Engineering_Mechanics_Colin_R_Ferguson_A_Solved_Ch1

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<p>Chapter 2</p><p>Heat Engine Cycles</p><p>2.1) An engine cylinder contains 7 × 10−5 kg of fuel with a heat of combustion, qc, of 45,000 kJ/kg. The</p><p>volume V1 at top dead center is 0.15×10−3 m3, and the volume V2 at bottom dead center is 1.50×10−3</p><p>m3. The air-fuel ratio is 16:1, and the mixture temperature T1 at the start of compression is 300 K.</p><p>Modeling the compression and combustion as an ideal gas (γ = 1.4, cv = 0.71 kJ/kg K) Otto cycle, (a)</p><p>what is the maximum temperature T3 and pressure P3 , (b) What is the pressure P1 at the start of</p><p>compression?</p><p>3</p><p>2</p><p>1</p><p>4</p><p>Volume (V/V2)</p><p>P</p><p>re</p><p>s</p><p>s</p><p>u</p><p>re</p><p>(</p><p>P</p><p>/P</p><p>1</p><p>)</p><p>Figure 2.1: Problem 2-1 P-V plot</p><p>a) The heat input is given by:</p><p>Qin = mfqc = mqin =</p><p>(</p><p>7.0× 10−5</p><p>)</p><p>(45,000)</p><p>Qin = 3.15kJ</p><p>The compression ratio is given by:</p><p>r =</p><p>V1</p><p>V2</p><p>=</p><p>1.50× 10−3</p><p>0.15× 10−3</p><p>r = 10</p><p>1</p><p>2 CHAPTER 2. HEAT ENGINE CYCLES</p><p>Process 1-2: isentropic compression</p><p>T2 = T1r</p><p>γ−1 = (300) (10)</p><p>1.4−1</p><p>T2 = 753K</p><p>Process 2-3: constant volume heat addition</p><p>T3 =</p><p>Qin</p><p>mcv</p><p>+ T2</p><p>The mass of gas in the cylinder:</p><p>m = ma +mf = mf (1 +AF ) =</p><p>(</p><p>7× 10−5</p><p>)</p><p>(1 + 16)</p><p>m = 1.19× 10−3 kg</p><p>The maximum temperature:</p><p>T3 =</p><p>Qin</p><p>mcv</p><p>+ T2 =</p><p>3.15</p><p>(</p><p>1.19× 10−3</p><p>)</p><p>(0.71)</p><p>+ 753</p><p>T3 = 4481K</p><p>The maximum pressure:</p><p>P3 =</p><p>mRT3</p><p>V3</p><p>=</p><p>(</p><p>1.19× 10−3</p><p>)</p><p>(0.287) (4481)</p><p>0.15× 10−3</p><p>P3 = 10,203kPa</p><p>b) Working backwards</p><p>P2 =</p><p>P3T2</p><p>T3</p><p>=</p><p>(10,203) (753)</p><p>4481</p><p>P2 = 1714kPa</p><p>P1 =</p><p>P2</p><p>rγ</p><p>=</p><p>1714</p><p>101.4</p><p>P1 = 68.2kPa</p><p>(Slightly throttled from Patm = 101 kPa)</p><p>3</p><p>2.2) The Lenoir air cycle is composed of three processes: 1-2 constant volume heat addition, 2-3 isentropic</p><p>expansion, and 3-1 constant pressure heat rejection. This cycle is named after Jean Lenoir (1822-1900),</p><p>a Belgian engineer who developed an internal combustion engine in 1858. It is a cycle in which com-</p><p>bustion occurs without compression of the mixture. (a) Draw the Lenoir cycle on p − V and T − s</p><p>diagrams, (b) Assuming the working fluid is an ideal gas with constant properties, derive an expression</p><p>for the thermal efficiency of the Lenoir air cycle, and (c) Compare the Lenoir cycle thermal efficiency</p><p>to the Otto cycle efficiency for standard atmospheric pressure and temperature, Qin = 1000 J, r = 8,</p><p>m = 1.0 g, and γ = 1.4.