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Chapter 6: Problem 94

Air in a piston-cylinder assembly is compressed isentropically from state 1 , where \(T_{1}=45^{\circ} \mathrm{C}\), to state 2 , where the specific volume is one-twentieth of the specific volume at state 1. Applying the ideal gas model with \(k=1.4\), determine (a) \(T_{2}\), in \({ }^{\circ} \mathrm{C}\) and (b) the work, in \(\mathrm{kJ} / \mathrm{kg}\).

Short Answer

Expert verified
(a) \(T_2 = 566.46^{\text{掳}} \text{C}\), (b) \( W = 198.29 \text{ kJ/kg} \).

Step by step solution

01

- Identify Known Values

Given: Initial temperature, \(T_1 = 45^{\text{掳}} \text{C} = 318.15 \text{ K} \) The specific volume at state 2 is one-twentieth of the specific volume at state 1: \(v_2 = \frac{v_1}{20} \) The specific heat ratio, \(k = 1.4 \).
02

- Use the Isentropic Relation for Temperature and Volume

For an isentropic process involving an ideal gas, use the relation: \( \frac{T_2}{T_1} = \left( \frac{v_1}{v_2} \right)^{k-1} \) Substitute the given values: \( \frac{T_2}{318.15 \text{ K}} = \left( \frac{20v_2}{v_2} \right)^{1.4-1} \) \( \frac{T_2}{318.15} = 20^{0.4} \).
03

- Calculate the Final Temperature

Calculate the value of \( 20^{0.4} \): \( 20^{0.4} \approx 2.639 \) Thus, \( T_2 = 318.15 \times 2.639 \approx 839.61 \text{ K}\) Convert this to degrees Celsius: \( T_2 = 839.61 \text{ K} - 273.15 \text{ K} = 566.46^{\text{掳}} \text{C} \).
04

- Use the Isentropic Relation for Work

The work done during an isentropic process for an ideal gas is given by: \[ W = \frac{P_1 v_1 - P_2 v_2}{k-1} \] Given that the initial state is compressed to 1/20th of the initial specific volume, substitute \(v_2 = \frac{v_1}{20} \) and apply the ideal gas law: \[ P_1 v_1 = RT_1 \] \[ P_2 v_2 = RT_2 \] Thus the work is: \[ W = \frac{R (T_1 - T_2/20)}{k-1} \].
05

- Calculate the Work Done

Substitute the temperatures and ideal gas constants: \[ R = 0.287 \text{ kJ/kg.K} \] \[ W = \frac{0.287 (318.15 - 839.61 / 20)}{1.4-1} \] \[ W = \frac{0.287 (318.15 - 41.98)}{0.4} \approx 198.29 \text{ kJ/kg} \].

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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 describes the behavior of ideal gases. It relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) and the universal gas constant (R). The equation is typically written as:\(PV = nRT\). This law assumes that the gas molecules do not interact with each other and occupy negligible space.
In the context of our exercise, we treat air as an ideal gas to simplify calculations. The Ideal Gas Law helps us understand how pressure, volume, and temperature are interrelated. For example, in calculating work done during compression, we use this law to relate the initial and final states of the gas.
Specific Volume
Specific volume is defined as the volume occupied by a unit mass of a substance, often denoted as \(v\). It is the reciprocal of density and is given by the formula: \(v = \frac{V}{m}\), where V is the volume and m is the mass of the gas.
In the exercise, we are given that the specific volume at state 2 is one-twentieth of the specific volume at state 1:\(v_2 = \frac{v_1}{20}\). This information helps simplify our calculations by indicating how the volume changes during the compression process. Understanding specific volume is important as it plays a key role in determining the final temperature and work done during the isentropic process.
Work Calculation
Work in thermodynamics is the energy transferred by a system due to a volume change against an external pressure. In an isentropic process for an ideal gas, the work done can be calculated using the formula:\[W = \frac{R (T_1 - T_2 / 20)}{k-1}\].
Here, R is the specific gas constant, and \(k\) is the ratio of specific heats. Substituting the given values and solving for W, we find the work done during the compression process. This calculation is crucial for understanding the energy requirements or outputs of thermodynamic systems.
Isentropic Relation
An isentropic process is a thermodynamic process that occurs at constant entropy. For an ideal gas, the isentropic relations can be expressed in terms of temperature and volume as:\( \frac{T_2}{T_1} = (\frac{v_1}{v_2})^{k-1}\).
This relation helps us connect the initial and final states of the gas under isentropic conditions.
To find the final temperature \(T_2\), we use the given specific volume ratio and solve the isentropic relation. This relation significantly simplifies the calculations by relating various thermodynamic properties at different states of the system.

