Engineering_Mechanics_Colin_R_Ferguson_A_Solved_Ch5

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<p>Chapter 5</p><p>Intake and Exhaust Flow</p><p>5.1) If an engine has a bore of 0.1 m, stroke of 0.08 m, inlet flow effective area of 4.0× 10−4 m2, and inlet</p><p>temperature of 320 K, what is the maximum speed it is intended to be operated while maintaining</p><p>good volumetric efficiency?</p><p>The Mach Index Z should be ≤ 0.6, so:</p><p>Up ≤</p><p>Ai · ci</p><p>1.3 · b2</p><p>The speed of sound in the inlet flow is</p><p>ci = (γRTi)</p><p>1/2</p><p>= [(1.4)(287)(320)]</p><p>1/2</p><p>= 358.6m/s</p><p>so</p><p>Up ≤</p><p>(</p><p>4.0× 10−4</p><p>)</p><p>(358.6)</p><p>(1.3)(1)2</p><p>= 11.0m/s</p><p>Solving for the speed</p><p>N =</p><p>Up</p><p>2s</p><p>=</p><p>11</p><p>(2)(0.08)</p><p>= 68.9 rev/s</p><p>N = 4138 rpm</p><p>1</p><p>2 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.2) Explain how unburned fuel can appear in the exhaust during the intake and exhaust strokes.</p><p>Fuel air mixtures can flow into the exhaust system via a short-circuit during the intake and exhaust</p><p>valve overlap period if the pressure conditions are such that</p><p>Pintake port > Pcylinder > Pexhaust port</p><p>Such conditions can exist with superchargers.</p><p>3</p><p>5.3) Combustion gases (γ= 1.3, R = 280 J/kg K) exit through the exhaust port of a two stroke engine</p><p>during blowdown. The exhaust port geometry can be modeled as a converging nozzle with a port</p><p>diameter of 2 cm. The cylinder gases are initially at 200 kPa and 393 K, and Patm = 101 kPa. What</p><p>is the initial velocity and mass flow rate of the exhaust flow ?</p><p>The initial pressure ratio is Po/Patm = 200/101 = 1.98</p><p>The critical pressure ratio for choked flow is</p><p>Po</p><p>P</p><p>=</p><p>[</p><p>1 +</p><p>(</p><p>γ − 1</p><p>2</p><p>)</p><p>M2</p><p>]γ/(γ−1)</p><p>=</p><p>(</p><p>γ + 1</p><p>2</p><p>)γ/(γ−1)</p><p>= 1.83</p><p>(5.1)</p><p>Therefore the initial flow will be choked and the flow sonic at the exit, with M = 1. The exit</p><p>temperature Te is</p><p>To</p><p>Te</p><p>= 1 +</p><p>(</p><p>γ − 1</p><p>2</p><p>)</p><p>M2</p><p>= 1.15</p><p>(5.2)</p><p>The initial exit velocity Ve is equal to the stagnation sound speed co</p><p>Ve = co = (γRTo)</p><p>1/2 = (1.3× 280× 342)1/2 = 352m/s (5.3)</p><p>The stagnation density is</p><p>ρo =</p><p>Po</p><p>RTo</p><p>=</p><p>200000</p><p>(280) (393)</p><p>= 1.817 kg/m</p><p>3</p><p>(5.4)</p><p>The density ratio ρo/ρe is</p><p>ρo</p><p>ρe</p><p>=</p><p>(</p><p>To</p><p>T</p><p>)γ/(γ−1)</p><p>=</p><p>393</p><p>342</p><p>= 1.58</p><p>(5.5)</p><p>So ρe = 1.817/1.58 = 1.15 kg/m3</p><p>The mass flow rate, ṁ, through the port is</p><p>ṁ = ρeAeVe = (1.15)(π/4)(0.022)(352) = 0.127 kg/s (5.6)</p><p>4 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.4) It was explained in the chapter that because of the pressure drop across a valve, it is advantageous to</p><p>close the intake valve after bottom dead center. Use the same logic to explain why exhaust valves are</p><p>closed after top dead center and what the effect of engine speed is on the residual fraction.