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Chapter 8: Problem 43

Air enters the compression process of a diesel cycle at \(120 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\). The pressure after compression is \(8 \mathrm{MPa}\) and \(1500 \mathrm{~kJ} / \mathrm{kg}\) is added during combustion. What are \((a)\) the cutoff ratio, \((b)\) the thermal efficiency, and ( \(c\) ) the MEP?

Short Answer

Expert verified
Cutoff ratio: 4.084, Thermal efficiency: 55.8%, MEP: 49.4 kPa.

Step by step solution

01

Identify Known Values

Given: Initial pressure, \( P_1 = 120 \text{ kPa} \), Initial temperature, \( T_1 = 15^{\text{\circ}} \text{C} = 288 \text{ K} \), Final pressure after compression, \( P_2 = 8 \text{ MPa} = 8000 \text{ kPa} \), Heat added during combustion, \( q_{\text{add}} = 1500 \text{ kJ/kg} \).
02

Calculate Compression Ratio (\(r\))

Use the isentropic relation for compression: \[ r = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}} \] With \(k = 1.4\) for air: \[ r = \left( \frac{8000}{120} \right)^{\frac{0.4}{1.4}} \] Calculating this, \[ r \approx 18.08 \]
03

Calculate the Cutoff Ratio (\(r_c\))

The cutoff ratio is the volume ratio after combustion to before combustion. Use the relation: \[ q_{\text{add}} = c_v (T_3 - T_2) \] With specific heat at constant volume \( c_v = 0.718 \text{ kJ/kg.K} \), \[ T_2 = T_1 \cdot r^{k-1} \] \[ T_2 = 288 \cdot 18.08^{0.4} \approx 708.75 \text{ K} \] Thus, \[ T_3 = T_2 + \frac{q_{\text{add}}}{c_v} = 708.75 + \frac{1500}{0.718} \approx 2895 \text{ K} \] Then, \[ r_c = \frac{T_3}{T_2} \approx \frac{2895}{708.75} \approx 4.084 \]
04

Calculate Thermal Efficiency (\(\eta_{th}\))

Use the efficiency formula for a Diesel cycle: \[ \eta_{th} = 1 - \frac{1}{r^{k-1}} \left( \frac{r_c^{k} - 1}{k (r_c - 1)} \right) \] Plug in the values, \[ \eta_{th} = 1 - \frac{1}{18.08^{0.4}} \left( \frac{4.084^{0.4} - 1}{1.4 (4.084 - 1)} \right) \] Calculating this, \[ \eta_{th} \approx 0.558 \text{ or } 55.8\% \]
05

Calculate Mean Effective Pressure (MEP)

Use the MEP formula: \[ \text{MEP} = \frac{W_{net}}{V_{d}} \] Where \( W_{net} = q_{add} \cdot \eta_{th} = 1500 \text{kJ/kg} \cdot 0.558 \approx 837 \text{kJ/kg} \), and \( V_{d} \) is the displacement volume: \( V_{d} = r - 1 \). Therefore, \[ \text{MEP} = \frac{837}{18.08-1} \approx 49.4 \text{kPa} \]

