This set of Bioprocess Engineering Questions and Answers for Experienced people focuses on “Energy Balance Numericals Without Reactions”.

1. Skim milk is prepared by the removal of some of the fat from whole milk. This skim milk is found to contain 90.5% water, 3.5% protein, 5.1% carbohydrate, 0.1% fat and 0.8% ash. If the original milk contained 4.5% fat, Calculate its composition assuming that fat only was removed to make the skim milk and that there are no losses in processing.

a) Fat = 4.5%, Water = 86.5%, protein = 3.3%, carbohydrate = 4.9%, ash = 0.8%

b) Fat = 4.5%, Water = 83.5%, protein = 3.0%, carbohydrate = 4.5%, ash = 0.9%

c) Fat = 4.6%, Water = 80.5%, protein = 3.5%, carbohydrate = 4.0%, ash = 0.9%

d) Fat = 4.6%, Water = 81.5%, protein = 3.3%, carbohydrate = 4.5%, ash = 0.8%

View Answer

Explanation: Basis: 100 kg of skim milk.

This contains, therefore, 0.1kg of fat. Let the fat which was removed from it to make skim milk be x kg.

Total original fat = (x + 0.1) kg

Total original mass = (100 + x) kg

and as it is shown that the original fat content was 4.5% so,

(x+0.1)/(100+x) = 0.045

where = x+0.1 = 0.045(100+x)

x = 4.6 kg

So the composition of the whole milk is then fat = 4.5%, water = 90.5/104.6 = 86.5%, protein = 3.5/104.6 = 3.3%, carbohydrate = 5.1/104.6 = 4.9% and ash = 0.8%.

2. If 35,000kg of whole milk containing 4% fat is to be separated in a 6 hour period into skim milk with 0.45% fat and cream with 45% fat, what are the flow rates of the two output streams from a continuous centrifuge which accomplishes this separation? (Basis 1 hour’s flow of whole milk.)

a) 5678 kg/hr

b) 5368 kg/hr

c) 2567 kg/hr

d) 2578 kg/hr

View Answer

Explanation: Mass in: Total mass = 35000/6 = 5833 kg.

Fat = 5833 x 0.04 = 233 kg.

And so Water plus solids-not-fat = 5600 kg.

Mass out: Let the mass of cream be x kg then its total fat content is 0.45x. The mass of skim milk is (5833 – x) and its total fat content is 0.0045 (5833 – x).

Material balance on fat: Fat in = Fat out

5833 x 0.04 = 0.0045(5833 – x) + 0.45x. And so x = 465 kg.

So that the flow of cream is 465 kg / hr and skim milk (5833 – 465) = 5368 kg/hr.

3. A textile dryer is found to consume 4 m^{3} / hr of natural gas with a calorific value of 800 kJ/mole. If the throughput of the dryer is 60 kg of wet cloth per hour, drying it from 55% moisture to 10% moisture, estimate the overall thermal efficiency of the dryer taking into account the latent heat of evaporation only.

Assuming the natural gas to be at standard temperature and pressure at which 1 mole occupies 22.4 liters and Latent heat of evaporation = 2257 kJ/K.

a) 40%

b) 50%

c) 58%

d) 48%

View Answer

Explanation: 60 kg of wet cloth contains

60 x 0.55 kg water = 33 kg moisture

and 60 x (1-0.55) = 27 kg bone dry cloth.

As the final product contains 10% moisture, the moisture in the product is 27/9 = 3 kg

And so Moisture removed / hr = 33 – 3 = 30 kg/hr

Latent heat of evaporation = 2257 kJ/K

Heat necessary to supply = 30 x 2257 = 6.8 x 10^{4} kJ/hr

Assuming the natural gas to be at standard temperature and pressure at which 1 mole occupies 22.4 litres.

Rate of flow of natural gas = 4 m^{3} / hr = (4 x 1000)/22.4 = 179 moles/hr

Heat available from combustion = 179 x 800 = 14.3 x 10^{4} kJ/hr

Approximate thermal efficiency of dryer = heat needed / heat used

= 6.8 x 10^{4} / 14.3 x 10^{4} = 48%.

4. It is desired to freeze 10,000 loaves of bread each weighing 0.75 kg from an initial room temperature of 18°C to a final temperature of -18°C. The bread-freezing operation is to be carried out in an air-blast freezing tunnel. It is found that the fan motors are rated at a total of 80 horsepower and measurements suggest that they are operating at around 90% of their rating, under which conditions their manufacturer’s data claims a motor efficiency of 86%. If 1 ton of refrigeration is 3.52 kW, estimate the maximum refrigeration load imposed by this freezing installation assuming that fans and motors are all within the freezing tunnel insulation and the heat-loss rate from the tunnel to the ambient air has been found to be 6.3 kW.

Extraction rate from freezing bread (maximum) = 104 kW

a) 46 tons of refrigeration

b) 40 tons of refrigeration

c) 56 tons of refrigeration

d) 50 tons of refrigeration

View Answer

Explanation: Extraction rate from freezing bread (maximum) = 104 kW

Fan rated horsepower = 80

Now 0.746 kW = 1 horsepower and the motor is operating at 90% of rating,

And so (fan + motor) power = (80 x 0.9) x 0.746 = 53.7 kW

With motors + fans in tunnel

Heat load from fans + motors = 53.7 kW

Heat load from ambient = 6.3 kW

Total heat load = (104 + 53.7 + 6.3) kW = 164 kW

= 46 tons of refrigeration.

