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Steam enters a well-insulated turbine operating at steady state at \(5 \mathrm{MPa}\) with a specific enthalpy of \(3000 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(9 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.1 \mathrm{MPa}\), specific enthalpy is \(2540 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(12 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in \(\mathrm{kW}\).

Short Answer

Expert verified
The power developed by the turbine is approximately 5460.49 kW.

Step by step solution

01

- Identify Given Data

List all known values from the problem.1. Inlet pressure: \(P_1 = 5 \text{ MPa}\)2. Inlet specific enthalpy: \(h_1 = 3000 \text{ kJ/kg}\)3. Inlet velocity: \(v_1 = 9 \text{ m/s}\)4. Exit pressure: \(P_2 = 0.1 \text{ MPa}\)5. Exit specific enthalpy: \(h_2 = 2540 \text{ kJ/kg}\)6. Exit velocity: \(v_2 = 100 \text{ m/s}\)7. Mass flow rate: \( \dot{m} = 12 \text{ kg/s}\)8. Neglecting potential energy effects.
02

- Apply the Steady-State Energy Equation

Use the steady-flow energy equation for a turbine to calculate the power developed.The energy equation can be represented as:\[\dot{Q} - \dot{W} = \dot{m}(h_2 - h_1 + \frac{v_2^2}{2} - \frac{v_1^2}{2})\]Since the turbine is well-insulated, \( \dot{Q} = 0 \).Substitute the known values:\[- \dot{W} = 12 \text{ kg/s} \times [(2540 \text{ kJ/kg} - 3000 \text{ kJ/kg}) + \frac{100^2 \text{ m}^2/\text{s}^2}{2 \times 1000 \text{ kJ/kg}} - \frac{9^2 \text{ m}^2/\text{s}^2}{2 \times 1000 \text{ kJ/kg}}]\]
03

- Simplify and Calculate

Simplify the expressions inside the bracket: \[2540 - 3000 + \frac{100^2}{2000} - \frac{9^2}{2000}\]Compute each term:\[2540 - 3000 = -460 \text{ kJ/kg}\]\[\frac{100^2}{2000} = \frac{10000}{2000} = 5 \text{ kJ/kg}\]\[\frac{9^2}{2000} = \frac{81}{2000} = 0.0405 \text{ kJ/kg}\]
04

- Total Specific Enthalpy Change

Combine the terms to find the total specific enthalpy change:\[-460 + 5 - 0.0405 = -455.0405 \text{ kJ/kg}\]
05

- Calculate the Power Developed

Plug the specific enthalpy change into the energy equation to find the power developed, in kW:\[- \dot{W} = 12 \text{ kg/s} \times (-455.0405 \text{ kJ/kg})\]Since \( \dot{W} \) is the power developed,\[\dot{W} = 5460.486 \text{ kW}\]

