# Heat Transfer Operations Questions and Answers – Packed Beds – Heat Transfer Coefficients

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This set of Heat Transfer Operations Question Bank focuses on “Packed Beds – Heat Transfer Coefficients”.

1. 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 $$\frac{1}{hi} = \frac{1}{hwall}+\frac{1}{hbed}$$.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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.

7. 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

8. Calculate hBED for the given parameters
Re = 12000
Pr = 0.47
Kg = 15W/mK
a) 8.535 KW/m2K
b) 0.5 KW/m2K
c) 8.35 KW/m2K
d) 5.35 KW/m2K

Explanation: $$\frac{Ke}{Kg}$$=5+0.1RePr=5+0.1×12000×0.47=569, Ke = 15×569 = 8.535 KW/m2K.

9. Calculate hwall
Re = 12000
Pr = 0.47
Kg = 15W/mK
a) 728 KW/m2K
b) 828 KW/m2K
c) 800 KW/m2K
d) 528 KW/m2K

Explanation: hwallDp/Kg=1.94Re0.5Pr0.33=1.94(12000)0.5(0.47)0.33=165.65 or hwall=165.65×15/0.003=828 KW/m2K.

10. What is the Nusselt number for the wall side if give Reynolds number = 12000 and Prantl number = 0.47?
a) 165.75
b) 175.65
c) 167.65
d) 165.65

Explanation: Nu=1.94 Re0.5 Pr0.33=1.94(12000)0.5(0.47)0.33=165.65.

11. What is the term X in the calculation of Nusselt Number for the wall side?
Nu=X Re0.5 Pr0.33
a) 0.027
b) 1.94
c) 0.0035
d) 0.023

Explanation: The correct expression for the nusselt number is Nu=1.94 Re0.5 Pr0.33, where Re is the Reynolds Number and Pr is the Prantl Number.

12. The relations established for the calculation of heat transfer coefficients is applicable for which of the following?
a) Spherical pellets
b) Ring shaped pellets
c) Cylindrical Pellets
d) Any shaped pellet
Explanation: The relations $$\frac{Ke}{Kg}$$=5+0.1RePr and Nu=1.94 Re0.5 Pr0.33 are more or less accurate to pellets of any possible shape, be it cylindrical, spherical, ring shaped e.t.c.