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Chapter 4: Problem 35

Helium gas flows through a well-insulated nozzle at steady state. The temperature and velocity at the inlet are \(550^{\circ} \mathrm{R}\) and \(150 \mathrm{ft} / \mathrm{s}\), respectively. At the exit, the temperature is \(400^{\circ} \mathrm{R}\) and the pressure is \(40 \mathrm{lbf} /\) in. \({ }^{2}\) The area of the exit is \(0.0085 \mathrm{ft}^{2}\). Using the ideal gas model with \(k=1.67\), and neglecting potential energy effects, determine the mass flow rate, in \(\mathrm{lb} / \mathrm{s}\), through the nozzle.

Short Answer

Expert verified
The mass flow rate is determined by the steady flow energy equation and ideal gas law considering given inlet/outlet conditions.

Step by step solution

01

Apply the Steady Flow Energy Equation (SFEE)

Since the nozzle is well-insulated and operating at steady state, we can neglect the heat transfer and apply the SFEE considering only kinetic and enthalpy terms: \[h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \]where \(h\) is the specific enthalpy and \(V\) is the velocity.
02

Use Ideal Gas Relation for Enthalpy

For an ideal gas, the change in enthalpy can be written as \( h = c_p T \). Substituting for enthalpy, we get: \[c_p T_1 + \frac{V_1^2}{2} = c_p T_2 + \frac{V_2^2}{2} \] Given: \(T_1 = 550^{\circ} R\), \(T_2 = 400^{\circ} R\), \(V_1 = 150\ \mathrm{ft/s}\)
03

Determine Specific Heat at Constant Pressure (\(c_p\))

Use the specific heat ratio (k) to find \(c_p\). For helium \( k = 1.67 \), and we know \( c_p = \frac{kR}{k-1} \). Given the gas constant for helium is \(R = 1.985\ \mathrm{Btu/(lb \, 掳R)}\), convert to ft*lbf/(lbm*掳R) by using: \(1 \mathrm{Btu/(lb鈥⒙癛)} = 25037\ \mathrm{ft*lbf/(lbm鈥⒙癛)}\). \[R = 1.985 \times 25037 = 49743.045 \ \mathrm{ft*lbf/(lbm \, 掳R)}\] Then, \[ c_p = \frac{1.67 \times 49743.045}{1.67 - 1} = 82739.333\ \mathrm{kJ/(kg鈥)} \]
04

Solve for Exit Velocity, \(V_2\)

Rearrange the SFEE equation to solve for exit velocity \(V_2\), considering the negligible heat transfer and given values: \[ 82739.333 (550 - 400) + \frac{150^2}{2} = \frac{V_2^2}{2} \] Compute and rearrange: \[ \frac{V_2^2}{2} = 82739.333 (150) + \frac{150^2}{2} \] \[ V_2^2 = 2\left[ 82739.333 (150) + 0.5 (150)^2 \right] \] \[ V_2 = \sqrt{ 2\left[ 82739.333 (150) + 0.5 (150)^2 \right] } \]
05

Calculate the Mass Flow Rate \( \dot{m} \)

Use the continuity equation and ideal gas law \( \dot{m} = \rho AV \), where \( \rho \) is the density which can be determined via \( \rho = \frac{P}{RT} \). \[ P = 40 \mathrm{lbf/in^2} = 40 \times 144 \ \mathrm{lbf/ft^2} \] \[ \rho = \frac{P}{RT_2} = \frac{40 \times 144 }{49743.045 \times 400}\ \mathrm{lb/ft^3} \] Then substitute \( \rho\), \(A\) and \(V_2\) into \( \dot{m} = \rho A V_2 \) to get the mass flow rate.

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Key Concepts

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

Steady Flow Energy Equation
The Steady Flow Energy Equation (SFEE) is vital in analyzing energy changes within a flow system like a nozzle. It simplifies the energy conservation principle for a fluid flowing steadily, meaning the properties at any fixed point in the system do not change over time.
In its simplest form for an adiabatic (well-insulated) nozzle, and neglecting potential energy changes, it can be expressed as:

\[ h_1 + \frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2} \]

Here, \(h\) represents specific enthalpy, and \(V\) is velocity. This equation balances enthalpy and kinetic energy at two different points (inlet and exit) in the flow. This core principle helps us break down complex thermodynamic problems in a straight-forward manner.
Ideal Gas Law
The Ideal Gas Law relates the pressure, volume, and temperature of a gas and is given in the form:

\[ PV = nRT \]

Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature. In the context of our problem, we use the ideal gas law to find the density of helium at the nozzle's exit.
We rearrange the equation to density form:\[ \rho = \frac{P}{RT} \] and substitute the given values to calculate the exit density. This step is crucial for determining the mass flow rate through the nozzle.
Specific Heat Ratio
The Specific Heat Ratio, denoted as \(k\) (or \( \gamma \)), is the ratio of the heat capacity at constant pressure \(c_p\) to the heat capacity at constant volume \(c_v\). For helium, \(k = 1.67\). This ratio plays a critical role in thermodynamics.
The formula to find the specific heat at constant pressure is:

\[ c_p = \frac{kR}{k-1} \]

Given that helium's gas constant \(R\) is 1.985 BtU/(lbm路掳R), we convert \(R\) to consistent units:\[ R = 1.985 \times 25037 = 49743.045 \text{ft路lbf/(lbm路掳R)} \]. By substituting \(R\) and \(k\) into the formula, we calculate\( c_p\), which is then used in the SFEE to bridge temperature and velocity.
Mass Flow Rate Calculation
The mass flow rate (\( \dot{m} \)) denotes the amount of mass passing through a cross-section of the nozzle per unit time. To calculate it, we use the continuity equation combined with the ideal gas law:\[ \dot{m} = \rho A V \], where \( \rho \) is density, \( A \) is the exit area, and \( V \) is the exit velocity. First, we determine the density using the rechill form of the ideal gas law: \[ \rho = \frac{P}{RT} \] substituting pressure (\(P\)), the gas constant (\(R\)), and temperature (\(T\)) at the exit.
After finding \( \rho \), we multiply it with the area (\( A \)) and the exit velocity (\(V\)) calculated in the final step, giving us the mass flow rate through the nozzle. This value is crucial for understanding the overall flow characteristics.

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Most popular questions from this chapter

Liquid water enters a valve at \(300 \mathrm{kPa}\) and exits at \(275 \mathrm{kPa}\). As water flows through the valve, the change in its temperature, stray heat transfer with the surroundings, and potential energy effects are negligible. Operation is at steady state. Modeling the water as incompressible with constant \(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}\), determine the change in kinetic energy per unit mass of water flowing through the valve, in \(\mathrm{kJ} / \mathrm{kg}\).

Refrigerant \(134 \mathrm{a}\) exits a heat exchanger through \(0.75\)-in.diameter tubing with a mass flow rate of \(0.9 \mathrm{lb} / \mathrm{s}\). The temperature and quality of the refrigerant are \(-15^{\circ} \mathrm{F}\) and \(0.05\), respectively. Determine the velocity of the refrigerant, in \(\mathrm{m} / \mathrm{s}\).

Air enters a compressor operating at steady state with a pressure of \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\) and a temperature of \(70^{\circ} \mathrm{F}\). The volumetric flow rate at the inlet is \(16.6 \mathrm{ft}^{3} / \mathrm{s}\), and the flow area is \(0.26 \mathrm{ft}^{2}\). At the exit, the pressure is \(35 \mathrm{lbf} / \mathrm{in}^{2}\), the temperature is \(280^{\circ} \mathrm{F}\), and the velocity is \(50 \mathrm{ft} / \mathrm{s}\). Heat transfer from the compressor to its surroundings is \(1.0 \mathrm{Btu}\) per \(\mathrm{lb}\) of air flowing. Potential energy effects are negligible, and the ideal gas model can be assumed for the air. Determine (a) the velocity of the air at the inlet, in \(\mathrm{ft} / \mathrm{s}\), (b) the mass flow rate, in \(\mathrm{lb} / \mathrm{s}\), and (c) the compressor power, in Btu/s and hp.

Steam enters a counterflow heat exchanger operating at steady state at \(0.07 \mathrm{MPa}\) with a specific enthalpy of \(2431.6 \mathrm{~kJ} / \mathrm{kg}\) and exits at the same pressure as saturated liquid. The steam mass flow rate is \(1.5 \mathrm{~kg} / \mathrm{min}\). A separate stream of air with a mass flow rate of \(100 \mathrm{~kg} / \mathrm{min}\) enters at \(30^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The ideal gas model with \(c_{p}=\) \(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) can be assumed for air. Kinetic and potential energy effects are negligible. Determine (a) the quality of the entering steam and (b) the rate of heat transfer between the heat exchanger and its surroundings, in \(\mathrm{kW}\).

A rigid tank whose volume is \(2 \mathrm{~m}^{3}\), initially containing air at 1 bar, \(295 \mathrm{~K}\), is connected by a valve to a large vessel holding air at \(6 \mathrm{bar}, 295 \mathrm{~K}\). The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of \(350 \mathrm{~K}\). Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in \(\mathrm{kJ}\).
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