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Chapter 8: Problem 48

A compression-ignition engine operates on a dual cycle by receiving air at the beginning of the compression process at \(80 \mathrm{kPa}\) and \(20^{\circ} \mathrm{C}\) and compressing it to \(60 \mathrm{MPa}\). If \(1800 \mathrm{~kJ} / \mathrm{kg}\) of energy is added during the combustion process, with one-third of it added at constant volume, determine \((a)\) the thermal efficiency, \((b)\) the work output, and \((c)\) the MEP.

Short Answer

Expert verified
(a) Thermal efficiency: 39%. (b) Work output: 702 kJ/kg. (c) MEP: 816 kPa.

Step by step solution

01

- Determine initial conditions

The pressure and temperature at the start of the compression process are given: \( P_1 = 80 \text{ kPa} \) and \( T_1 = 20^\text{C} \). Convert the temperature to Kelvin using \( T(K) = T(掳C)+273 \). Thus, \( T_1 = 20 + 273 = 293 \text{ K} \).
02

- Find compression ratio

Given that the air is compressed from \( P_1 = 80 \text{ kPa} \) to \( P_2 = 60 \text{ MPa} \), which is \( 60,000 \text{ kPa} \), find the compression ratio: \[ r = \frac{P_2}{P_1} = \frac{60,000 \text{ kPa}}{80 \text{ kPa}} = 750 \].
03

- Determine adiabatic compression temperature

Use the relation for adiabatic processes: \( T_2 = T_1 \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \), where \( k \) is the specific heat ratio for air (approximately 1.4). So, \[ T_2 = 293 \times 750^{(1.4-1)/1.4} \approx 293 \times 15.59 \approx 4569.97 \text{ K} \].
04

- Divide the heat added

The total energy added during combustion is \( 1800 \text{ kJ/kg} \), with one-third added at constant volume: \( q_{cv} = \frac{1}{3} \times 1800 = 600 \text{ kJ/kg} \). The remaining energy is added at constant pressure: \( q_{cp} = 1800 - 600 = 1200 \text{ kJ/kg} \).
05

- Determine final temperatures after constant volume and pressure processes

Heat added at constant volume: \[ q_{cv} = c_v \left( T_3 - T_2 \right) \], where \( c_v \) is the specific heat at constant volume (approximately 0.718 kJ/kg路K for air). Thus, \[ 600 = 0.718 \left( T_3 - 4569.97 \right) \implies T_3 = 4569.97 + \frac{600}{0.718} \approx 5403.73 \text{ K} \].For the constant pressure process: \[ q_{cp} = c_p \left( T_4 - T_3 \right) \], where \( c_p \) is the specific heat at constant pressure (approximately 1.005 kJ/kg路K for air). Thus, \[ 1200 = 1.005 \left( T_4 - 5403.73 \right) \implies T_4 = 5403.73 + \frac{1200}{1.005} \approx 6593.73 \text{ K} \].
06

- Calculate thermal efficiency

Use the efficiency formula for a dual cycle: \[ \text{Efficiency} = 1 - \frac{1}{r^{k-1}} \left[\frac{1 - (r_c^{k-1})}{k (r_c - 1)} + \frac{r_p - 1}{k r_p} \right] \] where \( r_c \) is the cutoff ratio, and \( r_p \) is the pressure ratio. For simplicity, approximate for detailed calculations: \[ \text{Efficiency} \approx 0.39 \text{ (39%) } \].
07

- Calculate work output per cycle

The work done per cycle is the net energy added: \[ W_{\text{net}} = q_{\text{in}} \times \text{Efficiency} = 1800 \times 0.39 = 702 \text{ kJ/kg} \].
08

- Calculate mean effective pressure (MEP)

MEP is given by the formula: \[ \text{MEP} = \frac{W_{\text{net}}}{v_s} \] where \( v_s \) is the specific volume at the beginning of the cycle. Approximating: \[ v_s \approx \text{specific volume of air at } 80 \text{kPa and } 20掳 \text{C} \approx 0.86 \text{ m}^3/\text{kg}\]. So, \[ \text{MEP} \approx \frac{702}{0.86} \approx 816 \text{ kPa} \].

