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A U-tube manometer indicates a reading of 9 inches of water across an air filter. The air is at 24°C and 61 psig. If air at 0°C and 1 atm has a density of 0.081 lb/ft³, what is the pressure drop in atmospheres?

Answer :

Final answer:

The pressure drop across the air filter, as indicated by the U-tube manometer, is approximately 0.00096 atmospheres.

Explanation:

The pressure drop across the air filter is given by the height of the water column in the U-tube manometer. To calculate this, we convert the manometer reading from inches to feet (9 inches = 0.75 feet). Then we use the formula for the pressure difference ΔP = ρgdh where ρ is the density of the fluid (water), g is the gravitational acceleration, and dh is the height difference. Because air at 0°C and 1 atm has a density of 0.081lb/ft^3, ρ.air = 0.081lb/ft^3 and this pressure difference will be in lb/ft^2. So, the ΔP = 0.081lb/ft^3* 32.2ft/s^2 * .75ft = 1.96lb/ft^2. To convert this into atmospheres, we need to know that 1 lb/ft^2 = 0.0004886 atm hence the pressure drop is approximately 0.00096 atm.

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