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Key Concepts

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

steady-state energy equation
To understand the power calculation of a steam turbine, we first need to grasp the concept of the steady-state energy equation. This equation helps us understand the balance of energy within the system, accounting for changes in various energy forms. For a turbine, the steady-state energy equation can be written as: \[\[\begin{equation} \dot{Q} - \dot{W} = \dot{m} \left( h_2 - h_1 + \frac{v_2^2}{2} - \frac{v_1^2}{2} \right) \end{equation}\]\] Here, \(\dot{Q}\) represents the heat transfer rate, \(\dot{W}\) the work rate, \(\dot{m}\) the mass flow rate, \(\text{h}\) the specific enthalpy, and \(\text{v}\) the velocity. In a well-insulated turbine, there's little to no heat transfer with the surroundings, allowing us to set \(\dot{Q} = 0\). By understanding this energy flow, we can use it to calculate the specific enthalpy variations and ultimately find the power output.
specific enthalpy
Specific enthalpy is a key term within thermodynamics that represents the total energy of a fluid per unit mass, combining its internal energy, pressure, and volume. It is crucial in the context of steam turbines, as the change in specific enthalpy across the turbine dictates the work done by the system. In the turbine entry, steam specific enthalpy \(\text{h}_1\) was given as \3000\ \text{kJ/kg} while at the exit, it was \2540\ \text{kJ/kg}. These values allow us to calculate the energy changes effectively.
turbine efficiency
Turbine efficiency denotes how effectively a turbine converts the energy in the steam into mechanical energy. It is defined as the ratio of actual work produced by the turbine to the ideal work based on isentropic processes. Efficiency can be impacted by factors such as energy losses due to friction, heat transfer, and internal energy dissipation. By understanding these elements, we gain insights into the overall performance and reliability of the turbine. While the exercise itself does not explicitly calculate efficiency, it is always a central consideration in real-world applications.
mass flow rate
The mass flow rate represents the amount of mass passing through a given point per unit time, a critical parameter in power calculations involving turbines. It's typically denoted as \(\dot{m}\), and in this exercise, is given as \22\ \text{kg/s}. A higher mass flow rate indicates more steam flows through the turbine, increasing the potential for power output. When used within our steady-state energy equation, it ensures we account for the total energy change over time, not merely per unit mass. Its impact is direct: higher flow rates result in more considerable power for the given enthalpy difference.

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Most popular questions from this chapter

Steam enters a horizontal pipe operating at steady state with a specific enthalpy of \(3200 \mathrm{~kJ} / \mathrm{kg}\) and a mass flow rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). At the exit, the specific enthalpy is \(1900 \mathrm{~kJ} / \mathrm{kg}\). If there is no significant change in kinetic energy from inlet to exit, determine the rate of heat transfer between the pipe and its surroundings, in \(\mathrm{kW}\).

Generating power by harnessing ocean tides and waves is being studied across the globe. Underwater turbines develop power from tidal currents. Wave-power devices develop power from the undulating motion of ocean waves. Although tides and waves long have been used to meet modest power generation needs, investigators today are aiming at large-scale power generation systems. Some see the oceans as providing a potentially unlimited renewable source of power. Critically evaluate the viability of tidaland wave power, considering both technical and economic issues. Write a report, including at least three references.

In a steam power plant, steam flows steadily from the boiler to the turbine through a \(0.25 \mathrm{~m}\) diameter pipe. Steam conditions at the boiler end are found to be \(P=4.5 \mathrm{Mpa}\), \(T=400^{\circ} \mathrm{C}, h=3204.7 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.06475 \mathrm{~m}^{3} / \mathrm{kg}\). At the turbine end, steam conditions are found to be \(P=4 \mathrm{MPa}\), \(T=393^{\circ} \mathrm{C}, h=3196.4 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.073 \mathrm{~m}^{3} / \mathrm{kg}\). There is a heat loss of \(6 \mathrm{~kJ} / \mathrm{kg}\) from the pipeline. Compute the steam flow rate.

A \(0.5 \mathrm{~m}^{3}\) tank initially contains air at \(300 \mathrm{kPa}, 350 \mathrm{~K}\). Air slowly escapes from the tank until the pressure drops to \(100 \mathrm{kPa}\). The air that remains in the tank undergoes a process described by \(p v^{1.3}=\) constant. For a control volume enclosing the tank, determine the heat transfer, in kJ. Assume ideal gas behavior with constant specific heats.

A well-insulated turbine operating at steady state develops \(23 \mathrm{MW}\) of power for a steam flow rate of \(40 \mathrm{~kg} / \mathrm{s}\). The steam enters at \(360^{\circ} \mathrm{C}\) with a velocity of \(35 \mathrm{~m} / \mathrm{s}\) and exits as saturated vapor at \(0.06\) bar with a velocity of \(120 \mathrm{~m} / \mathrm{s}\). Neglecting potential energy effects, determine the inlet pressure, in bar.

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