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Key Concepts

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

Cutoff Ratio Calculation
The cutoff ratio in a Diesel cycle represents the proportion of the cylinder volume after combustion to the volume before combustion. It's a critical parameter in Diesel engines because it gives insight into the amount of heat added during the cycle.
To calculate the cutoff ratio \(r_c\), we can follow these steps:
  • First, identify all known values. From the exercise: initial pressure \(P_1 = 120 \text{kPa}\), initial temperature \(T_1 = 288 \text{K}\), and final pressure after compression \(P_2 = 8000 \text{kPa}\).
  • Calculate the compression ratio \(r\) using the isentropic relation for compression: \[ r = \left( \frac{P_2}{P_1} \right)^{\frac{k-1}{k}}\] where \(k = 1.4\) for air. This gives us \[ r \approx 18.08 \]
  • Determine the temperature at the end of the compression process \(T_2\): \[ T_2 = T_1 \cdot r^{k-1} = 288 \cdot 18.08^{0.4} \approx 708.75 \text{K} \]
  • From the heat added during combustion \(q_{\text{add}} = 1500 \text{kJ/kg}\), calculate the temperature after combustion \(T_3\): \[ T_3 = T_2 + \frac{q_{\text{add}}}{c_v} = 708.75 + \frac{1500}{0.718} \approx 2895 \text{K} \]
  • Finally, determine the cutoff ratio: \[ r_c = \frac{T_3}{T_2} = \frac{2895}{708.75} \approx 4.084 \]
Understanding this process is crucial as \(r_c\) influences the engine's performance and efficiency significantly.
Thermal Efficiency Calculation
The thermal efficiency \(\eta_{\text{th}}\) of a Diesel cycle measures how effectively the engine converts the heat energy from fuel into useful work.
To calculate thermal efficiency, use the efficiency formula for a Diesel cycle:
  • First, summarize the known values: compression ratio \(r = 18.08\), cutoff ratio \(r_c = 4.084\), and \(k = 1.4 \).
  • Apply the efficiency formula for a Diesel cycle: \[ \eta_{th} = 1 - \frac{1}{r^{k-1}} \left( \frac{r_c^k - 1}{k (r_c - 1)} \right) \]
  • Substitute in the values: \[ \eta_{th} = 1 - \frac{1}{18.08^{0.4}} \left( \frac{4.084^{0.4} - 1}{1.4 \cdot (4.084 - 1)} \right) \]
  • After performing the calculations, you should find that: \[ \eta_{th} \approx 0.558 \text{ or 55.8\text{%}} \]
This efficiency percentage tells us how much of the input energy (fuel) is successfully converted into mechanical energy. The higher the thermal efficiency, the more effective the engine.
Mean Effective Pressure (MEP)
Mean Effective Pressure (MEP) is a parameter in internal combustion engines that illustrates the average pressure applied to the engine pistons during a cycle.
To determine MEP, we follow these steps:
  • Identify the net work done per cycle, \(W_{\text{net}}\), which equals the product of heat added and thermal efficiency: \[ W_{\text{net}} = q_{\text{add}} \cdot \eta_{th} = 1500 \text{kJ/kg} \cdot 0.558 \approx 837 \text{kJ/kg} \]
  • Calculate the displacement volume, \(V_d\), which is the difference in volume ratio before and after compression: \[ V_d = r - 1 = 18.08 - 1 \approx 17.08 \]
  • Now, determine MEP using the formula: \[ \text{MEP} = \frac{W_{\text{net}}}{V_d} = \frac{837}{17.08} \approx 49.4 \text{kPa} \]
This parameter is crucial because it indicates the engine's ability to generate work. Higher MEP values typically signify better performance and more efficient power output.

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Most popular questions from this chapter

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 $$ 8400 \dot{m}_{f}=25260 \quad \therefore \dot{m}_{f}=3.01 \mathrm{~kg} / \mathrm{s} $$

A diesel engine intakes atmospheric air at \(60^{\circ} \mathrm{F}\) and adds \(800 \mathrm{Btu} / \mathrm{lbm}\) of energy. If the maximum pressure is 1200 psia, calculate \((a)\) the cutoff ratio, \((b)\) the thermal efficiency, and \((c)\) the power output for an airflow of \(0.2 \mathrm{lbm} / \mathrm{s}\). (a) The compression process is isentropic. The temperature at state 2 (see Fig. 8.9) is calculated to be $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(520)\left(\frac{1200}{14.7}\right)^{0.2857}=1829^{\circ} \mathrm{R} $$ The temperature at state 3 is found from the first law as follows: $$ q_{\text {in }}=C_{p}\left(T_{3}-T_{2}\right) \quad 800=(0.24)\left(T_{3}-1829\right) \quad \therefore T_{3}=5162^{\circ} \mathrm{R} $$ The specific volumes of the three states are $$ v_{1}=\frac{R T_{1}}{P_{1}}=\frac{(53.3)(520)}{(14.7)(144)}=13.09 \mathrm{ft}^{3} / \mathrm{lbm} \quad v_{2}=\frac{R T_{2}}{P_{2}}=\frac{(53.3)(1829)}{(1200)(144)}=0.5642 \mathrm{ft}^{3} / 1 \mathrm{bm} $$ $$ v_{3}=\frac{R T_{3}}{P_{3}}=\frac{(53.3)(5162)}{(1200)(144)}=1.592 \mathrm{ft}^{3} / \mathrm{lbm} $$ The cutoff ratio is then \(r_{c}=v_{3} / v_{2}=1.592 / 0.5642=2.822\). (b) The compression ratio is \(r=v_{1} / v_{2}=13.09 / 0.5642=23.20\). The thermal efficiency can now be calculated, using (8.25): $$ \eta=1-\frac{1}{r^{k-1}} \frac{r_{c}^{k}-1}{k\left(r_{c}-1\right)}=1-\frac{1}{(23.2)^{0.4}} \frac{(2.822)^{1.4}-1}{(1.4)(2.822-1)}=0.6351 \quad \text { or } \quad 63.51 \% $$ (c) \(\dot{W}_{\text {out }}=\eta \dot{Q}_{\text {in }}=\eta \dot{m} q_{\text {in }}=[(0.6351)(0.2)(800)](778)=79,060 \mathrm{ft}-\mathrm{lbf} / \mathrm{s} \quad\) or \(\quad 143.7 \mathrm{hp}\).

