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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 Orifice Piston L=0.3m; d = 0.08m Vp/Vc=0.1 and Vb/Vc=0.9 Where V₁= 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 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.

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 Orifice Piston L=0.3m; d = 0.08m Vp/Vc=0.1 and Vb/Vc=0.9 Where V₁= 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 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.

Elements Of Electromagnetics
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: 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).
Transcribed Image Text: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: 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).
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
Transcribed Image Text: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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