Re-discuss questions 24 and 25 from the 2012 case study at 2 p.m
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24. As shown in the figure: the height difference between the liquid level in the distillation column and the inlet of the reboiler is H (m). The height difference between the inlet of the reboiler and the liquid level is 3 m. The density of the liquid at the reboiler inlet is ρL = 738 kg/m3, while the homogeneous density of both the reboiler and the rising pipe is ρ = 18.38 kg/m3. When H = 4 m, what is the total static head loss in the reboiler and the rising pipe? A 50.7 B 0.72 C 29 D 1.26 Solution: △P=ρLgH=738×9.81×4=28959Pa≈29KPa. 25. Based on the conditions given in question 24, what is the height difference H (in meters) between the liquid level in the methanol distillation column and the inlet to the reboiler when the total resistance of the system is 40 kPa? A 5.74 B 0.22 C 0.55 D 8.43 Solution: ρLgH’ – ρg(H+3) = ∑hf = 40×103; thus, H’ = 5.74 m. Question: Question 24 asks for the total static head loss in the reboiler and the rising pipe, so the formula from Question 25 should be used. For Question 25, the installation height of the reboiler should be considered as the driving force to overcome all the resistances in the system, so the method used in Question 24 should be applied. However, swapping them around doesn’t yield a solution. Please ask experts to help discuss this ；
24. Let the pressure inside the tower be P0, and the pressure at the reboiler inlet be P1; then the total static head loss across the reboiler and the rising pipe is △P = P1 – P0. Applying Bernoulli’s equation between the 0-0 level and the 1-1 level gives: P0 + ρlgZ0 = P1 + ρlu²/2 + hf1. Since ρlu²/2 + ρghf1 ≈ 0, it follows that ΔP = P1 – P0 = ρgH = 738 × 9.81 × 4 = 28959 Pa ≈ 29 KPa. Similarly, applying Bernoulli’s equation between the outlet of the rising pipe and the inlet of the reboiler yields: P2 + ρggZ2 + ρgu²/2 + hf2 = P1 + gρu²/2. With Z2 = H + 3hf2 ≈ ∑hf, we get P2 = P0, and thus ΔP = P1 – P0 = ρgH. Substituting the values, we have ρLgH’ – ρg(H + 3) = ∑hf = 40000; solving this equation gives H’ = 5.74 m