Title CHE654_2012_Homework4_Solutions.pdf Heat Transfer Heat Volume Kilogram Ordinary Differential Equation 175.9 KB 31
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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