5. Refer to Q4 and estimate the maximum refrigeration load imposed by this freezing installation assuming the fans but not their motors are in the tunnel.

a) 50.5 tons of refrigeration

b) 40.5 tons of refrigeration

c) 44.5 tons of refrigeration

d) 55.5 tons of refrigeration

View Answer

Explanation: With motors outside, the motor inefficiency = (1 – 0.86) does not impose a load on the refrigeration

Total heat load = (104 + [0.86 x 53.7] + 6.3)

= 156 kW

= 44.5 tons of refrigeration.

6. Water is pumped from a storage tank through a tube of 3.0 cm inside diameter at the rate of 0.001 m^{3} /s. What is the kinetic energy per kg water in the tube?

a) 2.00 J/kg

b) 1.00 J/kg

c) 5.00 J/kg

d) 0.05 J/kg

View Answer

Explanation: Kinetic energy E

_{k}= mv

^{2}, tube dia. D = 3.0 cm, m = 1 kg.

Cross-section area of tube A = \(\frac{1}{4}\) πD

^{2}= \(\frac{1}{4}\) π(3.0/100 m)

^{2}= 7.0686×10

^{-4}m

^{2}

Average velocity of water v = Q/A = 0.001 m

^{3}/(7.0686×10

^{-4}m

^{2}) = 1.415 m/s

KE per kg = Ek/m = 1/2 v

^{2}= 1/2 (1.415 m/s)

^{2}= 1.00 m

^{2}/s

^{2}= 1.00 J/kg.

7. Water is pumped from a storage tank (tank 1) to another tank (2) which is 40 ft above tank. Calculate the potential energy increase with each lb of water pumped from tank 1 to tank2.

a) 119.544 J/kg

b) 120.678 J/kg

c) 122.500 J/kg

d) 190.600 J/kg

View Answer

Explanation: PE increase per lb = E

_{p}/ m = gh = (9.806 m/s

^{2}) (40 m/3.2808) = 119.54 m

^{2}/s

^{2}= 119.544 J/kg.

8. Refer to Q7, and Express the answer in btu/lb.

a) 0.0421 btu/lb

b) 0.0532 btu/lb

c) 0.0514 btu/lb

d) 0.0432 btu/lb

View Answer

Explanation: PE increase per lb = E

_{p}/ m = gh = (9.806 m/s

^{2}) (40 m/3.2808) = 119.54 m

^{2}/s

^{2}= 119.544 J/kg = (119.544 btu/1055.6)/2.2046 lb = 0.0514 btu/lb.

9. Concentrated fermentation liquid containing 20% (w/w) gluconic acid from an evaporator has a flow rate of 2000 kg/h and a temperature 90 °C. It needs to be cooled to 6 °C in a heat exchanger with cooling water. The cooling water has a flow rate 2700 kg/h and an initial temperature 2 °C. If the cooling water leaves the heat exchanger at 50 °C, what is the rate of heat loss from gluconic acid solution to the surroundings? Assume the heat capacity of gluconic acid is 0.35 cal/g-°C^{-1}.

a) 69390.84 kJ/h

b) 65780.56 kJ/h

c) 67890.67 kJ/h

d) 65432.10 kJ/h

View Answer

Explanation: Heat lost by gluconic acid = heat gained by water [2000(0.2)×(0.35×4.184×1000/1000)×(90-6) + 2000×(0.8)×(376.92-25.2)]+Q

= 2700×(209.33-8.37)×611955.84+Q=542565

Q = -69390.84 kJ/h

Negative sign means heat loss by the system to surrounding.

Hence, heat loss to the surroundings is 69390.84 kJ/h.

10. How much work can be obtained from an adiabatic, continuous-flow turbine, if steam at 60 bar and 500^{o}C is used and the outlet stream is at 1 bar and 400^{o}C?

a) -144 kJ/kg

b) -155 kJ/kg

c) -166 kJ/kg

d) -177 kJ/kg

View Answer

Explanation: Basis: 1 kg steam

For superheated steam:

Inlet: (60 bar, 500°C) h1 = 3422 kJ/kg

Outlet: (1 bar, 400°C) h2 = 3278 kJ/kg

m_{in} = m_{out} = 1 kg

DE_{CV} + Sn_{i}(E_{k} + E_{p} + H)_{i} = Q + W_{S} (General form of the energy balance)

Now we make the following simplifying assumptions for this turbine:

▪ Adiabatic: Q = 0

▪ The inlet and outlet lines are not tremendously different in height: E_{p} = 0

▪ There is little difference in the velocity of the fluid in the inlet and outlet lines: E_{k} = 0

▪ There are no accumulation terms (steady-state operation with steady flow): DE_{CV} = 0

The simplified form of the energy balance is therefore:

H_{out} – H_{in} = W_{s} = m_{out}h_{out} – m_{in}h_{in} = m(h_{out} – h_{in})

Substituting in the values for the specific enthalpies gives:

W_{s} = m (h_{out} – h_{in}) or W_{s}/m = (3278 – 3422) kJ/kg = -144 kJ/kg.

**Sanfoundry Global Education & Learning Series – Bioprocess Engineering.**

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