Shell balances in cylindrical coordinates

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In the discussion about writing a shell balance for a cylindrical system, participants explore the appropriate method for dividing the conservation equation by the control volume's volume. The debate centers on why the division is done by 2π·ΔR·L instead of 2π·R·ΔR·L, which represents the volume of the control volume. It is noted that both methods are valid, but the choice may be based on convenience or aesthetic preference. However, one participant argues that this choice can affect the final answer obtained from integrating the resulting ordinary differential equation (ODE). The conversation emphasizes the importance of understanding the implications of these choices in transport phenomena.
Muhammad Waleed Khan
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I have a question regarding writing a shell balance for a cylindrical system with transport in one direction (in any area of transport phenomena). When we set up the conservation equation(say steady state), we multiply the flux and the area at the surfaces of our control volume and plug them into the eqn. Afterwards I believe that we divide the resulting equation by the volume of the control volume before taking the limit as DelX,DelZ or DelR approaches 0. In cartesian coordinates we divide this by delXdelYdelZ, but why in cylindrical CV do we divide by 2.pi.DelR.L instead of 2.pi.R.DelR.L (which is the volume of our CV)?
 
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Muhammad Waleed Khan said:
I have a question regarding writing a shell balance for a cylindrical system with transport in one direction (in any area of transport phenomena). When we set up the conservation equation(say steady state), we multiply the flux and the area at the surfaces of our control volume and plug them into the eqn. Afterwards I believe that we divide the resulting equation by the volume of the control volume before taking the limit as DelX,DelZ or DelR approaches 0. In cartesian coordinates we divide this by delXdelYdelZ, but why in cylindrical CV do we divide by 2.pi.DelR.L instead of 2.pi.R.DelR.L (which is the volume of our CV)?
Of course, you could also divide by ##2\pi r \Delta r L##. That won't change the final result. But, the authors must of thought it was more convenient and aesthetically pleasing to do it their way. Both ways are right.
 
Chestermiller said:
Of course, you could also divide by ##2\pi r \Delta r L##. That won't change the final result. But, the authors must of thought it was more convenient and aesthetically pleasing to do it their way. Both ways are right.
Yeah but it changes the final answer you get from integrating the resulting ODE.
 
Muhammad Waleed Khan said:
Yeah but it changes the final answer you get from integrating the resulting ODE.
Please show me how you think the answer will be different. Start out by showing me the two versions of the ODE that you get.
 

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