BIO3810 Assignment 1: Food Production Calculations and Analysis

This document presents a comprehensive solution to a food production assignment, covering various aspects of industrial engineering and food processing. The assignment includes detailed calculations involving unit conversions (BTU/lbm to kJ/kg, psig to kPa, and lb ft-1 s-1 to Pa.s), mass balances in juice concentration (evaporation and sugar addition), moisture adjustment and puffing calculations for grains, and process flow diagrams for expanded rice production. It also addresses calculations for mixing wet mashed potatoes with dried flakes, including water removal and raw potato requirements, and heat transfer calculations for canned peaches in an autoclave. Additionally, the solution explores energy efficiency calculations for a casein drying process, involving natural gas consumption and water evaporation. The solution provides step-by-step calculations, process flow diagrams, and final answers, making it a valuable resource for students studying food engineering and related fields. The assignment brief asks for conversion of units, calculations for juice concentration, and sugar addition, grain processing, and the use of process flow diagrams. It also includes questions about the mixing of mashed potatoes and dried flakes, heat transfer in a canning process, and energy efficiency in a drying process.

Running head: FOOD PRODUCTION
Industrial Engineering Food Processing
Name of the Student
Name of the University
Author note

1FOOD PRODUCTION
Question 1
a)
1 BTU = 1.05506 kJ
1 LBM = 0.453592Kg
Hence, 1 BTU/LBM = (1.05506/0.453592)kJ/Kg
= 2.326011 kJ/Kg
370 BTU/LBM = (370 x 2.326011) kJ/Kg = 860.624 kJ/Kg
b) 1 psig = 6.89476 kPa
28 psig= 28 x 6.89476 kPa
= 193.053 kPa
c)
Pa.s = Newton.second.mitre-2
= 1 kg.mitre.second-2.second.mitre-2
= 1 kg.second-1.mitre-1
1kg = 2.20462
Mitre = 3.28084 foot
1 kg.second-1.mitre-1 = (2.20462/3.28084) lb ft-1 s-1
12 kg.second-1.mitre-1 = 12 x (2.20462/3.28084) lb ft-1 s-1
= 8.064 lb ft-1 s-1

2FOOD PRODUCTION
Juice
3% solid , 97% water
Evaporation
(Water reduction)
Juice
12% solid, 88% water
Sugar added
(solid addition)
Juice
(12+2.5)% solid, (88-2.5)% water
Question 2
Initial pressed juice = 1000 kg
Amount of liquid = 1000 x 97/100 = 970kg
Let, x kg water is evaporated
Total quantity of juice = 1000-x kg
Amount of liquid 970-x kg
Hence,

3FOOD PRODUCTION
(970-x)/(1000-x) = 88/100
Or, 9700 - 100x = 8800 – 88x
Or, 12x= 900
Or, x = 900/12 = 75 kg
Quantity of water evaporated is 75 Kg from per 1000 kg pressed juice
Now per 1000 kg juice
Amount of solid = 1000 x 12/100 = 120kg
Let, quantity of added sugar is y
Total quantity of juice = 1000+y kg
Amount of solid = 120+y kg
Hence,
(120+y)/(1000+y) = 14.5/100
Or, 12000 + 100y = 14500 – 14.5x
Or, 85.5y = 2500
Or, y = 2500/85.5 = 29.24 kg
Quantity of sugar added 29.24 kg for each1000 kg pressed juice
Question 3

4FOOD PRODUCTION
Grain
89% solid, 11% water
Adjusted moisture on a dry basis (Water addition)
Grain
83.5% solid, 16.5% water
Puffing
(Water reduction)
Buckwheat
98% solid, 2% water
a)
Initial grain = 100 kg
Amount of water = 100 x 11/100 = 11kg
Let, x kg water added
Total quantity of grain = 100+x kg
Amount of water 11+x kg
Hence,

5FOOD PRODUCTION
(11+x)/(100+x) = 16.5/100
Or, 1100 + 100x = 1650 + 16.5x
Or, 83.5x= 550
Or, x = 550/83.5 = 6.587 kg
6.587 kg water should be added
b)
Conditioned grain = 100 kg
Amount of water = 16.5 Kg
Let, y kg water removed
Total quantity of grain = 100-y kg
Amount of water 16.5 - y kg
Hence,
(16.5-y)/(100-y) = 2/100
Or, 1650 - 100y = 200 - 200y
Or, 100 y= 1450
Or, y = 14.5 kg
14.5 kg is moisture is removed
c)
P1.V1/T1= P2.V2/T2
Or, P2=P1(V1.T2/V2.T2)
Since, during pressure the volume does change: V1=V2

