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Technical Module on
Design calculations of Milk chilling Centre
Refrigeration Load calculation
As, Heat Load, Q
𝑸 = 𝒎𝒄∆𝑻
Where,
Q- Heat required to remove to chill milk, Kcal/hr.
m- Mass flow rate of milk, Kg/hr.
c- Specific heat of milk, Kcal/ Kg°C
ΔT- temperature difference between incoming & chilled milk, °C
1 TR = 3028 Kcal
1HP = 641.41 Kcal/hr.
1 TR = 4.71 HP/hr.
Compressor
As,
1TR=4.71 HP
• Means if a compressor of 4.71 HP operates for 1 hr., it will
produce 1 Ton of refrigeration
So,
• For compressor of 70% efficiency
1 TR=6.72 HP
So 50 HP compressor (70%), will produce 7.44 TR if operates
for 1 hr.
Ice Bank tank (IBT) Design
Suppose chilling load (in one shift) = 145.51 TR
So we required 145.51 TR ice
As, 1 TR = 3028 Kcal
So,
145.51 TR = 145.51*3028
= 440604.28 Kcal
= mL
So,
m= 5507.55 Kgs of ice
Or 551 RMT of IBT coil
Where, L- latent heat of fusion of ice
= 80 Kcal/kg
Chiller Design
I. Specific gravity of milk = d1
II. Specific gravity of milk = d2
III. Chilling medium: ----
IV. Chilling medium inlet temperature, ti1°C
V. Chilling medium outlet temperature, to1°C
VI. Milk inlet temperature, ti2°C
VII. Milk outlet temperature, to2°C
VIII. Milk flow rate = M1 LPH
IX. Chilling medium flow rate = M2 LPH?
X. Specific heat of Milk = X1 Kcal/kg°C
XI. Specific heat of Brine = X2 Kcal/kg°C
So,
M1*d1*X1*(ti2-to2) = M2*d2*X2*(to1-ti1)
So, required chilling medium flow rate = M2= -----
LMTD- Log mean temperature difference
If,
 Chilling medium inlet temperature, ti1°C
 Chilling medium outlet temperature, to1°C
 Milk inlet temperature, ti2°C
 Milk outlet temperature, to2°C
Then,
𝐿𝑀𝑇𝐷 =
(ti2 − to1) − (to2 − ti1)
ln(
𝑡𝑖2 − 𝑡𝑜1
𝑡𝑜2 − 𝑡𝑖1)
Required nos. of Plates in chiller
As, Heat Load, Q
𝑸 = 𝑼𝑨(𝑳𝑴𝑻𝑫)
Where,
 U- Overall heat transfer coefficient, Kcal/m2 °C hr.
 A- Total effective heat transfer area, m2
 LMTD- Log mean temperature difference, °C
 Considering U (Overall heat transfer coefficient, Kcal/m2 °C hr.)
Required nos. of plates in chiller
𝑵 =
𝑻𝒐𝒕𝒂𝒍 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒂𝒓𝒆𝒂
𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒐𝒏𝒆 𝒑𝒍𝒂𝒕𝒆
Checklist
As mentioned in above slides, following are some essential information needed for
evaluation of MCCs on plant and machinery aspect to optimize chilling process.
1. Plate heat exchanger
 Make
 No. of Plates
 Plate’s dimension
 Make specification
 Milk inlet and outlet temperature
 Chilling medium inlet and outlet temperature
 Milk & chilling medium flow rate
 Plate overall heat transfer coefficient
2. Pump specification
 Make specification: Milk pump & Chilled water (or brine) pump
3. Compressor
 Make specification
4. IBT
 Storage capacity
 Evaporator coil length
Prepared By:
Mukesh Pratap Singh (NDS- DVC)
Contacts:
Mr. Basant Choudhary:
basant.choudhary@nddbdairyservices.com
Mr. Mukesh Pratap Singh:
mukesh.singh@nddbdairyservices.com
Thank you

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Technical Module on MCC Design calculation

  • 1. Technical Module on Design calculations of Milk chilling Centre
  • 2. Refrigeration Load calculation As, Heat Load, Q 𝑸 = 𝒎𝒄∆𝑻 Where, Q- Heat required to remove to chill milk, Kcal/hr. m- Mass flow rate of milk, Kg/hr. c- Specific heat of milk, Kcal/ Kg°C ΔT- temperature difference between incoming & chilled milk, °C 1 TR = 3028 Kcal 1HP = 641.41 Kcal/hr. 1 TR = 4.71 HP/hr.
