An air pump is used to fill a rigid gas bottle with air. The pump is connected to the gasbottle via a rigid pipe. The pump consists of a piston ofinternal diameter d=0.08 m and a cylinder with a strokeL=0.3m. Air enters the pump through an orifice located at thebottom of the cylinder to raise the pressure in the pump toatmospheric pressure. Air, drawn into the pump through theorifice, mixes completely and adiabatically with the cylindercontents. A valve allows the bottle's refill. The valve openswhen the pump pressure is equal to the gas bottle. The valveis always closed when the pump pressure is below the gasbottle pressure. Pressure drop across the valve can beneglected. The bottle, the connecting pipe and the pump are thermally insulated. Air inthe pump and connecting pipe undergoes isentropic compression and expansion processes.IF-oneThe working fluid (air) can be considered a perfect gas with R=287 J/KgK, cp=1005 J/KgKand y=1.4. Initially the piston is located at the bottom of the cylinder, the valve is openand pressure and temperature are 1bar and 298K in the pump, the bottle and thesurrounding ambient air.Air transferring to the gas bottle mixes completely and adiabatically on entering.Determine:1) The initial masses of air in the bottle (mb), cylinder (mc) and connecting pipe(mp).2) The mass in the bottle (mb2), in the connecting pipe (mp2) and the mass that hasbeen transferred to the bottle after the first compression stroke. Notes:a)b)c)A simplified sketch of the thermodynamic system showing cylinder/pistonat initial conditions can be represented as:Gas bottleValveCylinderL=0.3m; d = 0.08m Vp/Vc=0.1 and Vb/Vc=0.9Where V₂= connecting pipe volume, Vc= pump cylinder volume, V₁= bottle volumeFirst cycle of the pump comprises:First compression 1-2: Isentropic compression (valve is open; initialvolume being Vc+Vb+Vp)First expansion 2-3: Isentropic expansion. Bottle valve closedFilling of pump and connecting pipe 3-4. AdiabaticCylinderOrificeTo solve point 5 you would need the following result deduced by applyingthe First Law.External to the pump, air is always at atmospheric pressure (pa). As air is drawn into thecylinder, the boundary indicated in the figure below with a dashed line is drawn inwards.W = -PistonVolume VaAir (ma) drawninto cylinder.The work done on the "complete" system, including the air being drawn in, will be:- [ Padv = -Pa (0-V₁) = PavaThe process is adiabatic, hence Q=0 so that:W + Q = U₂ - U₁Applying the first law to the complete system constituted by the cylinder content and airdrawn in, we can write:Pava = (ma+mp2) cvT4 - mp2CvTc3 - macvTaBut using the perfect gas equation pV=mRT and R+Cv=Cp we obtain:(ma+mp2) CvT₁ = mp2CvTc3 + map Ta

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Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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An air pump is used to fill a rigid gas bottle with air. The pump is connected to the gas
bottle via a rigid pipe. The pump consists of a piston of
internal diameter d=0.08 m and a cylinder with a stroke
L=0.3m. Air enters the pump through an orifice located at the
bottom of the cylinder to raise the pressure in the pump to
atmospheric pressure. Air, drawn into the pump through the
orifice, mixes completely and adiabatically with the cylinder
contents. A valve allows the bottle's refill. The valve opens
when the pump pressure is equal to the gas bottle. The valve
is always closed when the pump pressure is below the gas
bottle pressure. Pressure drop across the valve can be
neglected. The bottle, the connecting pipe and the pump are thermally insulated. Air in
the pump and connecting pipe undergoes isentropic compression and expansion processes.
I
F-one
The working fluid (air) can be considered a perfect gas with R=287 J/KgK, cp=1005 J/KgK
and y=1.4. Initially the piston is located at the bottom of the cylinder, the valve is open
and pressure and temperature are 1bar and 298K in the pump, the bottle and the
surrounding ambient air.
Air transferring to the gas bottle mixes completely and adiabatically on entering.
Determine:
1) The initial masses of air in the bottle (mb), cylinder (mc) and connecting pipe(mp).
2) The mass in the bottle (mb2), in the connecting pipe (mp2) and the mass that has
been transferred to the bottle after the first compression stroke.

Notes:
a)
b)
c)
A simplified sketch of the thermodynamic system showing cylinder/piston
at initial conditions can be represented as:
Gas bottle
Valve
Cylinder
L=0.3m; d = 0.08m Vp/Vc=0.1 and Vb/Vc=0.9
Where V₂= connecting pipe volume, Vc= pump cylinder volume, V₁= bottle volume
First cycle of the pump comprises:
First compression 1-2: Isentropic compression (valve is open; initial
volume being Vc+Vb+Vp)
First expansion 2-3: Isentropic expansion. Bottle valve closed
Filling of pump and connecting pipe 3-4. Adiabatic
Cylinder
Orifice
To solve point 5 you would need the following result deduced by applying
the First Law.
External to the pump, air is always at atmospheric pressure (pa). As air is drawn into the
cylinder, the boundary indicated in the figure below with a dashed line is drawn inwards.
W = -
Piston
Volume Va
Air (ma) drawn
into cylinder.
The work done on the "complete" system, including the air being drawn in, will be:
- [ Padv = -Pa (0-V₁) = Pava
The process is adiabatic, hence Q=0 so that:
W + Q = U₂ - U₁
Applying the first law to the complete system constituted by the cylinder content and air
drawn in, we can write:
Pava = (ma+mp2) cvT4 - mp2CvTc3 - macvTa
But using the perfect gas equation pV=mRT and R+Cv=Cp we obtain:
(ma+mp2) CvT₁ = mp2CvTc3 + map Ta

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4) Calculate the air temperature and pressure in the pump at the end of the following expansion stroke, just before the piston uncovers the orifice (orifice dimensions are neglectable).
5) Calculate the mass and temperature of air in the pump (i.e. the cylinder and
connecting pipe) after the orifice has been exposed and the pressure has risen to atmospheric pressure (bottle valve is now closed).
7) Calculate the piston position and the air temperature in the pump during the next compression stroke at the time when the valve to the gas bottle opens again.
8) Calculate the pressure in the bottle when no more air can be transferred by the pump. (This will be after an undetermined number of pumping cycles).

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