Download CHE654_2012_Homework4_Solutions.pdf PDF

TitleCHE654_2012_Homework4_Solutions.pdf
TagsHeat Transfer Heat Volume Kilogram Ordinary Differential Equation
File Size175.9 KB
Total Pages31
Document Text Contents
Page 30

30

10 = c3 + 0.3c4 � c4 = 18.5183 � c1 = 11.4817




T2 = 11.4817e

-0.3t – 0.06667t2 + 4.4445t + 18.5183




Check: At t = 0, T2 = 11.4817 + 18.5183 = 30 °C � Correct


When Tank empties, M1 = 300 – 10t = 0 � t = 30 minutes


Therefore, at t = 30 min: T2 = 91.85 °°°°C




(b) First calculate the time for liquid in Tank 1 to reach 150 °C:


T1 = 200 – 0.2(30 – t)

2 = 150 � t = 14.19 minutes


When calculating this time, I treated the heat transfer area as being constant. In
reality, it is not because the liquid moves up and down although there is no boiling
yet. If you take into account the non-constant area, the answer is 14.94 minutes, but
the math is somewhat complex. This gives z1 = 0.4442 m.



Energy balance after boiling occurs in Tank 1:


dM1 = –10 – v but v = q/λ
dt

= –10 – UAT(200 – T1)/λ

= –10 – UAT(200 – T1)/λ

= –10 – (25)(4V/D + πD2/4 )(200 – 150)/1500

= –10 – (25)(6.6667V+0.28274)(50)/1500 = –10.23562 – 5.5556V

But φ = M1/V or M1 = φV

V(t = 14.94 min) = 0.4442π(0.3)2 = 0.1256 m3

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