Subtopic: Fluidization of granular solids – pneumatic, bucket, screw and belt conveying
Question Statement:
A vertical bucket elevator lifts $30.0\text{ tonnes/hour}$ of wheat grain through a net vertical height of $25.0\text{ m}$. The power required to scoop (dig) the grain in the boot is $12\%$ of the power required for vertical lifting. If the mechanical drive transmission efficiency is $80\%$, the total motor power required in $\text{kW}$ is ________ (round off to two decimal places).
Answer Key:
2.86
Step-by-Step Solution & Explanation
Multi-Approach Solution:
**Method 1: Component Power Balance**
1. Mass flow rate of grain:
$\dot{m} = \frac{30000\text{ kg}}{3600\text{ s}} = 8.3333\text{ kg/s}$
2. Net vertical lifting power:
$P_{\text{lift}} = \dot{m} g H = 8.3333\text{ kg/s} \times 9.81\text{ m/s}^2 \times 25.0\text{ m} = 2043.75\text{ W} = 2.04375\text{ kW}$
3. Boot scooping power ($12\%$):
$P_{\text{dig}} = 0.12 \times 2.04375\text{ kW} = 0.24525\text{ kW}$
4. Total shaft power:
$P_{\text{shaft}} = P_{\text{lift}} + P_{\text{dig}} = 2.04375 + 0.24525 = 2.2890\text{ kW}$
5. Motor electrical power requirement with transmission efficiency $\eta = 0.80$:
$P_{\text{motor}} = \frac{P_{\text{shaft}}}{\eta} = \frac{2.2890\text{ kW}}{0.80} = 2.86125\text{ kW} \approx 2.86\text{ kW}$
**Method 2: Multiplier Shortcut**
$P_{\text{motor}} = \frac{\dot{m} g H \times 1.12}{\eta} = \frac{8.3333 \times 9.81 \times 25 \times 1.12}{0.80 \times 1000} = 2.861\text{ kW}$