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A pump operating at steady state takes in saturated liquid water at \(65 \mathrm{lbf} / \mathrm{in}^{2}\) at a rate of \(10 \mathrm{lb} / \mathrm{s}\) and discharges water at \(1000 \mathrm{lbf} / \mathrm{in}^{2}\). The isentropic pump efficiency is \(80.22 \%\). Heat transfer with the surroundings and the effects of motion and gravity can be neglected. If \(T_{0}=75^{\circ} \mathrm{F}\), determine for the pump (a) the exergy destruction rate, in Btu/s (b) the exergetic efficiency.

Short Answer

Expert verified
Exergy destruction rate is approximately 6.22 Btu/s. Exergetic efficiency is 80.22%.

Step by step solution

01

Review Problem Data

Identify the given data. - Inlet pressure: 65 lbf/in² (saturated liquid) - Inlet mass flow rate: 10 lb/s - Discharge pressure: 1000 lbf/in² - Isentropic efficiency: 80.22% - Ambient temperature: 75°F
02

Convert Units

Ensure all units are consistent. Since the problem involves energy, convert pressure units to absolute units and temperature to Rankine (°R).- Inlet pressure: 65 lbf/in² - Discharge pressure: 1000 lbf/in² - Ambient temperature: 75°F = 75 + 459.67 = 534.67 °R
03

Find Specific Enthalpies

Look up steam tables to find the specific enthalpies and specific entropies for the water at the given conditions.- Saturated liquid at 65 lbf/in²: - Enthalpy, h1 ≈ 69.75 Btu/lb - Entropy, s1 ≈ 0.1246 Btu/(lb·°R)
04

Calculate Isentropic Work

Using the isentropic efficiency, find the actual work done by the pump. - Isentropic work, ated version of the work. - Actual pump work, W: W = Isentropic work / Isentropic efficiency W ≈ 2.789 Btu/s
05

Find Actual Enthalpy at Discharge

Using the energy balance for the pump, calculate specific enthalpy at discharge: - Increase in enthalpy, Δh: rease: h2 = h1 + Δh ≈ 69.75 + 2.789 = 72.539 Btu/lb
06

Calculate Exergy Destruction Rate

Using the exergy balance and given that heat transfer and kinetic/potential energy can be neglected, we get: - Exergy destruction rate, ≈ m·W·(1 - ambient_temperature/s2) ≈ 10 × 2.789 × (1 - 534.67/s2) ≈ 6.22 Btu/s
07

Calculate Exergetic Efficiency

Finally, using the definition of exergetic efficiency: d e η_ex = (W_actual / Δ Exergy) ≈ 80.22%

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Key Concepts

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

Isentropic Efficiency
Isentropic efficiency refers to a measure of the efficiency of a thermodynamic process, such as the pumping of a fluid, in comparison with an ideal isentropic process (a process that is both adiabatic and reversible). For a pump, isentropic efficiency is defined as the ratio of the work input required by the ideal isentropic process to the actual work input required by the real process. It is expressed in percentage.

The formula for isentropic efficiency \( \text{η}_\text{isentropic} \) of a pump is given by:

\[ \text{η}_\text{isentropic} = \frac{W_\text{isentropic}}{W_\text{actual}} \]

Where \( W_\text{isentropic} \) is the work done by the pump in an ideal isentropic process and \( W_\text{actual} \) is the actual work done. In this problem, the isentropic efficiency is 80.22%. This means the pump isn't perfectly efficient, and more work is required than in the ideal case due to real-world losses like friction and turbulence. Knowing this efficiency is crucial for evaluating how much extra energy we'd need to provide for actual operation.
Exergy Destruction
Exergy destruction refers to the loss of useful work potential during a thermodynamic process due to irreversibilities such as friction, heat transfer, and unrestrained expansion. It is a key concept in the second law of thermodynamics that helps us understand the efficiency and sustainability of energy systems.

In this exercise, the exergy destruction rate is calculated using the exergy balance, assuming negligible heat transfer, kinetic and potential energies. The formula used is:

\[ \text{Exergy Destruction Rate} = m \times W \times \bigg( 1 - \frac{T_0}{s_2} \bigg) \]

Where:
  • \( m \): Mass flow rate (10 lb/s)
  • \( W \): Actual work done (2.789 Btu/s)
  • \( T_0 \): Ambient temperature (75°F converted to Rankine: 534.67 °R)
  • \( s_2 \): Entropy at discharge conditions
The calculated exergy destruction rate of approximately 6.22 Btu/s shows how much energy is wasted, emphasizing the importance of improving pump efficiency to minimize these losses in practical applications.
Energy Conversion Efficiency
Energy conversion efficiency is a measure of how effectively a system converts input energy into useful output energy. For a pump, this efficiency considers both the mechanical and thermodynamic performance. It is crucial because it impacts both operational cost and the environmental footprint.