</p><p>The p− V and T − s diagrams are shown below:</p><p>3</p><p>2</p><p>1</p><p>Volume</p><p>P</p><p>re</p><p>s</p><p>s</p><p>u</p><p>re</p><p>Figure 2.2: Problem 2-2 p− V plot</p><p>3</p><p>2</p><p>1</p><p>Entropy</p><p>T</p><p>e</p><p>m</p><p>p</p><p>e</p><p>ra</p><p>tu</p><p>re</p><p>Figure 2.3: Problem 2-2 T − s plot</p><p>4 CHAPTER 2. HEAT ENGINE CYCLES</p><p>Process 1-2: Constant volume heat addition</p><p>T2 =</p><p>Q1−2</p><p>mcv</p><p>+ T1</p><p>=</p><p>1000</p><p>(1)(0.72)</p><p>+ 298</p><p>= 1687K</p><p>Process 2-3: Isentropic expansion</p><p>T3 = T2 (1/r)</p><p>γ−1</p><p>= 1687(1/8)0.4</p><p>= 734K</p><p>The Lenoir cycle thermal efficiency is</p><p>η =</p><p>Wout</p><p>Qin</p><p>= 1−</p><p>Qout</p><p>Qin</p><p>= 1−</p><p>mcp(T3 − T1)</p><p>mcv(T2 − T1)</p><p>= 1− γ</p><p>(T3 − T1)</p><p>(T2 − T1)</p><p>= 1− 1.4</p><p>(734− 298)</p><p>(1687− 298)</p><p>= 0.56</p><p>The Otto cycle efficiency is η = 1− r1−γ= 0.565, slightly greater than the Lenoir cycle.</p><p>5</p><p>2.3) (a) Show for an Otto cycle that T3/T2 = T4/T1. (b) Derive the Otto cycle efficiency equation, Equation</p><p>(2.10).</p><p>3</p><p>2</p><p>1</p><p>4</p><p>Volume (V/V2)</p><p>P</p><p>re</p><p>s</p><p>s</p><p>u</p><p>re</p><p>(</p><p>P</p><p>/P</p><p>1</p><p>)</p><p>Figure 2.4: Problem 2-3 P-V plot</p><p>a) From the cycle analysis of Section 2.2; for an isentropic compression:</p><p>T2</p><p>T1</p><p>= rγ−1</p><p>T4</p><p>T3</p><p>=</p><p>(</p><p>1</p><p>r</p><p>)γ−1</p><p>(</p><p>T2</p><p>T1</p><p>)</p><p>·</p><p>(</p><p>T4</p><p>T3</p><p>)</p><p>= 1</p><p>T3</p><p>T2</p><p>=</p><p>T4</p><p>T1</p><p>b) The thermal efficiency is</p><p>η = 1−</p><p>Qin</p><p>Qout</p><p>Qout = mc (T4 − T3)</p><p>Qin = mc (T3 − T2)</p><p>Substituting we obtain</p><p>η = 1−</p><p>Qin</p><p>Qout</p><p>= 1−</p><p>T4 − T1</p><p>T3 − T2</p><p>= 1−</p><p>T1</p><p>(</p><p>T4</p><p>T1</p><p>− 1</p><p>)</p><p>T2</p><p>(</p><p>T3</p><p>T2</p><p>− 1</p><p>)</p><p>η = 1−</p><p>T1</p><p>T2</p><p>6 CHAPTER 2. HEAT ENGINE CYCLES</p><p>2.4) Derive the Otto and Diesel cycle imep equation, Equation (2.11).</p><p>The imep is defined as:</p><p>imep =</p><p>W</p><p>Vd</p><p>W = η ·Qin</p><p>r =</p><p>Vc + Vd</p><p>Vc</p><p>−→ Vc =</p><p>Vd</p><p>r − 1</p><p>V1 = Vc + Vd = Vd</p><p>(</p><p>r</p><p>r − 1</p><p>)</p><p>imep</p><p>P1</p><p>=</p><p>(</p><p>1</p><p>P1</p><p>)(</p><p>η ·Qin</p><p>V1</p><p>)(</p><p>r</p><p>r − 1</p><p>)</p><p>7</p><p>2.5) For equal maximum temperature and heat input, which cycle will be more efficient, the Diesel or Otto?</p><p>Prove your answer by comparing the two cycles on the T − s diagram. The two cycles should have a</p><p>common state corresponding to the start of compression.</p><p>The two cycles are plotted on the Figure below, labeled as Diesel (d) and Otto (o). The common start</p><p>of compression is labeled point 1. If the Diesel cycle is assumed to have a greater compression ratio</p><p>than the Otto cycle, then T2d > T2o. The constant pressure heat addition line for the Diesel cycle has</p><p>a smaller slope than the constant volume heat addition line for the Otto cycle, since constant pressure</p><p>processes will have a smaller temperature rise than constant volume processes for the same heat input.</p><p>The Figure indicates that s3d</p><p>and 320 K, the non-dimensional heat input</p><p>Qin/P1V1 = 30, and γ=1.3. Find the thermal efficiency and the values of α and β.</p><p>The limited pressure parameters are α and β</p><p>α =</p><p>1</p><p>rγ</p><p>P3</p><p>P1</p><p>=</p><p>(</p><p>1</p><p>171.3</p><p>)(</p><p>8000</p><p>101</p><p>)</p><p>α = 1.99</p><p>β = 1 +</p><p>γ − 1</p><p>αγ</p><p>[</p><p>Qin</p><p>P1V1</p><p>·</p><p>1</p><p>r γ−1</p><p>−</p><p>α− 1</p><p>γ − 1</p><p>]</p><p>= 1 +</p><p>1.3− 1</p><p>(1.99)(1.3)</p><p>[</p><p>30</p><p>(</p><p>1</p><p>170.3</p><p>)</p><p>−</p><p>(1.99− 1)</p><p>(1.33− 1)</p><p>]</p><p>β = 2.104</p><p>η = 1−</p><p>1</p><p>rγ−1</p><p>[</p><p>αβγ</p><p>− 1</p><p>α− 1 + αγ(β − 1)</p><p>]</p><p>= 1−</p><p>1</p><p>170.3</p><p>[</p><p>(1.99)(2.104)1.3 − 1</p><p>(1.99− 1) + (1.3)(1.99)(2.104− 1)</p><p>]</p><p>η = 0.53</p><p>13</p><p>2.11) (a) Derive the equation for the Miller cycle efficiency, Equation (2.23). (b) Derive the equation for the</p><p>Miller cycle imep, Equation (2.24).