6FOOD PRODUCTION
P2=P1(T2/T1)
T1 is room temperature, that is 298.15K
T2 is described temperature that is 2000C=473.15K
P2 = P1(473.15/298.15) = P1 (1.587)
Standard atmospheric pressure is P1= 101325Pa
P2=101325(1.587) = 160798Pa
Or, P2 = 1.6 MPa (approx.)
Therefore the pressure will be 1.6 MPa
Question 4
Process flow chart of expanded rice making in a local industry

7FOOD PRODUCTION
Question 5
a)
Ratio of mixing wet mashed potatoes with dried flex = 95:5
Percentage of wet mashed potato 95%
Percentage of dried flakes 5%
Water content in mashed potato is 82%
Water content in flakes is 3%
Water content in mixture due to mashed potato is 95% x 82% = 77.8%
Water content in mixture due to dried flakes is 5% x 3% = 1.5%
Total water content in mixture of wet mashed potatoes and dried flakes = 77.8 + 1.5 = 79.3%
Solid content before = 100 – 79.3 = 20.7%
After drying the water will be 3%
Solid content after= 100 – 3% = 97 kg
Le, m amount of water is evaporated
Before evaporation the total amount (100+m) kg
Total solid content was (100 + m) x 20.7%
Total solid content after drying = 97Kg
(100 + m) x (20.7/100) = 97
Or, m = (97 x 100/20.7) – 100
Or, m = 368.56
Total amount of water removed is 368.6 Kg

8FOOD PRODUCTION
Wet mashed potato
Mix wet mashed potato with flakes
95:5
Granulator
Drum Dryer
(Water reduction)
Flakes
97% solid, 3% water
Potato flakes
b)
Ratio of mixing wet mashed potatoes with dried flex = 95:5
Percentage of wet mashed potato 95%
Percentage of dried flakes 5%
Water content in mashed potato is 82%
Water content in flakes is 3%
Water content in mixture due to mashed potato is 95% x 82% = 77.8%

9FOOD PRODUCTION
Water content in mixture due to dried flakes is 5% x 3% = 1.5%
Total water content in mixture of wet mashed potatoes and dried flakes = 77.8 + 1.5 = 79.3%
The moisture content of the granulated paste fed to the dryer is 79.3%
c)
Amount of water removed 368.6 Kg
Granulated pest 368.6 Kg + 100kg = 468.6 kg
Ratio of mixing wet mashed potatoes with dried flex = 95:5
Considering the fact that 82% water is present in raw potato as well it can be said that
Masked potato = 468.6 x (95/100) = 445.17 kg
let, the weight of the potato is m kg
8.5% of the raw potato weight is lost on peeling
After peeling the potato the quantity will be = m x (91.5/100) = 465.17 kg
Or, m x (91.5/100) = 445.17
Or, m= 445.17 x (100/91.5)
Or, m = 486.524
The amount of raw potatoes needed toproduce 100 kg of dried flakes is 486.524 kg
d)
Moisture level of potato = 82%
Solid level of potato = 18%
Total solid level = 87.574 kg
Price 1 ton price =$200

10FOOD PRODUCTION
Price of 907.185 kg = $200
Price of 87.574 kg = (200/907.185) x 87.574 = $19.3
For Moisture level of potato = 85%
Solid level of potato = 15%
Total solid level in 486.524 kg= 72.978 kg
Price 1 ton price =$200
Price of 907.185 kg = $200
Price of 72.978 kg = (200/907.185) x 72.978 = $16
Question 6
Number of cans 1000
Weight of each can 60 g
Weight of all cans 60kg
Specific heat of the can metal is 0.50 kJ/kg.K.
Lost temperature during the cooling process of 120 to 40 = (120-40) = 800C
Lost head from can metals during cooling process = (0.50 x 80 x 60) = 2400 kJ
Peach per can 0.45Kg
Total weight of peach = (1000 x 0.45) = 450Kg
Percentage of water 93%

11FOOD PRODUCTION
Autoclave
1000 cans of peaches
Heating 120C
Cooled Autoclave
1000 cans of peaches
Cooling Water
15 C
Leaving Water
35C
Percentage of peach 7%
Total weight of water = (450*93)/100 = 418.5 Kg
Total weight of peach = (450*7)/100 = 31.5 kG
Specific heat of water = 4.19 kJ/kgK
Lost head from water during cooling process = (4.19 x 80 x 418.5) = 140281.2 kJ
Heat content of the autoclave walls above 40°C is 1.6 x 104 kJ
Heat released by peach 140281.2 kJ- 166.4kJ = 140114.8 kJ
Specific head of peach = 140114.8/ (80 x 31.5)= 55.610
Question 7
Calorific value of natural gas 890 kJ/mole
Molar mas of natural gas is 19 g/mole
890kj/mole = 890 x (1/19) x 1000