WABT Weighted Average Bed Temperature Reactor Formula Example

Overview

Engineering reference data for WABT Weighted Average Bed Temperature Reactor Formula Example in thermodynamics.

Key Formulas

First Law

$\Delta U = Q - W$

Energy is conserved — heat added minus work done.

Ideal Gas Law

$PV = nRT$

Relates pressure, volume, and temperature of an ideal gas.

Heat Transfer

$Q = mc\Delta T$

Sensible heat transfer.

Carnot Efficiency

$\eta = 1 - T_C/T_H$

Maximum efficiency between two temperatures.

Variables

Symbol	Description	Unit
$U$	Internal energy	J
$Q$	Heat	J
$W$	Work	J
$P$	Pressure	Pa
$V$	Volume	m³
$T$	Temperature	K

Calculation Examples

To illustrate the application of WABT calculations in practical scenarios, here are three detailed examples based on common reactor configurations.

Example 1: Single Reactor with Two Thermocouples

This is the simplest case where the reactor has one catalytic bed with temperature indicators only at the inlet ($T_1$) and outlet ($T_2$).

• Formula: Since there's one bed, $N = 1$ and the weight fraction $W_{c} = 1$. The WABT is: $$WABT = \frac{T_1 + 2 \times T_2}{3}$$
• Given Data: $T_1 = 350\,°C$, $T_2 = 395\,°C$
• Calculation: $$WABT = \frac{350 + 2 \times 395}{3} = \frac{1140}{3} = 380\,°C$$

Example 2: Three Reactors in Series

Here, each reactor (bed) has its own inlet and outlet temperatures, and the catalyst bulk density varies, requiring weighted averaging.

• Step 1: Calculate Individual WABTs For each bed $i$, using $WABT_i = \frac{T_{in,i} + 2 \times T_{out,i}}{3}$:

  • Bed 1: $T_{in,1} = T_1 = 345\,°C$, $T_{out,1} = T_2 = 380\,°C$ → $WABT_1 = \frac{345 + 2 \times 380}{3} = \frac{1105}{3} \approx 368.33\,°C$
  • Bed 2: $T_{in,2} = T_3 = 370\,°C$, $T_{out,2} = T_4 = 390\,°C$ → $WABT_2 = \frac{370 + 2 \times 390}{3} = \frac{1150}{3} \approx 383.33\,°C$
  • Bed 3: $T_{in,3} = T_5 = 385\,°C$, $T_{out,3} = T_6 = 395\,°C$ → $WABT_3 = \frac{385 + 2 \times 395}{3} = \frac{1175}{3} \approx 391.67\,°C$

• Step 2: Determine Weight Fractions Catalyst weight per bed is calculated from bed volume and bulk density ($\rho$). Assume volumes: Bed 1 = 18 m³, Bed 2 = 30 m³, Bed 3 = 30 m³.

  • Weight in Bed 1: $18 \times 550 = 9900\,kg$
  • Weight in Bed 2: $30 \times 800 = 24000\,kg$
  • Weight in Bed 3: $30 \times 750 = 22500\,kg$
  • Total weight: $9900 + 24000 + 22500 = 56400\,kg$
  • Weight fractions: $W_{c,1} = \frac{9900}{56400} \approx 0.18$, $W_{c,2} = \frac{24000}{56400} \approx 0.42$, $W_{c,3} = \frac{22500}{56400} \approx 0.40$

• Step 3: Compute Global WABT

  $$WABT = (W_{c,1} \times WABT_1) + (W_{c,2} \times WABT_2) + (W_{c,3} \times WABT_3)$$ $$WABT = (0.18 \times 368.33) + (0.42 \times 383.33) + (0.40 \times 391.67) \approx 384\,°C$$

Example 3: Single Reactor with Multiple Thermocouple Chains

This case involves a reactor with two thermocouple chains at different radial positions, allowing for more precise temperature averaging across catalyst layers.

• Given: Temperature readings $T_1$ to $T_{16}$ as provided, with catalyst layer densities: top layer ($1/6$ of bed) with $\rho_1 = 550\,kg/m³$, and remaining five layers ($5/6$ of bed) with $\rho_2 = 800\,kg/m³$. Bed volume is 18 m³.

• Step 1: Calculate Weight Fractions for Each Layer Assume each layer has equal volume fraction of $1/6$ of the bed.

  • Weight in top layer: $(18 \times \frac{1}{6}) \times 550 = 3 \times 550 = 1650\,kg$
  • Weight in each of the other five layers: $(18 \times \frac{1}{6}) \times 800 = 3 \times 800 = 2400\,kg$
  • Total weight: $1650 + 5 \times 2400 = 13650\,kg$
  • Weight fractions: $W_{c,1} = \frac{1650}{13650} \approx 0.12$, $W_{c,2} = W_{c,3} = W_{c,4} = W_{c,5} = W_{c,6} = \frac{2400}{13650} \approx 0.176$

• Step 2: Average Temperatures at Each Level At each height, there are two thermocouples (e.g., $T_2$ and $T_9$ for the top layer). The inlet and outlet temperatures for each layer are averages of these pairs.

  • For layer 1: $T_{in,1} = \frac{T_2 + T_9}{2}$, $T_{out,1} = \frac{T_3 + T_{10}}{2}$
  • Similar for layers 2 to 6.

• Step 3: Compute Layer WABTs and Global WABT Use $WABT_i = \frac{T_{in,i} + 2 \times T_{out,i}}{3}$ for each layer, then weight them. Using the given temperatures, the calculated global WABT is approximately $395.3\,°C$.

• Comparison Note: Using only the inlet ($T_1 = 351\,°C$) and outlet ($T_{16} = 411\,°C$) temperatures in the simple formula gives $WABT = \frac{351 + 2 \times 411}{3} = 391.0\,°C$, while a simple average of $T_1$ and $T_{16}$ is $381\,°C$. The detailed method with multiple thermocouples provides a more accurate representation of the actual catalyst bed temperature, which is crucial for monitoring catalyst activity and deactivation over time.

References

• Original Source