</p><p>We need to get the η and imep as a function of λ and γ. Using a state by state cycle analysis</p><p>1-2 T2 = T1rc</p><p>γ−1</p><p>2-3 Qin = mcv(T3 − T2)</p><p>3-4 T4 = T3re</p><p>1−γ</p><p>4-1 Qout = mcv(T4 − T5) +mcp(T5 − T1)</p><p>a) Miller cycle efficiency derivation:</p><p>λ =</p><p>re</p><p>rc</p><p>=</p><p>V5</p><p>V3</p><p>V1</p><p>V2</p><p>=</p><p>V5</p><p>V1</p><p>=</p><p>P5V5</p><p>P1V1</p><p>=</p><p>RT5</p><p>RT1</p><p>=</p><p>T5</p><p>T1</p><p>−→ T5 = λT1</p><p>η = 1−</p><p>Qout</p><p>Qin</p><p>=</p><p>1−mcv(T4 − T5) +mcp(T5 − T1)</p><p>Qin</p><p>=</p><p>1− (T4 − T5) + γ(T5 − T1)</p><p>Qin</p><p>mcv</p><p>T3 = T2 +</p><p>Qin</p><p>mcv</p><p>= T1rc</p><p>γ−1 +</p><p>Qin</p><p>mcv</p><p>T4 = T1λ</p><p>1−γ +</p><p>Qin</p><p>mcv</p><p>re</p><p>1−γ</p><p>Solving for efficiency:</p><p>η = 1− re</p><p>1−γ</p><p>−</p><p>[</p><p>λ1−γ</p><p>− λ+ γ(λ− 1)</p><p>Qin</p><p>cvT1</p><p>]</p><p>= 1− (λrc)</p><p>1−γ</p><p>−</p><p>[</p><p>λ1−γ</p><p>− λ(1 − γ)− γ</p><p>Qin</p><p>cvT1</p><p>]</p><p>η = 1− (λrc)</p><p>1−γ</p><p>−</p><p>[</p><p>λ1−γ</p><p>− λ(1− γ)− γ</p><p>Qin</p><p>P1V1</p><p>(γ − 1)</p><p>]</p><p>b) Miller cycle imep derivation:</p><p>imep =</p><p>W</p><p>Vd</p><p>W = η ·Qin</p><p>Vd = V5 − V2 = re ·</p><p>V1</p><p>rc</p><p>−</p><p>V1</p><p>rc</p><p>= V1</p><p>(</p><p>λrc − 1</p><p>rc</p><p>)</p><p>imep</p><p>P1</p><p>=</p><p>(</p><p>Qin</p><p>P1V1</p><p>)(</p><p>rc</p><p>λrc − 1</p><p>)</p><p>(η)</p><p>14 CHAPTER 2. HEAT ENGINE CYCLES</p><p>2.12) For Otto and Miller cycles that have equal compression ratios, rc = 10, what are the respective thermal</p><p>efficiencies and non-dimensional imeps? Assume that the parameter, λ is equal to 1.5 for the Miller</p><p>cycle, the specific heat ratio γ = 1.3, and Qin/P1V1 = 30.</p><p>a) The Otto thermal efficiency is:</p><p>ηotto = 1− rc</p><p>1−γ = 1− (10)−0.3</p><p>ηotto = 0.50</p><p>The Miller thermal efficiency is:</p><p>ηmiller = 1− (λrc)</p><p>1−γ</p><p>−</p><p>(</p><p>λ1−γ</p><p>− λ(1− γ)− γ</p><p>γ − 1</p><p>)(</p><p>P1V1</p><p>Qin</p><p>)</p><p>= 1− (15)−0.3</p><p>−</p><p>(</p><p>1.5−0.3</p><p>− 1.5(−0.3)− 1.3</p><p>(0.3)(30)</p><p>)</p><p>ηmiller = 0.55</p><p>The Miller cycle has a 10% greater efficiency</p><p>b) The Otto imep is:</p><p>(</p><p>imep</p><p>P1</p><p>)</p><p>otto</p><p>= ηotto</p><p>(</p><p>Qin</p><p>P1V1</p><p>)(</p><p>rc</p><p>rc − 1</p><p>)</p><p>= (0.50)(30)</p><p>(</p><p>10</p><p>9</p><p>)</p><p>(</p><p>imep</p><p>P1</p><p>)</p><p>otto</p><p>= 16.6</p><p>The Miller imep is:</p><p>(</p><p>imep</p><p>P1</p><p>)</p><p>miller</p><p>= ηmiller</p><p>(</p><p>Qin</p><p>P1V1</p><p>)(</p><p>rc</p><p>λrc − 1</p><p>)</p><p>= (0.55)(30)</p><p>(</p><p>10</p><p>14</p><p>)</p><p>(</p><p>imep</p><p>P1</p><p>)</p><p>miller</p><p>= 11.8</p><p>The Miller cycle has a 29% lower imep.</p><p>15</p><p>2.13) Develop a complete expansion cycle model in which the expansion stroke continues until the pressure</p><p>is atmospheric. Derive an expression for the efficiency in terms of γ, α = V4/V3, and β = V1/V4.