EX40HC – Tutorial 3 Building basic dynamic simulations with Simul ink Problem 1. Consider Problem 1 of Tutorial 2 again, where we have a stirred tank heater with no inlet or outlet flow (see scheme below). The aim of the process is to heat the fluid in the tank until it reaches the desired temperature ππ via heating that is provided by a coil, inside which steam condenses at a temperature πππ π π π π π π π π π . Assuming that the system is initially at steady-state determine the following: a) Using the mathematical model already developed in Tutorial 2, verify that the dynamic response of the fluid temperatureππ(π‘π‘) in relation to the steam temperature πππ π π π (t) can be described by a first order lag process. Determine the numerical values of the time constant and of the gain of the process and comment on their values and units. Assuming that at time π‘π‘=0 the steam temperature increases by 10oC (step change) develop a Simulink model for this process and plot the time profile of the heater temperature versus time to verify the findings of Tutorial 2. (Answers: ππππ=837π π ,πΎπΎππ=1 Β°CΒ°C) b) Assume now that due to poor insulation there are non-negligible heat losses to the surroundings. The surrounding temperature πππΈπΈ(π‘π‘) can be considered constant and equal to 25oC. Revise accordingly the mathematical model of a) and verify whether the dynamic response of the fluid temperatureππ(π‘π‘) in relation to the steam temperature πππ π π π (π‘π‘) can still be described by a first order lag process. Determine the numerical values of the time constant and of the gain of the process and comment on their values and units, particularly in relation to those obtained in a). Calculate the time necessary for the temperature in the vessel to increase by 9oC, if the steam temperature increases again via a step change of 10oC at time π‘π‘=0, and verify your findings via an updated Simu link model. Provide an updated plot of the time profile of the heater temperature vs time and compare the response with that obtained in a). (Answers:ππππ=761π π , πΎπΎππ=0.91Β°CΒ°C, π‘π‘=3505π π )