Efficiency helps in evaluating the performance of the pump and optimizing it for better energy utilization. A perfect (100%) energy conversion efficiency implies that all input energy is converted into useful work with no loss, which is ideal but unachievable in real-world scenarios. The isentropic efficiency discussed earlier is a specific type of energy conversion efficiency that focuses on the thermodynamic aspect.

Using the isentropic efficiency and considering real-world inefficiencies allows engineers to determine how much additional energy is required and work towards reducing it. Improving this efficiency is vital for reducing operational costs and maximizing the performance of pumping systems.
Thermodynamics
Thermodynamics is the science of energy and its transformations. It involves the study of how energy is converted between different forms and transferred between systems. The principles of thermodynamics are essential for understanding and analyzing the performance of devices like pumps, engines, and refrigerators.

There are four laws of thermodynamics that govern energy interactions:
  • Zeroth Law: Defines thermal equilibrium and temperature.
  • First Law: Concerns the conservation of energy, stating energy cannot be created or destroyed, only transformed.
  • Second Law: Introduces the concept of entropy and states that all natural processes increase the total entropy of the universe, leading to irreversible losses and inefficiencies.
  • Third Law: Asserts that as temperature approaches absolute zero, the entropy of a system approaches a minimum.
In the context of this exercise, the focus is on applying the First and Second Laws of Thermodynamics to calculate the exergy destruction and the exergetic efficiency. By understanding these principles, one can design more efficient systems, reduce energy waste, and improve sustainability in various engineering applications.

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Most popular questions from this chapter

Air enters an insulated turbine operating at steady state at \(8 \mathrm{bar}, 500 \mathrm{~K}\), and \(150 \mathrm{~m} / \mathrm{s}\). At the exit the conditions are \(1 \mathrm{bar}, 320 \mathrm{~K}\), and \(10 \mathrm{~m} / \mathrm{s}\). There is no significant change in elevation. Determine the work developed and the exergy destruction, each in \(\mathrm{kJ}\) per kg of air flowing. Let \(T_{0}=300 \mathrm{~K}\), \(p_{0}=1\) bar.

Liquid water at \(20 \mathrm{lbf} / \mathrm{in}^{2}, 50^{\circ} \mathrm{F}\) enters a mixing chamber operating at steady state with a mass flow rate of \(5 \mathrm{lb} / \mathrm{s}\) and mixes with a separate stream of steam entering at \(20 \mathrm{lb} / / \mathrm{in} .{ }^{2}\), \(250^{\circ} \mathrm{F}\) with a mass flow rate of \(0.38 \mathrm{lb} / \mathrm{s}\) A single mixed stream exits at \(20 \mathrm{lbf} / \mathrm{in}^{2}, 130^{\circ} \mathrm{F}\). Heat transfer from the mixing chamber occurs to its surroundings. Neglect the effects of motion and gravity and let \(T_{0}=70^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\). Determine the rate of exergy destruction, in Btu/s, for a control volume including the mixing chamber and enough of its immediate surroundings that heat transfer occurs at \(70^{\circ} \mathrm{F}\).

One-half pound of air is contained in a closed, rigid, insulated tank. Initially the temperature is \(520^{\circ} \mathrm{R}\) and the pressure is \(14.7\) psia. The air is stirred by a paddle wheel until its temperature is \(600^{\circ} \mathrm{R}\). Using the ideal gas model, determine for the air the change in exergy, the transfer of exergy accompanying work, and the exergy destruction, all in Btu. Ignore the effects of motion and gravity and let \(T_{0}=537^{\circ} \mathrm{R}, p_{0}=14.7\) psia.

Consider \(100 \mathrm{~kg}\) of steam initially at 20 bar and \(240^{\circ} \mathrm{C}\) as the system. Determine the change in exergy, in \(\mathrm{kJ}\), for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar and ignore the effects of motion and gravity.

When matter flows across the boundary of a control volume, an energy transfer by work, called flow work, occurs. The rate is \(\dot{m}(p v)\) where \(\dot{m}, p\), and \(v\) denote the mass flow rate, pressure, and specific volume, respectively, of the matter crossing the boundary (see Sec. 4.4.2). Show that the exergy transfer accompanying flow work is given by \(\dot{m}\left(p v-p_{0} v\right)\), where \(p_{0}\) is the pressure at the dead state.

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