mass m0
(ii) how many strokes are required to obtain 1% of the original mass?


so assuming mr is the mass of air after rth stroke i am assuming m2 must be the 2nd stroke =m1(15%) +m2,m3 must be m2 +m1 so we have a number series going on.

if mo is total air mass ie mo =100% can i express m2 and m3 as a percentage of mo ie m2 must equal 30% and m3 must equal 45% of mo respectively?not sure thats accurate enough.

qii. i will need to think more about,if every stroke assuming its in equilibrium removes 15% of air every time 100/15 =approx 7 strokes required to get to this point maybe thats also too simplistic.


apologies if i have got this in the wrong forum!

If the pump removes 15% then it leaves behind 85%.
So, if the original mass of air is m0, the mass left after 1 stroke m1 = m0 x 0.85.
The mass left after 2 strokes: m2 = m1 x 0.85 = m0 x 0.85^2, etc...
and so m3 = .... I'll leave that up to you!
This forms a geometrice progression.
So the mass left after r strokes mr = m0 x 0.85^r.

For the second part, your approach is incorrect. You need to find r such that mr = 0.01 x m0, ie you need to find r such that 0.85^r = 0.01. Hint: That will involve logs!!

A vacuum pump removes 15% of the air that is in a container at the beginning of a stroke. If mr is the mass of air in the container after the r th stroke:

(i) express m2 and m3 in terms of the original mass m0

(ii) how many strokes are required to obtain 1% of the original mass?

so assuming mr is the mass of air after rth stroke i am assuming m2 must be the 2nd stroke =m1(15%) +m2,m3 must be m2 +m1 so we have a number series going on.

if mo is total air mass ie mo =100% can i express m2 and m3 as a percentage of mo ie m2 must equal 30% and m3 must equal 45% of mo respectively?not sure thats accurate enough.

qii. i will need to think more about,if every stroke assuming its in equilibrium removes 15% of air every time 100/15 =approx 7 strokes required to get to this point maybe thats also too simplistic.
----------------------------------------------------------

your assuming the vacuum takes out 15 percent of the full tank. it actually removes 15% of whats left.
e.g
M=(100% full tank),
M-Mx15% will give you the air left after first stroke,
which can be written as Mx0.85, now on the second stroke you remove another 15% hence (Mx0.85)x0.85.... etc
now for r strokes the mass (should really be volume) of air in the tank is given by Mx0.85^r

(i) express m2 and m3 in terms of the original mass m0

m2 is the mass of air after the second stroke hence m0x0.85^2, since m0 = M
m3 = m0x0.85^3

(ii) how many strokes are required to obtain 1% of the original mass?

now this part requires logs so i doubt you can do this.
but heres how it goes:

Mx0.85^r = 1%, where M is a full tank (100%)

1x0.85^r=1/100

Log(base(0.85)) 1/100 = r

r=28.3.....

so it would take about 29 strokes to get 1% fuel.

many thanks for reply and "words of encouragement", its obviously a bit more complicated than i anticipated,then you prob guessed that.

i shall have to try and redo this lesson again in the hope i can understand it better, hate logs and powers why are they so complicated?.