Question

Air is to be moved up through a grain bed at the rate of 2 \(\text{m}^3/\text{s}\) against a pressure of 375 Pa. Determine the input power required. Assume 75% efficiency.

• A. 500 W
• B. 1000 W
• C. 750 W
• D. 75 W

Hint:

Always remember that input power is higher than theoretical air power due to efficiency losses:
\[ P_{\text{input}} = \frac{\text{Flow} \times \text{Pressure}}{\text{Efficiency}} \] Dividing by 0.75 is equivalent to multiplying by \(4/3\).
\[ 750 \times \frac{4}{3} = 1000\text{ W} \]

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Fanning air through a grain bed requires electrical energy to overcome the static pressure resistance offered by the pore paths of the grain stack.

Step 2: Key Formula or Approach:
1. The theoretical air power (\(P_{\text{air}}\)) is:
\[ P_{\text{air}} = Q \times \Delta p \] where:
\(Q\) = volumetric air flow rate (\(\text{m}^3/\text{s}\))
\(\Delta p\) = static pressure difference (Pa)
2. The mechanical input power (\(P_{\text{input}}\)) required by the fan motor is:
\[ P_{\text{input}} = \frac{P_{\text{air}}}{\eta} \] where \(\eta\) is the fan efficiency.

Step 3: Detailed Explanation:
Given values:
- Flow rate (\(Q\)) = \(2\text{ m}^3/\text{s}\)
- Static pressure (\(\Delta p\)) = \(375\text{ Pa}\)
- Efficiency (\(\eta\)) = \(75\% = 0.75\)
First, calculate the theoretical power required to move the air:
\[ P_{\text{air}} = 2 \times 375 = 750\text{ W} \] Now, calculate the actual electrical input power needed using the efficiency factor:
\[ P_{\text{input}} = \frac{750}{0.75} = 1000\text{ W} \]

Step 4: Final Answer:
The correct option is 2, which corresponds to 1000 W.