# Heat Transfer Operations Questions and Answers – Packed Beds – Pressure Drop Calculations

This set of Heat Transfer Operations Multiple Choice Questions & Answers focuses on “Packed Beds – Pressure Drop Calculations”.

1. The Ergun equation is used to calculate the ___________
a) Pressure drop in packed beds
b) Reynolds number in packed columns
c) Overall heat transfer coefficients
d) Number of particles

Explanation: The Ergun Equation given as f = ΔP/L DP/ρϑS2 (∈3/1-∈) is fundamentally used to calculate the pressure drop in packed bed of given length and particle size, for a non-laminar flow regime.

2. In the Ergun equation, what do you understand by the term X?
f = ΔP/L X/ρϑS2(∈3/(1-∈))
a) Diameter of the tube
b) Diameter of the pellets
c) Effective diameter
d) Equivalent diameter of the tube

Explanation: f= ΔP/L DP/ρϑS2(∈3/(1-∈)) is the Ergun equation for non-laminar flow regime, here DP is the Particle diameter or the pellet diameter and not the equivalent one.

3. If we have very small dp/di < 0.01, then we may use ____________
a) Ergun Equation
b) Kozeny Carman Equation
c) Sieder Tate Equation
d) Arrhenius Equation

Explanation: When we have very small particle size such as dp/di = 0.01, the flow regime more or less becomes laminar, hence the relation for laminar flow is Kozeny Carman equation.

4. What is the Reynolds number for the following setup?
Density = 990 Kg/m3
Superficial velocity = 10m/s
DP = 3mm
Porosity = 0.14
Viscosity = 2.5×10-5 kg/ms
a) 2.76×103
b) 2×105
c) 0.76×105
d) 2.76×105

Explanation: The Reynolds number for packed bed is defined as Re = $$\frac{\rho v_s D_P}{(1-\in)\mu} = \frac{990×2×0.003}{(1-0.14)×2.5×10^{-5}}$$ = 2.76×105.

5. What is the friction factor for the given parameters for a non laminar packed bed setup?
Density = 990 Kg/m3
Superficial velocity = 10m/s
DP = 3mm
Porosity = 0.14
Viscosity = 2.5×10-5kg/ms
a) 1.751
b) 1.740
c) 1.752
d) 1.749

Explanation: The Reynolds number for packed bed is defined as Re = $$\frac{\rho v_s D_P}{(1-\in)\mu} = \frac{990×2×0.003}{(1-0.14)×2.5×10^{-5}}$$ = 2.76×105.
Now the Friction Factor is f=150/Re +1.75 = 150/2.76×105+1.75 = 1.751.
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6. What is the pressure drop per unit length for the given parameters for a laminar packed bed setup?
Sphericity = 0.6
Superficial velocity = 10m/s
DP = 3mm
Porosity = 0.4
Viscosity = 2.5×10-5kg/ms
a) -15.625Pa/m
b) -15.625KPa/m
c) -1.875Pa/m
d) -1.875KPa/m

Explanation: The pressure drop per unit length of the bed is defined as $$\frac{∆P}{L} = \frac{-180μ}{S^2 D_P^2} (\frac{(1-∈)^2}{∈^3}) ϑ_S$$=$$-180×2.5\frac{\frac{10^{-5}}{0.6^2}}{0.003^2}×(\frac{(1-0.14)^2}{0.14^3})×2$$ = -15625 Pa/m.

7. If we have a tube of diameter 30mm and particles of diameter 2mm, the length of the tube is 10cm and the number of pellets is 100. Then what is the porosity?
a) 0.45
b) 0.95
c) 0.99
d) 0.65

Explanation: Porosity = Vv/VT = (VT – Vp×N)/VT = $$\frac{\frac{\pi}{4}×0.03^2×0.1-100×\frac{4\pi}{3×8}×0.002^3}{\frac{\pi}{4}×0.03^2×0.1}$$ = 0.994.

8. If we have a tube of diameter 30mm and particles of diameter 3mm, the length of the tube is 10cm and the number of pellets is 100. The flow rate is 5cm3/s, then what is the superficial velocity?
a) 0.5cm/s
b) 0.45cm/s
c) 0.7cm/s
d) 0.8cm/s

Explanation: Superficial velocity = Q/A = flow rate/cross section = $$\frac{5}{\frac{\pi}{4}×3^2}$$=0.7cm/s.