</p><p>a) With reference to Figure 2.7, the thermal efficiency is defined as</p><p>η = 1−</p><p>Qout</p><p>Qin</p><p>= 1−</p><p>cp(T4 − T1)</p><p>cv(T3 − T2)</p><p>Since heat addition is at constant volume and heat rejection is at constant pressure</p><p>η = 1− γ</p><p>[</p><p>T4(1−</p><p>T1</p><p>T4</p><p>)</p><p>T3(1−</p><p>T2</p><p>T3</p><p>)</p><p>]</p><p>For the isentropic expansion from 1 to 2 and from 3 to 4</p><p>T2</p><p>T1</p><p>=</p><p>(</p><p>V1</p><p>V2</p><p>)γ−1</p><p>=</p><p>(</p><p>V1</p><p>V4</p><p>·</p><p>V4</p><p>V2</p><p>)γ−1</p><p>= (βα)</p><p>γ−1</p><p>T4</p><p>T3</p><p>=</p><p>(</p><p>V3</p><p>V4</p><p>)γ−1</p><p>=</p><p>(</p><p>1</p><p>α</p><p>)γ−1</p><p>Simplifying</p><p>T2</p><p>T1</p><p>·</p><p>T4</p><p>T3</p><p>= βγ−1</p><p>T2</p><p>T3</p><p>= βγ−1</p><p>(</p><p>T1</p><p>T4</p><p>)</p><p>Since P1 =P4, from the ideal gas law</p><p>T1</p><p>V1</p><p>=</p><p>T4</p><p>V4</p><p>−→</p><p>T1</p><p>T4</p><p>=</p><p>V1</p><p>V4</p><p>= β</p><p>So the thermal efficiency can be expressed as</p><p>η = 1−</p><p>( γ</p><p>αγ−1</p><p>) (1− β)</p><p>(1− βγ)</p><p>16 CHAPTER 2. HEAT ENGINE CYCLES</p><p>2.14) If a four-cylinder, four-stroke engine with a 0.1 m bore and an 0.08 m stroke operating at 2000 rpm</p><p>has the same heat/mass loss parameters as Example 2.3, how much indicated power (kW) would it</p><p>produce? What if it were a two-stroke engine?</p><p>The engine speed ω =N * 2π/60 = 2000 rev/min * 2π/60 = 209.4 rad/s. Using the FiniteHeatMassLoss.m</p><p>program as shown below, imep/P1=11.55, so imep= 11.55 e+5 kPa.</p><p>The displacement volume Vd = nc ∗ (π/4) b</p><p>2s = 4(π/4) (0.1)2(0.08) = 2.51e− 3 m3.</p><p>The indicated power Ẇi is</p><p>Ẇi = imepVd N/2</p><p>= (11.55e+ 5)(2.51e− 3)</p><p>2000</p><p>2 60</p><p>= 48, 320W</p><p>= 48 kW</p><p>If the engine is two stroke, the indicated power would double to 96 kW.</p><p>The program input is:</p><p>function [ ] = FiniteHeatMassLoss( )</p><p>% Gas cycle heat release code with and w/o heat transfer</p><p>% data structure for engine parameters</p><p>clear();</p><p>thetas = -20; % start of heat release (deg)</p><p>thetad = 40; % duration of heat release (deg)</p><p>r =10; % compression ratio</p><p>gamma = 1.4; % gas const</p><p>Q = 20.; % dimensionless total heat release</p><p>h = 0.2; % dimensionless ht coefficient</p><p>tw = 1.2; % dimensionless cylinder wall temp</p><p>beta = 1.5; % dimensionless volume</p><p>a = 5; % weibe parameter a</p><p>n = 3; % weibe exponent n</p><p>omega =209.4; % engine speed rad/s</p><p>c = 0.8; % mass loss coeff</p><p>The program output is:</p><p>Weibe Heat Release with Heat and Mass Loss</p><p>Theta_start = -20.00</p><p>Theta_dur = 40.00</p><p>P_max/P1 = 85.34</p><p>Theta @P_max = 11.0</p><p>P_tdc/P_max = 0.72</p><p>Net Work/P1V1 = 10.40</p><p>Heat Loss/P1V1 = 4.45</p><p>Mass Loss/m = 0.024</p><p>Efficiency = 0.520</p><p>Eff./Eff. Otto = 0.864</p><p>Imep/P1 = 11.55</p><p>17</p><p>2.15) Using the program BurnFraction.m, and assuming that a = 5, the beginning of heat addition is -10◦ ,</p><p>and the duration of heat addition is 40◦ , (a) Plot the Weibe heat release fraction curve for the following</p><p>form factor values: n = 2, 3, and 4. (b) At what crank angle is 0.10, 0.50, and 0.90 of the heat released?