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Chapter 7: Problem 17

Determine the exergy, in \(\mathrm{kJ}\), of the contents of a \(2-\mathrm{m}^{3}\) storage tank, if the tank is filled with (a) air as an ideal gas at \(400^{\circ} \mathrm{C}\) and \(0.35\) bar. (b) water vapor at \(400^{\circ} \mathrm{C}\) and \(0.35\) bar. Ignore the effects of motion and gravity and let \(T_{0}=17^{\circ} \mathrm{C}\), \(p_{0}=1 \mathrm{~atm} .\)

Short Answer

Expert verified
Exergy of air: 170.13 kJ, exergy of water vapor: 735.94 kJ

Step by step solution

01

Define known parameters

Given:- Volume of tank, V = 2 m鲁- Temperature of air and water vapor, T = 400掳C = 673K- Pressure, p = 0.35 bar = 35 kPa- Reference environment temperature, T鈧� = 17掳C = 290K- Reference environment pressure, p鈧� = 1 atm = 101.325 kPa
02

Determine the properties of air

For air as an ideal gas:- The gas constant for air, R = 0.287 kJ/kg路K- Use the ideal gas equation of state to find the mass of air:givwn ideal gas equation:\[ pV = mRT \]Rearrange to solve for m:\[ m = \frac{pV}{RT} \]\[ m = \frac{(35 kPa)(2 m鲁)}{(0.287 kJ/kg路K)(673 K)} \]\[ = 0.36 kg \]
03

Calculate specific exergy of air

Use the specific exergy formula for an ideal gas:\[ e_a = c_p (T - T鈧�) - T鈧�R \ln \frac{p}{p鈧�} \]with for air, \( c_p =1.005 kJ/kg路K \), thus we get specifc exergy:e_a = (1.005 kJ/kg路K) (673 K - 290 K) - (290 K) (0.287 kJ/kg路K) \ln \left(\frac{35}{101.325}\right)Setting an approximation for this natural logarithm \(\ln \left(\frac{35}{101.325}\right)\approx \ln 0.35 \approx -1.05 so we set into:e_a = 1.005*383 - 290*0.287*(-1.05)e鈧� = 385.215 kJ/kg +87.3525kJ/kg\)\approx 472.57 kJ/kg
04

Determine exergy of air

The exergy (蠄) of air in the tank is obtained by multiplying the specific exergy by the mass (m):\[ \psi = e_a \times m \]Substitute e鈧� and m:\[ \psi = 472.57 \times 0.36 \]\[ \approx 170.13 kJ \]
05

Determine properties of water vapor

For water vapor at 400掳C and 0.35 bar, interpolate properties from steam tables to find specific enthalpy (h) and entropy (s):h 鈮� 3150 kJ/kgs 鈮� 8.1 kJ/kg路K. Also, properties of reference environment for water at 290K is: h鈧� =419.17 kJ/kg and s鈧� = 1.359 kJ/kg路k
06

Calculate specific exergy of water vapor

Use the specific exergy formula for water vapor:蠄 = (h - h鈧�) - T鈧�(s - s鈧�)蠄 = (3150 - 419.17) - 290(8.1 - 1.359)蠄 鈮� 2730.83 - 290脳(6.741) 蠄 =2730.83-1994.89 蠄 鈮� 735.94 kJ/kg
07

Determine exergy of water vapor

again we set exergy to extexic value related with water vapor as per volume relation using PV=mRT we obtain mass m of water vapor by setting properties interpolations as : assume

