Question

Air enters a nozzle steadily at 50 psia, \(140^{\circ} \mathrm{F}\), and \(150 \mathrm{ft} / \mathrm{s}\) and leaves at 14.7 psia and \(900 \mathrm{ft} / \mathrm{s}\). The heat loss from the nozzle is estimated to be 6.5 Btu/lbm of air flowing. The inlet area of the nozzle is \(0.1 \mathrm{ft}^{2} .\) Determine (a) the exit temperature of air and ( \(b\) ) the exit area of the nozzle.

Short answer

Answer: (a) The exit temperature of air is 60.5°F. (b) The exit area of the nozzle is 0.0111 ft².

Step-by-step solution

Calculate the inlet enthalpy and entropy.

To find these values, we need to apply the ideal gas law and find the specific volume, enthalpy, and entropy of the inlet air. We are given the pressure and temperature at the inlet: \(P_1 = 50\) psia and \(T_1 = 140^\circ F\). We also need the specific heat capacities of air, \(c_p = 0.240\) Btu/lbm·\(^\circ\)F, and \(c_v = 0.171\) Btu/lbm·\(^\circ\)F, and the gas constant, \(R = 53.3\) ft·lbf/lbm·\(^\circ\)F. First, we can convert the temperatures to Rankine: \(T_1 = 140^\circ F + 459.67^\circ R = 599.67^\circ R\). Specific volume, \(v_1 = \dfrac{R \times T_1}{P_1} = \dfrac{53.3 \times 599.67}{50 \times 144} = 4.418 \,\mathrm{ft^3/lbm}\) Enthalpy, \(h_1 = c_p \times T_1 = 0.240 \times 599.67 = 143.92 \,\mathrm{Btu/lbm}\) Entropy, \(s_1 = c_v \cdot \ln \left( \dfrac{T_1}{T_{ref}}\right) - R \cdot \ln \left( \dfrac{P_1}{P_{ref}}\right)\). Choose a reference temperature \(T_{ref} = 492.62^\circ R\), and use absolute pressure: \(P_{ref} = 14.696\) psia. So, \(s_1 = 0.171 \cdot \ln \left( \dfrac{599.67}{492.62}\right) - 53.3 \cdot \ln \left( \dfrac{50}{14.696}\right) = 0.1625 \,\mathrm{Btu/lbm·R}\)

Apply the steady flow energy equation (SFEE).

We are given the heat loss per lbm of air flowing, \(Q = -6.5\) Btu/lbm. The SFEE is \(-W + Q = \Delta h + \dfrac{\Delta V^2}{2gc}\), where \(W\) is the work done, \(gc\) is the gravitational constant (\(gc = 32.174\) ft/lbmlbf), and \(\Delta V\) represents the change in velocity of the air. Since the process is steady and involves only changes in pressure and velocity, we can assume \(W = 0\). Rearrange the SFEE for exit enthalpy: \(h_2 = h_1 + \dfrac{V_1^2 - V_2^2}{2gc} + Q = 143.92 + \dfrac{150^2 - 900^2}{2 \cdot 32.174} - 6.5 = 124.84 \,\mathrm{Btu/lbm}\)

Find the exit temperature.

To find the exit temperature, we use the equation for specific enthalpy: \(h_2 = c_p \times T_2\). Rearrange this equation for \(T_2\): \(T_2 = \dfrac{h_2}{c_p} = \dfrac{124.84}{0.240} = 520.17^\circ R\). Convert back to Fahrenheit: \(T_2 = 520.17^\circ R - 459.67^\circ F = 60.5^\circ F\) (a) The exit temperature of air is \(60.5^\circ F\).

Calculate the mass flow rate.

To find the mass flow rate, we will use the equation for mass flow rate: \(\dot{m} = \rho_1 \times A_1 \times V_1\), where \(\rho_1\) represents the inlet air density, and \(A_1\) represents the inlet area. First, we need to find \(\rho_1\): \(\rho_1 = \dfrac{1}{v_1} = \dfrac{1}{4.418} = 0.226\,\mathrm{lbm/ft^3}\) Now, we can find the mass flow rate: \(\dot{m} = 0.226 \times 0.1 \times 150 = 3.392 \,\mathrm{lbm/s}\)

Calculate the exit area.

