Batch Vacuum Pan Design Calculation | Sugarprocesstech

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

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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 m²/m³

Formula = S/ V = 6.6

Srike volume = Weight/density of massecuite = 80 MT / 1.42 MT / m³ = 56.33 m³ ≈ 57 m³ = 570 HL

Heating surface required = 57 x 6.6 = 376 m²

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 m²
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:

$\text{Formula: Area of the tube plate} = N \times \frac{\sqrt{3}}{2} (P)^2 \times 1.2 + \text{Down take area}$

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 m² = N x ( Π /4) x ID²

A = 1643 x 0.785 x ( 0.09875)² = 12.58 m²

Area of the down take ( A_D )in m² = 12.58 / 2.5 = 5.032 m²

$\text{Dia of the down take} = \sqrt{1.27 \times A_{D}} = 2.53 \, \text{m} = 2500 \, \text{mm}$

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

$\text{Dia of the tube plate (D_{TP}) in m} = \sqrt{1.27 \times A}$

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	M³
3	Volume of downtake	Q2 = 0.785 x D2 x D2 x H1	3.93	M³
4	Volume of the bottom ring	Q3 = 0.785 x D1 x D1 x H2	1.46	M³
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	M²
7	A2	0.785 x (D3)²	3.80	M²
8	Volume of the bottom cone	Q4 = h/3 (A1+A2+√A1A2)	9.20	M³
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	M³
11	Graining Volume	Q1+Q2+Q3+Q4 – Q5	23.68	M³
			42.0	%

5. Srike height:

Formula : Strick height ( Hs) in m = Volume of the massecuite above the tube plate in m³ / ( 0.785 x ID² of the vapour space in m)

Here

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

= 57 m³ – 23.68 m³ = 33.32 m³

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)

$\text{Dia of the vapour dome in meters } = \sqrt{ \frac{4}{\pi} \times \frac{\text{Volume of Vapour}}{\text{Velocity of Vapour}} }$

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

$\text{So volume of the vapour} = 376 \times 60 \times 11.04 = 249062 \, \text{m}^3/\text{hr} \left( \text{m}^2 \times \frac{\text{kg}}{\text{m}^2 \times \text{hr}} \times \frac{\text{m}^3}{\text{kg}} \right)$

= 249062 / 3600 = 69.18 m³ /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

$\text{Formula: Height of the top cone } H_{TC} = \frac{D_{TP} - D_{d}}{2} \times \tan \alpha$

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

$\text{Vapour Inlet pipe dia} = \sqrt{ \frac{4}{\pi} \times \frac{\text{Volume of Vapour}}{\text{Velocity of Vapour}} }$

S.No	Description	Value	UOM
1	Heating surface of the pan	376	m²
2	Evaporation rate	60	kgs/m²/hr
3	Pan Inlet Vapour Temperature	94	^oC
4	Pan Outlet Vapour Temperature	52	^oC
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	M³/kg
2	Specific volume of Pan Outlet vapour	10.98	M³/kg
3	Volume of the pan inlet vapour	12.853	M³/sec
4	Volume of the pan outlet vapour	68.81	M³/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 cm² area for 10 m² heating surface of the vacuum pan

SO Cross section area of non-condensable gases pipeline in cm² = Heating surface in m²/10

= 376 /10 = 37.6 cm²

$\text{Dia of each NCG pipeline} = \sqrt{ \frac{\text{Total area of the NCG in m}^2/\text{sec}}{0.785 \times \text{No. of extraction points}} }$

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 = $\sqrt{\frac{\text{Volume of the massecuite}}{0.785 \times \text{Time required for massecuite discharge in sec} \times \text{Velocity of the massecuite}}}$

Strike Volume = 56.33 m³
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.

$\text{Dia of each condensate line} = \sqrt{ \frac{\text{Volume of condensate in m}^3/\text{sec}}{0.785 \times \text{Velocity of condensate} \times \text{No. of condensate extraction points}} }$

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 m³ /sec

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

13. Calendria shell thickness

$\text{Formula: Calendria shell thickness (ts) in mm} = \frac{P \times D_{i}}{(2 \times f \times J) - P} + C \cdot A$

Here

P = Hydraulic test pressure in kg/cm² = 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/cm² = 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

$\text{Tube plate thickness (tp) in mm} = F \times G \times \frac{0.25 \times P}{f} + C \cdot A$

$F = \sqrt{ \frac{K}{f + 3K} }$

$K = \frac{E_{s} \times t_{s} \times (D_{o} - t_{s})}{E_{t} \times t_{t} \times N_{t} \times (D_{t} - t_{t})}$

Here

C . A = corrosion allowance in mm = 1.5 mm
f = Allowable stress in kg/cm²= 1400kg/cm²
P = Design pressure in kg/cm²= 2.72 kg/cm²
Es = Modulus factor for MS sheet in kg/cm²= 2.1 x 10⁶ Kg/cm²
Et = Modulus factor for SS sheet in kg/cm²= 1.9 x 10⁶ Kg/cm²
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.
$K = \frac{2.1 \times 10^6 \times 18 \times (6100 - 18)}{1.9 \times 10^6 \times 1.625 \times 1643 \times (102 - 1.625)}$
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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