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure, volume, and temperature of an ideal gas. This law is essential for our exercise with air in a storage tank. The equation is:\[ pV = mRT \]where:
  • p is the pressure,
  • V is the volume,
  • m is the mass,
  • R is the specific gas constant,
  • T is the temperature.
We can rearrange this equation to find the mass of air inside the tank:\[ m = \frac{pV}{RT} \]Plugging in the known values (p = 35 kPa, V = 2 m鲁, R = 0.287 kJ/kg路K, T = 673 K), we get m = 0.36 kg. This mass calculation is crucial for further steps in exergy analysis.
Specific Exergy
Specific exergy is the measure of the work potential of a system's unit mass when it interacts with a reference environment. For an ideal gas like air, the specific exergy formula is given by:\[ e = c_p (T - T_0) - T_0R \ln \left( \frac{p}{p_0} \right) \]where:
  • c_p is the specific heat at constant pressure,
  • T is the system's temperature,
  • T_0 is the reference temperature,
  • R is the specific gas constant,
  • p is the system's pressure,
  • p_0 is the reference pressure.
For air at 400掳C and 0.35 bar with reference conditions of 17掳C and 1 atm, we determined:\[ e_{air} = 1.005 \cdot 383 - 290 \cdot 0.287 \cdot \ln \left( \frac{35}{101.325} \right) \]Simplifying:\[ e_{air} = 385.215 + 87.3525 \approx 472.57 \text{ kJ/kg} \]This specific exergy value is multiplied by the mass of air to find the total exergy.
Steam Tables
Steam tables are invaluable tools in thermodynamics, especially for water vapor calculations. They provide various properties such as specific enthalpy (h) and specific entropy (s) at different temperatures and pressures. For our exercise:
  • Specific enthalpy at 400掳C and 0.35 bar: \( h \approx 3150 \text{ kJ/kg} \)
  • Specific entropy at 400掳C and 0.35 bar: \( s \approx 8.1 \text{ kJ/kg路K} \)
For reference conditions (17掳C or 290K):
  • \( h_0 = 419.17 \text{ kJ/kg} \)
  • \( s_0 = 1.359 \text{ kJ/kg路K} \)
These interpolated values are used to calculate the specific exergy for water vapor.
Reference Environment State
The reference environment state is crucial for exergy calculations. It represents the 'dead state' where a system is in equilibrium with its surroundings, meaning it has zero exergy. For our calculations, we use:
  • Reference temperature, \( T_0 = 17掳C = 290K \)
  • Reference pressure, \( p_0 = 1 \text{ atm} = 101.325 kPa \)
These conditions are the baseline to assess the work potential of our system. Specific enthalpy and entropy values at this state are also used for accurate exergy measurements. They simplify the formulas and provide the necessary constants to determine the system's deviation from equilibrium, reflecting its potential to perform work.

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Most popular questions from this chapter

Refrigerant 134 a enters a water-jacketed compressor operating at steady state at \(10^{\circ} \mathrm{C}, 3.2\) bar and exits at \(70^{\circ} \mathrm{C}, 10\) bar. Cooling water enters as a separate stream at \(20^{\circ} \mathrm{C}\) and exits at \(32^{\circ} \mathrm{C}\) with no significant change in pressure. The refrigerant mass flow rate is \(1.63 \mathrm{~kg} / \mathrm{s}\), and the power input to the compressor is \(55.2 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Assuming no heat transfer between the outer surface of the water jacket and the surroundings, and neglecting the effects of motion and gravity (a) determine the mass flow rate of cooling water, in \(\mathrm{kg} / \mathrm{s}\), and (b) perform a full exergy accounting, in \(\mathrm{kW}\), based on the compressor power input and comment. Let \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar.

Air at 1 bar, \(17^{\circ} \mathrm{C}\), and a mass flow rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) enters an insulated compressor operating at steady state and exits at 3 bar, \(147^{\circ} \mathrm{C}\). Determine the power required by the compressor and the rate of exergy destruction, each in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=17^{\circ} \mathrm{C}\), \(p_{0}=1\) bar.

Determine the specific exergy, in \(\mathrm{kJ} / \mathrm{kg}\) of (a) saturated water vapor at \(100^{\circ} \mathrm{C}\). (b) saturated liquid water at \(4^{\circ} \mathrm{C}\). (c) ammonia at \(-40^{\circ} \mathrm{C}, 40 \mathrm{kPa}\). In each case, consider a fixed mass at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=100 \mathrm{kPa}\)

A compressor operating at steady state takes in \(1 \mathrm{~kg} / \mathrm{s}\) of air at 1 bar and \(25^{\circ} \mathrm{C}\) and compresses it to 8 bar and \(160^{\circ} \mathrm{C}\). The power input to the compressor is \(230 \mathrm{~kW}\), and heat transfer occurs from the compressor to the surroundings at an average surface temperature of \(50^{\circ} \mathrm{C}\). (a) Perform a full exergy accounting of the power input to the compressor. (b) Devise and evaluate an exergetic efficiency for the compressor. (c) Evaluating exergy at 8 cents per \(\mathrm{kW} \cdot \mathrm{h}\), determine the hourly costs of the power input, exergy loss associated with heat transfer, and exergy destruction. Neglect the effects of motion and gravity. Let \(T_{0}=25^{\circ} \mathrm{C}\), \(p_{\text {? }}=1\) bar.

Determine the specific exergy, in \(\mathrm{kJ} / \mathrm{kg}\), at \(0.01^{\circ} \mathrm{C}\) of water as a (a) saturated vapor, (b) saturated liquid, (c) saturated solid. In each case, consider a fixed mass at rest and zero elevation relative to an exergy reference environment for which \(T_{0}=20^{\circ} \mathrm{C}, p_{0}=1\) bar.
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