To find the exit area, \(A_2\), we need to find the exit density, \(\rho_2\). We can use the isentropic relations since the exit entropy is the same as the inlet entropy, \(s_1 = s_2\). Density ratio: \(\dfrac{\rho_2}{\rho_1} = \left(\dfrac{T_2}{T_1}\right)^{1/(\gamma - 1)} \Longrightarrow \rho_2 = \rho_1 \times \left(\dfrac{T_2}{T_1}\right)^{1/(\gamma - 1)}\) Here, \(\gamma\) is the ratio of specific heat capacities: \(\gamma = \dfrac{c_p}{c_v} = \dfrac{0.240}{0.171} = 1.4037\) Next, we can find \(\rho_2\): \(\rho_2 = 0.226 \times \left(\dfrac{520.17}{599.67}\right)^{1/(1.4037 - 1)} = 0.340\,\mathrm{lbm/ft^3}\) Finally, we can find the exit area, \(A_2\): \(A_2 = \dfrac{\dot{m}}{\rho_2 \times V_2} = \dfrac{3.392}{0.340 \times 900} = 0.0111 \,\mathrm{ft^2}\) (b) The exit area of the nozzle is \(0.0111 \,\mathrm{ft^2}\).

Key concepts

Steady Flow Energy Equation

Understanding the steady flow energy equation (SFEE) is crucial when studying nozzle thermodynamics. It depicts the principle that the total energy within a controlled volume with a steady flow remains consistent. This equation is often written as \( -W + Q = \Delta h + \dfrac{\Delta V^2}{2gc} \), where \(W\) is work done by the system, \(Q\) is the heat transfer, \(\Delta h\) is the change in specific enthalpy, \(\Delta V\) is the change in velocity, and \(gc\) is the gravitational constant.

For a nozzle, where no work is done on or by the fluid (\(W = 0\)), and considering any heat transfer (\(Q\)), the equation simplifies and becomes a tool to connect enthalpy with the kinetic energy of the fluid entering and exiting the system. This relationship allows us to determine various states of the fluid such as exit temperature and velocity by using given inlet conditions and the estimated heat loss or gain.

Specific Enthalpy

Specific enthalpy is a property that combines a substance's internal energy with its work capability due to pressure and volume. It is represented by \(h\) and expressed in units of energy per unit mass. For an ideal gas, specific enthalpy can be calculated by multiplying the specific heat capacity at constant pressure (\(c_p\)) with the temperature, as \(h = c_p \times T\).

This concept is integral in thermodynamics and particularly in nozzle operation analysis, where we often need to calculate the enthalpy at various points in the flow. This step is pivotal in leading us to understand the temperature changes and flow characteristics from the nozzle's entrance to its exit, as seen in our exercise where the exit temperature is found using the specific enthalpy.

Isentropic Process

An isentropic process is an idealized thermodynamic process that is both reversible and adiabatic, meaning there is no heat transfer into or out of the system, and no entropy change occurs. The term 'isentropic' literally means 'constant entropy', with entropy being a measure of a system's disorder.

In reality, truly isentropic processes are rare because some heat transfer and frictional effects are usually present. However, in our nozzle problem, we assume that the flow is isentropic to simplify our calculations. This assumption allows us to use the entropy at the inlet and assume it's equal to the entropy at the exit. Applying isentropic relations provides insights into the density and temperature changes within the nozzle.

Mass Flow Rate

The mass flow rate, denoted as \(\dot{m}\), represents the rate at which mass passes through a given surface area. It is a critical concept in the analysis of fluid systems and is expressed as \(\dot{m} = \rho \times A \times V\), where \(\rho\) is the fluid density, \(A\) is the cross-sectional area, and \(V\) is the velocity of the fluid.

In our exercise, determining the mass flow rate at the nozzle's entrance gives us insight into how much mass is passing through per unit of time, which in turn helps to calculate the exit area of the nozzle given the mass flow rate is constant in a steady flow process. These calculations hinge on continuity and the conservation of mass within the boundary of the nozzle.