</p><p>a.) The problem data is entered into the program BurnFraction.m and the resulting cumulative and</p><p>rate of burn fraction plots are shown below.</p><p>−10 0 10 20 30</p><p>0</p><p>0.2</p><p>0.4</p><p>0.6</p><p>0.8</p><p>1</p><p>Crank Angle (deg)</p><p>C</p><p>um</p><p>ul</p><p>at</p><p>iv</p><p>e</p><p>B</p><p>ur</p><p>n</p><p>F</p><p>ra</p><p>ct</p><p>io</p><p>n</p><p>n=2</p><p>n=3</p><p>n=4</p><p>Figure 2.6: Problem 2-15 Burn fraction plot</p><p>−10 0 10 20 30</p><p>0</p><p>0.5</p><p>1</p><p>1.5</p><p>2</p><p>2.5</p><p>Crank Angle (deg)</p><p>B</p><p>ur</p><p>n</p><p>R</p><p>at</p><p>e</p><p>(1</p><p>/d</p><p>eg</p><p>)</p><p>n=2</p><p>n=3</p><p>n=4</p><p>Figure 2.7: Problem 2-15 Rate of burn fraction plot</p><p>b.) As the form factor increases, the crank angles for the 0.1, 0.5, and 0.9 burn fractions all increase.</p><p>For n=2, the 0.1 fraction is at -3.9 degrees, the 0.5 fraction at 4.9 degrees, and the 0.9 fraction at</p><p>17.5 degrees. For n=3, the 0.1 fraction is at 1.3 degrees, the 0.5 fraction at 11.0 degrees, and the 0.9</p><p>fraction at 21.1 degrees. For n=4, the 0.1 fraction is at 3.4 degrees, the 0.5 fraction at 14.6 degrees,</p><p>18 CHAPTER 2. HEAT ENGINE CYCLES</p><p>and the 0.9 fraction at 23.1 degrees.</p><p>The program, modified for Problem 2.15 is:</p><p>function [ ]=BurnFraction2_15( )</p><p>% this program computes and plots the cumulative burn fraction</p><p>% and the instantanous burnrate</p><p>clear();</p><p>a = 5; % Weibe efficiency factor</p><p>thetas = -10; % start of combustion</p><p>thetad = 40; % duration of combustion</p><p>theta=linspace(thetas,thetas+thetad,100); %crankangle theta vector</p><p>dum=(theta-thetas)/thetad; % theta diference vector</p><p>i1=0;</p><p>n = 2; % Weibe form factor</p><p>temp=-a*dum.^n;</p><p>xb1=1.-exp(temp); %burn fraction</p><p>dxb1=n*a*(1-xb1).*dum.^(n-1); %element by element vector multiplication</p><p>n = 3; % Weibe form factor</p><p>temp=-a*dum.^n;</p><p>xb2=1.-exp(temp); %burn fraction</p><p>dxb2=n*a*(1-xb2).*dum.^(n-1);</p><p>n = 4; % Weibe form factor</p><p>temp=-a*dum.^n;</p><p>xb3=1.-exp(temp); %burn fraction</p><p>dxb3=n*a*(1-xb3).*dum.^(n-1);</p><p>%get heat release angles</p><p>for i=1:100</p><p>test=xb3(i);</p><p>if test > 0.1 i1 = i; break</p><p>end</p><p>end</p><p>%get heat release angles</p><p>for i=1:100</p><p>test=xb3(i);</p><p>if test > 0.5 i2 = i; break</p><p>end</p><p>end</p><p>%get heat release angles</p><p>for i=1:100</p><p>test=xb3(i);</p><p>if test > 0.9 i3 = i; break</p><p>end</p><p>end</p><p>fprintf (’crank angle for 0.1 burn fraction is %6.1f degrees \n’, theta(i1));</p><p>fprintf (’crank angle for 0.5 burn fraction is %6.1f degrees \n’, theta(i2));</p><p>fprintf (’crank angle for 0.9 burn fraction is %6.1f degrees \n’, theta(i3));</p><p>%plot results</p><p>plot(theta,xb1,’-’,theta,xb2,’--’,theta,xb3,’:’,’linewidth’,2);</p><p>set(gca, ’fontsize’, 18,’linewidth’,2);</p><p>%grid</p><p>xlabel(’Crank Angle (deg)’,’fontsize’, 18);</p><p>ylabel(’Cumulative Burn Fraction’,’fontsize’, 18);</p><p>legend(’n=2’,’n=3’,’n=4’);</p><p>19</p><p>figure();</p><p>plot(theta,dxb1,’-’,theta,dxb2,’--’,theta,dxb3,’:’,’linewidth’,2);</p><p>set(gca, ’fontsize’, 18,’linewidth’,2);</p><p>%grid</p><p>xlabel(’Crank Angle (deg)’,’fontsize’, 18);</p><p>ylabel(’Burn Rate (1/deg)’,’fontsize’, 18);</p><p>legend(’n=2’,’n=3’,’n=4’);</p><p>end</p><p>20 