</p><p>Exhaust valves are closed after top dead center for two main reasons. The first reason is the fi-</p><p>nite opening and closing time of cam actuated poppet valves. Due to the mechanical inertia of the</p><p>valve train, and to avoid float, there is a limit on the acceleration and deceleration of the valve. This</p><p>implies that there is a minimum closing time required to close a poppet valve. To maximize the open</p><p>duration time of the valve during the exhaust stroke, it is therefore necessary to extend the closing</p><p>time past top dead center.</p><p>The second reason is the effect of finite engine speed. In the limit of zero engine speed, there would</p><p>be no pressure drop across the exhaust valve during the exhaust stroke. Thus after top dead center</p><p>when the piston begins to move downward, if the exhaust valve is open, there would be a flow reversal,</p><p>since the motion of the piston would pull the exhaust gas from the exhaust manifold to the cylinder.</p><p>As the engine speed increases, the pressure drop across the exhaust valve will increase, and the angle</p><p>of flow reversal will occur later and later past top dead center. The duration between top dead center</p><p>and the onset of flow reversal can be used to extend the exhaust process.</p><p>The residual fraction will be a minimum when the onset of flow reversal is coincident with the valve</p><p>closing. If the valve closing occurs earlier, then additional residuals will remain in the cylinder, in-</p><p>creasing the residual fraction. If the valve closing occurs later, then exhaust products will return to</p><p>the cylinder from the exhaust manifold, also increasing the residual fraction.</p><p>5</p><p>5.5) Suppose an engine were constructed with variable valve timing, thus ensuring optimum timing at all</p><p>speeds. Explain how the volumetric efficiency would depend on speed for wide-open throttle operation</p><p>with short pipes and Z As, the minimum flow area is the constant seat area As.</p><p>For d = 22 mm, Ac = Ad = 380 mm2. For d = 16 mm, Ac = Ad = 201 mm2. So for the two inlet valve</p><p>configuration, the total area is 2 x 201 = 402 mm2, a 6% increase, relative to the single inlet valve at</p><p>d = 22 mm.</p><p>As stated on p. 142, the use of multiple valves increases the valve area per unit piston area, and hence</p><p>the speed at which the engine power becomes flow limited. Heads are often wedge-shaped or domed to</p><p>increase the valve area to piston area, so that intake valve area to piston area ratios of up to 0.5 can</p><p>be obtained.</p><p>9</p><p>5.9) If the inlet Mach index in each case in Figure 5.11 is held to Zi = 0.6 and ci = 400 m/s, Ai =</p><p>0.35ni(π/4) d</p><p>2</p><p>i where ni = number of intake valves, then what would be the maximum piston speed in</p><p>each case?</p><p>Up,max =</p><p>Ai · ci</p><p>1.3 · b2</p><p>=</p><p>(0.35)(ni)</p><p>(</p><p>π</p><p>4</p><p>) (</p><p>di</p><p>2</p><p>)</p><p>(ci)</p><p>1.3 · b2</p><p>= 85.48(ni)</p><p>(</p><p>di</p><p>b</p><p>)2</p><p>a)</p><p>Up,max = (85.48)(0.194) = 16.4m/s</p><p>b)</p><p>Up,max = (85.48)(0.240) = 20.3m/s</p><p>c)</p><p>Up,max = (85.48)(0.218) = 18.4m/s</p><p>10 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.10) A four-stroke four cylinder square (bore = stroke) engine has a displacement volume of 5 L and</p><p>operates at 3000 rpm. The intake air temperature is 350 K, the intake manifold length is 1.25 m long,</p><p>and C̄f = 0.38. (a) For a Mach index Z = 0.6, what is the mean piston speed and average</p><p>effective</p><p>intake valve flow area ? (b) At what engine speed would the intake manifold be ”tuned” for increased</p><p>intake mass flow ?