CE3106 Example 2 Solutions (Q2 and Q3) PDF

Preview

Summary

Examples...

Description

2. Chen’s method: hcb = fc hfc + fs hnb

- please be prepared to explain it

We have a corrected convective coefficient (fchfc) and corrected nucleate coefficient (fshnb) Forster-Zuber equation is used for the nucleate pool boiling: h nb

 k 0L.79C 0pL.45 0L.49  = 0.00122 0.5 0.29 0.24 0.24  (Tw − Ts )0.24 (pw − ps )0.75    L   V 

  0.1350.79  29700.45  3800.49  hnb = 0.00122 0.5 0.29 0.24 0.24  0 . 024 0 . 00025 435000 7 . 6       (145 − 133)0.24 (750000 − 620000)0.75

hnb = 4078 W/m2K Now we go to the calculations of the convective part, considering that we have a single (boiling) liquid phase in flow: Total hydrocarbon flow = 54000 / 3600 = 15 kg/s = 15 / 250 = 0.06 kg/s I (att: Mass flow per tube per tube) Mass flow per unit area (mass flux) =

0.06 / ( x 0.0212 / 4) = 173.2 kg/m2s (both phases – L and V (vapour) Mass fraction of vapour = 0.2 Thus, we have the respective Re number for a single (boiling liquid) phase in a flow (1− x )Gdi (1 − 0.2)  173.2  0.021 = = 11639  8000 −4 L 2.5  10 (turbulent flow)

ReL =

Prandtl number  LCpL 2.5  10 −4  2970 PrL =

kL

=

0.135

= 5.5

Colburn equation (year 1,2): kL 0.135 0.8  116390.8  5.50.33 = ReL PrL0.33 = 0.023  d 0.021 = 464.4 W / m2 K hfc = 0.023

Lochkhard-Martinelli parameter (turbulent-turbulent flow): 1  x  =  X tt  1 − x 

0.9

 L   V

  

0.5

 V   L

  

0.1

 0.2  =   1 − 0.2 

0.9

0.5

 380     7.6 

 1.2  10 −5     −4   2 . 5 10  

Therefore from the appropriate chart, the correction of the convective component of the heat transfer coefficient is fc = 3.5 (enhancement due to two-phase flow)

0.1

= 1.499

Correction for the nucleate boiling term (suppression) (fs) : ReLfc1.25 = 55719, so from chart at slide 50 fs = 0.53 Therefore, the effective heat transfer coefficient for forced convective boiling:

hcb = fc hfc + fs hnb = ( 3.5 464.4) + (0.53 4078) = 3787 W / m 2K 3. Now we have a vertical thermosyphon reboiler and the rate or flow (circulation) of process fluid is not known since it is now a natural circulation, due to density difference – no pump for the hydrocarbon mixture (process fluid). In Q2 we had to have a pump, otherwise we would not be able to provide the flow given (54000 kg/h). Reduced temperature = = (133C + 273C) / (353C + 273C) = 0.649  0.65 Mean temperature difference = 160C − 133C = 27C (133C is the Tsat temperature for the hydrocarbon mixture and 160C is the saturation steam temperature, both given in Example 2). Heat flux from Frank and Prickett chart  25000 W/m2 (graph given separately).

Steam mass flow = 2700 / 3600 = 0.75 kg/s Therefore heat exchanged = 0.75 x 2083 = 1562 kW (Mass flow x heat of vaporisation) Therefore total tube surface area = 1562000 / 25000 = = 62.5 m2 (be careful with dimensions) For 250 tubes of 25 mm outside diameter: Tube length = 62.5 / (250 x  x 0.025) = 3.18 m...