This article discussed 80 MT capacity **batch vacuum pan design** calculation of the Heating surface, Number of tubes, Dia of the downtake and tube plate, Graining volume, Dia of the vapour doom, Height of the top cone, Dia of vapour inlet and outlet pipelines , Pipeline dia of noxious gases, Dia of condensate pipeline, Dia of Massecuite discharge, Calendria shell thickness and tube plate thicknesses

## Batch Vacuum Pan Calculation in sugar industry | Crystallization process

Capacity required for* crystallization process *equipment like *supply tanks*, Capacity of *Batch/continuous pans,* condensers, molasses conditioners, spray pond, crystallizers capacity .. etc.

Please go through the below link for complete information regarding the above topics

Capacity Calculation of Pan Section

Basic design aspects of vacuum pan Click Here

### Design of 80 MT capacity Batch Pan

**1. Heating Surface :**

The heating surface is calculated based on **S/V ratio**

For batch pan with 3rd vapour has a heating medium it will be* 6.6 to 6.7 m ^{2}/m^{3}*

Formula = S/ V = 6.6

Srike volume = Weight/density of massecuite = 80 MT / 1.42 MT / m^{3} = 56.33 m^{3} ≈ 57 m^{3} = 570 HL

Heating surface required = 57 x 6.6 = 376 m^{2}

**2. Number of tubes**

Generally, the specifications of tubes for batch vacuum pans as follows

*102 mm OD / 16 g ( 1.625 mm) / 800 mm length*

Formula : S = Π x D x L x N

Here

- S = heating surface area of the pan in m
^{2} - D = Mean dia of tube in m ( OD – thickness of tube )
- D = 102 – 1.625 = 100.375 mm
- L = Effective Length of the tube in m ( Total length – 2 x Tube plate thickness – 2 x Tube expansion elevation and projection of the tube )
- L = 800 -( 2 x 32) – (2 x 5) = 800 – 64 – 10 = 726 mm
- N = Number of the tubes

So N = 376 / ( 3.14 x 0.726 x 0.100375 ) = 1643 nos.

**3. Dia of the Downtake and tube plate:**

- Here N = Number of the tubes
- P = Pitch of the tube in m

Considered 20% extra area to arrange vapour distribution in calendria of batch vacuum pan

Here first we calculate the required dia for down take

It is calculated based on the circulation ratio of the pan and it is generally maintained minimum 2.5

Circulation Ratio = Cross Section Areqa of the Tubes / Area of the Downtake

Total cross-section area of the tubes (A) in m^{2} = N x ( Π /4) x ID^{2}

A = 1643 x 0.785 x ( 0.09875)^{2} = 12.58 m^{2}

Area of the down take ( A_{D} )in m^{2} = 12.58 / 2.5 = 5.032 m^{2}

**Pitch of the Tube**

Legment of the tubes for vacuum pan = 16 mm

So Pitch of the tube = OD + legment + Tube tolerance + Hole tolerance

Pitch of the tube P = 102 + 16 + 0.5 + 0.1 = 118.6 mm

Area of the tube plate A in m2= 1643 x 0.866 x (0.1186 )2 x 1. 2 + 5.032

A = 24.02 + 5.032 = 29.05 m2

D_{TP} = 6100 mm

Generally, dia of the downtake is maintained at 40% on tube plate dia for proper circulation of massecuite in pan.

Here 6100 x 40 % = 2440 mm ( As per the above calculation it is 2500 mm. So it is OK )

**4. Graining volume of the batch vacuum pan**

Please go through the below link for complete information regarding this topic

The Concept of Graining Volume of the Batch Pan

Sl.no. |
Description |
Formula |
Values |
UOM |

Input Data | ||||

1 | Capacity of pan | 80 | T | |

2 | No. of tubes | N | 1643 | nos. |

3 | Tube thickness | t | 1.6 | mm |

4 | Tube Length | H1 | 800 | mm |

5 | Tube OD | OD | 102 | mm |

6 | Dia of pan | D1 | 6100 | mm |

7 | Dia of the down take | D2 | 2500 | mm |

8 | Dia of bottom inverted cone | D3 | 2200 | mm |

9 | Height of the bottom ring | H2 | 50 | mm |

10 | Angle of bottom cone | α | 18 | Deg |

11 | Angle of bottom inverted cone | Φ | 35 | Deg |

Graining Volume Calculation |
||||

1 | ID of the tube | ID = OD – 2t | 98.8 | |

2 | Volume of massecuite in tubes | Q1 = 0.785 x ID x ID x H1 x N | 10.07 | M^{3} |

