This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Power Requirements for Mixing”.
1. “Power required for turbulent flow is independent of the viscosity of the fluid but proportional to fluid density”, this statement is applicable to which type of regime?
a) Laminar regime
b) Turbulent regime
c) Transition regime
d) Streamline regime
View Answer
Explanation: In turbulent regime, Power required for turbulent flow is independent of the viscosity of the fluid but proportional to fluid density. The turbulent regime is fully developed at Rei > 103 or 104 for most small impellers in baffled vessels. For the same impellers in vessels without baffles, the power curves are somewhat different, Without baffles, turbulence is not fully developed until Rei > 105; even then the value of Np’ is reduced to between 1/2 and 1/10 that with baffles.
2. Pseudoplastic consume less power than Newtonian fluids?
a) True
b) False
View Answer
Explanation: The laminar region extends to higher Reynolds numbers in pseudoplastic fluids than in Newtonian systems. At Rei below 10 and above 200 the results for Newtonian and non-Newtonian fluids are essentially the same; in the intermediate range, pseudoplastic liquids consume less power than Newtonian fluids. Flow patterns in pseudoplastic and Newtonian fluids differ significantly. Even when there is high turbulence near the impeller in pseudoplastic systems, the bulk liquid may be moving very slowly and consuming relatively little power.
3. Power consumption is less in ungassed fluids.
a) True
b) False
View Answer
Explanation: Liquids into which gas is sparged have reduced power requirements. Gas bubbles decrease the density of the fluid. The presence of bubbles also affects the hydrodynamic behaviour of fluid around the impeller. Large gas-filled cavities develop behind the stirrer blades in aerated liquids; these cavities reduce the resistance to fluid flow and decrease the drag coefficient of the impeller. With sparging, the power consumed could be reduced to as little as half the ungassed value, depending on gas flow rate.
4. Fluid height in the tank is H = 16 m and Diameter of the tank, D = 14 m. Calculate the slurry volume in the tank.
a) 2450 m3
b) 2463 m3
c) 3450 m3
d) 3463 m3
View Answer
Explanation: Fluid Height in Tank , H = 16 m and Diameter of tank, D = 14 m
Slurry volume in tank = π * D2*H/4 = π * (14)2*16 / 4 = 2463 m3.
5. If the diameter of an agitator impeller is assumed to be 33% of a tank diameter of 14 meters, what would the impeller diameter be?
a) 4.62 m
b) 4.50 m
c) 4.60 m
d) 4.52 m
View Answer
Explanation: Agitator Impeller Diameter, D = 33 % of tank diameter = 14 * 33% m = 4.62 m.
6. What is the degree of agitation if bulk fluid velocity is 23.55 ft/min? (For 6 ft/min., degree of agitation = 1 and Degree of agitation varies from 0 to 10).
a) 5
b) 6
c) 4
d) 3
View Answer
Explanation: Degree of Agitation = bulk fluid velocity / 6
= 23.55/6 = 3.93 ∼ 4
7. It is proposed to mix a batch of water at 80 deg F in an agitated vessel. Determine the agitator speed in appropriate units for Reynolds number calculation for agitation speed of 60 rpm. Use an approximate loss of efficiency of 20% for the motor, gearbox, and bearings. Equipment details are given below:
Equipment Data: Vessel Diameter (T) = 9.0 ft. Impeller Diameter (D) = 3.0 ft. Impeller type = Flat blade turbine. # of blades = 6 Blade width (W) =7 inches. Blade length (L) = 9 inches. Number of Baffles = 4; equally spaced. Baffle width (B) = 11 inches. Liquid Depth (H) = 9.0 feet. Impeller height above vessel bottom (Z) = 3 feet.
a) 2, 400 rph
b) 4, 500 rph
c) 3, 500 rph
d) 3, 600 rph
View Answer
Explanation: N = revolutions per hour (rph). The agitation speed is given in the example as 60 revolutions per minute, which when multiplied by 60 minutes/hr yields:
N = 60 rpm * 60 = 3,600 rph.
8. Determine the Reynolds number for an agitated liquid with an impeller diameter of 3 feet, an agitator speed of 3,600 rph, fluid density of 64 lbs/cu. ft., and fluid viscosity of 2.12 lb/hr-ft.
a) 977,035
b) 900,045
c) 877,035
d) 800,045
View Answer
Explanation: NRe = Reynolds number for agitated liquids = ( D2 * N * ρB / μB).
NRe = (3.02 * 3,600 * 64.0 / 2.12)
NRe = 977,035.
9. What is the agitator speed in revolutions per second (rps) if the agitation speed is 60 rpm?
a) 3 rps
b) 1 rps
c) 6 rps
d) 4 rps
View Answer
Explanation: Convert rpm to rps by dividing by 60:
N = 60 rpm / 60 = 1 rps.
10. Using a Power Number of 6, calculate the uncorrected power required for an agitated liquid with density 64 lbs/cu. ft. and an impeller diameter of 3 ft, if the agitator speed is 1 rps. (Use Gc = 32.2 ft-lb/sec).
a) 5.3 Hp
b) 6.3 Hp
c) 5.6 Hp
d) 6.6 Hp
View Answer
Explanation: The correlations calculate the delivered Power to the fluid. This must be corrected back to the agitator motor and account for efficiency losses in the motor, gearbox, and bearings.
PDelivered = (NP ρB * N3 * DA5) / GC
PDelivered = 6 * 64 * 13 * 35 / 32.2
PDelivered = 2,898 ft-lb/sec
or using 550 ft-lb/sec per Hp,
PDelivered = 5.3 Hp.
11. If the motor efficiency is reduced by 30%, what is the required motor power to deliver 5.3 Hp to the fluid?
a) 5.3 Hp
b) 6.3 Hp
c) 5.6 Hp
d) 6.6 Hp
View Answer
Explanation: PMotor = PDelivered / ( 1 – Losses )
PMotor = 5.3 / ( 1 – 0.20 )
PMotor = 6.6 Hp.
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