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(Solved): The level-flow system in Figure \( Q 1 \) is to be analyzed. The flow into the system, \( F_{0} \), ...




The level-flow system in Figure \( Q 1 \) is to be analyzed. The flow into the system, \( F_{0} \), is independent of the sys
The level-flow system in Figure \( Q 1 \) is to be analyzed. The flow into the system, \( F_{0} \), is independent of the system pressures. The feed is entirely liquid, and the first vessel is closed and has a non-soluble gas in the space above the nonvolatile liquid. The flows \( F_{1} \) and \( F_{2} \) depend only on the pressure drops, because the restrictions in the pipes are fixed. The system is isothermal. 1- Derive the differential equations that represent the dynamic response of \( h_{1} \) and \( h_{2} \). Solve the differential equations using Matlab (ode23 or ode45), and plot the response of \( h_{1} \) and \( h_{2} \) 2- Derive the linearized model in a deviated format for this system response to a step change in \( \mathrm{F}_{0} \) from 10 to \( 20 \mathrm{~m}^{3} / \mathrm{min} \). 3- Solve the equations using Matlab or Simulink to show the response of \( h_{1} \) and \( h_{2} \), and, without specific numerical values, 4- Sketch the dynamic responses using Matlab or Simulink. 5- At \( 7 \mathrm{~min} \), what will be height of the first and second tank. 6- What are the cases where the second tank will overflow? State at least one possible case. Density of liquid \( = \), density of gas \( =1.204 \) \( \mathrm{kg} / \mathrm{m}^{3} \), The height and diameter of the first tank is \( 7 \mathrm{~m} \) and \( 2 \mathrm{~m} \) respectively, The height and diameter of the second tank is \( 5 \mathrm{~m} \) and \( 3 \mathrm{~m} \). The Molecular weight is \( 28.97 \mathrm{~kg} / \mathrm{kmol} \) The steady state values are \( h_{1, w}=2 \mathrm{~m}, h_{2, n} \) \( =3 \mathrm{~m}, \mathrm{~T}=25^{\circ} \mathrm{C}, \mathrm{F}_{0}=10 \mathrm{~m}^{2} / \mathrm{min} \) Assume (realistic assumptions) for any missing data.


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Solution Assuming you used wing tire balance on both * dli al = fo- c3JH noco dhi dt dhi finta = ç 3 51
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