3 | Volume of downtake | Q2 = 0.785 x D2 x D2 x H1 | 3.93 | M^{3} |

4 | Volume of the bottom ring | Q3 = 0.785 x D1 x D1 x H2 | 1.46 | M^{3} |

5 | Height of the bottom cone | h 1 = [(D1 – D3)/2 ] x TAN α | 633.59 | mm |

6 | A1 | 0.785 x (D1)^{2} |
29.21 | M^{2} |

7 | A2 | 0.785 x (D3)^{2} |
3.80 | M^{2} |

8 | Volume of the bottom cone | Q4 = h/3 (A1+A2+√A1A2) | 9.20 | M^{3} |

9 | Height of the bottom inverted cone | h2 = [( D3)/2 ] x TAN Φ | 770.23 | mm |

10 | Volume of the inverted cone | Q5 = 1/3 x 0.785 x (D3)^{2} h2 |
0.98 | M^{3} |

11 | Graining Volume | Q1+Q2+Q3+Q4 – Q5 | 23.68 | M^{3} |

42.0 | % |

**5. Srike height:**

Formula : Strick height ( Hs) in m = Volume of the massecuite above the tube plate in m^{3} / ( 0.785 x ID^{2} of the vapour space in m)

Here

Volume of the massecuite above the tube plate = Strike volume – graining volume

= 57 m^{3} – 23.68 m^{3} = 33.32 m^{3}

ID of the vapour space = ID of the tube plate = 6100 mm – ( 2 x thickness of the vapour shell)

= 6100 – (2 x 16) = 6068 mm

Hs = 33.32 / [0.785 x (6.068)^{2}] = 1.152 m = 1150 mm

#### 6. Dia of the Vapour Dome (D_{d})

Here Volume of the vapour = Heating surface of pan **x** Evaporation rate **x** Specific volume of the vapour at 700 mm of Hg vacuum

The evaporation rate of the batch vacuum pans depends on type of the massecuites

**The average evaporation rate in Batch Pans is considered as follows**

- For ‘A ‘ Massecuite – 60 kg/m² /hr ,
- For ‘B’ Massecuite – 55 kg/m² /hr
- And for ‘ C’ Massecuite – 50 kg/m² /hr

= 249062 / 3600 = 69.18 m^{3} /sec

Vapour velocity through the doom to be considered as 25 m/sec

D_{d} = √ 1.27 x ( 69.18 / 25 ) = 1.874 m ≈ 1900 mm

**7. Height of the top cone part**

Here D_{TP } = ID of the tube plate in mm = 6100 – (2 x 16) = 6068 mm

Dd = Dia of the top doom = 1900 – (2 x 18) = 1864 mm

∝ = Angle of the cone = 18^{o} ( Generally, it is in the range of 18^{o} to 20^{o} )

HTC = [ (6068 – 1864) /2 ] x Tan 18^{o}

= 2102 x 0.325 = 683 mm

#### 8. Vapour inlet and outlet line dia

S.No |
Description |
Value |
UOM |

1 | Heating surface of the pan | 376 | m^{2} |

2 | Evaporation rate | 60 | kgs/m^{2}/hr |

3 | Pan Inlet Vapour Temperature | 94 | ^{o}C |

4 | Pan Outlet Vapour Temperature | 52 | ^{o}C |

5 | Pan Inlet Vapour Velocity | 35 | m/sec |

6 | Pan Outlet Vapour Velocity | 55 | m/sec |

Result | |||

1 | Specific volume of Pan inlet vapour | pan boiling calculation2.0510 | M^{3}/kg |