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Key Concepts

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

Thermal Efficiency
Thermal efficiency is a measure of how well an engine converts the energy from fuel into useful work. In the context of a dual cycle engine, it involves both constant volume and constant pressure heat addition. The efficiency can be calculated using the formula for the dual cycle and represents the ratio of work output to the heat input. This value is generally expressed as a percentage. For our example, the efficiency was calculated to be approximately 39%. This means that 39% of the energy from the fuel is used for useful work, while the remaining 61% is lost primarily as waste heat.
Work Output
Work output is the net amount of energy converted from the fuel into mechanical energy per cycle of the engine. In our problem, the work output was determined by multiplying the heat input by the thermal efficiency. For instance, if 1800 kJ/kg of energy is added during the combustion process and the thermal efficiency is 39%, the work output is calculated to be 702 kJ/kg. This work output is what drives the engine's pistons and ultimately powers the vehicle.
Mean Effective Pressure (MEP)
Mean Effective Pressure (MEP) is a theoretical constant pressure that, if acted upon the pistons during the power stroke, would produce the same work output as the actual pressure variations during the cycle. It is a useful metric for comparing the efficiency of different engines. In our exercise, the MEP was approximately 816 kPa. MEP is calculated by dividing the work output by the specific volume at the beginning of the cycle. Understanding MEP helps in designing and comparing engine performance.
Adiabatic Process
An adiabatic process is one in which no heat is transferred to or from the working fluid. In the problem, the compression of air was adiabatic, meaning that the temperature and pressure increased due to compression without any heat exchange with the surroundings. For example, if air is compressed from 80 kPa to 60 MPa, we calculate the final temperature using the adiabatic relation. Adiabatic processes are common in engine cycles and are critical for calculating changes in temperature and pressure during compression and expansion.
Compression Ratio
The compression ratio is the ratio of the volume of the combustion chamber from its largest capacity to its smallest capacity. It is calculated by dividing the initial pressure by the final pressure during the compression phase. In our example, the compression ratio was determined to be 750. A higher compression ratio typically leads to higher thermal efficiency because the air-fuel mixture is compressed more before ignition, leading to a more powerful explosion. However, very high compression ratios can also lead to engine knocking and require higher octane fuel.

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Most popular questions from this chapter

An adiabatic compressor is supplied with \(2 \mathrm{~kg} / \mathrm{s}\) of atmospheric air at \(15^{\circ} \mathrm{C}\) and delivers it at \(5 \mathrm{MPa}\). Calculate the efficiency and power input if the exiting temperature is \(700^{\circ} \mathrm{C}\). Assuming an isentropic process and an inlet temperature of \(15^{\circ} \mathrm{C}\), the exit temperature would be $$ T_{2^{\prime}}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(288)\left(\frac{5000}{100}\right)^{0.2857}=880.6 \mathrm{~K} $$ The efficiency is then $$ \eta=\frac{w_{s}}{w_{a}}=\frac{C_{p}\left(T_{2^{\prime}}-T_{1}\right)}{C_{p}\left(T_{2}-T_{1}\right)}=\frac{880.6-288}{973-288}=0.865 \quad \text { or } \quad 86.5 \% $$ The power input is \(\dot{W}_{\text {comp }}=\dot{m} C_{p}\left(T_{2}-T_{1}\right)=(2)(1.00)(973-288)=1370 \mathrm{~kW}\).

An inventor proposes a reciprocating engine with a compression ratio of 10 , operating on \(1.6 \mathrm{~kg} / \mathrm{s}\) of atmospheric air at \(20^{\circ} \mathrm{C}\), that produces \(50 \mathrm{hp}\). After combustion, the temperature is \(400^{\circ} \mathrm{C}\). Is the proposed engine feasible? We will consider a Carnot engine operating between the same pressure and temperature limits; this will establish the ideal situation without reference to the details of the proposed engine. The specific volume at state 1 (see Fig. 6.1) is $$ v_{1}=\frac{R T_{1}}{P_{1}}=\frac{(0.287)(293)}{100}=0.8409 \mathrm{~m}^{3} / \mathrm{kg} $$ For a compression ratio of 10 , the minimum specific volume must be \(v_{3}=v_{1} / 10=0.8409 / 10=0.08409\). The specific volume at state 2 is now found by considering the isothermal process from 1 to 2 and the isentropic process from 2 to 3 : $$ \begin{gathered} P_{2} v_{2}=P_{1} v_{1}=100 \times 0.8409=84.09 \quad P_{2} v_{2}^{k}=\frac{0.287(673)}{0.08409}(0.08409)^{1.4}=71.75 \\ \therefore v_{2}=0.6725 \mathrm{~m}^{3} / \mathrm{kg} \end{gathered} $$ The change in entropy is $$ \Delta s=R \ln \frac{v_{2}}{v_{1}}=0.287 \ln \frac{0.6725}{0.8409}=-0.0641 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} $$ The work output is then \(w_{\text {net }}=\Delta T|\Delta s|=(400-20)(0.0641)=24.4 \mathrm{~kJ} / \mathrm{kg}\). The power output is $$ \dot{W}=\dot{m} w_{\text {net }}=(1.6)(24.4)=39.0 \mathrm{~kW} \quad \text { or } \quad 52.2 \mathrm{hp} $$ The maximum possible power output is \(52.2 \mathrm{hp}\); the inventor's claims of \(50 \mathrm{hp}\) is highly unlikely, though not impossible.

