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

Question

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 &#34;complete&#34; 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

Accepted Answer

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