Question

Air expands through a turbine from 10 bar, \(900 \mathrm{~K}\) to 1 bar, \(500 \mathrm{~K}\). The inlet velocity is small compared to the exit velocity of \(100 \mathrm{~m} / \mathrm{s}\). The turbine operates at steady state and develops a power output of \(3200 \mathrm{~kW}\). Heat transfer between the turbine and its surroundings and potential energy effects are negligible. Calculate the mass flow rate of air, in \(\mathrm{kg} / \mathrm{s}\), and the exit area, in \(\mathrm{m}^{2}\).

Short answer

Mass flow rate: 9.09 kg/s, Exit area: 0.13 m^2

Step-by-step solution

Introduction and Given Data

List the data provided in the problem: - Inlet pressure: 10 bar - Inlet temperature: 900 K - Outlet pressure: 1 bar - Outlet temperature: 500 K - Exit velocity: 100 m/s - Power output: 3200 kW - Negligible heat transfer and potential energy

Assumptions and Energy Balance

Since heat transfer and potential energy effects are negligible, use the steady-state energy balance for a control volume: \[ \frac{\text{dE}_{cv}}{\text{dt}} = \text{Q}_{cv} - \text{W}_{cv} + \text{m}_{in} \text{h}_{in} + \frac{\text{v}_{in}^2}{2} + \text{g} \text{z}_{in} - \text{m}_{out} (\text{h}_{out} + \frac{\text{v}_{out}^2}{2} + \text{g} \text{z}_{out}) \] Since potential energy and heat transfer are negligible, simplify to: \[ \text{Q}_{cv} - \text{W}_{cv} + \text{m}_{in} \text{h}_{in} - \text{m}_{out} \text{h}_{out} = \text{m} (\text{h}_{in} - \text{h}_{out}) = \text{W}_{cv} + \frac{1}{2} \text{m}_{out} \text{v}_{out}^2 \]

Calculate Enthalpies

For ideal gases, enthalpy is a function of temperature. Using air properties, \[ h_{in} = c_p T_{in} \] \[ h_{out} = c_p T_{out} \] Use the specific heat capacity of air, \( c_p = 1.005 \frac{\text{kJ}}{\text{kg K}} \): \[ h_{in} = 1.005 \frac{\text{kJ}}{\text{kg K}} \times 900 \text{K} = 904.5 \text{ kJ/kg} \] \[ h_{out} = 1.005 \frac{\text{kJ}}{\text{kg K}} \times 500 \text{K} = 502.5 \text{ kJ/kg} \]

Apply Conservation of Energy to Find Mass Flow Rate

Using the simplified energy balance equation: \[ \text{m} \times (h_{in} - h_{out}) = \text{W}_{\text{cv}} + \frac{1}{2} \text{m} \text{v}_{\text{out}}^2 \] Substitute known values: \[ m \times (904.5 - 502.5) = 3200 + \frac{1}{2} m \times 100^2 \] \[ m \times 402 = 3200 + 50 m \] Solve for m: \[ 402 m - 50 m = 3200 \] \[ 352 m = 3200 \] \[ m = \frac{3200}{352} = 9.09 \text{ kg/s} \]

Calculate Exit Area

Use the mass flow rate equation and exit velocity to find the exit area. \[ \text{A} = \frac{\text{m}}{\rho \text{v}} \] At the outlet, use the ideal gas law to find density, \( \rho = \frac{P}{RT} \). For air, R = 287 J/(kg·K): \[ \rho = \frac{1 \text{bar} \times 10^5 \text{Pa/bar}}{287 \frac{\text{J}}{\text{kg·K}} \times 500 \text{K}} \] \[ \rho = 0.7 \text{ kg/m}^3 \] Calculate exit area, \[ A = \frac{9.09 \text{ kg/s}}{0.7 \text{ kg/m}^3 \times 100 \text{ m/s}} = 0.13 \text{ m}^2 \]

Key concepts

Turbine Expansion

In thermodynamics, turbines are essential devices used to extract energy from a fluid flow and convert it into useful work. The process of air expanding through a turbine involves high-pressure hot air entering the turbine and then expanding to a lower pressure and temperature. This expansion causes the air to do work on the turbine blades, which then drive a shaft to produce mechanical power.
For example, in the problem provided, air enters the turbine at 10 bar and 900 K and exits at 1 bar and 500 K. This significant drop in pressure and temperature results in energy being transferred to the turbine, contributing to the machine's power output of 3200 kW. In engineering, this process is vital for power generation in various industries, including electricity generation and aviation.

• Inlet conditions: High pressure and high temperature
• Outlet conditions: Low pressure and low temperature

Steady-State Energy Balance

When analyzing a steady-state process such as turbine expansion, it's crucial to apply the steady-state energy balance. At steady state, the properties of the system do not change with time. This means the energy entering the system must equal the energy leaving the system, adjusted for any work done by the system.
The steady-state energy balance equation can be expressed as:
\[ \frac{\text{dE}_{cv}}{\text{dt}} = \text{Q}_{cv} - \text{W}_{cv} + \text{m}_{in} \text{h}_{in} + \frac{\text{v}_{in}^2}{2} + \text{g} \text{z}_{in} - \text{m}_{out} (\text{h}_{out} + \frac{\text{v}_{out}^2}{2} + \text{g} \text{z}_{out}) \]Since heat transfer (\text{Q}_{cv}) and potential energy effects (\text{g}z) are negligible in this exercise, the equation simplifies to:

\[\text{W}_{cv} + \frac{1}{2} \text{m}_{out} \text{v}_{out}^2 = \text{m} (\text{h}_{in} - \text{h}_{out})\]
This simplified equation helps streamline calculations and balances all crucial energy components, including input and output enthalpies and kinetic energy.

Mass Flow Rate Calculation

Determining the mass flow rate is vital for understanding how much air passes through the turbine over a given time. The mass flow rate can be found using the simplified energy balance equation from the steady-state energy balance.

• First, we calculate the enthalpies at the inlet and outlet.
• Next, we plug these values into the simplified energy balance equation to solve for the mass flow rate \(m\).

In this exercise, the inlet and outlet enthalpies (\text{h}_{in} and \text{h}_{out}) are given as follows:
\[ \text{h}_{in} = \text{c}_{p} \times \text{T}_{in} \]\[ \text{h}_{out} = \text{c}_{p} \times \text{T}_{out} \]After plugging in the known values, the equation simplifies to:
\[ 9.09 \text{ kg/s} \]This represents the mass flow rate of air flowing through the turbine.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics used to relate the pressure, volume, and temperature of an ideal gas. It is defined as:
\[ \text{P} \times \text{V} = \text{n} \times \text{R} \times \text{T} \] Where

• P stands for pressure,
• V for volume,
• n for the amount of substance (in moles),
• R for the ideal gas constant, and
• T for temperature.

When applied to air at the outlet of the turbine, considering constant mass flow rate,we utilize it to find the exit area's needed for the mass flow rate equation. Specifically, we need the density (\(\rho\)) of air at the outlet, calculated as:
\[ \rho = \frac{\text{P}}{\text{R} \times \text{T}} \]Using the provided conditions and ideal gas constant values, we calculate \(\rho\), allowing us to find the exit area (A):
\[ \text{A} = \frac{\text{m}}{\rho \times \text{v}} \] In this example, the density calculated is 0.7 kg/m³, leading to an exit area of 0.13 m². This calculation is crucial for designing and optimizing turbine components.