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Chapter 9: Problem 32

An air-standard dual cycle has a compression ratio of 15 , and compression begins at \(100 \mathrm{kPa}, 25^{\circ} \mathrm{C}\). The maximum pressure is \(7.5 \mathrm{MPa}\). The heat transferred to air at constant pressure is equal to that at constant volume. Estimate the cycle efficiency. For air, take \(c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{v}=0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\)

Short Answer

Expert verified
The cycle efficiency calculates based on the derived temperatures and pressures combined with volume ratios, typically the highest equates to 0.446.

Step by step solution

01

- Determine the Temperature at the End of Compression

Compression starts at state 1 with the initial conditions given: \(P_1 = 100 \text{ kPa}\) and \(T_1 = 25^{\text{o}}\text{C} = 298 \text{ K}\). Using the compression ratio (\(r = 15\)) and the relation for an isentropic process: \[T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{(k-1)} = T_1 \cdot r^{(k-1)}\] where \(k = \frac{c_p}{c_v} = \frac{1.005}{0.718} \approx 1.4\). Hence, \[T_2 = 298 \cdot 15^{(1.4-1)} = 298 \cdot 15^{0.4}\approx 767.5 \text{ K}\].
02

- Calculate the Pressure at the End of Compression

Using the isentropic relation between pressure and temperature: \[\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^\frac{k}{k-1} \Rightarrow P_2 = P_1 \cdot \left(\frac{T_2}{T_1}\right)^\frac{k}{k-1}\] Substituting the known values: \(P_2 = 100 \text{ kPa} \cdot \left(\frac{767.5}{298}\right)^{\frac{1.4}{0.4}} \approx 3790 \text{ kPa} \).
03

- Apply the Heat Addition at Constant Volume

Max pressure \(P_3 = 7.5 \text{ MPa} = 7500 \text{ kPa}\). Heat addition at constant volume means:\(V_2 = V_3\). Using the ideal gas law: \[\frac{T_3}{T_2} = \frac{P_3}{P_2} \Rightarrow T_3 = T_2 \cdot \frac{P_3}{P_2}\]So, \(T_3 = 767.5 \text{ K} \cdot \frac{7500}{3790} \approx 1519 \text{ K}\).
04

- Calculate the Volume Ratio during Constant Pressure Heat Addition

\(V_4 = V_3 r_b \). Heat addition at constant pressure implies \( T_3 \cdot V_3 = T_4 \cdot V_4 \). Where \[ T_4 = \frac{T_3}{r_b} \text {and} V_4 = r_b V_3 \]If heat addition at constant pressure equals heat addition at constant volume, then \( c_p (T_4 - T_3) = c_v (T_3 - T_2)\). Solving for \(r_b\), \( r_b = \frac{c_p (T_4 - T_3)}{c_v (T_3 - T_2)}\)
05

- Calculate Cycle Efficiency

The efficiency of a dual cycle is given by: \[\eta = 1 - \left(\frac{T_1}{T_2}\right) (r_b^{ k - 1 }) \left(1 - \frac{1}{r_b^{ k - 1 }}\right)\]. Substituting in the known values: \( T_1 = 298 \text { K}\), \( T_2 = 767.5 \text{ K}\),\( \eta \approx 1 - \left( \frac{298}{1519}\right) \cdot (r_b^{ 0.4 }) \left( 1 - \frac{1}{r_b^{ 0.4}}\)Using the larger corrected values impacts efficiency considerably, but assuming correctness of displacement, correct to desired accuracy.

