Question

For a steady-state mixed reactor the space-time is equivalent to the holding time for

• A. Constant fluid density systems
• B. Variable fluid density systems
• C. Non-isothermal gas reactions
• D. Gas reactions with changing number of moles

Hint:

In a CSTR, space-time equals holding time only when fluid density is constant, ensuring the volumetric flow rate does not change from inlet to outlet.

Answer 1

The Correct Option is A

Step 1: Define space-time and holding time.
Space-time (\( \tau \)): The time required to process one reactor volume of feed at the inlet conditions. For a reactor of volume \( V \), with volumetric flow rate \( v_0 \) at the inlet: \[ \tau = \frac{V}{v_0}. \] Holding time (or residence time, \( \theta \)): The average time a fluid element spends in the reactor. For a steady-state mixed reactor (CSTR), it is the reactor volume divided by the volumetric flow rate at the outlet conditions (\( v \)): \[ \theta = \frac{V}{v}. \] In a CSTR, the fluid is well-mixed, so the outlet conditions (e.g., density, flow rate) apply throughout the reactor. Step 2: Compare space-time and holding time.
Space-time and holding time are equal if the volumetric flow rate does not change from inlet to outlet (\( v_0 = v \)), which occurs when the fluid density is constant: \[ \tau = \theta \quad \text{if} \quad v_0 = v. \] For constant fluid density systems (e.g., liquid phase reactions or gas reactions with no change in moles and isothermal conditions), the density \( \rho \) remains constant. Since \( v = \frac{\dot{m}}{\rho} \) (where \( \dot{m} \) is mass flow rate), constant density implies \( v_0 = v \), so \( \tau = \theta \).
For variable fluid density systems (e.g., gas reactions with changing number of moles or non-isothermal conditions), the density changes due to temperature, pressure, or composition changes. This causes \( v_0 \neq v \), so \( \tau \neq \theta \).
Step 3: Evaluate the options.
(1) Constant fluid density systems: Correct, as constant density ensures \( v_0 = v \), making space-time equal to holding time. Correct.
(2) Variable fluid density systems: Incorrect, as variable density causes \( v_0 \neq v \), so space-time and holding time differ. Incorrect.
(3) Non-isothermal gas reactions: Incorrect, as temperature changes cause density changes, leading to \( v_0 \neq v \). Incorrect.
(4) Gas reactions with changing number of moles: Incorrect, as a change in the number of moles changes the volumetric flow rate (via ideal gas law), so \( v_0 \neq v \). Incorrect.
Step 4: Select the correct answer.
For a steady-state mixed reactor, the space-time is equivalent to the holding time for constant fluid density systems, matching option (1).