Question

2. In the constant volume, stirred-tank heater shown in the figure below, the flow rates and material properties are held constant. The tank contains a heating element, which is initially off and the tank is the same temperature as the inlet stream. When t=0, the heater is turned on to a constant heating rate Q. For this material, assume C = C = Cp = Cy. m,Ti P,V,C, T(t) ń, r(t) o a. Write a transient thermal energy balance to determine the dynamic process model in terms of the symbols given on the flow chart above. b. State the initial value constraint. Solve the ODE and apply the IVC to determine the function T(t). d. Liquid density is 900 kg/ k] m3, heat capacity is 3.00 “Y/kg m is equal to 50.0 kg, T; is 300 K, V is 0.2 mº, and initially T=Tį. At t=0, the heating rate 0 (t > 0) c. ·K min' is instantly increased to 7000 kJ min What is the temperature in the tank after 4 min?

Answer 1

) im To Sol Given Given min Moor Cp Curl IPvc TH P- Densih V=volume Heat Rate = 8 in TLE). o o Energy Inlet tenp= Ti outlet Temp=T. Balenie on System. (In)-(out) + Con Accumu! Ti as Referent temp mc (T:-T;) - mc (T-T;) + (1 = PVC di D-r$ 7- Ti) - Pvca] Toking dt. dt (3) 1-on iT- Ti At to

30-mill 7-Ti) - Pracolo Pue di tric(T-T:) = 8 dj + me t = 6 + Ti mc. at at PVC PVC PNC di + im T=0 & Tim dt PV PVC evi Solving Lot T(:f)= Sl S + Tim) (J. F.elt tc. PVC pava ) (7 rely Jif meat emt The*). 5642 ) left 474 t(e) = (huet ma te mt e pv Tvtc. T= D me + no ti T:) me t ce ev. At t-o T=Ti nc =- me. T- (1-é nou) + Ti ml

Noy Г-оол, C-33]-к. msova/ 1,- у — 0°m2 Тіг дор Ő - 7000 W/m Он t- Чm.. - Эроо - Soxy 1 - е So12 бор.0° *Зоо 1- Т. чь-66 О• 34 | Зоо Temp After Lemin Т- 331я и

using given data and enery balance we can find the required model and temp

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