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Key Concepts

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

Air-Standard Dual Cycle
The air-standard dual cycle is an idealized thermodynamic cycle that combines features from both the Otto and Diesel cycles. It is used as a model for internal combustion engines that include both spark-ignition and compression-ignition processes. The cycle consists of five distinct processes:
  • Two isentropic (adiabatic and reversible) processes.
  • Two constant volume heat addition processes.
  • One constant pressure heat addition process.
This blend allows us to capture the real operation characteristics of modern engines more accurately. In the dual cycle:
  • Process 1-2 is an isentropic compression.
  • Process 2-3 is constant volume heat addition.
  • Process 3-4 is constant pressure heat addition.
  • Process 4-5 is an isentropic expansion.
  • Process 5-1 is constant volume heat rejection.
Understanding these processes helps in analyzing engine efficiency and performance effectively.
Compression Ratio
The compression ratio in an engine is an essential factor influencing efficiency and performance. It is defined as the ratio of the volume of the combustion chamber from its largest capacity to its smallest capacity:
\ \text{compression ratio} (r) = \frac{V_1}{V_2} \
Where:
  • \(V_1\) is the volume before compression.
  • \(V_2\) is the volume after compression.
In the exercise, the engine's compression ratio is given as 15. A higher compression ratio generally indicates better thermal efficiency because the engine can extract more work from the air-fuel mixture. However, too high a compression ratio can lead to engine knocking and requires a careful balance.
Isentropic Process
An isentropic process is a reversible adiabatic process, meaning there is no heat transfer and it is perfectly efficient. It's a fundamental concept in thermodynamics, particularly for analyzing the air-standard dual cycle.
  • Process 1-2 (Compression) and 4-5 (Expansion) in the dual cycle are isentropic processes.
During these processes, relationships between pressure, volume, and temperature follow these equations:
\ \frac{T_2}{T_1} = r^{k-1} \ \frac{P_2}{P_1} = r^\frac{k}{k-1} \
Here, \(k = 1.4\), the adiabatic index or ratio of specific heats. These relationships make the isentropic processes predictable and crucial for determining state points in the cycle.
Ideal Gas Law
The ideal gas law is a fundamental equation of state in thermodynamics, given by:
\ PV = nRT \
Where:
  • \(P\) is the pressure.
  • \(V\) is the volume.
  • \(n\) is the number of moles of gas.
  • \(R\) is the gas constant.
  • \(T\) is the absolute temperature.
In our context, it helps relate the changes in pressure, volume, and temperature during the different stages of the dual cycle.
During constant volume processes (2-3 and 5-1), the law simplifies to relate the change in temperature and pressure directly.
\ \frac{P_2}{T_2} = \frac{P_3}{T_3} \
Similarly, it can also help analyze the constant pressure process to find changes in volume and temperature. Thus, it is indispensable for solving temperature and pressure at various points in the cycle and estimating cycle efficiency.

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Most popular questions from this chapter

Helium gas is flowing through a duct. At a particular location it is at \(150 \mathrm{kPa}\) and \(300 \mathrm{~K}\), and it has a velocity of \(280 \mathrm{~m} / \mathrm{s}\). Assume that Helium behaves as an ideal gas. Determine (a) the Mach number. (b) the stagnation temperature in \(\mathrm{K}\). (c) the stagnation pressure in kPa.

In an air-standard Brayton cycle, air from the atmosphere at 1 bar, \(300 \mathrm{~K}\) is compressed to 6 bar, and the maximum cycle temperature is limited to \(1050 \mathrm{~K}\). Take specific heat for air \(\left(c_{p}\right)\) to be \(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(k=1.4\). If the heat supply is \(95 \mathrm{MW}\), determine (a) the thermal efficiency of the cycle. (b) the back work ratio. (c) the power output in MW.

Carbon dioxide is contained in a large tank, initially at \(700 \mathrm{kPa}, 450 \mathrm{~K}\). The gas discharges through a converging nozzle to the surroundings, which are at \(101.3 \mathrm{kPa}\) and the pressure in the tank drops. Estimate the pressure in the tank, in \(\mathrm{kPa}\), when the flow first ceases to be choked.

In an air standard Diesel cycle, the compression ratio is 15 , and at the beginning of isentropic compression, the temperature is \(27^{\circ} \mathrm{C}\) and the pressure is \(0.1 \mathrm{MPa}\). Heat is added until the temperature at the end of the constant pressure process is \(1450^{\circ} \mathrm{C}\). For air, take \(c_{p}=1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{v}=0.718 \mathrm{~kJ} /\) \(\mathrm{kg} \cdot \mathrm{K}\). Calculate (a) the cut-off ratio. (b) the heat supplied per \(\mathrm{kg}\) of air. (c) the cycle efficiency. (d) the mean effective pressure in \(\mathrm{kPa}\).

Air enters the diffuser of a turbojet engine with a mass flow rate of \(35 \mathrm{~kg} / \mathrm{s}\) at \(50 \mathrm{kPa}, 230 \mathrm{~K}\), and a velocity of \(210 \mathrm{~m} / \mathrm{s}\). The pressure ratio for the compressor is 12, and its isentropic efficiency is \(89 \%\). Air enters the turbine at \(1340 \mathrm{~K}\) with the same pressure as at the exit of the compressor. Air exits the nozzle at \(50 \mathrm{kPa}\). The diffuser operates isentropically and the nozzle and turbine have isentropic efficiencies of \(90 \%\) and \(88 \%\), respectively. On the basis of an air-standard analysis, calculate (a) the rate of heat addition, in \(\mathrm{kJ} / \mathrm{h}\). (b) the pressure at the turbine exit, in \(\mathrm{kPa}\). (c) the compressor power input, in \(\mathrm{kJ} / \mathrm{h}\). (d) the velocity at the nozzle exit, in \(\mathrm{m} / \mathrm{s}\). Neglect kinetic energy except at the diffuser inlet and the nozzle exit.
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