Chapter 4: Problem 159
A hot-air home heating system takes \(500 \mathrm{ft}^{3} / \mathrm{min}\) \((\mathrm{cfm})\) air at 14.7 psia, \(65 \mathrm{~F}\) into a furnace, heats it to \(130 \mathrm{~F},\) and delivers the flow to a square duct \(0.5 \mathrm{ft}\) by \(0.5 \mathrm{ft}\) at 15 psia. What is the velocity in the duct?
Short Answer
Step by step solution
Understand the Given Parameters
Convert Temperatures to Absolute Scale
Calculate Initial and Final Volumetric Flow Rates
Calculate the Cross-sectional Area of the Duct
Calculate the Velocity in the Duct
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ideal Gas Law
- \( P \) is the pressure of the gas,
- \( V \) is the volume it occupies,
- \( n \) represents the number of moles of gas,
- \( R \) is the ideal gas constant,
- \( T \) is the absolute temperature of the gas.
In practical applications like the heating system problem, the Ideal Gas Law can be used to find changes in gas behavior under varying conditions. For example, when a gas is heated in a duct, its temperature changes, which affects its pressure and volume proportionally. This principle allows us to relate initial and final states of the gas to determine volumetric flow rates and duct velocity.
Volumetric Flow Rate
To determine the flow rate before and after heating in our problem, we used the Ideal Gas Law in a form suited for changes in state: \[ \frac{Q_{1}}{T_{1}} \cdot P_{1} = \frac{Q_{2}}{T_{2}} \cdot P_{2} \] where:
- \( Q_{1} \) = initial volumetric flow rate,
- \( T_{1} \) = initial temperature in Rankine,
- \( P_{1} \) = initial pressure,
- \( Q_{2} \) = final volumetric flow rate,
- \( T_{2} \) = final temperature in Rankine,
- \( P_{2} \) = final pressure.
By rearranging and solving this equation, you can find the final flow rate \( Q_{2} \), revealing how heating affects the volume of air delivered by the system.
Duct Velocity
The velocity \( V \) can be calculated using the formula:\[ V = \frac{Q_{2}}{A} \] where \( Q_{2} \) is the volumetric flow rate we found earlier, and \( A \) is the cross-sectional area of the duct.
In the problem, we determine \( A \) for a square duct using the formula for an area of a square \( (\text{side})^2 \). Given a duct that is \( 0.5 \) ft by \( 0.5 \) ft, the area \( A \) equals \( 0.25 \, \text{ft}^2 \). Using this area, along with the volume flow rate post-heating \( Q_{2} \), allows us to solve for the duct velocity, telling us how fast the air is moving once it's been heated.
Temperature Conversion
The conversion from Fahrenheit to Rankine is straightforward and follows the formula:\[ T_{R} = T_{F} + 459.67 \] where:
- \( T_{R} \) is the temperature in Rankine,
- \( T_{F} \) is the temperature in Fahrenheit.