An adiabatic compressor receives \(20 \mathrm{~m}^{3} / \mathrm{min}\) of air from the atmosphere at \(20^{\circ} \mathrm{C}\) and compresses it to \(10 \mathrm{MPa}\). Calculate the minimum power requirement. SOLUTION An isentropic compression requires the minimum power input for an adiabatic compressor. The outlet temperature for such a process is $$ T_{2}=T_{1}\left(\frac{P_{2}}{P_{1}}\right)^{(k-1) / k}=(293)\left(\frac{10000}{100}\right)^{0.2857}=1092 \mathrm{~K} $$ To find the mass flux, we must know the density. It is \(\rho=P / R T=100 /(0.287)(293)=1.189 \mathrm{~kg} / \mathrm{m}^{3}\). The mass flux is then (the flow rate is given) \(\dot{m}=\rho(A V)=(1.189)(20 / 60)=0.3963 \mathrm{~kg} / \mathrm{s}\). The minimum power requirement is now calculated to be $$ \dot{W}_{\text {comp }}=\dot{m}\left(h_{2}-h_{1}\right)=\dot{m} C_{p}\left(T_{2}-T_{1}\right)=(0.3963)(1.00)(1092-293)=317 \mathrm{~kW} $$

Air at \(90 \mathrm{kPa}\) and \(15^{\circ} \mathrm{C}\) is supplied to an ideal cycle at intake. If the compression ratio is 10 and the heat supplied is \(300 \mathrm{~kJ} / \mathrm{kg}\), calculate the efficiency and the maximum temperature for \((a)\) a Stirling cycle, and (b) an Ericsson cycle. (a) For the constant-temperature process, the heat transfer equals the work. Referring to Fig. 8.13, the first law gives $$ q_{\text {out }}=w_{1-2}=R T_{1} \ln \frac{v_{1}}{v_{2}}=(0.287)(288) \ln 10=190.3 \mathrm{~kJ} / \mathrm{kg} $$ The work output for the cycle is then \(w_{\text {out }}=q_{\text {in }}-q_{\text {out }}=300-190.3=109.7 \mathrm{~kJ} / \mathrm{kg}\). The efficiency is $$ \eta=\frac{w_{\text {out }}}{q_{\text {in }}}=\frac{109.7}{300}=0.366 $$ The high temperature is found from $$ \eta=1-\frac{T_{L}}{T_{H}} \quad \therefore T_{H}=\frac{T_{L}}{1-\eta}=\frac{288}{1-0.366}=454 \mathrm{~K} $$ (b) For the Ericsson cycle of Fig. 8.14, the compression ratio is \(v_{4} / v_{2}\). The constant-temperature heat addition \(3 \rightarrow 4\) provides $$ q_{\text {in }}=w_{3-4}=R T_{4} \ln \frac{v_{4}}{v_{3}} \quad \therefore 300=(0.287) T_{4} \ln \frac{v_{4}}{v_{3}} $$ The constant-pressure process \(2 \rightarrow 3\) allows $$ \frac{T_{3}}{v_{3}}=\frac{T_{2}}{v_{2}}=\frac{288}{v_{4} / 10} $$ The constant-pressure process \(4 \rightarrow 1\) demands $$ \frac{T_{4}}{v_{4}}=\frac{T_{1}}{v_{1}}=\frac{P_{1}}{R}=\frac{90}{0.287}=313.6 $$ Recognizing that \(T_{3}=T_{4}\), the above can be combined to give $$ 300=(0.287)\left(313.6 v_{4}\right) \ln \frac{v_{4}}{v_{3}} \quad v_{3}=0.1089 v_{4}^{2} $$ The above two equations are solved simultaneously by trial and error to give $$ v_{4}=3.94 \mathrm{~m}^{3} / \mathrm{kg} \quad v_{3}=1.69 \mathrm{~m}^{3} / \mathrm{kg} $$ Thus, from the compression ratio, \(v_{2}=v_{4} / 10=0.394 \mathrm{~m}^{3} / \mathrm{kg}\). The specific volume of state 1 is $$ v_{1}=\frac{R T}{P_{1}}=\frac{(0.287)(288)}{90}=0.9184 \mathrm{~m}^{3} / \mathrm{kg} $$ The heat rejected is then $$ q_{\mathrm{out}}=R T_{1} \ln \frac{v_{1}}{v_{2}}=(0.287)(288) \ln \frac{0.9184}{0.394}=70.0 \mathrm{~kJ} / \mathrm{kg} $$ The net work for the cycle is \(w_{\text {out }}=q_{\text {in }}-q_{\text {out }}=300-70.0=230 \mathrm{~kJ} / \mathrm{kg}\). The efficiency is then \(\eta=w_{\text {out }} / q_{\text {in }}=\) \(230 / 300=0.767\). This allows us to calculate the high temperature: $$ \eta=1-\frac{T_{L}}{T_{H}} \quad 0.767=1-\frac{288}{T_{H}} \quad \therefore T_{H}=1240 \mathrm{~K} $$

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.
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