A cylindrical air duct in an air conditioning system has a length of 5.5 m and a radius of `7.2 xx 10^(-2)` m . A fan forces air (`eta = 1.8 xx 10^(-5) Pa -s`) through the duct , such that the air in a room (volume = `280 m^3`) is replenished every ten minutes . Determine the difference in pressure between the ends of the air duct .

Options

A. 14 Pa
B. 4.4 Pa
C. 1.1 Pa
D. 0.47 Pa

Answer

To determine the difference in pressure between the ends of the air duct, we can follow these steps:

### Step 1: Calculate the Volume Flow Rate
The volume of air in the room is given as \(280 \, m^3\) and it is replenished every 10 minutes. We need to convert this time into seconds to find the flow rate in cubic meters per second.

\[ \text{Time} = 10 \, \text{minutes} = 10 \times 60 \, \text{seconds} = 600 \, \text{seconds} \]

Now, we can calculate the volume flow rate \(Q\):

\[ Q = \frac{\text{Volume}}{\text{Time}} = \frac{280 \, m^3}{600 \, s} = 0.4667 \, m^3/s \approx 0.467 \, m^3/s \]

### Step 2: Use the Hagen-Poiseuille Equation
The Hagen-Poiseuille equation relates the volume flow rate \(Q\) to the pressure difference \(\Delta P\) across a cylindrical duct:

\[ Q = \frac{\Delta P \cdot \pi r^4}{8 \eta L} \]

Where:
- \(r\) is the radius of the duct,
- \(\eta\) is the dynamic viscosity,
- \(L\) is the length of the duct.

### Step 3: Substitute Known Values
We have:
- \(Q = 0.467 \, m^3/s\)
- \(\eta = 1.8 \times 10^{-5} \, Pa \cdot s\)
- \(L = 5.5 \, m\)
- \(r = 7.2 \times 10^{-2} \, m\)

Now, substitute these values into the equation:

\[ 0.467 = \frac{\Delta P \cdot \pi (7.2 \times 10^{-2})^4}{8 \cdot (1.8 \times 10^{-5}) \cdot 5.5} \]

### Step 4: Calculate \(r^4\)
First, calculate \(r^4\):

\[ (7.2 \times 10^{-2})^4 = 2.016 \times 10^{-7} \, m^4 \]

### Step 5: Substitute and Rearrange for \(\Delta P\)
Now substitute \(r^4\) back into the equation:

\[ 0.467 = \frac{\Delta P \cdot \pi (2.016 \times 10^{-7})}{8 \cdot (1.8 \times 10^{-5}) \cdot 5.5} \]

Calculate the denominator:

\[ 8 \cdot (1.8 \times 10^{-5}) \cdot 5.5 = 7.92 \times 10^{-5} \]

Now, we can rearrange to solve for \(\Delta P\):

\[ \Delta P = \frac{0.467 \cdot 7.92 \times 10^{-5}}{\pi (2.016 \times 10^{-7})} \]

### Step 6: Calculate \(\Delta P\)
Calculating the numerator:

\[ 0.467 \cdot 7.92 \times 10^{-5} = 3.694 \times 10^{-5} \]

Now calculate \(\Delta P\):

\[ \Delta P = \frac{3.694 \times 10^{-5}}{\pi (2.016 \times 10^{-7})} \approx \frac{3.694 \times 10^{-5}}{6.339 \times 10^{-7}} \approx 58.2 \, Pa \]

### Final Step: Conclusion
Thus, the pressure difference between the ends of the air duct is approximately:

\[ \Delta P \approx 58.2 \, Pa \]