An interesting little (simple) brain teaser for you – no matter whether what discipline or type of engineering professional you are.

Dear Colleagues

An interesting little (simple) brain teaser for you – no matter whether what discipline or type of engineering professional you are.

The Challenge

A tank at atmospheric pressure contains 1 kg of air. The tank is then pressurized with an additional 4 kgs of air. What is the resultant gauge pressure (in bars) in the tank after this 4 kgs of air has been added?

Note that Absolute pressure = gauge pressure + atmospheric pressure

A. 1 bar
B. 3 bar
C. 4 bar
D. 5 bar

You could use the Universal Gas Law in your deliberations:

PV = nRT

Where V is volume; P is absolute pressure; n is the number of moles (‘molecules’); R is a constant and T is the temperature.

Assume atmospheric pressure is 1 bar (the initial pressure) – naturally this is a number which could change depending on what altitude you are at.

Solution (suggested – you could do it in a myriad of different ways)

There is initially 1kg of air occupying V0 initial volume.

With 4 kgs of air added; there is now a total of 5 kg of air in the tank (thus the final number of moles or particles is effectively 5 times that of the initial number).

The volume (V0) and temperature (TO) of the tank still stays the same.

Thus:

P0 V0  = n0 R T0 (initial state)
P1  V0  =  n1 R T0  (final state where n1 = 5 x n0)
Thus – dividing each side of equation 1 and 2 we get:

 P1  V0 / P0 V0  =  5 x n0 R T0  /  n0 R T0

P1  V0 / P0 V0  =  5 x n0 R T0  /  n0 R T0

Thus P1 = 5 P0

Hence, the final pressure P1 = 5 bar absolute or 4 bar gauge pressure (thus answer C above)


Do you agree?

My appreciation to David Spitzer (the flow guru) for his contribution.

Albert Einstein remarked: I never teach my pupils. I only attempt to provide the conditions in which they can learn.

Yours in engineering learning

Steve

Mackay’s Musings – 19th July’16 #609
780, 293 readers – www.idc-online.com/blogs/stevemackay

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