1. Which one is the correct temperature profile for high dp/di Packed bed setup?
a) Laminar temperature profile
b) Turbulent Temperature profile
c) Laminar near the walls and turbulent at the centre
d) Turbulent near the walls and laminar at the centre
Explanation: When we have a high Dp/Di, which means the pellet size is large, the flow becomes completely laminar near the walls and turbulent at the bed as the void spaces are more near the walls.
2. Which one is the correct temperature profile for very low dp/di packed bed setup?
a) Laminar temperature profile
b) Turbulent Temperature profile
c) Laminar near the walls and turbulent at the centre
d) Turbulent near the walls and laminar at the centre
Explanation: When we have very low Dp/Di, which means the pellet size is very small, the flow becomes completely laminar inside the bed as the void spaces come very close to one another
3. The renowned equation used to calculate the pressure drop in a non-ideal packed bed is known as______________
a) Sieder Tate Equation
b) Kozeny Carman Equation
c) Ergun Equation
d) Arrhenius Equation
Explanation: The Ergun equation expresses the friction factor as the Reynolds number and is commonly used to calculate the pressure drop in packed beds having non laminar flow regime.
4. To predict the heat transfer for different particles or different tube size, the heat transfer coefficient can be represented as \(\frac{1}{hi} = \frac{1}{hwall}+\frac{1}{hbed}\).
a) True
b) False
Explanation: There are two types of heat transfer coefficients existing on the packed bed, they are the coefficients on the wall side and the other in bed side. If we represent these two resistances as a series, we have overall coefficient as \(\frac{1}{hi} = \frac{1}{hwall}+\frac{1}{hbed}\).
5. What is the term Ke in the expression for hBED?
hBED = \(\frac{4Ke}{r}\)
a) Effective Thermal Conductivity
b) Gas Thermal Conductivity
c) Pellet’s Thermal Conductivity
d) Wall Thermal Conductivity
Explanation: The expression given in the question is used to calculate the Bed heat transfer coefficient, and to calculate it we require the effective thermal heat transfer coefficient which considers the flow rate and the gas thermal coefficient into account to be calculated.
6. Calculate the heat transfer coefficient of bed if the effective thermal conductivity is 15W/mK and 30mm diameter tube.
a) 4 KW/m2K
b) 5 KW/m2K
c) 6 KW/m2K
d) 7 KW/m2K
Explanation:: The expression for hBED = \(\frac{4Ke}{r}\), hence hBED = \(\frac{4×15}{0.015}\) = 4 KW/m2K.
7. The effective thermal conductivity of a packed bed is almost ____ the Thermal Conductivity of the gas.
a) Six
b) Five
c) Seven
d) Eight
Explanation: The relation between effective and gas thermal heat transfer coefficient can be well represented as \(\frac{Ke}{Kg}\) = 5+0.1RePr, where Re is the Reynolds Number and Pr is the Prantl Number. Hence for very small value of Re and Pr, we can assume the relation to be \(\frac{Ke}{Kg}\)=5.
8. The relation between effective thermal conductivity of a packed bed and the thermal conductivity of the gas is?
a) \(\frac{Ke}{Kg}\)=5+0.1RePr
b) \(\frac{Ke}{Kg}\)=5
c) \(\frac{Ke}{Kg}\)=5+0.1Re
d) \(\frac{Ke}{Kg}\)=5+0.1RePr0.33
Explanation:The relation between effective and gas thermal heat transfer coefficient can be well represented as \(\frac{Ke}{Kg}\)=5+0.1RePr, where Re is the Reynolds Number and Pr is the Prantl Number.
9. In the given relation, they Reynolds number is calculated with the ______ and the Prantl Number for the _______
\(\frac{Ke}{Kg}\)=5+0.1RePr
a) Particle diameter, Gas
b) Gas Void, Particle
c) Particle diameter, Particle
d) Gas Void, Gas
Explanation: For the relation, \(\frac{Ke}{Kg}\)=5+0.1RePr, we calculate the Reynolds number with the particle diameter because it is the particle that acts as the obstruction to the flow and the Prantl Number for the gas side.
10. For the Calculation of hwall, identify the correct relation.
a) Sieder Tate Equation
b) hwallDp/Ke=1.94 Re0.5 Pr0.33
c) hwallDp/Kg=1.94 Re0.5 Pr0.33
d) hwallDt/Ke=1.94 Re0.5 Pr0.33
Explanation: As the flow is non linear and non laminar, we cannot use the Sieder Tate Equation for this operation, hence we use this formula given as:hwallDp/Kg=1.94 Re0.5 Pr0.33