</p><p>a) The displacement volume is</p><p>Vd = nc</p><p>π</p><p>4</p><p>b2s = nc</p><p>π</p><p>4</p><p>b3</p><p>The bore and stroke are</p><p>b = s =</p><p>(</p><p>4 · Vd</p><p>nc · π</p><p>)1/3</p><p>=</p><p>(</p><p>4 · 5× 10−3</p><p>4 · π</p><p>)1/3</p><p>= 0.117m</p><p>The mean piston speed up is</p><p>up = 2(s)(N) = 2(0.117)</p><p>(</p><p>3000</p><p>60</p><p>)</p><p>= 11.6m/s</p><p>The average effective flow area Ai is</p><p>Ai =</p><p>1.3 · b2 · up</p><p>ci</p><p>=</p><p>(1.3)(0.117)2(11.6)</p><p>(1.4 · 287 · 350)1/2</p><p>= 5.5× 10−4m2</p><p>The intake valve area Av is</p><p>Av =</p><p>Ai</p><p>Cf</p><p>=</p><p>5.5× 10−4</p><p>0.38</p><p>= 1.45× 10−3m2</p><p>b)</p><p>Nt =</p><p>a · c0</p><p>Lt</p><p>=</p><p>(7.5)(1.4 · 287 · 350)</p><p>1/2</p><p>1.25</p><p>= 2250 rpm</p><p>11</p><p>5.11) Compare the predicted resonant tuning rpm Nt (Equation 5.60) of a Helmholtz resonator model with</p><p>the simple acoustic tuning rpm of Equation 5.58 and also the experimental results for maximum vol-</p><p>umetric efficiency ev plotted in Figure 5.17. Assume Di is equal to the inlet pipe diameter. Make a</p><p>table of the tuning rpm versus tuning inlet pipe length for the five cases shown in Figure 5.17. Assume</p><p>b = 83 mm, s = 106 mm, Di = 0.05 m, r = 9, To = 300 K.</p><p>The Helmholtz effective volume is</p><p>Veff =</p><p>Vd</p><p>2</p><p>(</p><p>r + 1</p><p>r − 1</p><p>)</p><p>=</p><p>π</p><p>2 · 4</p><p>(0.083)2(0.106)</p><p>(</p><p>9 + 1</p><p>9− 1</p><p>)</p><p>= 3.58× 10−4m3</p><p>The sound speed with γ = 1.4 is</p><p>c0 = (γ · R · T0)</p><p>1/2</p><p>= (1.4 · 287 · 300)</p><p>1/2</p><p>= 347m/s</p><p>The acoustic tuning equation is</p><p>Nt =</p><p>7.5c0</p><p>Li</p><p>=</p><p>2604</p><p>Li</p><p>The Helmholtz tuning equation is</p><p>Nt =</p><p>15</p><p>π</p><p>(c0)</p><p>( π</p><p>4D</p><p>2</p><p>i</p><p>veff · Li</p><p>)1/2</p><p>=</p><p>15</p><p>π</p><p>(347)</p><p>( π</p><p>4 (0.05)</p><p>2</p><p>(3.58× 10−4) · Li</p><p>)</p><p>1/2</p><p>=</p><p>3880</p><p>L</p><p>1/2</p><p>i</p><p>The following table compares the predictions of the tuning equation with experimental results for</p><p>maximum volumetric efficiency. Neither of the models reproduces the complex dependence of the vol-</p><p>umetric efficiency on engine speed and pipe length. Both of the models do indicate that as the engine</p><p>speed increases, the tuned pipe length decreases, one scaling as L−1, and the other as L</p><p>− 1/2</p><p>L Experiment N (rpm) Helmholtz N (rpm) Acoustic N (rpm)</p><p>0 3900 ∞ ∞</p><p>0.203 5000 8611 12,827</p><p>0.381 4500 6285 6834</p><p>0.546 4000 5250 4769</p><p>0.826 3500 4268 3152</p><p>12 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.12) Three camshafts are available for an engine. The valve maximum lift and intake and exhaust opening</p><p>and closing angles in degrees relative to top dead center (tdc) and bottom dead center (bdc) are</p><p>tabulated below .