1.07 The effect of the separation of the plates

Let us return to our previous analogy, of an air pump transferring air between two rigid vessels. Rather than using two isolated chambers, let us consider two chambers that are formed in a single rigid vessel, separated by an elastic membrane. (The initial volumes of the two chambers are the same as in the previous example). We again apply a constant force to the pump handle and begin to transfer air between the two chambers. Now, however, due to the build up of pressure in chamber B and the suction from chamber A, the membrane will flex. This increases the volume of chamber B, and reduces the volume of chamber A. The result of this action, is that the pressure increase in chamber B and the suction in chamber A, will be less than it was, for the rigid isolated chambers, when the same amount of air is transferred.

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Comparing with the point at which the pump handle stopped turning in the previous example, (with two seperate rigid vessels). Considering chamber B. The same amount of air has now been transferred into chamber B, but the volume of the chamber has now increased. Therefore the air is less compressed and so the pressure is lower. Considering chamber A. The same amount of air has been removed as before but now the volume of chamber A has reduced. Therefore there is less space for the remaining gas to expand into, and so the pressure does not drop as much as before (i.e. there is less suction). Therefore the back pressure from chamber B and the suction from chamber A, are not great enough at this point to stop the pump handle turning. Therefore the pump handle will continue turning for longer and will transfer more air than before until the limiting pressures are reached.