Soal Dan Jawaban Internalcombustionengines

SOAL DAN JAWABAN INTERNAL COMBUSTION ENGINES II SOAL: 1. A five-cylinder, in-line engine has an 8.l5-cm bore, a 7.82-cm stroke, and a connecting rod length of 15.4 cm. Each piston has a skirt length of 6.5 cm and a mass of 0.32 kg. At a certain engine speed and crank angle, the instantaneous piston speed is 8.25 m/sec, and clearance between the piston and cylinder wall is 0.004 mm. SAE 10W-30 motor oil is used in the engine, and at the temperature of the piston-cylinder interface the dynamic viscosity of the oil is 0.006 N-sec/m2. The engine performs a power stroke in the cylinder described and the crank angle is as shown in Fig. 1. At this point, pressure in the cylinder is 3200 kPa and the compressive force in the connecting rod is 8.1 kN. Calculate the friction force on one piston and the thrust force on the cylinder wall at this condition.

Fig. 1. 2. A four-cylinder, two-stroke cycle engine, with a 2.65-liter displacement and crankcase compression, is running at 2400 RPM and an air-fuel ratio of 16.2:1. At this condition the trapping efficiency is 72%, relative charge is 87%, and the exhaust residual from the previous cycle in each cylinder is 7%. Oil is added to the intake air flow such that the input fuel-to-oil ratio is 50:1. Calculate: 1. rate of oil use 2. rate of unburned oil added to the exhaust flow 3. A diesel engine is fitted with a turbocharger, which comprises a radial compressor driven by a radial exhaust gas turbine. The air is drawn into the compressor at a pressure of 0.95 bar and at a temperature of 15°C, and is delivered to the engine at a pressure of 2.0 bar. The engine is operating on a gravimetric air/fuel ratio of 18 : 1, and the exhaust leaves the engine at a temperature of 600°C and at a pressure of 1.8 bar; the turbine exhausts at 1.05 bar. The isentropic efficiencies of the compressor and turbine are 70 per cent and 80 per cent, respectively. Using the values cpair = 1.01 kJ/kg.K, γair = 1.4 and cpex = 1.15kJ/kg K, γex = 1.33 Calculate: (i) the temperature of the air leaving the compressor (ii) the temperature of the gases leaving the turbine (iii) the mechanical power loss in the turbocharger expressed as a percentage of the power generated in the turbine.

4. Compare the cooling effect of fuel evaporation on charge temperature in a turbocharged spark ignition engine for the following two cases: (a) the carburettor placed before the compressor (b) the carburettor placed after the compressor The specific heat capacity of the air and the latent heat of evaporation of the fuel are both constant. For the air/fuel ratio of 12.5 : 1, the evaporation of the fuel causes a 25 Κ drop in mixture temperature. The compressor efficiency is 70 per cent for the pressure ratio of 1.5, and the ambient air is at 15°C. Assume the following property values: - for air cp = 1.01 kJ/kg K, γ = 1.4 - for air/fuel mixture cp = 1.05 kJ/kg K, γ = 1.34 Finally, compare the compressor work in both cases.

JAWABAN: Soal 1: A five-cylinder, in-line engine has an 8.l5-cm bore, a 7.82-cm stroke, and a connecting rod length of 15.4 cm. Each piston has a skirt length of 6.5 cm and a mass of 0.32 kg. At a certain engine speed and crank angle, the instantaneous piston speed is 8.25 m/sec, and clearance between the piston and cylinder wall is 0.004 mm. SAE 10W-30 motor oil is used in the engine, and at the temperature of the piston-cylinder interface the dynamic viscosity of the oil is 0.006 N-sec/m2. The engine performs a power stroke in the cylinder described and the crank angle is as shown in Fig. 1. At this point, pressure in the cylinder is 3200 kPa and the compressive force in the connecting rod is 8.1 kN. Calculate the friction force on one piston and the thrust force on the cylinder wall at this condition. Data: Bore B = 8.15 cm = 0.0815 m Stroke S = 7.82 cm Connecting rod length L = 15.4 cm Piston skirt length h = 6.5 cm Piston mass m = 0.32 kg Instantaneous piston speed ΔU = 8.25 m/sec Clearance between the piston and cylinder wall Δy = 0.004 mm = 0.000004 m Dynamic viscosity of the oil µ = 0.006 N-sec/m2 Pressure in the cylinder P = 3200 kPa Compressive force in the connecting rod Fr = 8.1 kN = 8100 N

Solution: Shear stresses between the piston and cylinder wall: τs = μ(dU/dy) = μ(ΔU/Δy) = (0.006 N-sec/m2)[(8.25 m/sec)/(0.000004 m)] = 12,375N/m2 Contact area between the piston and cylinder wall: A = πBh = π(0.0815 m)(0.065 m) = 0.0166 m2 Friction force on the piston: Ff = τsA = (12,375 N/m2)(0.0166 m2) = 205 N