</p><p>CAM IO IC EO EC LIFT</p><p>(before tdc) (after bdc) (before bdc) (after tdc) (mm)</p><p>Factory 30 60 60 30 9.5</p><p>A 26 66 66 26 11.4</p><p>B 22 62 62 22 10.3</p><p>Draw a sketch of the three cam timing diagrams. Discuss the effects these different cams might have,</p><p>including duration and overlap effects.</p><p>The duration and overlap of each cam is listed in the table below:</p><p>Timing (Degrees)</p><p>Cam Overlap Duration of Intake Duration of Exhaust</p><p>θcc - θio θic - θio θcc - θco</p><p>Factory 60 270 270</p><p>A 52 272 272</p><p>B 44 264 264</p><p>• Factory</p><p>– This has the largest overlap, so it will have the largest short-circuiting</p><p>– The intake closing is the earliest, so there may be fuel starvation at high RPM</p><p>• Cam A</p><p>– The exhaust valve opens earliest, so it will have the shortest effective expansion stroke</p><p>– The intake closing is the latest, so it will have good performance at high RPM</p><p>– The duration of intake and exhaust is the largest, also giving good performance at high RPM</p><p>• Cam B</p><p>– The overlap is the smallest, so it will have the least short-circuiting</p><p>– The exhaust closing is the earliest, so the residual fraction will be the highest. This will</p><p>increase the initial pressure, and decrease the combustion temperature.</p><p>– The duration of intake and exhaust is the smallest, which will hinder performance at high</p><p>RPM</p><p>13</p><p>5.13) Derive an expression for the volumetric efficiency of a supercharged engine, using an analysis similar</p><p>to the derivation of Equation 5.51.</p><p>With a supercharged engine, since the intake manifold pressure is always greater than the cylinder</p><p>pressure, there is no flow of residual gas from the cylinder to the intake manifold. The start of intake</p><p>of fresh mixture (is) is coincident with the opening of the intake valve (io). Also, because of the super-</p><p>charging, there is flow from the cylinder into the exhaust port, until the exhaust valve is closed. The</p><p>intake process enthalpy flow integral is</p><p>ic</p><p>∫</p><p>io</p><p>[</p><p>(ṁcpT )in − (ṁcpT )out</p><p>]</p><p>dt</p><p>The mass inducted mi is</p><p>mi =</p><p>ic</p><p>∫</p><p>io</p><p>ṁindt</p><p>And the mass flow from the cylinder into the exhaust mov during intake is</p><p>mov =</p><p>ic</p><p>∫</p><p>io</p><p>ṁovdt</p><p>Assuming constant temperature intake and exhaust flow, the intake enthalpy flow integral becomes</p><p>cpTimi − cpTovmov</p><p>The first law, Equation 5.45 is</p><p>PicVic − PioVio</p><p>γ − 1</p><p>=</p><p>∫</p><p>−Pdv − cpTovmov + cpTimi +Q</p><p>The volumetric efficiency ev is</p><p>ev =</p><p>mi</p><p>ρiVd</p><p>Upon substitution of the first law, an using the ideal gas relation</p><p>ρiVdcpTi = PiVd</p><p>(</p><p>γ − 1</p><p>γ</p><p>)</p><p>The volumetric efficiency can be written</p><p>ev =</p><p>(</p><p>1</p><p>γ</p><p>)</p><p>PicVic − PioVio</p><p>PiVd</p><p>+</p><p>(</p><p>γ − 1</p><p>γ</p><p>)</p><p>ic</p><p>∫</p><p>io</p><p>PdV</p><p>PiVd</p><p>−</p><p>(</p><p>γ − 1</p><p>γ</p><p>)</p><p>Q</p><p>PiVd</p><p>+</p><p>(</p><p>Tov</p><p>Ti</p><p>)(</p><p>mov</p><p>ρi</p><p>vd</p><p>)</p><p>The main difference from the throttled case derive in th text is the positive sign of the mov term due</p><p>to the outflow from the cylinder into the exhaust. Also, the Tov/Ti ∼ 1. For the throttled case, the flow</p><p>is in the reverse direction from the exhaust to the lower pressure cylinder.