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Chapter 5: Problem 167

An air-conditioning system requires airflow at the main supply duct at a rate of \(130 \mathrm{m}^{3} / \mathrm{min}\). The average velocity of air in the circular duct is not to exceed \(8 \mathrm{m} / \mathrm{s}\) to avoid excessive vibration and pressure drops. Assuming the fan converts 80 percent of the electrical energy it consumes into kinetic energy of air, determine the size of the electric motor needed to drive the fan and the diameter of the main duct. Take the density of air to be \(1.20 \mathrm{kg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
Answer: The size of the electric motor needed to drive the fan is approximately 104.52 W, and the diameter of the main duct is approximately 0.588 m.

Step by step solution

01

Calculate the cross-sectional area of the duct

We have the airflow rate, which is 130 m³/min, and we know that the average air velocity in the duct should not exceed 8 m/s. Using these values, we can calculate the cross-sectional area (A) of the duct using the formula: A = Airflow Rate / Air Velocity Convert airflow rate to m³/s: 130 m³/min × (1 min / 60 s) = 130/60 m³/s Now calculate the cross-sectional area of the duct: A = (130/60 m³/s) / 8 m/s = 130 / (60 * 8) m² A ≈ 0.2717 m²
02

Calculate the diameter of the duct

We know the cross-sectional area of the duct (A) from Step 1, and since the duct is circular, we can use the formula for the area of a circle to calculate its diameter (D): A = π (D/2)² We now solve for the diameter (D): D = 2 × √(A/π) Substituting the value of A: D = 2 × √(0.2717 m² / π) ≈ 0.588 m So, the diameter of the main duct is approximately 0.588 m.
03

Calculate the mass flow rate of the air

We can calculate the mass flow rate (m_dot) of the air using the formula: m_dot = A × Air Velocity × Air Density Substitute the values for A, Air Velocity, and Air Density: m_dot = 0.2717 m² × 8 m/s × 1.20 kg/m³ ≈ 2.613 kg/s The mass flow rate of the air is approximately 2.613 kg/s.
04

Calculate the kinetic energy of the air

We can calculate the kinetic energy (KE) of the air using the formula: KE = (1/2) × m_dot × Air Velocity² Substitute the values for m_dot and Air Velocity: KE = (1/2) × 2.613 kg/s × (8 m/s)² ≈ 83.616 kg·m²/s² The kinetic energy of the air is approximately 83.616 kg·m²/s².
05

Calculate the electrical power required

We can use the given efficiency of the fan to calculate the electrical power (P) required to drive it: P = KE / (Efficiency) Considering that the fan converts 80% of the electrical energy into kinetic energy, the efficiency is 0.8: P = 83.616 kg·m²/s² / 0.8 ≈ 104.520 kg·m²/s² The electrical power required to drive the fan is approximately 104.52 kg·m²/s².
06

Determine the size of the electric motor needed

To determine the size of the electric motor, we will need to convert the electrical power required, which is in kg·m²/s², to watts (W), knowing that 1 kg·m²/s² is equivalent to 1 W: Electric motor size ≈ 104.52 W As a result, the size of the electric motor needed to drive the fan is approximately 104.52 W, and the diameter of the main duct is approximately 0.588 m.

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Most popular questions from this chapter

A piston-cylinder device initially contains \(1.2 \mathrm{kg}\) of air at \(700 \mathrm{kPa}\) and \(200^{\circ} \mathrm{C}\). At this state, the piston is touching on a pair of stops. The mass of the piston is such that 600 -kPa pressure is required to move it. A valve at the bottom of the tank is opened, and air is withdrawn from the cylinder. The valve is closed when the volume of the cylinder decreases to 80 percent of the initial volume. If it is estimated that \(40 \mathrm{kJ}\) of heat is lost from the cylinder, determine \((a)\) the final temperature of the air in the cylinder, (b) the amount of mass that has escaped from the cylinder, and \((c)\) the work done. Use constant specific heats at the average temperature.

Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at \(2 \mathrm{MPa}\) pressure. The exhaust gases enter the heat exchanger at \(400^{\circ} \mathrm{C}\) at a rate of \(32 \mathrm{kg} / \mathrm{min}\) while water enters at \(15^{\circ} \mathrm{C}\). The heat exchanger is not well insulated, and it is estimated that 10 percent of heat given up by the exhaust gases is lost to the surroundings. If the mass flow rate of the exhaust gases is 15 times that of the water, determine ( \(a\) ) the temperature of the exhaust gases at the heat exchanger exit and ( \(b\) ) the rate of heat transfer to the water. Use the constant specific heat properties of air for the exhaust gases.

In large gas-turbine power plants, air is preheated by the exhaust gases in a heat exchanger called the regenerator before it enters the combustion chamber. Air enters the regenerator at \(1 \mathrm{MPa}\) and \(550 \mathrm{K}\) at a mass flow rate of \(800 \mathrm{kg} / \mathrm{min}\). Heat is transferred to the air at a rate of \(3200 \mathrm{kJ} / \mathrm{s}\). Exhaust gases enter the regenerator at \(140 \mathrm{kPa}\) and \(800 \mathrm{K}\) and leave at \(130 \mathrm{kPa}\) and \(600 \mathrm{K}\). Treating the exhaust gases as air, determine ( \(a\) ) the exit temperature of the air and \((b)\) the mass flow rate of exhaust gases.

Steam enters a long, horizontal pipe with an inlet diameter of \(D_{1}=16 \mathrm{cm}\) at \(2 \mathrm{MPa}\) and \(300^{\circ} \mathrm{C}\) with a velocity of \(2.5 \mathrm{m} / \mathrm{s}\). Farther downstream, the conditions are \(1.8 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C},\) and the diameter is \(D_{2}=14 \mathrm{cm} .\) Determine (a) the mass flow rate of the steam and ( \(b\) ) the rate of heat transfer.

An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at \(7^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) at a rate of \(0.55 \mathrm{m}^{3} / \mathrm{s}\) while warm air enters at \(34^{\circ} \mathrm{C}\) and 105 kPa. The air leaves the room at \(24^{\circ} \mathrm{C}\). The ratio of the mass flow rates of the hot to cold air streams is \(1.6 .\) Using variable specific heats, determine \((a)\) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room.
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