2 | Specific volume of Pan Outlet vapour | 10.98 | M^{3}/kg |

3 | Volume of the pan inlet vapour | 12.853 | M^{3}/sec |

4 | Volume of the pan outlet vapour | 68.81 | M^{3}/sec |

5 | Pan Inlet vapour Line Dia | 0.684 | mtrs |

Say | 700 | mm | |

6 | Pan Outlet vapour Line Dia | 1.262 | mm |

Say | 1300 | mm |

#### 9. **Vapour space height** (Free space above the massecuite level up to top cone )

Generally, the Vapour space required above the massecuite level is minimum 1500 mm.

**10. Noxious gases connections/ Non-condensable gases (NCG)**

Generally, The removal of non-condensable gases requires a 1 cm^{2} area for 10 m^{2 } heating surface of the vacuum pan

SO Cross section area of non-condensable gases pipeline in cm^{2} = Heating surface in m^{2 }/10

= 376 /10 = 37.6 cm^{2}

Here considered 6 nos. of non-condensable connections.

So Dia of the each non condensable gases pipe line = √ 376 / (0.785 x 6 ) = 2.82 cm ≈ 32 mm

Non condensable gas connections = 32 mm x 6 nos.

**11. Dia of massecuite discharge**

Dia of the massecuite discharge =

- Strike Volume = 56.33 m
^{3} - Time required for massecuite discharge = 10 to 15 min.
- Velocity of massecuite = 0.15 m/sec
- Dia of the massecuite discharge = √ 56.33 / ( 0.785 x 12 x 60 x 0.15 ) = 0.815 m ≈ 800 mm

**12. Dia of Condensate piping**

Generally, for 80 MT pan condensate withdrawal points required minimum 2 nos. It is better to go with 3 nos. of connections.

Volume of the condensate = [Heating surface X Evap. Rate ] / [ Density of water x 3600].

= 376 x 60 / ( 1 x 3600 )

= 22.560 / 3600 = 0.00627 m^{3} /sec

Dia of the each condensate line = √ 0.00627 / ( 0.785 x 1 x 3 ) = 0.052 ≈ 80 mm

**13. Calendria shell thickness**

Here

- P = Hydraulic test pressure in kg/cm
^{2 }= 3 kg/cm2 - Di = ID of the Calendria in mm = 6100 – ( 2 x 16) = 6068 mm ( Here 16 mm considered for shell thickness for calculation purpose)
- F = Allowable stress in kg/cm
^{2 }= For Mild steel it sis considered as 1400 kg/cm2 - J = Welding Joint efficiency in mm = 0.75 mm
- C= corrosion allowance in mm = 3.0 mm
- t s ={ 3.0 x 6068 / [ ( 2 x 1400 x 0.75 ) – 3 ] } + 3 = 12 mm

But according to standard specifications calendric and body shell plate thickness for the 80 MT pan is 18 mm and for the bottom saucer is 25 mm

**14. Tube Plate thickness**

Here

- C . A = corrosion allowance in mm = 1.5 mm
- f = Allowable stress in kg/cm
^{2 }= 1400kg/cm^{2} - P = Design pressure in kg/cm
^{2 }= 2.72 kg/cm^{2} - Es = Modulus factor for MS sheet in kg/cm
^{2 }= 2.1 x 10^{6}Kg/cm^{2} - Et = Modulus factor for SS sheet in kg/cm
^{2 }= 1.9 x 10^{6}Kg/cm^{2} - G = ID of the shell in mm = 6068 mm
- ts = Thickness of the shell in mm = 18 mm
- tt = Thickness of tube in mm = 1.625 mm
- do= OD of the tube in mm = 102 mm
- Do = OD of the calendria sheet in mm = 6100 mm
- Nt = Number of tubes = 1643 nos.
- K = 0.4515
- F = √ 0.4515 / [ 2 + 3(0.4515)] = 0.3669
- tp = 0.3669 x 6068 x √ (0.25 x 2.75 /1400) + 1.5 = 50.56 mm

According to standard specification, 36 mm thickness is sufficient for 80 MT tube plate.

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