</p><p>14 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.14) A supercharger has an isentropic efficiency of 0.75 and consumes 20 kW. If the volumetric flowrate</p><p>of standard air into the supercharger is 250 L/s, what is the air temperature, pressure, and density</p><p>exiting the supercharger ? Assume standard inlet conditions.</p><p>From the energy equation, the isentropic work wi is</p><p>wi = cp (T1 − T2s)</p><p>The mass airflow rate through the supercharger is</p><p>ṁa = ρiv̇i =</p><p>Pi</p><p>RTi</p><p>(v̇i) =</p><p>101,000</p><p>287 · 298</p><p>(</p><p>250× 10−3</p><p>)</p><p>= 0.295 kg/s</p><p>The isentropic work is related to the power</p><p>wi = ηcẆact = ηc</p><p>(</p><p>Ẇact</p><p>ṁa</p><p>)</p><p>= (0.75)</p><p>(</p><p>−20</p><p>0.295</p><p>)</p><p>= −50.8 kJ/kg</p><p>so the outlet isentropic temperature T2s is</p><p>T2s = T1 −</p><p>wi</p><p>cp</p><p>= 298−</p><p>−50.8</p><p>1.0</p><p>= 349K</p><p>The actual outlet temperature, T2 is</p><p>T2 = T1 +</p><p>T2s − T1</p><p>ηc</p><p>= 298 +</p><p>349− 298</p><p>0.75</p><p>= 366K</p><p>The outlet actual pressure P2 is</p><p>P2 = P1</p><p>(</p><p>T2s</p><p>T1</p><p>)</p><p>γ</p><p>γ−1</p><p>= 101</p><p>(</p><p>349</p><p>298</p><p>)</p><p>1.4</p><p>0.4</p><p>= 175 kPa</p><p>and outlet density</p><p>ρ2 =</p><p>P2</p><p>RT2</p><p>=</p><p>175</p><p>(0.287)(366)</p><p>= 1.67 kg/m3</p><p>15</p><p>5.15) Develop Equation 5.65 for the work done in an isentropic compression.</p><p>The energy equation for an open system with negligible kinetic and potential energy change is</p><p>q1→2 − w1→2 = h2 − h1</p><p>for an isentropic process q1−2 = 0</p><p>for an ideal gas h2 − h1 = cp (T2 − T1) Therefore</p><p>w1→2,isentropic = cp (T1 − T2)</p><p>Also for an isentropic process</p><p>T2</p><p>T1</p><p>=</p><p>(</p><p>P2</p><p>P1</p><p>)</p><p>γ−1</p><p>γ</p><p>so</p><p>w1→2,isentropic = cpT1</p><p>(</p><p>1−</p><p>T2</p><p>T1</p><p>)</p><p>= cpT1</p><p>(</p><p>1−</p><p>(</p><p>P2</p><p>P1</p><p>)</p><p>γ−1</p><p>γ</p><p>)</p><p>16 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>5.16) The airflow into a four-stroke 3.5 L engine operating at 3000 rpm with a volumetric efficiency of 0.75</p><p>is to be supercharged to 145 kPa from ambient Po, To conditions. An intercooler cools the compressed</p><p>air to 325 K. If the supercharger isentropic efficiency is 0.60, what is the power consumption of the</p><p>supercharger?</p><p>The isentropic work ws into the supercharger is</p><p>ws = cpT1</p><p>[</p><p>(</p><p>P2</p><p>P1</p><p>)</p><p>γ−1</p><p>γ</p><p>− 1</p><p>]</p><p>= (1.0)(300)</p><p>[</p><p>(</p><p>145</p><p>101</p><p>)</p><p>0.4</p><p>1.4</p><p>− 1</p><p>]</p><p>= 32.65 kJ/kg</p><p>The actual work into the supercharger, wact is</p><p>wact =</p><p>ws</p><p>ηc</p><p>=</p><p>32.65</p><p>0.60</p><p>= 54.42 kJ/kg</p><p>The air mass flowrate through the supercharger and engine is</p><p>ṁa = evρivd</p><p>N</p><p>2</p><p>ρi =</p><p>Pi</p><p>RTi</p><p>=</p><p>145,000</p><p>287 · 325</p><p>= 1.55 kg/m3</p><p>ṁa = (0.75)(1.55)(3.5× 10−3)</p><p>(</p><p>3000</p><p>2 · 60</p><p>)</p><p>= 0.102 kg/s</p><p>The power consumption of the supercharger is</p><p>ẇ = ṁawact = (0.102)(54.42) = 5.54kW</p><p>17</p><p>5.17) A Roots supercharger map is given in Figure 5.28. Match (i.e., find the resultant pressure ratio) this</p><p>supercharger to a 2.0 liter, 4-stroke engine with the following volumetric efficiencies.