The energy balance equation over a tubular reactor under transient conditions is
A. An ordinary non-linear differential equation
B. An algebric differential equation
C. A linear partial differential equation
D. A non-linear partial differential equation
Answer: Option D
Solution (By Examveda Team)
In chemical reaction engineering, the energy balance for a tubular reactor under transient (unsteady-state) conditions involves time-dependent and position-dependent variables such as temperature and concentration.This results in a balance that includes both time and spatial derivatives, making it a partial differential equation (PDE).
Due to the presence of temperature-dependent properties (like reaction rate constants, heat capacities, and heat generation terms), the equation typically contains non-linear terms.
Therefore, the energy balance under these conditions is expressed as a non-linear partial differential equation.
Option A: An ordinary non-linear differential equation is incorrect because it doesn't account for spatial variation.
Option B: An algebraic differential equation is incorrect because the energy balance is not algebraic — it involves derivatives.
Option C: A linear partial differential equation is incorrect because the equation is usually non-linear due to reaction kinetics and temperature dependencies.
Correct answer: A non-linear partial differential equation.
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Comments (2)
A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these

The correct answer is:
D. A non-linear partial differential equation
✅ Explanation:
In a tubular reactor under transient conditions, we perform an unsteady-state energy balance, considering both:
Time variation → introduces
∂
𝑇
∂
𝑡
∂t
∂T
Axial position → introduces
∂
𝑇
∂
𝑧
∂z
∂T
or
∂
2
𝑇
∂
𝑧
2
∂z
2
∂
2
T
The reaction term often involves non-linear dependence on temperature due to Arrhenius kinetics, such as:
𝑟
(
𝑇
)
=
𝑘
0
𝑒
−
𝐸
/
𝑅
𝑇
r(T)=k
0
e
−E/RT
This makes the overall energy balance equation non-linear.
General form of the energy balance:
𝜌
𝐶
𝑝
∂
𝑇
∂
𝑡
+
𝜌
𝐶
𝑝
𝑢
∂
𝑇
∂
𝑧
=
𝑘
∂
2
𝑇
∂
𝑧
2
±
(
−
Δ
𝐻
𝑟
)
𝑟
(
𝑇
,
𝐶
)
ρC
p
∂t
∂T
+ρC
p
u
∂z
∂T
=k
∂z
2
∂
2
T
±(−ΔH
r
)r(T,C)
Where:
𝑇
T: temperature (function of
𝑧
z and
𝑡
t)
𝑢
u: velocity
𝑟
(
𝑇
,
𝐶
)
r(T,C): rate of reaction (non-linear)
𝜌
,
𝐶
𝑝
ρ,C
p
: density, specific heat
𝑘
k: thermal conductivity
Δ
𝐻
𝑟
ΔH
r
: heat of reaction
Summary:
Partial: because it involves multiple independent variables (time and position)
Non-linear: due to exponential temperature dependence in reaction rate
👉 Correct answer: D. A non-linear partial differential equation
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