Examveda

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)

  1. Atanu Chatterjee
    Atanu Chatterjee:
    4 months ago

    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

  2. Dhruv Singh
    Dhruv Singh:
    7 years ago

    Explain it?

Related Questions on Chemical Reaction Engineering

A first order gaseous phase reaction is catalysed by a non-porous solid. The kinetic rate constant and the external mass transfer co-efficients are k and $${{\text{k}}_{\text{g}}}$$ respectively. The effective rate constant (keff) is given by

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. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$