CHAPTER 2. HEAT ENGINE CYCLES</p><p>2.16) Using the program FiniteHeatRelease.m, determine the effect of heat release duration on the net</p><p>work, power, mean effective pressure, and thermal efficiency for a four stroke engine with heat release</p><p>durations of 40, 30, 20, 10, and 5 degrees. Assume that the total heat addition Qin = 2500 J, the start</p><p>of heat release θs remains constant at -10◦ atdc, a = 5, n = 3, and γ = 1.4. The engine bore and stroke</p><p>are 0.095 m, the compression ratio is 9:1, and engine speed is 3000 rpm.</p><p>The table below shows the effect of heat release duration. For a heat release model, the work, imep, and</p><p>efficiency will increase as the energy release duration decreases, and the cycle approaches a constant</p><p>volume Otto cycle. The Figure shows the greater peak pressure with the shorter 20 degree duration of</p><p>Engine 1 relative to the longer 40 degree duration of Engine 2.</p><p>However, if the heat release duration occurs only during the compression stroke, as for the 10 and 5</p><p>degree duration points, this will not be the case.</p><p>θd Wi/P1V1 imep/P1 η</p><p>+40 18.75 21.09 0.568</p><p>+30 19.09 21.47 0.578</p><p>+20 19.25 21.65 0.583</p><p>+10 19.21 21.61 0.582</p><p>+5 19.11 21.50 0.579</p><p>Table for Problem 2.16</p><p>−200 −100 0 100 200</p><p>0</p><p>20</p><p>40</p><p>60</p><p>80</p><p>100</p><p>120</p><p>140</p><p>Theta (deg)</p><p>P</p><p>re</p><p>ss</p><p>ur</p><p>e</p><p>(b</p><p>ar</p><p>)</p><p>Engine 1</p><p>Engine 2</p><p>Figure 2.8: Problem 2-16 Pressure profile plot</p><p>The input to and output of the program FiniteHeatRelease.m is shown below.</p><p>function[ ]=FiniteHeatRelease( )</p><p>% Gas cycle heat release code for two engines</p><p>% engine parameters</p><p>clear();</p><p>thetas(1,1)= -10; % Engine1 start of heat release (deg)</p><p>thetas(2,1)= -10; % Engine2 start of heat release (deg)</p><p>thetad(1,1) = 20; % Engine1 duration of heat release (deg)</p><p>21</p><p>thetad(2,1) = 40; % Engine2 duration of heat release (deg)</p><p>r=9; %compression ratio</p><p>gamma= 1.4; %gas const</p><p>q= 33.0; % dimensionless total heat release Qin/P1V1</p><p>a= 5; %weibe parameter a</p><p>n= 3; %weibe exponent n</p><p>The program output is</p><p>Engine 1 Engine 2</p><p>Theta_start -10.00 -10.00</p><p>Theta_dur 20.00 40.00</p><p>P_max/P_1 130.55 95.33</p><p>Theta_max 8.0 20.0</p><p>Net Work/P1V1 19.25 18.75</p><p>Efficiency 0.583 0.568</p><p>Eff. Ratio 0.997 0.972</p><p>Imep/P1 21.65 21.09</p><p>22 CHAPTER 2. HEAT ENGINE CYCLES</p><p>2.17) If a four cylinder unthrottled Otto cycle engine is to generate 100 kW at an engine speed of 2500 rpm,</p><p>what should its bore and stroke be? Assume a square block engine with equal bore and stroke and a</p><p>compression ratio of 10:1. The total heat addition Qin = 2200 J, the start of heat release θs remains</p><p>constant at -15◦ atdc, the combustion duration is 40◦ , a = 5, n = 3, and γ = 1.4. Use the single</p><p>cylinder program FiniteHeatRelease.m, and solve for a power output of 25 kW.</p><p>Since the program FiniteHeatRelease.m is non dimensional, we choose Qin/P1V1= 20, as that is</p><p>the value used in Example 2.3. Please note that this solution does not use the Qin value given. The</p><p>program input and output is shown below, resulting in Wi/P1V1= 11.85.