</p><p>N (rpm) ev(%)</p><p>1000 68</p><p>2000 68</p><p>3000 75</p><p>4000 76</p><p>5000 73</p><p>6000 70</p><p>Find the power required to drive the supercharger at each condition as well as the outlet temperature.</p><p>Choose a compressor speed</p><p>Nc equal to twice the engine speed N .</p><p>Following the procedure outlined on page 155, one assumes a pressure ratio, and then reads the</p><p>compressor efficiency ηc and mass flow rate ṁc from the compressor map. The outlet temperature T2</p><p>is</p><p>T2 = T1</p><p>[</p><p>1 +</p><p>(P2/P1)</p><p>(γ−1)/γ − 1</p><p>ηc</p><p>]</p><p>The outlet density from the compressor is obtained from the ideal gas law:</p><p>ρ2 =</p><p>P2</p><p>RT2</p><p>The mass flow rate through the engine is thus</p><p>ṁe = ev ρ2 Vd N/2</p><p>The calculated engine mass flowrate is compared with the compressor mass flowrate, and iterated until</p><p>the difference is less than a certain percentage, say 10%. A program to compute the engine mass</p><p>flowrate, compressor outlet temperature, power consumption, and compare with the given compressor</p><p>mass flowrate is given below. The specific heat ratio and average specific heat were assumed to be γ</p><p>= 1.40 and cp = 1.01 kJ/kg-K, respectively. Sample output of the program:</p><p>Engine flow(kg/s)= 0.0945 Supercharger flow =0.0900, Diff(\%) = 5.0</p><p>Compressor power(kW)= 8.11, Outlet Temperature (K) = 387</p><p>The Table below shows the iterative procedure for three representative engine speeds.</p><p>For Ne = 1000 rpm and ev = 0.68:</p><p>P2/P1 ηc ṁc ṁe Diff T2 Ẇc</p><p>(kg/s) (kg/s) (%) (K) (kW)</p><p>1.3 0.40 0.0120 0.0140 20 356 0.70</p><p>1.0 0.43 0.0170 0.0137 20 317 0.33</p><p>1.25 0.42 0.0140 0.0143 2 345 0.66</p><p>For Ne = 3000 rpm and ev = 0.75:</p><p>P2/P1 ηc ṁc ṁe Diff T2 Ẇc</p><p>(kg/s) (kg/s) (%) (K) (kW)</p><p>1.3 0.55 0.050 0.050 0.1 340 2.13</p><p>For Ne = 6000 rpm and ev = 0.70:</p><p>18 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>P2/P1 ηc ṁc ṁe Diff T2 Ẇc</p><p>(kg/s) (kg/s) (%) (K) (kW)</p><p>1.3 0.37 0.105 0.088 16 361 6.65</p><p>1.4 0.40 0.10 0.09 8 373 7.59</p><p>1.5 0.41 0.090 0.0945 5 387 8.11</p><p>Note how the outlet temperature and compressor power required increases as the engine speed and</p><p>thus the engine/compressor mass airflow increases.</p><p>The program to compute the supercharger matching is given below:</p><p>% program to compute supercharger matching</p><p>% enter efficiency and flowrate from map for a given dP</p><p>dP= 1.50; % pressure ratio</p><p>nc=0.41; % supercharger efficiency</p><p>mdotc= 0.09; % supercharger flowrate (kg/s)</p><p>N=6000; %engine speed (rpm)</p><p>ev= 0.70; % volumetric efficiency of engine</p><p>Vd= 2e-3; % engine displacement volume (m^3)</p><p>cp= 1.01; % cp of air at average T = 350 K</p><p>T1= 298; %inlet T (K)</p><p>P1= 100000; %inlet P (Pa)</p><p>gamma = 1.4;</p><p>gam= (gamma -1) /gamma;</p><p>R = 287;</p><p>%compute P2, T2</p><p>P2= P1*dP;</p><p>T2= T1*(1+ (dP^gam -1)/nc);</p><p>rho2= P2/(R*T2);</p><p>mdote = ev*rho2*Vd*N/60/2;</p><p>diff = (mdote-mdotc)/mdotc *100;</p><p>power=mdotc*cp*(T2-T1); %(kW)</p><p>fprintf(’Engine flow(kg/s)=%6.4f Supercharger flow=%6.4f,Diff = %6.1f \n’,...