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Chapter 7: Problem 67

A stream of hot water at \(300^{\circ} \mathrm{F}, 500 \mathrm{lbf} / \mathrm{in}^{2}\), and a velocity of \(20 \mathrm{ft} / \mathrm{s}\) is obtained from a geothermal supply. Determine the specific flow exergy, in Btu/lb. The velocity is relative to the exergy reference environment for which \(T_{0}=77^{\circ} \mathrm{F}, p_{0}=\) \(1 \mathrm{~atm}\). Neglect the effect of gravity.

Short Answer

Expert verified
-42.829 Btu/lb

Step by step solution

01

Understand the Given Data

Identify and list the given data:- Temperature of the hot water: \(T = 300^{\circ} \mathrm{F}\)- Pressure of the hot water: \(P = 500 \mathrm{lbf} / \mathrm{in}^{2}\)- Velocity of the hot water: \(v = 20 \mathrm{ft} / \mathrm{s}\)- Reference environment temperature: \(T_{0}=77^{\circ} \mathrm{F}\)- Reference environment pressure: \(p_{0} = 1 \mathrm{atm}\)
02

Convert Temperature Units

Convert the temperatures from Fahrenheit to Rankine, as the thermodynamic calculations are easier in absolute temperature units:\[T_{\text{R}} = T + 459.67\]So for the given stream, \[T_{\text{stream}} = 300 + 459.67 = 759.67 \text{R}\]And for the environment, \[T_{0} = 77 + 459.67 = 536.67 \text{R}\]
03

Determine Specific Flow Exergy Formula

Recall the specific flow exergy equation for a flow stream relative to a reference environment:\[\psi = (h - h_0) - T_0(s - s_0) + \frac{v^2}{2}\]where \(h\) and \(s\) are the specific enthalpy and entropy at the given state, and \(h_0\) and \(s_0\) are the specific enthalpy and entropy at the reference state.
04

Lookup Specific Enthalpy and Entropy

Utilize the steam tables to find the specific enthalpy \(h\) and specific entropy \(s\) of water at given conditions:For \(T = 300^{\circ} \mathrm{F}\) and \(P = 500 \mathrm{lbf}/\mathrm{in}^2\):\(h = 275.15 \text{ Btu/lb}\), \(s = 0.6407 \text{ Btu/lb-R}\)For the reference state \(T_{0} = 77^{\circ} \mathrm{F}\) and \(p_{0} = 1 \text{ atm}\):\(h_0 = 49.35 \text{ Btu/lb}\), \(s_0 = 0.1305 \text{ Btu/lb-R}\)
05

Plug Values into the Exergy Formula

Substitute the values into the specific flow exergy formula:\[\psi =(275.15 - 49.35) - 536.67(0.6407 - 0.1305) + \frac{(20)^2}{2 \times 32.174}\]Simplify each term step-by-step.
06

Simplify and Calculate

Perform the calculations:\[(275.15 - 49.35) = 225.8 \text{ Btu/lb}\]\[536.67 (0.6407 - 0.1305) = 274.839 \text{ Btu/lb}\]\[\frac{(20)^2}{2 \times 32.174} = 6.21 \text{ ft}^2/\text{s}^2 \approx 6.21 \text{ Btu/lb}\]\[\psi = 225.8 - 274.839 + 6.21\]\[\psi = 225.8 + 6.21 - 274.839 = -42.829 \text{ Btu/lb}\]

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Key Concepts

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

Thermodynamic Properties
In thermodynamics, properties such as temperature, pressure, enthalpy, and entropy are crucial for analyzing energy systems. These properties help us determine the state of a system and its ability to do work.
  • **Temperature** measures the thermal energy within a substance.
  • **Pressure** is the force per unit area applied by the substance.
  • **Enthalpy (h)** represents the total heat content of a system at constant pressure.
  • **Entropy (s)** is a measure of the disorder or randomness in a system.

Understanding these properties allows us to evaluate the energy interactions within the system. For instance, in calculating the specific flow exergy of a hot water stream, knowing the temperature and pressure helps us determine the specific enthalpy and entropy from steam tables.
Enthalpy and Entropy
Enthalpy and entropy are key concepts in thermodynamics used to describe the energy and disorder within a system, respectively.

**Enthalpy (h):** This is the measure of total heat content in a system. It includes internal energy plus the product of pressure and volume.

**Entropy (s):** This is the measure of disorder or randomness. Higher entropy indicates more disorder.

In the context of our exercise:
  • Specific enthalpy and entropy values were obtained from steam tables for both the given state (300掳F, 500 lbf/in虏) and the reference environment (77掳F, 1 atm).
  • These values allow us to calculate the energy available to do work, which is necessary for determining specific flow exergy.

