/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Steam enters a turbine operating... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 47

Steam enters a turbine operating at steady state with a mass flow of \(10 \mathrm{~kg} / \mathrm{min}\), a specific enthalpy of \(3100 \mathrm{~kJ} / \mathrm{kg}\), and a velocity of \(30 \mathrm{~m} / \mathrm{s}\). At the exit, the specific enthalpy is \(2300 \mathrm{~kJ} / \mathrm{kg}\) and the velocity is \(45 \mathrm{~m} / \mathrm{s}\). The elevation of the inlet is \(3 \mathrm{~m}\) higher than at the exit. Heat transfer from the turbine to its surroundings occurs at a rate of \(1.1 \mathrm{~kJ}\) per \(\mathrm{kg}\) of steam flowing. Let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the power developed by the turbine, in \(\mathrm{kW}\).

Short Answer

Expert verified
The power developed by the turbine is approximately 133.06 kW.

Step by step solution

01

- Understand the Energy Balance Equation

The energy balance for the turbine in steady-state includes the change in enthalpy, kinetic energy, potential energy, and heat transfer. The relationship is given by the equation:\[\frac{\text{dE}}{\text{dt}} = \text{mass flow rate} \times \bigg[\text{Specific Enthalpy (in) - Specific Enthalpy (out)} + \frac{V^2_{\text{in}} - V^2_{\text{out}}}{2} + g (z_{\text{in}} - z_{\text{out}})\bigg] - \text{Heat Loss} \]
02

- Convert Mass Flow Rate to kg/s

Given the mass flow rate is in kg/min, convert it to kg/s.\[\text{Mass flow rate} = \frac{10 \text{ kg/min}}{60 \text{ s/min}} = \frac{1}{6} \text{ kg/s}\]
03

- Apply the Energy Balance Equation

Substitute the given values into the energy balance equation. Remember to convert the heat transfer per kg to a total value by multiplying by the mass flow rate.\[\text{Power} = \frac{1}{6} \text{ kg/s} \times \bigg[ (3100 \text{ kJ/kg} - 2300 \text{ kJ/kg}) + \frac{(30 \text{ m/s})^2 - (45 \text{ m/s})^2}{2 \times 1000} + 9.81 \text{ m/s}^2 \times (3 \text{ m}) \bigg] - \frac{1}{6} \text{ kg/s} \times 1.1 \text{ kJ/kg} \]
04

- Calculate Each Term of the Equation

Break down the equation into individual components:1. Change in Enthalpy:\[3100 \text{ kJ/kg} - 2300 \text{ kJ/kg} = 800 \text{ kJ/kg} \]2. Change in Kinetic Energy:\[\frac{(30 \text{ m/s})^2 - (45 \text{ m/s})^2}{2 \times 1000} = \frac{900 - 2025}{2000} = -0.5625 \text{ kJ/kg}\]3. Change in Potential Energy:\[9.81 \text{ m/s}^2 \times 3 \text{ m} = 29.43 \text{ J/kg} = 0.02943 \text{ kJ/kg}\]4. Heat Transfer Loss:\[\frac{1.1 \text{ kJ/kg}}{6} = 0.18333 \text{ kJ/s} = 0.18333 \text{ kW}\]
05

- Calculate the Total Power Developed

Combine all values to find the total power developed by the turbine:\[\text{Power} = \frac{1}{6} \text{ kg/s} \times \bigg[800 - 0.5625 + 0.02943\bigg] - 0.18333 \text{ kW}\]\[\text{Power} = \frac{1}{6} \times 799.46693 - 0.18333 \text{ kW} = 133.24449 - 0.18333 \text{ kW} \]\[\text{Power} \text{ developed} = 133.06116 \text{ kW} \]

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Key Concepts

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

energy balance equation
Understanding the energy balance equation is crucial for calculating the power developed by a turbine. This equation takes into account the mass flow rate, the change in specific enthalpy, kinetic energy, potential energy, and heat transfer.
The relationship is given by the equation:
\[\frac{\text{dE}}{\text{dt}} = \text{mass flow rate} \times \bigg[\text{Specific Enthalpy (in) - Specific Enthalpy (out)} + \frac{V^2_{\text{in}} - V^2_{\text{out}}}{2} + g (z_{\text{in}} - z_{\text{out}})\bigg] - \text{Heat Loss}\]

In this formula:
  • Mass Flow Rate: The amount of mass passing through the turbine per unit of time.
  • Specific Enthalpy (in) - Specific Enthalpy (out): The difference in the steam's enthalpy between the inlet and outlet.
  • \(\frac{V^2_{\text{in}} - V^2_{\text{out}}}{2}\): This represents the change in kinetic energy per unit mass.
  • \(g (z_{\text{in}} - z_{\text{out}})\): The change in potential energy per unit mass due to the difference in elevation.
  • Heat Transfer Loss: The energy lost due to heat transferring out of the turbine.
By balancing all these factors, we can determine the net power developed by the turbine.
enthalpy change
Enthalpy change is one of the vital components in determining the power developed by a turbine. Enthalpy is a measure of the total energy of a thermodynamic system, which includes both internal energy and the energy required to displace its environment.

The change in enthalpy \(\Delta H\) is calculated by:
\[\Delta H = H_{\text{in}} - H_{\text{out}}\]

In the given exercise:
  • The specific enthalpy at the inlet \(H_{\text{in}}\) is \(3100 \ \text{kJ/kg}\).
  • The specific enthalpy at the outlet \(H_{\text{out}}\) is \(2300 \ \text{kJ/kg}\).
The change in enthalpy is:
\[\Delta H = 3100 \ \text{kJ/kg} - 2300 \ \text{kJ/kg} = 800 \ \text{kJ/kg}\]
This value is critical, as it represents the significant portion of energy transformation occurring within the turbine.
kinetic energy
Kinetic energy reflects the energy a body has because of its motion. In a turbine scenario, the steam's velocity change contributes to the overall energy balance.