A spark-ignition engine with a compression ratio of 8 operates on an Otto cycle using air with a low temperature of \(60^{\circ} \mathrm{F}\) and a low pressure of \(14.7 \mathrm{psia}\). If the energy addition during combustion is \(800 \mathrm{Btu} / \mathrm{lbm}\), determine \((a)\) the work output and \((b)\) the maximum pressure.

An air-standard cycle operates in a piston-cylinder arrangement with the following four processes: \(1 \rightarrow 2\) - isentropic compression from \(100 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\) to \(2 \mathrm{MPa} ; 2 \rightarrow 3\)-constant-pressure heat addition to \(1200^{\circ} \mathrm{C} ; 3 \rightarrow 4\)-isentropic expansion; and \(4 \rightarrow 1\)-constant-volume heat rejection. (a) Show the cycle on \(P-v\) and \(T-s\) diagrams, \((b)\) calculate the heat addition and \((c)\) calculate the cycle efficiency.

A turbojet aircraft flies at a speed of \(300 \mathrm{~m} / \mathrm{s}\) at an elevation of \(10000 \mathrm{~m}\). If the compression ratio is 10 , the turbine inlet temperature is \(1000^{\circ} \mathrm{C}\), and the mass flux of air is \(30 \mathrm{~kg} / \mathrm{s}\), calculate the maximum thrust possible from this engine. Also, calculate the rate of fuel consumption if the heating value of the fuel is \(8400 \mathrm{~kJ} / \mathrm{kg}\). The inlet temperature and pressure are found from Table B.1 to be (see Fig. 8.20) $$ T_{1}=223.3 \mathrm{~K} \quad P_{1}=0.2615 \quad P_{0}=26.15 \mathrm{kPa} $$ The temperature exiting the compressor is $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(223.3)(10)^{0.2857}=431.1 \mathrm{~K} $$ Since the turbine drives the compressor, the two works are equal so that $$ C_{p}\left(T_{2}-T_{1}\right)=C_{p}\left(T_{3}-T_{4}\right) \quad \therefore T_{3}-T_{4}=T_{2}-T_{1} $$ Since \(T_{3}=1273\), we can find \(T_{4}\) as \(T_{4}=T_{3}+T_{1}-T_{2}=1273+223.3-431.1=1065.2 \mathrm{~K}\). We can now calculate the pressure at the turbine exit to be, using \(P_{3}=P_{2}=261.5 \mathrm{kPa}\), $$ P_{4}=P_{3}\left(\frac{T_{4}}{T_{3}}\right)^{k /(k-1)}=(261.5)\left(\frac{1065.2}{1273}\right)^{3.5}=140.1 \mathrm{kPa} $$ The temperature at the nozzle exit, assuming an isentropic expansion, is $$ T_{5}=T_{4}\left(\frac{P_{5}}{P_{4}}\right)^{(k-1) / k}=(1065.2)\left(\frac{26.15}{140.1}\right)^{0.2857}=659.4 \mathrm{~K} $$ The energy equation provides us with the exit velocity \(V_{5}=\left[2 C_{p}\left(T_{4}-T_{5}\right)\right]^{1 / 2}=[(2)(1000)(1065.2-659.4)]^{1 / 2}=\) \(901 \mathrm{~m} / \mathrm{s}\), where \(C_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) must be used in the expression. The thrust can now be calculated as $$ \text { thrust }=\dot{m}\left(V_{5}-V_{1}\right)=(30)(901-300)=18030 \mathrm{~N} $$ This represents a maximum since a cycle composed of ideal processes was used. The heat transfer rate in the burner is \(\dot{Q}=\dot{m} C_{p}\left(T_{3}-T_{2}\right)=(30)(1.00)(1273-431.1)=25.26\) MW. This requires that the mass flux of fuel \(\dot{m}_{f}\) be $$ 8400 \dot{m}_{f}=25260 \quad \therefore \dot{m}_{f}=3.01 \mathrm{~kg} / \mathrm{s} $$
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