</p><p>Therefore,</p><p>V1 =</p><p>Wi</p><p>P1 11.85</p><p>=</p><p>1200</p><p>105 11.85</p><p>= 1.01× 10−3m3</p><p>Since V1 = Vd r/(r − 1),</p><p>b =</p><p>r − 1</p><p>r</p><p>V1</p><p>4</p><p>π</p><p>=</p><p>10− 1</p><p>10</p><p>1.01× 10−3 4</p><p>π</p><p>= 0.105m</p><p>function [ ]=FiniteHeatRelease( )</p><p>% Gas cycle heat release code for two engines</p><p>% engine parameters</p><p>clear();</p><p>thetas(1,1)= -15; % Engine1 start of heat release (deg)</p><p>thetas(2,1)= -15; % Engine2 start of heat release (deg)</p><p>thetad(1,1) = 40; % Engine1 duration of heat release (deg)</p><p>thetad(2,1) = 40; % Engine2 duration of heat release (deg)</p><p>r=10; %compression ratio</p><p>gamma= 1.4; %gas const</p><p>q= 20; % dimensionless total heat release Qin/P1V1</p><p>a= 5; %weibe parameter a</p><p>n= 3; %weibe exponent n</p><p>Engine 1 Engine 2</p><p>Theta_start -15.00 -15.00</p><p>Theta_dur 40.00 40.00</p><p>P_max/P_1 78.98 78.98</p><p>Theta_max 15.0 15.0</p><p>Net Work/P1V1 11.85 11.85</p><p>Efficiency 0.592 0.592</p><p>Eff. Ratio 0.984 0.984</p><p>Imep/P1 13.16 13.16</p><p>23</p><p>2.18) Develop a four-stroke Diesel cycle model (along the lines used in Example 2.4).Compute the net cycle</p><p>efficiency and the net imep for an engine with r = 22, γ = 1.3, Ti = 300K, Pi = 101kPa, Pi/Pe =</p><p>0.98,M = 29, and qin = 2090 kJ/kggas.</p><p>The program FourStrokeOtto.m was modified into FourStrokeDiesel.m for this problem. The con-</p><p>stant volume heat addition was changed to a constant pressure heat addition, and the efficiency η</p><p>redefined accordingly, as shown below. The output is given below. With a compression ratio r of 22,</p><p>the residual fraction f = 0.015, and the thermal efficiency η = 0.51.</p><p>Four Stroke Diesel Cycle</p><p>State 1 2 3 4</p><p>Pressure (kPa): 101.0 5616.6 5616.6 443.3</p><p>Temperature (K): 309.9 783.2 2443.8 1360.2</p><p>Ideal Thermal Eff.= 0.513 Net Thermal Eff.= 0.513</p><p>Exhaust Temp. (K)= 971.2 Volumetric Eff.= 1.00</p><p>Residual Fraction 0.015 Net Imep (bar)= 12.56</p><p>% Four stroke Diesel cycle model</p><p>% input parameters</p><p>Ti = 300; % inlet temperature, K</p><p>Pi = 101; % inlet pressure, kPa</p><p>Pe = 103; % exhaust pressure, kPa</p><p>r = 22; % compression ratio</p><p>qin = 2090; % heat input, kJ/kg (mixture)</p><p>gamma = 1.3; % ideal gas specific heat ratio</p><p>R = 0.287; % gas constant kJ/kg K</p><p>cp= 1.24; %const vol specific heat, kJ/kg K</p><p>f=0.05;% guess value of residual fraction f</p><p>Te = 1000; % guess value of exhaust temp, K</p><p>tol=0.0001; % tolerance for convergence</p><p>err = 2*tol; %error initialization</p><p>gam=(gamma -1)/gamma;</p><p>while (err > tol) %while loop for cycle calc</p><p>%intake stroke</p><p>T1=(1-f)*Ti + f*(1 - (1- Pi/Pe)*gam)*Te;</p><p>P1=Pi;</p><p>%isentropic compression</p><p>P2=P1*r^gamma;</p><p>T2=T1*r^(gamma-1);</p><p>%const p heat addition</p><p>T3=T2 + qin*(1-f)/cp;</p><p>P3=P2;</p><p>%isentropic expansion</p><p>beta = T3/T2;</p><p>P4=P3*(beta/r)^gamma;</p><p>T4=T3*(beta/r)^(gamma-1);</p><p>%isentropic blowdown</p><p>P5=Pe;</p><p>T5=T4*(P4/Pe)^(-gam);</p><p>%const p exhaust stroke</p><p>Te=T5;</p><p>fnew=(1/r)*(Pe/P4)^(1/gamma); %new residual fraction</p><p>err=abs(fnew-f)/fnew;</p><p>f=fnew;</p><p>24 CHAPTER 2. HEAT ENGINE CYCLES</p><p>end</p><p>%cycle parameters</p><p>eta= 1 - 1/r^(gamma-1)*(beta^gamma -1)/(gamma*(beta-1));</p><p>imep = P1*(qin*(1-f)/(R*T1))/(1-(1/r))*eta;</p><p>pmep=Pe-Pi;</p><p>etanet= eta*(1-pmep/imep);</p><p>imepnet= (imep-pmep)/100.;</p><p>voleff=1-(Pe/Pi -1)/(gamma*(r-1));</p><p>%output calcs</p><p>fprintf(’ \nFour Stroke Diesel Cycle \n’)</p><p>fprintf(’State 1 2 3 4 \n’);</p><p>fprintf(’Pressure (kPa): %6.1f %6.1f %6.1f %6.1f \n’,P1,P2,P3,P4);</p><p>fprintf(’Temperature (K): %6.1f %6.1f %6.1f %6.1f \n’,T1,T2,T3,T4);</p><p>fprintf(’Ideal Thermal Eff.