Formulas:
  • Specific enthalpy (h) and specific entropy (s) from the state and reference environment are utilized in the exergy calculation.
  • Example values: h = 275.15 Btu/lb and s = 0.6407 Btu/lb-R for the hot water.
Reference Environment
The reference environment acts as a baseline to measure the exergy of a system. It is typically defined by standard temperature and pressure conditions:
  • For this problem, the reference environment was specified as 77掳F and 1 atm.
  • In absolute temperature units, these values are converted to Rankine (R): 77掳F becomes 536.67 R.

**Why is the reference environment important?**
  • It allows us to measure the exergy or the maximum useful work that can be extracted as the system comes into equilibrium with the environment.
  • Specific enthalpy and entropy values at these conditions (h鈧� = 49.35 Btu/lb and s鈧� = 0.1305 Btu/lb-R) are used in the exergy calculation.

In simpler terms, the reference environment provides a consistent baseline so we can compare different energy states under the same conditions.

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Most popular questions from this chapter

A thermal reservoir at \(1000 \mathrm{~K}\) is separated from another thermal reservoir at \(350 \mathrm{~K}\) by a \(1 \mathrm{~cm}\) by \(1 \mathrm{~cm}\) square-cross section rod insulated on its lateral surfaces. At steady state, energy transfer by conduction takes place through the rod. The rod length is \(L\), and the thermal conductivity is \(0.5\) \(\mathrm{kW} / \mathrm{m}+\mathrm{K}\). Plot the following quantities, each in \(\mathrm{kW}\), versus \(L\) ranging from \(0.01\) to \(1 \mathrm{~m}\) : the rate of conduction through the rod, the rates of exergy transfer accompanying heat transfer into and out of the rod, and the rate of exergy destruction. Let \(T_{0}=300 \mathrm{~K}\).

Water at \(T_{1}=100^{\circ} \mathrm{F}, p_{1}=30 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) enters a counterflow heat exchanger operating at steady state with a mass flow rate of \(100 \mathrm{lb} / \mathrm{s}\) and exits at \(T_{2}=200^{\circ} \mathrm{F}\) with closely the same pressure. Air enters in a separate stream at \(T_{3}=540^{\circ} \mathrm{F}\) and exits at \(T_{4}=140^{\circ} \mathrm{F}\) with no significant change in pressure. Air can be modeled as an ideal gas and stray heat transfer can be ignored. Determine (a) the mass flow rate of the air, in \(\mathrm{lb} / \mathrm{s}\), and (b) the rate of exergy destruction within the heat exchanger, in Btu/s. Ignore the effects of motion and gravity and let \(T_{0}=60^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

Air is compressed in an axial-flow compressor operating at steady state from \(27^{\circ} \mathrm{C}, 1\) bar to a pressure of \(2.1\) bar. The work required is \(94.6 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. Heat transfer from the compressor occurs at an average surface temperature of \(40^{\circ} \mathrm{C}\) at the rate of \(14 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing. The effects of motion and gravity can be ignored. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=\) 1 bar. Assuming ideal gas behavior, (a) determine the temperature of the air at the exit, in \({ }^{\circ} \mathrm{C}\), (b) determine the rate of exergy destruction within the compressor, in kJ per \(\mathrm{kg}\) of air flowing, and (c) perform a full exergy accounting. in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of air flowing, based on work input.

Steam at \(450 \mathrm{lbf} / \mathrm{in}^{2}, 700^{\circ} \mathrm{F}\) enters a well-insulated turbine operating at steady state and exits as saturated vapor at a pressure \(p\). (a) For \(p=50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\), determine the exergy destruction rate, in Btu per lb of steam expanding through the turbine, and the turbine exergetic and isentropic efficiencies. (b) Plot the exergy destruction rate, in Btu per lb of steam flowing, and the exergetic efficiency and isentropic efficiency, each versus pressure \(p\) ranging from 1 to \(50 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) Ignore the effects of motion and gravity and let \(T_{0}=70^{\circ} \mathrm{F}\), \(p_{0}=1 \mathrm{~atm} .\)

Determine the specific exergy, in Btu, of one pound mass of (a) saturated liquid Refrigerant \(134 a\) at \(-5^{\circ} \mathrm{F}\) - (b) saturated vapor Refrigerant \(134 a\) at \(140^{\circ} \mathrm{F}\). (c) Refrigerant \(134 \mathrm{a}\) at \(60^{\circ} \mathrm{F}, 20 \mathrm{lbf} / \mathrm{in} .^{2}\) (d) Refrigerant \(134 \mathrm{a}\) at \(60^{\circ} \mathrm{F}, 10 \mathrm{lbf} / \mathrm{in}^{2}{ }^{2}\) In each case, consider a fixed mass at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=\) \(60^{\circ} \mathrm{F}, p_{0}=15 \mathrm{lbt} / \mathrm{in}^{2}\)
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