2.docx - OBJECTIVE The purpose of this experiment is to analyze the relation between the time and pressure in a first-order fluid system The analysis of

2.docx - OBJECTIVE The purpose of this experiment is to...

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OBJECTIVE The purpose of this experiment is to analyze the relation between the time and pressure in a first-order fluid system. The analysis of the natural response of the system will allow the comparison with the experimental response using a simulation response. INTRODUCTION The following system is first-order fluid system in which the experimental data will be produced from. This system is composed of an air pressure tube where an initial pressure comes from, a valve V that allows and keep this air pressure into the right side of the system, a rigid tank TA that is connected to a capillary tube resistor RS, which discharge the pressure, and a pressure transducer that measures the decrease in pressure overtime due to the discharged air. Figure 1: Experimental system The system works in a way that the mass flow rate is: dM dt = ρ ( q ¿ - q out ) (1) Where,
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dM dt is the mass flow rate ρ is the density inside the tank q ¿ is the incoming flow rate, which is zero in this system q out is the outgoing flow rate The previous equation (1) can be reformulated with the incoming flow rate dM dt = ρ q Where, q out is now written as - q (2) The mass flow rate can further be calculated knowing that the mass is equal to the pressure times the volume of the tank. Then doing a derivative over time on each side, which results in a product rule on the right hand side, the equation becomes: dM dt = ρ dV dt + V dt (3) Considering that the volume in the tank is static, equation (3) can be written as dM dt = V dt (4) The equation (1) can be substituted in equation (4) to give ρ q = V dt (5) Knowing that the fluid is air, the adiabatic process can be analyzed to relate volume flow to air density. The next equation considers the thermodynamic process to relate pressure and density together. ρ = dp ɣ ( p + p a ) (6)
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Where, Ɣ is the ratio of the fluid's specific heat p is the atmospheric pressure p a is the gauge pressure The ratio of specific heat is done due to the fact that the thermodynamic process while discharging occurs is unknown. Then it is important to consider both isothermal and isentropic process and their specific heat. Also, the fluid is air, so the ratio is between the isothermal and isentropic values will be the specific heat’s value is 1.4. Isolating 1 dt in equation (5) and substituting the inverse of equation (6) into equation (5) produces: dp dt = q Ɣ ( p + p a ) V (7) This equation can be rewritten including the fluid’s capacitance in terms of fluid specific heat, absolute pressure, and volume.
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