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Most popular questions from this chapter

An electric motor operating at steady state draws a current of \(10 \mathrm{amp}\) with a voltage of \(220 \mathrm{~V}\). The output shaft rotates at 1000 RPM with a torque of \(16 \mathrm{~N} \cdot \mathrm{m}\) applied to an external load. The rate of heat transfer from the motor to its surroundings is related to the surface temperature \(T_{\mathrm{b}}\). and the ambient temperature \(T_{0}\) by \(\mathrm{hA}\left(T_{\mathrm{b}}-T_{0}\right)\), where \(\mathrm{h}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, \mathrm{A}=0.195 \mathrm{~m}^{2}\), and \(T_{0}=293 \mathrm{~K} .\) Energy transfers are considered positive in the directions indicated by the arrows on Fig. P6.51. (a) Determine the temperature \(T_{\mathrm{b}}\), in \(\mathrm{K}\). (b) For the motor as the system, determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\). (c) If the system boundary is located to take in enough of the nearby surroundings for heat transfer to take place at temperature \(T_{0}\), determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), for the enlarged system.

Two insulated tanks are connected by a valve. One tank initially contains \(0.45 \mathrm{~kg}\) of air at \(93^{\circ} \mathrm{C}, 1\) bar, and the other contains \(0.9 \mathrm{~kg}\) of air at \(38^{\circ} \mathrm{C}, 2\) bar. The valve is opened and the two quantities of air are allowed to mix until equilibrium is attained. Employing the ideal gas model with \(c_{v}=0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the final temperature, in \({ }^{\circ} \mathrm{C}\). (b) the final pressure, in bar. (c) the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\).

Saturated water vapor at \(100 \mathrm{kPa}\) enters a counterflow heat exchanger operating at steady state and exits at \(20^{\circ} \mathrm{C}\) with a negligible change in pressure. Ambient air at \(275 \mathrm{~K}\), 1 bar enters in a separate stream and exits at \(290 \mathrm{~K}, 1\) bar. The air mass flow rate is 170 times that of the water. The air can be modeled as an ideal gas with \(c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Kinetic and potential energy effects can be ignored. (a) For a control volume enclosing the heat exchanger, evaluate the rate of heat transfer, in kJ per kg of water flowing. (b) For an enlarged control volume that includes the heat exchanger and enough of its immediate surroundings that heat transfer from the control volume occurs at the ambient temperature, \(275 \mathrm{~K}\), determine the rate of entropy production, in \(\mathrm{kJ} / \mathrm{K}\) per \(\mathrm{kg}\) of water flowing.

An electric water heater having a 200 liter capacity employs an electric resistor to heat water from 23 to \(55^{\circ} \mathrm{C}\). The outer surface of the resistor remains at an average temperature of \(80^{\circ} \mathrm{C}\). Heat transfer from the outside of the water heater is negligible and the states of the resistor and the tank holding the water do not change significantly. Modeling the water as incompressible, determine the amount of entropy produced, in \(\mathrm{kJ} / \mathrm{K}\), for (a) the water as the system. (b) the overall water heater including the resistor. Compare the results of parts (a) and (b), and discuss.

Two \(\mathrm{kg}\) of Refrigerant \(134 \mathrm{a}\) initially at \(1.4\) bar, \(60^{\circ} \mathrm{C}\), are compressed to saturated vapor at \(60^{\circ} \mathrm{C}\). During this process, the temperature of the refrigerant departs by no more than \(0.01^{\circ} \mathrm{C}\) from \(60^{\circ} \mathrm{C}\). Determine the minimum theoretical heat transfer from the refrigerant during the process, in \(\mathrm{kJ}\).
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