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Chapter 4: Problem 4

Air modeled as an ideal gas enters a combustion chamber at \(20 \mathrm{lbf} / \mathrm{in}^{2}\) and \(70^{\circ} \mathrm{F}\) through a rectangular duct, \(5 \mathrm{ft}\) by \(4 \mathrm{ft}\). If the mass flow rate of the air is \(830,000 \mathrm{lb} / \mathrm{h}\), determine the velocity, in \(\mathrm{ft} / \mathrm{s}\).

Short Answer

Expert verified
The velocity is approximately 111.82 ft/s.

Step by step solution

01

- Convert Pressure to Absolute Pressure

Add the atmospheric pressure to the given gauge pressure to convert it to absolute pressure. However, the given pressure is already in absolute terms, so no conversion is needed. \[P_1 = 20 \, \text{lbf/in}^2\]
02

- Convert Temperature to Rankine

Convert the given temperature from Fahrenheit to Rankine using the relationship: \[T_{1R} = T_{1F} + 459.67\]Substitute the given temperature: \[T_{1R} = 70 + 459.67 = 529.67 \, \text{R} \]
03

- Calculate the Specific Volume

Use the ideal gas law equation to find the specific volume of the air: \[(P_1 v_1 = R T_1)\]Where:\(P_1 = 20 \, \text{lbf/in}^2 = 20 \, \times 144 \, \text{lbf/ft}^2\)\(R = 53.35 \, \text{ft} \, \text{lbf/(lb} \, \text{R)}\)\(T_1 = 529.67 \, \text{R}\).Solve for specific volume \(v_1\): \[v_1 = \frac{R T_1}{P_1} = \frac{53.35 \, \text{ft} \, \text{lbf/(lb} \, \text{R)} \, \times 529.67 \, \text{R}}{20 \, \times 144 \, \text{lbf/ft}^2} \approx 9.7 \, \text{ft}^3/\text{lb}\]
04

- Calculate the Volume Flow Rate

Find the volume flow rate \(Q\) using the mass flow rate \(\dot{m}\) and the specific volume \(v_1\):\[\dot{m} = 830,000 \, \text{lb/h} \, \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 230.56 \, \text{lb/s}\]\[Q = \dot{m} \, v_1 = 230.56 \, \text{lb/s} \, \times 9.7 \, \text{ft}^3/\text{lb} \approx 2236.432 \, \text{ft}^3 / \, \text{s}\]
05

- Calculate the Duct Area

Calculate the cross-sectional area \(A\) of the duct:\[A = 5 \, \text{ft} \, \times 4 \, \text{ft} = 20 \, \text{ft}^2\]
06

- Determine the Velocity

Use the volume flow rate \(Q\) and the cross-sectional area \(A\) to find the velocity \(v\):\[v = \frac{Q}{A} = \frac{2236.432 \, \text{ft}^3/\text{s}}{20 \, \text{ft}^2} \approx 111.82 \, \text{ft/s}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Conversion
When working with gases, it is very important to use absolute pressure rather than gauge pressure. Absolute pressure includes atmospheric pressure. For example, if you have a gauge pressure of 20 lbf/in虏, you normally would add the atmospheric pressure (14.7 lbf/in虏 at sea level). However, in this case, the pressure is already provided in absolute terms, so no conversion is necessary. Remember, using absolute pressure ensures accurate calculations in equations like the Ideal Gas Law.
Temperature Conversion
Temperature conversion is crucial for thermodynamic calculations. In this example, we convert from Fahrenheit to Rankine. The conversion formula is straightforward: \(T_{R} = T_{F} + 459.67\). Given the temperature is 70掳F, the Rankine conversion is: \(70 + 459.67 = 529.67 \text{R}\). This new temperature in Rankine will be used in further calculations like the Ideal Gas Law.
Specific Volume
Specific volume denotes the volume occupied by a unit of mass of a substance. To find the specific volume of air, the Ideal Gas Law is used: \(P \times v = R \times T\). Here, \(P = 20 \text{ lbf/in}^2 = 20 \times 144 \text{ lbf/ft}^2\), \(R = 53.35 \text{ ft lbf/(lb R)}\), and \(T = 529.67 \text{ R}\). Solving for specific volume, \ v = \frac{R \times T}{P} = \frac{53.35 \times 529.67}{20 \times 144} \, you get approximately 9.7 ft鲁/lb. This value is essential for finding other properties like volume flow rate.
Volume Flow Rate
Volume flow rate represents the volume of fluid passing through a section per unit time. Using the mass flow rate and specific volume, we can find it: \(Q = \dot{m} \times v\). Here, \(\dot{m} = 830,000 \text{ lb/h} \times \frac{1 \text{ h}}{3600 \text{ s}} = 230.56 \text{ lb/s}\) and specific volume \(v = 9.7 \text{ ft}^3/\text{lb}\). So, \ Q = 230.56 \text{ lb/s} \times 9.7 \text{ ft}^3/\text{lb} \, which is approximately 2236.432 ft鲁/s. Knowing the volume flow rate helps in determining the velocity through the duct.
Duct Cross-Sectional Area
The cross-sectional area of the duct is a critical part of calculating gas velocity. The duct dimensions are provided as 5 ft by 4 ft. Therefore, the area is: \(A = 5 \text{ ft} \times 4 \text{ ft} = 20 \text{ ft}^2\). With this area and the previously calculated volume flow rate, velocity can be determined using the formula \ v = \frac{Q}{A} \. Here: \(v = \frac{2236.432 \text{ ft}^3/\text{s}}{20 \text{ ft}^2} = 111.82 \text{ ft/s}\). This final velocity tells you how fast the gas is moving through the duct.