9. What is the relation between superficial velocity and average fluid velocity in a packed bed?
a) ϑ=φu
b) ϑ=u/φ
c) ϑ=φ0.14u
d) ϑ=φ$$^\frac{\vartheta}{u}u$$

Explanation: The correct relation between the superficial velocity and the average velocity is given as ϑ=φu, whereϑ is the porosity of the bed. The average fluid velocity is the cumulative mean of the velocity of fluid through the voids which is difficult to measure.

10. In the Formula for porosity, what is the meaning of the term V?
∅ = $$\frac{V}{V_T}$$
a) Volume of Void
b) Volume of Pure Solid
c) Total Volume
d) Volume of the tube material

Explanation: The porosity or the void fraction is the measure of the void area of the bed, hence the definition is
∅ = $$\frac{V}{V_T}$$
Where V is the void volume, which is the volume excluding particles.

11. Which one of the following is the correct expression for Friction Factor in Packed Bed Columns?
a) f = 150/Re
b) f = 150/Re+1.75
c) f = 180/Re+1.5
d) f = 100/Re+0.0035

Explanation: The friction factor for a packed bed column for non laminar flow is given by the Ergun equation which can be represented as f = 150/Re+1.75.

12. Calculate the pressure drop in the following setup.
Density = 990 Kg/m3
Superficial velocity = 10m/s
DP = 3mm
Porosity = 0.14
Viscosity = 2.5×10-5 kg/ms
Column length = 1m
a) 7Pa
b) 724×106Pa
c) 7×106Pa
d) 72atm

Explanation: The Reynolds number for packed bed is defined as Re = $$\frac{\rho v_s D_P}{(1-\in)\mu} = \frac{990×2×0.003}{(1-0.14)×2.5×10^{-5}}$$ = 2.76×105.
Now the Friction Factor is = 150/Re +1.75 = 150/2.76×105+1.75 = 1.751, hence the formula for pressure drop is, f = $$\frac{∆P}{L} \frac{Dp}{\rho\vartheta_s^2}(\frac{\in^3}{1-\in}) or ∆P⁄L=[latex]\frac{f}{\frac{Dp}{\rho\vartheta_s^2}(\frac{\in^3}{1-\in})}$$ = $$\frac{1.751}{\frac{0.003}{990×2*2}×\frac{0.14^3}{1-0.14}}$$ = 724×106Pa.

13. A fully packed bed with small particle size has _________ pressure drop than a bed with larger particle size.
a) More
b) Less
c) Same
d) Negligible

Explanation: When we have fully packed bed, then the pressure drop experienced by the fluid is more for small particles as in that case the void space becomes very less because of the closed packed structure formed between the pellets.

14. What is the term Ɵ in the Kozeny Carman equation for laminar flow in packed beds?
$$\frac{∆P}{L} = \frac{-180μ}{S^2 D_P^2} (\frac{(1-∈)^2}{∈^3}) ϑ_S$$
a) Surface Area of pellets
b) Sphericity of Pellets
c) Void fraction
d) Density

Explanation: The give n term is Sphericity of the pellet. Sphericity is defined as the fraction/degree to which a particular shape resembles sphere.

15. Which one of the following is correct expression for sphericity of a particle?
Note: Vp = Volume of particle, Ap = Surface areas of the Particle.
a) S = $$\frac{6Vp}{Ap}$$
b) S = $$\frac{\pi^{\frac{1}{3}}(6Vp)^\frac{2}{3}}{Ap}$$
c) S = $$\frac{\pi^{\frac{1}{3}}(6Vp)^\frac{4}{3}}{Ap}$$
d) S = $$\frac{(6Vp)^\frac{2}{3}}{Ap}$$

Explanation: The sphericity is defined as the fraction/degree to which a particular shape resembles sphere. Hence the correct expression is S = $$\frac{\pi^{\frac{1}{3}}(6Vp)^\frac{4}{3}}{Ap}$$, where Vp = Volume of particle, Ap = Surface areas of the Particle.

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