Bioprocess Engineering Multiple Choice Questions 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
Answer: B
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
Answer: A
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
Answer: B
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
Answer: B
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. Refer to Q4, and estimate the Agitator impeller diameter D, assuming it to be 33% of tank diameter.
A. 4.62 m
B. 4.50 m
C. 4.60 m
D. 4.52 m
Answer: A
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
Answer: C
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
Answer: D
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. Refer to Q7, and determine the agitated batch liquid Reynolds number. (ρB = 64.0 lbs./cu. ft. and μB = 2.12 lb./hr-ft.)
A. 977,035
B. 900,045
C. 877,035
D. 800,045
Answer: A
Explanation: NRe = Reynolds number for agitated liquids = ( D2 * N * ρB / μB).
NRe = (3.02 * 3,600 * 64.0 / 2.12)
NRe = 977,035.
9. Refer to Q7 and Q8, and determine the agitator speed in appropriate units for power calculation.
A. 3 rps
B. 1 rps
C. 6 rps
D. 4 rps
Answer: B
Explanation: N = revolutions per sec (rps). The agitation speed is given in the example as 60 revolutions per minute, which when divided by 60 seconds/minute yields:
N = 60 rpm / 60 = 1 rps.
10. Refer to Q7, Q8 and Q9, and determine Power Number from the appropriate correlation, and calculate the uncorrected Power required. (Gc = 32.2).
A. 5.3 Hp
B. 6.3 Hp
C. 5.6 Hp
D. 6.6 Hp
Answer: A
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. Refer to Q7, 8, 9 and 10, and determine the motor Power required using estimated losses at 30%.
A. 5.3 Hp
B. 6.3 Hp
C. 5.6 Hp
D. 6.6 Hp