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Most popular questions from this chapter

A \(380-\mathrm{L}\) tank contains steam, initially at \(400^{\circ} \mathrm{C}, 3\) bar. A valve is opened, and steam flows out of the tank at a constant mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). During steam removal, a heater maintains the temperature within the tank constant. Determine the time, in s, at which \(75 \%\) of the initial mass remains in the tank; also determine the specific volume, in \(\mathrm{m}^{3} / \mathrm{kg}\), and pressure, in bar, in the tank at that time.

A rigid tank whose volume is \(0.5 \mathrm{~m}^{3}\), initially containing ammonia at \(20^{\circ} \mathrm{C}, 1.5\) bar, is connected by a valve to a large supply line carrying ammonia at 12 bar, \(60^{\circ} \mathrm{C}\). The valve is opened only as long as required to fill the tank with additional ammonia, bringing the total mass of ammonia in the tank to \(143.36 \mathrm{~kg}\). Finally, the tank holds a two-phase liquid-vapor mixture at \(20^{\circ} \mathrm{C}\). Determine the heat transfer between the tank contents and the surroundings, in \(\mathrm{kJ}\), ignoring kinetic and potential energy effects.

Steam with a quality of \(0.7\), pressure of \(1.5\) bar, and flow rate of \(10 \mathrm{~kg} / \mathrm{s}\) enters a steam separator operating at steady state. Saturated vapor at \(1.5\) bar exits the separator at state 2 at a rate of \(6.9 \mathrm{~kg} / \mathrm{s}\) while saturated liquid at \(1.5\) bar exits the separator at state 3 . Neglecting kinetic and potential energy effects, determine the rate of heat transfer, in \(\mathrm{kW}\), and its associated direction.

A rigid, well-insulated tank of volume \(0.9 \mathrm{~m}^{3}\) is initially evacuated. At time \(t=0\), air from the surroundings at 1 bar, \(27^{\circ} \mathrm{C}\) begins to flow into the tank. An electric resistor transfers energy to the air in the tank at a constant rate for 5 minutes, after which time the pressure in the tank is 1 bar and the temperature is \(457^{\circ} \mathrm{C}\). Modeling air as an ideal gas, determine the power input to the tank, in \(\mathrm{kW}\).

A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of \(0.42 \mathrm{gal} / \mathrm{min}\). The inlet pressure and temperature are \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\), and \(180^{\circ} \mathrm{F}\), respectively; at the exit the pressure is \(120 \mathrm{lbf} / \mathrm{in} .^{2}\) The pump requires \(1 / 35 \mathrm{hp}\) of power input. Water can be modeled as an incompressible substance with constant density of \(60.58 \mathrm{lb} / \mathrm{ft}^{3}\) and constant specific heat of \(1 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\). Neglecting kinetic and potential energy effects, determine the temperature change, in \({ }^{\circ} \mathrm{R}\), as the water flows through the pump. Comment on this change.
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