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Chapter 1: Problem 12

An equation for the frictional pressure loss \(\Delta\) p (inches \(\mathrm{H}_{2} \mathrm{O}\) ) in a circular duct of inside diameter \(d\) (in.) and length \(L\) (ft) for air flowing with velocity \(V\) (ft/min) is \\[ \Delta p=0.027\left(\frac{L}{d^{122}}\right)\left(\frac{V}{V_{0}}\right)^{1.82} \\] where \(V_{0}\) is a reference velocity cqual to \(1000 \mathrm{ft} / \mathrm{min}\). Find the units of the "constant" 0.027.

Short Answer

Expert verified
The units of the 'constant' 0.027 are \(\text{inches}^{121} \text{ of water} / \text{foot}\).

Step by step solution

01

Identify the Known Quantities and Their Units

Firstly, look at the equation and identify all the known quantities and their respective units. These include the frictional pressure loss \(\Delta p\) (in inches of water), the inside diameter \(d\) (in inches), the length \(L\) (in feet), and the velocity \(V\) (in feet per minute). Additionally, there is a reference velocity \(V_{0} = 1000\) ft/min.
02

Identify the Constant's Position in Equation

The next task is to identify where the 'constant' 0.027 appears in the equation. Recognize that it appears at the start, effecting all the quantities.
03

Analyze the Equation for Units

Look at the equation again noticing that the constant 0.027 and the rest of equation must together result in the units of pressure for balance. Break down the equation to isolate the constant. The equation becomes: \[0.027 = \Delta p / \left(\frac{L}{d^{122}}\right) \left(\frac{V}{V_{0}}\right)^{1.82}\]
04

Substitute the Quantities with Their Units

Substitute the quantities in the equation from Step 3 with their units. Hence obtain: \[0.027 = \text{(inches of water)} / \left(\frac{\text{foot}}{\text{inch}^{122}}\right) \left(\frac{\text{foot/minute}}{\text{foot/minute}}\right)^{1.82}\] Since foot/minute cancels out, simplify the expression to: \[0.027 = \text{(inches of water)} / \left(\frac{\text{foot}}{\text{inch}^{122}}\right)\]
05

Final Calculation for the Units of Constant

Finally, solving for the constant's unit, invert the above equation to get: \[0.027 \text{ unit} = (\text{inches of water}) \times (\text{inch}^{122} / \text{foot})\] Simplifying it further, noting that 1 foot is 12 inches, compute: \[0.027 \text{ unit} = (\text{inches of water}) \times (\text{inch}^{121} / \text{12}) = \text{inches}^{122} \text{ of water} / \text{12 inches}\] Hence, the units of constant is \(\text{inches}^{121} \text{ of water} / \text{foot}\).

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Most popular questions from this chapter

Some measurements on a blood sample at \(37^{\circ} \mathrm{C}\left(98.6^{\circ} \mathrm{F}\right)\) indicate a shearing stress of \(0.52 \mathrm{N} / \mathrm{m}^{2}\) for a corresponding rate of shearing strain of \(200 \mathrm{s}^{-1}\). Determine the apparent viscosity of the blood and compare it with the viscosity of water at the same temperature.

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