</p><p>mdote,mdotc, diff);</p><p>fprintf(’Compressor power(kW)=%5.2f, Outlet Temperature(K)=%5.0f \n’,power,T2);</p><p>19</p><p>5.18) A naturally aspirated four-cylinder, four-stroke gasoline engine has the following specifications.</p><p>Vd 2316 cm3</p><p>b 96 mm</p><p>s 80 mm</p><p>r 9.5</p><p>Ẇb 83 kW at N = 5400 rpm</p><p>A turbocharged version of the engine utilizes the compressor mapped in Figure 5.29. Estimate the</p><p>brake power of the turbocharged engine at N = 5400 rpm if the compressor ratio is P2/P1 = 1.5. What</p><p>is the compressor efficiency and rotational speed? What is the heat transfer to the inter-cooler? Make</p><p>the following assumptions.</p><p>• For the naturally aspirated (NA) engine</p><p>Inlet manifold conditions: Ti = 310 K, Pi = 1.0 bar, φ = 1.0.</p><p>Volumetric efficiency: ev = 0.84.</p><p>Mechanical efficiency: ηm = bmep/(imep)net = 0.90.</p><p>• For the turbocharged (TC) engine</p><p>Aftercooled gas temperature: Ti = 340 K. Volumetric efficiency: ev = 0.91.</p><p>Mechanical efficiency: ηm = bmep/(imep)net = 0.88.</p><p>• For a given engine speed and displacement, the indicated power is proportional to airflow rate:</p><p>Ẇi ≃ imepnet ≃ evPi/Ti</p><p>In practice, the compression ratio was lowered to 8.7 to avoid knock and the engine produced 117 kW</p><p>at 5280 rpm.</p><p>a.) The density of the mass flow entering the engine from the turbo charger is obtained from the ideal</p><p>gas law:</p><p>ρi =</p><p>Pi</p><p>RTi</p><p>=</p><p>(1.5)101000</p><p>(287) (340)</p><p>= 1.55 kg/m</p><p>3</p><p>The mass flow rate through the engine and turbocharger</p><p>ṁa = ev ρi Vd N/2 = (0.91)(1.55)(2.316× 10−3)(90/2) = 0.147 kg/s</p><p>At this air mass flowrate and pressure ratio, the compressor map indicates that the compressor efficiency</p><p>is about 0.65 and speed about 1660 rps (99600 rpm).</p><p>b.) The turbocharger outlet temperature T2 is</p><p>T2 = T1</p><p>[</p><p>1 +</p><p>(P2/P1)</p><p>(γ−1)/γ − 1</p><p>ηc</p><p>]</p><p>= 310</p><p>[</p><p>1 +</p><p>(1.5)(1.4−1)/1.4 − 1</p><p>0.65</p><p>]</p><p>= 368 K</p><p>The heat transfer to the inter-cooler is</p><p>Q = ṁacp(T2 − Ti)</p><p>= 0.147(1.01)(368− 340)</p><p>= 4.11 kW</p><p>20 CHAPTER 5. INTAKE AND EXHAUST FLOW</p><p>c.) The brake power is Ẇb = bmepVdN/2, and the mechanical efficiency ηm = bmep/imep, so the</p><p>ratio of the turbocharged engine brake power Ẇb,tc to the naturally aspirated engine power Ẇb,na is</p><p>Ẇb,tc</p><p>Ẇb,na</p><p>≃</p><p>(imep ηm)tc</p><p>(imep ηm)na</p><p>Since imep ≃ evPi/Ti,</p><p>Ẇb,tc</p><p>Ẇb,na</p><p>=</p><p>( evPi</p><p>Ti</p><p>ηm)tc</p><p>( evPi</p><p>Ti</p><p>ηm)na</p><p>=</p><p>(0.91 (1.5)</p><p>340 (0.88))tc</p><p>(0.84 (1.0)</p><p>310 (0.90))na</p><p>= 1.45</p><p>Therefore the turbocharged engine brake power Ẇb,tc = 1.45 Ẇb,na = 1.45 x 83 = 120 kW.</p>