Question

Air at \(600 \mathrm{kPa}\) and \(500 \mathrm{K}\) enters an adiabatic nozzle that has an inlet-to-exit area ratio of 2: 1 with a velocity of \(120 \mathrm{m} / \mathrm{s}\) and leaves with a velocity of \(380 \mathrm{m} / \mathrm{s}\). Determine (a) the exit temperature and \((b)\) the exit pressure of the air.

Short answer

(b) What is the exit pressure of the air in the adiabatic nozzle?

Step-by-step solution

Understand Given Values and Variables

Given values include: 1. Inlet pressure: \(P_1 = 600 \,\text{kPa}\) 2. Inlet temperature: \(T_1 = 500 \,\text{K}\) 3. Inlet velocity: \(v_1 = 120 \,\text{m/s}\) 4. Outlet velocity: \(v_2 = 380 \,\text{m/s}\) 5. Inlet-to-exit area ratio: \(A_1:A_2=2:1\) Unknown values to be calculated include exit temperature (\(T_2\)) and exit pressure (\(P_2\)).

Apply the Conservation of Mass Equation

Since the nozzle is adiabatic, we assume there is no mass added or removed, and conservation of mass holds true. For an adiabatic nozzle, the conservation of mass equation is: $$\dot{m} = \rho_1 A_1 v_1 = \rho_2 A_2 v_2$$ Here, \(\dot{m}\) is the mass flow rate, \(\rho_1\) and \(\rho_2\) are the air densities at the inlet and exit, and \(A_1\) and \(A_2\) are the inlet and exit areas. We can write this equation in terms of air densities at the inlet and exit: $$\frac{\rho_2}{\rho_1} = \frac{A_1}{A_2} = 2$$

Express Specific Volume in Terms of Pressure and Temperature

The specific volume of the air (\(v\)) can be expressed using the ideal gas law as: $$v = \frac{R T}{P}$$ Here, \(R\) is the specific gas constant for air: \(R = 287 \,\text{J/(kg·K)}\). The ratio of specific volumes at the exit and the inlet can be found by dividing the specific volume at the exit by the specific volume at the inlet: $$\frac{v_2}{v_1} = \frac{P_1 T_2}{P_2 T_1}$$

Apply the Conservation of Energy Equation

The conservation of energy equation for an adiabatic nozzle can be written as: $$h_1 + \frac{1}{2}v_1^2 = h_2 + \frac{1}{2}v_2^2$$ Note that to simplify calculations we can assume changes in potential energy are negligible. The specific enthalpy \(h\) can be expressed as: $$h = c_p T$$ Where \(c_p\) is the specific heat of air at constant pressure: \(c_p=1005 \,\text{J/(kg·K)}\). Substituting this expression for enthalpy gives: $$c_p T_1 + \frac{1}{2}v_1^2 = c_p T_2 + \frac{1}{2}v_2^2$$

Calculate Exit Temperature T2

Now, we can solve for the exit temperature \(T_2\) from the conservation of energy equation: $$T_2 = T_1 + \frac{1}{2 c_p}(v_1^2 - v_2^2)$$ Plugging in the given values, we get: $$T_2 = 500\,\text{K} + \frac{1}{2(1005\,\text{J/(kg·K)})}[(120 \,\text{m/s})^2 - (380 \,\text{m/s})^2]$$ $$T_2 = 407.28\,\text{K}$$

Calculate Exit Pressure P2

Next, we can calculate the exit pressure \(P_2\) using the ratio of specific volumes: $$\frac{P_1 T_2}{P_2 T_1} = 2$$ Rearranging and plugging in the given values and calculated exit temperature, we get: $$P_2=\frac{P_1 T_2}{2 T_1} = \frac{600\,\text{kPa} \cdot 407.28\,\text{K}}{2 \cdot 500\,\text{K}} = 244.97 \,\text{kPa}$$

Results

(a) The exit temperature of the air is \(T_2 = 407.28 \,\text{K}\). (b) The exit pressure of the air is \(P_2 = 244.97 \,\text{kPa}\).

Key concepts

Conservation of Mass

Understanding the principle of the conservation of mass is fundamental in studying fluid dynamics, especially when dealing with nozzles. This law states that the mass cannot be created or destroyed within an isolated system. In an adiabatic nozzle, air or fluid flows are isolated from mass exchange with the environment. Therefore, the mass flow rate entering the nozzle must equal the mass flow rate exiting it.

This conservation is mathematically represented in the equation \( \dot{m} = \rho_1 A_1 v_1 = \rho_2 A_2 v_2 \), where \( \dot{m} \) symbolizes the mass flow rate; \( \rho \) is the density of the fluid; \( A \) stands for the cross-sectional area of flow, and \( v \) is the velocity with which the fluid moves through the nozzle. If the inlet-to-exit area ratio changes, so must one or more of the variables of density or velocity to maintain the mass balance.

Ideal Gas Law

The ideal gas law serves as a cornerstone for understanding the behavior of gases under various conditions. It is elegantly represented by the equation \( PV = nRT \) or, in terms of specific volume, \( v = \frac{RT}{P} \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, \( T \) is the absolute temperature, and \( v \) is the specific volume.

In the context of a nozzle operating under adiabatic conditions, and assuming air behaves as an ideal gas, we apply this law to relate pressure and temperature with volume or density. The relationship guides us in predicting how the pressure and temperature of air will change as it flows through the nozzle and allows for the thorough analysis and calculation of these variables at the exit point.

Conservation of Energy

The conservation of energy principle—also known as the first law of thermodynamics—asserts that energy cannot be created or destroyed in an isolated system, but it can change forms. For an adiabatic nozzle, this translates to the total energy (enthalpy plus kinetic energy) remaining constant between the inlet and outlet.

The relevant formula for an adiabatic process with negligible changes in potential energy is \( h_1 + \frac{1}{2}v_1^2 = h_2 + \frac{1}{2}v_2^2 \). Here, \( h \) represents specific enthalpy and \( v \) symbolizes velocity. By manipulating this equation, we find that as the velocity of the fluid increases (as it often does when passing through a convergent nozzle), the enthalpy—and hence the temperature—must decrease if no heat is added or removed from the system. This result is crucial for determining other state variables like temperature and pressure at the nozzle's exit.

Specific Enthalpy

Specific enthalpy is a property of a substance that represents the total heat content per unit mass. It's a combination of internal energy plus the energy required to displace the environment to make room for the substance (pressure-volume work). For an ideal gas, the specific enthalpy depends only on temperature and can be calculated using the expression \( h = c_p T \) where \( c_p \) is the specific heat at constant pressure.

Within the conservation of energy framework for an adiabatic nozzle, specific enthalpy plays a pivotal role. Changes in enthalpy between the inlet and outlet directly correspond to changes in kinetic energy due to the adiabatic nature of the flow. Therefore, an increase in fluid speed is necessarily accompanied by a drop in thermal energy or temperature, as depicted in the cooling of the air as it speeds up from inlet to outlet in the given exercise. This concept is essential for realizing energy transformations in adiabatic processes and helps predict thermal properties of the exiting fluid based on starting conditions and flow dynamics.