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# 物理代写|热力学代写Thermodynamics代考|Conserving Mass in an Open System

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## 物理代写|热力学代写Thermodynamics代考|Conserving Mass in an Open System

The best way to begin every thermodynamic analysis is by defining a system. A system describes a region enclosed by an imaginary boundary (which may be fixed or flexible) that contains a mass or volume to use for analysis. A system that doesn’t allow mass to enter or leave is called a closed system. The mass inside a closed system is often called the control mass. A system that allows mass to enter and leave is called an open system. The volume of an open system is often called the control volume.

This chapter focuses on thermodynamic analysis using the conservation of mass and conservation of energy for open systems. Consenvation of mass means that the mass flow rate of material entering a system minus the mass flow rate leaving equals the mass that may accumulate within the system, as described by this equation:
$$\dot{m}{\text {in }}-\dot{m}{\text {out }}=\frac{d m_{3 \mathrm{~s}}}{d t}$$
When you see a “dot” over a variable like mass (mi), the dot means that the variable is on a rate basis or per unit time. The units for mass flow rate are kilograms per second.

Defining mass and volumetric flow rates
The size of the inlets and outlets of some open systems, such as nozzles and diffusers in jet engines, is important because they’re sized to take advantage of changes in kinetic energy. The mass flow rate entering or leaving an open system is related to the area of the opening $(A)$, the average fluid velocity normal to the inlet (V), and the fluid density $(\rho)$, as shown in this equation:
$$\dot{m}=\rho \mathrm{VA}$$
I use bold font for velocity (V) and italicized font for total volume $(V)$ to distinguish between these two variables throughout this book.

In this equation, the units of area are in square meters, velocity is in meters per second, and density is in kilograms per cubic meter.

The volumetric flow rate is related to the mass flow rate and is calculated either by using the average fluid velocity $(V)$ and the area $(A)$ of the opening, or by dividing the mass flow rate by the fluid density, as shown here:
$$\dot{V}=V A=\frac{\dot{m}}{\rho}$$
The units for volumetric flow rate are cubic meters per second.

## 物理代写|热力学代写Thermodynamics代考|Applying conservation of mass to a system

Here’s an example that shows you how to use the conservation of mass principle for an open system. Figure 6-1 shows a jet engine mounted on an aircraft. The system is defined by the dashed line around the engine. The system has two inlets, one for air and the other for fuel. The system has one outlet for exhaust.

The aircraft is flying at 250 meters per second. The air temperature is -50 degrees Celsius, and the pressure is 30 kilopascals. The air mass flow rate into the engine is 60 kilograms per second, and the fuel mass flow rate is 1 kilogram per second. The exhaust is 300 degrees Celsius and has a velocity of 1,000 meters per second. You can analyze this system to determine the volumetric flow rates into and out of the engine by following these steps:

Write the conservation of mass equation for the system.
No mass accumulates within the system, so the mass flow in equals the mass flow out.
$$\dot{m}{\text {air }}+\dot{m}{\text {tual }}=\dot{m}_{\text {estuast }}=(60+1) \mathrm{kg} / \mathrm{s}=61 \mathrm{~kg} / \mathrm{s}$$
To determine the volumetric flow rates, you need to find the gas density at the inlet and the exhaust.

Find the density $\rho_{\mathrm{in}}$ of the air at the inlet, using the ideal-gas-law equation.
$$\rho_{\mathrm{m}}=\frac{P}{R T_{\mathrm{im}}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(223 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.469 \mathrm{~kg} / \mathrm{m}^3$$

Find the density $\rho_{\text {ont }}$ of the exhaust at the exit, using the ideal-gas-law equation. Assume the exhaust has the properties of air.
$$\rho_{\text {out }}=\frac{P}{R T_{\text {ost }}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(573 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.182 \mathrm{~kg} / \mathrm{m}^3$$

Calculate the volumetric flow rate of air at the inlet, using the mass flow rate of the incoming air: $60 \mathrm{~kg} / \mathrm{s}$.
$$\dot{V}{\mathrm{in}}=\frac{\dot{m}{\text {air }}}{\rho_{\text {in }}}=\frac{60 \mathrm{~kg} / \mathrm{s}}{0.469 \mathrm{~kg} / \mathrm{m}^3}=128 \mathrm{~m}^3 / \mathrm{sec}$$

Calculate the volumetric flow rate of the exhaust at the exit, using the total mass flow rate of fuel plus air.
$$\dot{V}{\text {cut }}=\frac{\tilde{m}{\text {eathauf }}}{\rho_{\text {out }}}=\frac{61 \mathrm{~kg} / \mathrm{s}}{0.182 \mathrm{~kg} / \mathrm{m}^3}=335 \mathrm{~m}^3 / \mathrm{sec}$$
The volumetric flow rate depends on the density and mass flow rate of the air.

## 物理代写|热力学代写Thermodynamics代考|Conserving Mass in an Open System

$$\dot{m}{\text {in }}-\dot{m}{\text {out }}=\frac{d m_{3 \mathrm{~s}}}{d t}$$

$$\dot{m}=\rho \mathrm{VA}$$

$$\dot{V}=V A=\frac{\dot{m}}{\rho}$$

## 物理代写|热力学代写Thermodynamics代考|Applying conservation of mass to a system

$$\dot{m}{\text {air }}+\dot{m}{\text {tual }}=\dot{m}_{\text {estuast }}=(60+1) \mathrm{kg} / \mathrm{s}=61 \mathrm{~kg} / \mathrm{s}$$

$$\rho_{\mathrm{m}}=\frac{P}{R T_{\mathrm{im}}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(223 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.469 \mathrm{~kg} / \mathrm{m}^3$$

$$\rho_{\text {out }}=\frac{P}{R T_{\text {ost }}}=\frac{30 \mathrm{kPa}}{(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(573 \mathrm{~K})}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right)=0.182 \mathrm{~kg} / \mathrm{m}^3$$

$$\dot{V}{\mathrm{in}}=\frac{\dot{m}{\text {air }}}{\rho_{\text {in }}}=\frac{60 \mathrm{~kg} / \mathrm{s}}{0.469 \mathrm{~kg} / \mathrm{m}^3}=128 \mathrm{~m}^3 / \mathrm{sec}$$

$$\dot{V}{\text {cut }}=\frac{\tilde{m}{\text {eathauf }}}{\rho_{\text {out }}}=\frac{61 \mathrm{~kg} / \mathrm{s}}{0.182 \mathrm{~kg} / \mathrm{m}^3}=335 \mathrm{~m}^3 / \mathrm{sec}$$

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