The change in kinetic energy \(\Delta KE\) per unit mass can be calculated using the velocities at the inlet and outlet:
\[\Delta KE = \frac{V^2_{\text{in}} - V^2_{\text{out}}}{2}\]

For this exercise:
  • The velocity at the inlet \(V_{\text{in}}\) is \(30 \ \text{m/s}\).
  • The velocity at the outlet \(V_{\text{out}}\) is \(45 \ \text{m/s}\).
Calculating the change in kinetic energy:
\[\Delta KE = \frac{30^2 - 45^2}{2 \times 1000}\]
\[\Delta KE = \frac{900 - 2025}{2000} = -0.5625 \ \text{kJ/kg}\]
This negative value indicates that some kinetic energy was lost, which needs to be considered in the total power calculation.
potential energy
Potential energy in this context refers to the energy a unit mass of steam possesses due to its position in a gravitational field. The difference in height between the inlet and the outlet contributes to this.

The change in potential energy \(\Delta PE\) is given by:
\[\Delta PE = g (z_{\text{in}} - z_{\text{out}})\]

For the given problem:
  • The height at the inlet \(z_{\text{in}}\) is 3 meters higher than the outlet \(z_{\text{out}}\).

Using gravitational acceleration:\[g = 9.81 \ \text{m/s}^2\]The change in potential energy becomes:
\[\Delta PE = 9.81 \ \text{m/s}^2 \times 3 \ \text{m} = 29.43 \ \text{J/kg} = 0.02943 \ \text{kJ/kg}\]

This small gain in potential energy is a part of the overall energy transformation process in the system.
heat transfer
Heat transfer is another essential aspect to consider when calculating the turbine's power. This concept refers to the heat energy that is lost or gained by the system.

In the given problem, there is a heat loss from the turbine to its surroundings at a rate of \(1.1 \ \text{kJ/kg}\) of steam flowing.

The total heat transfer loss can be calculated by multiplying this rate by the mass flow rate.
  • Given mass flow rate \( \dot{m} \) is \(\frac{1}{6} \ \text{kg/s}\).
  • Heat loss per unit mass is \(1.1 \ \text{kJ/kg}\).
The total heat loss \( Q \) is:
\[Q = \frac{1.1}{6} \ \text{kJ/s} = 0.18333 \ \text{kW}\]
This value will be subtracted from the other terms in the energy balance equation to find the net power output of the turbine.

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Most popular questions from this chapter

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.

A pump is used to circulate hot water in a home heating system. Water enters the well-insulated pump operating at steady state at a rate of \(0.42 \mathrm{gal} / \mathrm{min}\). The inlet pressure and temperature are \(14.7 \mathrm{lbf} / \mathrm{in}^{2}\), and \(180^{\circ} \mathrm{F}\), respectively; at the exit the pressure is \(120 \mathrm{lbf} / \mathrm{in} .^{2}\) The pump requires \(1 / 35 \mathrm{hp}\) of power input. Water can be modeled as an incompressible substance with constant density of \(60.58 \mathrm{lb} / \mathrm{ft}^{3}\) and constant specific heat of \(1 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\). Neglecting kinetic and potential energy effects, determine the temperature change, in \({ }^{\circ} \mathrm{R}\), as the water flows through the pump. Comment on this change.

Air with a mass flow rate of \(2.3 \mathrm{~kg} / \mathrm{s}\) enters a horizontal nozzle operating at steady state at \(450 \mathrm{~K}, 350 \mathrm{kPa}\), and velocity of \(3 \mathrm{~m} / \mathrm{s}\). At the exit, the temperature is \(300 \mathrm{~K}\) and the velocity is \(460 \mathrm{~m} / \mathrm{s}\). Using the ideal gas model for air with constant \(c_{p}=1.011 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the area at the inlet, in \(\mathrm{m}^{2}\). (b) the heat transfer between the nozzle at its surroundings, in kW. Specify whether the heat transfer is to or from the air.

Refrigerant \(134 \mathrm{a}\) at a flow rate of \(0.5 \mathrm{lb} / \mathrm{s}\) enters a heat exchanger in a refrigeration system operating at steady state as saturated liquid at \(0^{\circ} \mathrm{F}\) and exits at \(20^{\circ} \mathrm{F}\) at a pressure of \(20 \mathrm{lbf} / \mathrm{in}^{2}\) A separate air stream passes in counterflow to the Refrigerant \(134 \mathrm{a}\) stream, entering at \(120^{\circ} \mathrm{F}\) and exiting at \(77^{\circ} \mathrm{F}\). The outside of the heat exchanger is well insulated. Neglecting kinetic and potential energy effects and modeling the air as an ideal gas, determine the mass flow rate of air,

Oil enters a counterflow heat exchanger at \(450 \mathrm{~K}\) with a mass flow rate of \(10 \mathrm{~kg} / \mathrm{s}\) and exits at \(350 \mathrm{~K}\). A separate stream of liquid water enters at \(20^{\circ} \mathrm{C}, 5\) bar. Each stream experiences no significant change in pressure. Stray heat transfer with the surroundings of the heat exchanger and kinetic and potential energy effects can be ignored. The specific heat of the oil is constant, \(c=2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). If the designer wants to ensure no water vapor is present in the exiting water stream, what is the allowed range of mass flow rates for the water, in \(\mathrm{kg} / \mathrm{s}\) ?
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