  • 3. Compressor As, 1TR=4.71 HP • Means if a compressor of 4.71 HP operates for 1 hr., it will produce 1 Ton of refrigeration So, • For compressor of 70% efficiency 1 TR=6.72 HP So 50 HP compressor (70%), will produce 7.44 TR if operates for 1 hr.
  • 4. Ice Bank tank (IBT) Design Suppose chilling load (in one shift) = 145.51 TR So we required 145.51 TR ice As, 1 TR = 3028 Kcal So, 145.51 TR = 145.51*3028 = 440604.28 Kcal = mL So, m= 5507.55 Kgs of ice Or 551 RMT of IBT coil Where, L- latent heat of fusion of ice = 80 Kcal/kg
  • 5. Chiller Design I. Specific gravity of milk = d1 II. Specific gravity of milk = d2 III. Chilling medium: ---- IV. Chilling medium inlet temperature, ti1°C V. Chilling medium outlet temperature, to1°C VI. Milk inlet temperature, ti2°C VII. Milk outlet temperature, to2°C VIII. Milk flow rate = M1 LPH IX. Chilling medium flow rate = M2 LPH? X. Specific heat of Milk = X1 Kcal/kg°C XI. Specific heat of Brine = X2 Kcal/kg°C So, M1*d1*X1*(ti2-to2) = M2*d2*X2*(to1-ti1) So, required chilling medium flow rate = M2= -----
  • 6. LMTD- Log mean temperature difference If,  Chilling medium inlet temperature, ti1°C  Chilling medium outlet temperature, to1°C  Milk inlet temperature, ti2°C  Milk outlet temperature, to2°C Then, 𝐿𝑀𝑇𝐷 = (ti2 − to1) − (to2 − ti1) ln( 𝑡𝑖2 − 𝑡𝑜1 𝑡𝑜2 − 𝑡𝑖1)
  • 7. Required nos. of Plates in chiller As, Heat Load, Q 𝑸 = 𝑼𝑨(𝑳𝑴𝑻𝑫) Where,  U- Overall heat transfer coefficient, Kcal/m2 °C hr.  A- Total effective heat transfer area, m2  LMTD- Log mean temperature difference, °C  Considering U (Overall heat transfer coefficient, Kcal/m2 °C hr.) Required nos. of plates in chiller 𝑵 = 𝑻𝒐𝒕𝒂𝒍 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒂𝒓𝒆𝒂 𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒉𝒆𝒂𝒕 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒐𝒏𝒆 𝒑𝒍𝒂𝒕𝒆
  • 8. Checklist As mentioned in above slides, following are some essential information needed for evaluation of MCCs on plant and machinery aspect to optimize chilling process. 1. Plate heat exchanger  Make  No. of Plates  Plate’s dimension  Make specification  Milk inlet and outlet temperature  Chilling medium inlet and outlet temperature  Milk & chilling medium flow rate  Plate overall heat transfer coefficient 2. Pump specification  Make specification: Milk pump & Chilled water (or brine) pump 3. Compressor  Make specification 4. IBT  Storage capacity  Evaporator coil length
  • 9. Prepared By: Mukesh Pratap Singh (NDS- DVC) Contacts: Mr. Basant Choudhary: basant.choudhary@nddbdairyservices.com Mr. Mukesh Pratap Singh: mukesh.singh@nddbdairyservices.com