Module 5. Pasteurizers

Lesson 20

SOLVING NUMERICAL

20.1 Problem

Required: Milk pasteurizer, capacity 12,000 1tr/h

                 Inlet temperature                 150C

                 Pasteurizing temperature     850C

                 Outlet temperature                40C

                 Holding time 15 sec.            50 ltr.

                 Regeneration 80%

                 Heat transfer area per plate = 0.375 m2

Overall heat transfer co-efficient for heating, regeneration and cooling and deep cooling are 2320, 2290, 2100 and 1800 (k.cal/m2 hr °C) respectively

Services Available :

                   24,000 l t r / h  hot water of 90°C

                   18,000 l t r / h well water of 11°C

                   24,000 l t r / h chilled water of 1 °C

Branch to centrifuge at about 45 °C. Calculate the number of plates and pressure drop. Make suitable assumptions. For simplification of calculations, take specific heat of milk also as 1.0 k.cal / kg oC.

Solution

Assumption: Neglecting the reduction in volume.

When calculating plate requirement of pasteurizers we normally do not take into account the separation of cream in the centrifuge, which is about 10 %.

Calculation

A regeneration efficiency of 80% means that 80% of the heat is regained in the regenerating section.

In this case 80 (85 – 15) = 56°C rise in temperature

                  100

 

Milk at 15 oC inlet temperature is raised by 56°C in regeneration section, and milk outlet temperature will then be 15 + 56 = 71°C

The  milk pasteurized at 85°C will be cooled down by the raw milk by 56°C and the milk entering the cooling section will be 85 – 56 = 29°C

 

So the complete temperature course will be:

15 – 71 – 85 – 29 – 15 - 4°C. (Assuming a minimum of 4°C temperature gradient between milk and well water)

Pasteurizing Section

Heating by 24,000 l t r / h water of 90 °C

The LMTD calculations are given below:

 

2

 

Applying heat balance X is calculated (temperature drop in hot water)

24000x1x(90-X) = 12000x1x(85-71)

X = 83

Equation for LMTD (Log Mean Temperature Difference)

           

           

            = 8o

The LMTD diagram can be represented as:

           

For heat exchangers heat transfer rate is given as:

            Q= UA

Where,

            Q = rate of heat transfer

            A = Area of heat transfer

             = Log mean temperature difference

 

Or

            Q= UaN ..........(1)

Where,

            A = area of one plate

            N = number of plates

And since,

            Q = m s Δt.......(2)

From equations 1 and 2

            UaN tLMTD = m s Δt

            N =             

Arrangement 3 × 5 / 2 × 8 = 32 plates.

Here,

3 × 5  is for milk (3 passes and 5 channels)

2 × 8 is for hot water (2 passes and 5 channels)

Regenerating Section

The LMTD calculations are given below:

3

We know,

           

 

Since, ∆t2 = ∆t1  = 14 therefore we take the average value,

            

              

            = 14

LMTD diagram can be represented as

           

We know,

            UaN  tLMTD = m s Δt

           

           

Arrangement   6 × 5 / 6 × 5 = 61

However, in the entering range 15 – 71 a branch to the centrifuge has to be provided as close as possible to 45 °C.

Each group means a rise in temperature of 

If we make a branch after 3 groups the temperature there will be approximately 43°C.

Arrangement reg. section   I:  3 × 5 / 3 × 5 = 31   plates

                      Reg. section II:  3 × 5 / 3 × 5 = 31   plates

Cooling

By means of 18.000 l t r / h well water of 11 °C.

4

                                                                                

            MwSwΔt = MmSmΔ

            18000 x 1 x (11-X) = 12000 x 1 x (29-15)

            X = 20o

              

              

            

             = 6.1o

 

LMTD diagram can be represented as:

           

                                       

             t = 6.1 °C

We Know,

            UaN tLTMD = m s Δt

 

           

Arrangement    3 × 6 / 3 × 6 = 37

Deep Cooling

By means of 24.000 l t r / h chilled water of 1°C.