= %6.3f Net Thermal Eff.= %6.3f \n’,eta, etanet);</p><p>fprintf(’Exhaust Temp. (K)= %6.1f Volumetric Eff.= %6.2f \n’,Te, voleff);</p><p>fprintf(’Residual Fraction %6.3f Net Imep (bar)= %6.2f \n’,f, imepnet);</p><p>Four Stroke Diesel Cycle</p><p>State 1 2 3 4</p><p>Pressure (kPa): 101.0 5616.6 5616.6 443.8</p><p>Temperature (K): 309.6 782.7 2443.8 1360.5</p><p>Ideal Thermal Eff.= 0.513 Net Thermal Eff.= 0.514</p><p>Exhaust Temp. (K)= 964.6 Volumetric Eff.= 1.00</p><p>Residual Fraction 0.014 Net Imep (bar)= 12.60</p><p>25</p><p>2.19) Using the program FourStrokeOtto.m, plot the effect of inlet throttling from 100 kPa to 25 kPa on the</p><p>peak pressure, P3, and the volumetric efficiency ηv. Assume the following conditions: Ti = 300K, r =</p><p>9, γ = 1.3, and qin = 2400 kJ/kggas.</p><p>The inlet pressure was varied as shown in the program input below, and the peak pressure and volu-</p><p>metric efficiency plotted below. There is a linear relationship between peak and inlet pressure. The</p><p>volumetric efficiency rises in a non-linear fashion from 0.62 to 1.00.</p><p>20 40 60 80 100</p><p>1000</p><p>2000</p><p>3000</p><p>4000</p><p>5000</p><p>6000</p><p>7000</p><p>8000</p><p>9000</p><p>Inlet Pressure (kPa)</p><p>P</p><p>ea</p><p>k</p><p>P</p><p>re</p><p>ss</p><p>ur</p><p>e</p><p>(k</p><p>P</p><p>a)</p><p>Figure 2.9: Problem 2-19 Peak pressure plot</p><p>20 40 60 80 100</p><p>0.65</p><p>0.7</p><p>0.75</p><p>0.8</p><p>0.85</p><p>0.9</p><p>0.95</p><p>1</p><p>Inlet Pressure (kPa)</p><p>V</p><p>ol</p><p>um</p><p>et</p><p>ric</p><p>E</p><p>ffi</p><p>ci</p><p>en</p><p>cy</p><p>Figure 2.10: Problem 2-19 Volumetric Efficiency plot</p><p>% Four stroke Otto cycle model</p><p>% input parameters</p><p>Ti = 300; % inlet temperature, K</p><p>Pi = 28; % inlet pressure, kPa</p><p>Pe = 100; % exhaust pressure, kPa</p><p>26 CHAPTER 2. HEAT ENGINE CYCLES</p><p>r = 9; % compression ratio</p><p>qin = 2400; % heat input, kJ/kg (mixture)</p><p>gamma = 1.3; % ideal gas specific heat ratio</p><p>R = 0.287; % gas constant kJ/kg K</p><p>cv= R/(gamma-1); %const vol specific heat, kJ/kg K</p><p>f=0.05;% guess value of residual fraction</p><p>f</p><p>Te = 1000; % guess value of exhaust temp, K</p><p>tol=0.0001; % tolerance for convergence</p><p>err = 2*tol; %error initialization</p><p>gam=(gamma -1)/gamma;</p><p>27</p><p>2.20) In Example 2.3, Te is the exhaust temperature during the constant pressure exhaust stroke. It is not</p><p>the same as the average temperature of the gases exhausted. Explain.</p><p>Gases at temperatures greater than Te are also exhausted during the blowdown phase. The average</p><p>temperature of the gases leaving the engine depend on the definition of average. A time average, a</p><p>mass flow rate average, and enthalpy average are all different, since the mass and enthalpy flow varies</p><p>with crank angle.</p>