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Chapter 9: Problem 130

Steam expands isentropically through a converging nozzle operating at steady state from a large tank at \(1.83\) bar, \(280^{\circ} \mathrm{C}\). The mass flow rate is \(2 \mathrm{~kg} / \mathrm{s}\), the flow is choked, and the exit plane pressure is 1 bar. Using steam table data as needed, determine the diameter of the nozzle, in cm, at locations where the pressure is \(1.5\) bar, and 1 bar, respectively.

Short Answer

Expert verified
The nozzle diameters at pressures 1.5 bar and 1 bar are determined by relating isentropic flow conditions and applying mass conservation and steam properties from tables.

Step by step solution

01

- Identify Initial Conditions

The steam expands isentropically from a large tank at pressure \( P_1 = 1.83 \) bar and temperature \( T_1 = 280^{\circ} \mathrm{C} \).
02

- Find Properties at Initial State

Using steam tables for superheated steam, find the specific enthalpy \( h_1 \) and specific entropy \( s_1 \) at \( P_1 = 1.83 \) bar and \( T_1 = 280^{\circ} \mathrm{C} \).
03

- Determine Properties Along the Isentropic Path

Since the process is isentropic, the specific entropy remains constant. Use the steam tables to find the specific enthalpy and temperature at \( P_2 = 1.5 \) bar and \( s_2 = s_1 \). Similarly, find the same properties at the exit pressure \( P_3 = 1 \) bar, where the flow is choked.
04

- Calculate Enthalpy Changes

Verify the values from the steam tables to find the specific enthalpies at \( P_2 = 1.5 \) bar and \( P_3 = 1 \) bar. Use the specific enthalpy formulas: \( h_2 \) and \( h_3 \) with \( s_2 = s_1 \) and \( s_3 = s_1 \).
05

- Apply Conservation of Mass and Energy

Use the mass flow rate equation and the isentropic flow relations to relate the velocities \( V_2 \) and \( V_3 \) with the given conditions and specific volumes obtained from steam tables.
06

- Calculate Nozzle Areas

Use the mass flow rate equation \( \dot{m} = \rho AV \) to relate the flow velocity, density, and cross-sectional area of the nozzle at pressures \( P_2 \) and \( P_3 \). Rearrange it to find the diameters \( D_2 \) and \( D_3 \).
07

- Determine Nozzle Diameters

Finally, convert the areas into diameters using the formula \( A = \frac{\pi D^2}{4} \). The nozzle diameters at \( P_2 = 1.5 \) bar and \( P_3 = 1 \) bar are then calculated and expressed in cm.

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Key Concepts

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

steam tables
Steam tables are incredibly useful tools in thermodynamics, especially when dealing with steam-based problems like isentropic nozzle flow. They contain a variety of thermodynamic properties such as specific enthalpy, specific entropy, temperature, and pressure for different states of water and steam. These tables help engineers and students quickly find needed properties without complex calculations. For instance, in the given exercise, the steam table helps find specific enthalpy (\(h_1\)) and specific entropy (\(s_1\)) at the initial state of 1.83 bar and 280掳C. Since the process is isentropic, the specific entropy remains constant, allowing us to use the steam tables again to find properties at intermediate and exit states, such as at pressures 1.5 bar and 1 bar.
mass flow rate
The mass flow rate is a critical parameter in fluid dynamics and thermodynamics that measures the amount of mass passing through a cross-sectional area per unit time. In the given problem, the mass flow rate is provided as 2 kg/s. This value remains constant throughout the nozzle due to the conservation of mass principle. It is essential for calculating other flow properties such as velocity and area. The mass flow rate equation used in this problem is \( \dot{m} = \rho A V \), where \( \rho \) is the density, \( A \) is the cross-sectional area, and \( V \) is the velocity of the steam. By understanding the mass flow rate, we can determine how the steam's velocity and nozzle diameter change at different pressures within the nozzle.
thermodynamic properties
Thermodynamic properties such as pressure, temperature, specific enthalpy, and specific entropy are key to solving problems involving isentropic nozzle flow. These properties describe the state of a thermodynamic system and provide insights into the energy transitions taking place. For example, specific enthalpy (\(h\)) is a measure of the total heat content per unit mass, and specific entropy (\(s\)) is a measure of disorder or randomness in a system. In the provided solution, keeping specific entropy constant due to the isentropic nature of the process simplifies the calculation and interrelation of different states. As the steam expands isentropically through the nozzle, its temperature and pressure change, affecting other properties like enthalpy, density, and velocity. Using these thermodynamic properties from steam tables, we can accurately determine the changing states and dimensions of the nozzle.

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Most popular questions from this chapter

Air enters the compressor of a cold air-standard Brayton cycle with regeneration, intercooling, and reheat at \(100 \mathrm{kPa}\), \(300 \mathrm{~K}\), with a mass flow rate of \(6 \mathrm{~kg} / \mathrm{s}\). The compressor pressure ratio is 10 , and the pressure ratios are the same across each compressor stage. The intercooler and reheater both operate at the same pressure. The temperature at the inlet to the second compressor stage is \(300 \mathrm{~K}\), and the inlet temperature for each turbine stage is \(1400 \mathrm{~K}\). The compressor and turbine stages each have isentropic efficiencies of \(80 \%\) and the regenerator effectiveness is \(80 \%\). For \(k=1.4\), calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the net power developed, in \(\mathrm{kW}\). (d) the rates of exergy destruction in the compressor and turbine stages as well as the regenerator, in \(\mathrm{kW}\), for \(T_{0}=300 \mathrm{~K}\).

Air enters a two-stage compressor operating at steady state at 1 bar, \(290 \mathrm{~K}\). The overall pressure ratio across the stages is 16 and each stage operates isentropically. Intercooling occurs at the pressure that minimizes total compressor work, as determined in Example 9.10. Air exits the intercooler at \(290 \mathrm{~K}\). Assuming ideal gas behavior with \(k=1.4\), determine (a) the intercooler pressure, in bar, and the heat transfer, in kJ per kg of air flowing. (b) the work required for each compressor stage, in kJ per \(\mathrm{kg}\) of air flowing.

Air is the working fluid in an Ericsson cycle. Expansion through the turbine takes place at a temperature of \(2000^{\circ} \mathrm{R}\). Heat transfer from the compressor occurs at \(520^{\circ} \mathrm{R}\). The compressor pressure ratio is 10 . Assuming the ideal gas model and ignoring kinetic and potential energy effects, determine (a) the net work, in Btu per lb of air flowing. (b) the thermal efficiency.

The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of \(2.5\). The state of the air at the beginning of compression is fixed by \(p_{1}=95 \mathrm{kPa}, T_{1}=22^{\circ} \mathrm{C}\), and \(V_{1}=\) \(3.17\) liters. Determine the net work per cycle, in \(\mathrm{kJ}\), the power developed by the engine, in \(\mathrm{kW}\), and the thermal efficiency, if the cycle is executed 1000 times per min.

Air enters the compressor of a regenerative gas turbine at \(14.5 \mathrm{lbf} / \mathrm{in}^{2}, 77^{\circ} \mathrm{F}\), and is compressed to \(60 \mathrm{lbf} / \mathrm{in}^{2}\) The air then passes through the regenerator and exits at \(1120^{\circ} \mathrm{R}\). The temperature at the turbine inlet is \(1700^{\circ} \mathrm{R}\). The compressor and turbine each have an isentropic efficiency of \(84 \%\). The net power developed is \(1000 \mathrm{hp}\). Using an air-standard analysis, calculate (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the regenerator effectiveness (d) the mass flow rate of the air, in lb/s.
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