5

  

            MmSmΔt = McScΔ

            12000 x 1 x (15-4) = 24000 x 1 x (X-1)

            X = 6.5o

              

              

              

            = 5.25o

LMTD diagram can be represented as:

                         t = 5.25 °C

            N =     12000 × 11 × 1           = 36

                0.375 × 1800 × 5.25

            Arrangement    3 × 6 / 3 × 6 = 37 

Pressure Drops

Milk                 15 × 1. 12 + 6 × 0.78   + 28 × 0, 1   + 3 (holding section) = 27. 3 mwc

Hot water          2 × 1. 75 + 3 × 0.19                                                        =   4. 1 mwc

Well water        3 × 1. 75 + 4 × 0. 11                                                       =   5. 7 mwc

Chilled water    3 × 3. 00 + 4 × 0. 19                                                        =   9. 8 mwc

 

20.2 PROBLEM :

Calculate the number of plates in each section of HTST pasteurizer, for the following data:

S. No.

Parameters

Milk

Hot Water

Chilled Water

1

Flow Rate

10,000lph

30,000 lph

30,000 lph

2

Inlet temperature

4°C

85°C

1°C

3

Outlet temperature

4°C

--

--

4

Overall heat Transfer coefficient ‘U’ k.cal/m2 hr °C

2320

2290

1800

 

 

Area of each plate is 0.375 m2; Regeneration Efficiency: 80%; Past. Temperature: 75°C

Soln:

Regeneration Section

Regeneration  80/100 = (x-4)/75-4

           X=60.8

Heat balance between raw milk and heated milk

            10,000 x 0.93 x (60.8-4) =10,000 x 0.93 x (75-x)

           X=18.2 °C

Note: Remember that rate of heat is different than the heat balance

Heating Section:

Heat gained by milk = Heat lost by hot water

            10,000 x 0.93 x (75-60.8) =30,000 x1 x (85-X)

            14.2/3 =   85-X           or                     X=80.3 °C

Chilling Section:

Heat lost by milk = Heat gained by chilled water    

           10,000 x 0.93 x (18.2-4) = 30,000 x 1x (X-1)

           X=5.7

Heat gained by raw milk in Regeneration section is rate of transfer in that section

10,000 x 0.93 x (60.8-4) = UA∆TLMTD

 

 

Fig. 20.1 Temperature profiles in various section of pasteurizer

 

       Regeneration Section:                      Chilling section                       Heating section

Balancing the heat exchanged to the Rate of heat transfer equation,

           

           

                =43.3 or 44 plates                              =11.26≈12                       = 29                

 

Answer: The number of plates required in Regeneration, heating and chilling sections are 44, 12 and 29 respectively.

Role of Correction Factor

The above calculations using LMTD, may not be the true representative of counter current flow, as the milk and service fluid are differing in the flow rates.  To control the pressures on both sides to be equal, the flow in the channels is made unequal.  When the milk and service fluid flow is in the ratio of the range of 0.66 to 1.5, it is possible to have equal number of passes on both sides of the exchangers making the flow to be counter current. When it becomes necessary to go for unequal passes due to large differences in flow rates, it becomes necessary to apply significant correction factor to take into account the non-counter current flow situation.

 

The extent of LMTD correction depends upon several factors, including the number of NTU/pass and number of channels per pass.  Correction factor is required also because, in the end zones of plates, heat is transferred only on one side, and with even number of plates, one fluid has an extra stream or passage over the other.

 

 ∆T LMTD cross flow = F X ∆F LMTD

           

Where,  

           

          


20.3 Problem

 

For calculating the time taken for bringing up to pasteurization temperature in a batch pasteurizer

 

Data:

Pasteurization Temperature, Tp

75

C

Conductivity of vat wall , k

13

Initial temperature of milk, To

4

C

Thickness of wall,  x

0.003

Temperature of Hot fluid

85

C

Co-eff of heat transfer, steam

5000

Volume of Milk

3

m3

Co-eff of heat transfer, milk

860

Density of milk

1030

kg/m3

 

 

 

 

Specific heat of milk

0.93

k.cal kg C

 

 

 

 

Area of heat transfer

9.5

m2

 

 

 

 

Overal Heat transfer co-eff

627.5258

k.cal/m2h C

1.008369 h

 

 

 

Time taken for heating, t =

61

min