Crank offset equals half of stroke length: R = S/2 = (7.82 cm)/2 = 3.91 cm. Angle between the connecting rod and the centerline of cylinder: tanϕ = R/L = 3.91/15.4 = 0.2539 ϕ = atan(0.2539) = 14.25o Find the acceleration term: ∑Fx = m.a = -Fr cosϕ + Fp - Ff m(dUp/dt) = -Fr cosϕ + P(π/4)B2 - Ff = -(8.1 kN) cos(14.25o) + (3200 kPa)(π/4)(0.0815 m)2 – (205 N) = 8638 N Find the thrust force: m(dUp/dt) = -Fr cosϕ + P(π/4)B2 - Ff Fr = [- m(dUp/dt) + P(π/4)B2 - Ff ]/cosϕ Ft = Fr sinϕ = 8100 N sin (14.25o) = or Ft = [-m(dUp/dt) + P(π/4)B2 – Ff] tanϕ = [-(8636 N) + (3200 kPa)(π/4)(0.0815 m)2 – (205 N)] tan(14.25o) =1994 N This force would be in the plane of the connecting rod on the major thrust side of the cylinder. Soal 2: A four-cylinder, two-stroke cycle engine, with a 2.65-liter displacement and crankcase compression, is running at 2400 RPM and an air-fuel ratio of 16.2:1. At this condition the trapping efficiency is 72%, relative charge is 87%, and the exhaust residual from the previous cycle in each cylinder is 7%. Oil is added to the intake air flow such that the input fuel-to-oil ratio is 50:1. Calculate: 1. rate of oil use 2. rate of unburned oil added to the exhaust flow Data: Number of cylinder = 4 Number of revolution per cycle n = 1 rev/cycle (2 stoke engine) Displacement Vd = 2.65 liter = 0.00265 m3 Speed N = 2400 RPM Air-fuel ratio AF = 16.2 : 1 Traffing Efficiency λte = 72% Relative Charge λrc = 87% (volumetric efficiency) Exhaust residual = 7% Fuel-to-oil ratio FO = 50:1 Air density ρa = 1.181 kg/m3 Solution: 1) Rate of oil use: The total charge trapped (mtc) in the engine with time: λrc = 𝑚̇tc /[Vdρa(N/n)] 𝑚̇tc = VdρaλrcN/n = (0.00265 m3/cycle)(1.181 kg/m3)(0.87)(2400/60 rev/sec)/(1 rev/cycle) = 0.01089 kg/sec With 7% exhaust residual, only 93% of the trapped charge is air-fuel (mmt):

𝑚̇mt = (0.1089 kg/sec)(0.93) = 0.1013 kg/sec Find the mass of air and fuel ingested (mmi): Trapping Efficiency λte = mmt/mmi 𝑚̇mi = 𝑚̇mt/λte = (0.1013 kg/sec)/(0.72) = 0.1407 kg/sec With AF = 16.2, the rate of fuel input is: 𝑚̇f = mmi/(AF +1) = (0.1407 kg/sc)/(16.2 + 1) = 0.00818 kg/sec Oil input: 𝑚̇oil = 𝑚̇f /FO = (0.00818 kg/sec)/(50) = 0.000164 kg/sec = 0.59 kg/hr ≈ 0.67 L/hr 2) Rate of unburned oil added to the exhaust flow Of the oil input, 72% is trapped in the cylinders and burned, and 28% enters the exhaust during valve overlap. Mass of unburned oil in the exhaust: 𝑚̇oil = (0.59 kg/hr)(0.28) = 0.17kg/hr Soal 3: A diesel engine is fitted with a turbocharger, which comprises a radial compressor driven by a radial exhaust gas turbine. The air is drawn into the compressor at a pressure of 0.95 bar and at a temperature of 15°C, and is delivered to the engine at a pressure of 2.0 bar. The engine is operating on a gravimetric air/fuel ratio of 18 : 1, and the exhaust leaves the engine at a temperature of 600°C and at a pressure of 1.8 bar; the turbine exhausts at 1.05 bar. The isentropic efficiencies of the compressor and turbine are 70 per cent and 80 per cent, respectively. Using the values cpair = 1.01 kJ/kg.K, γair = 1.4 and cpex = 1.15kJ/kg K, γex = 1.33 Calculate: (i) the temperature of the air leaving the compressor (ii) the temperature of the gases leaving the turbine (iii) the mechanical power loss in the turbocharger expressed as a percentage of the power generated in the turbine. Data: Intake pressure p1 = 0.95 bar Intake temperature T1 = 15°C = 15 + 273 = 288 K Output pressure p2 = 2.0 bar Air/fuel ratio = 18 : 1 Engine exhaust temperature T3 = 600°C = 600 + 273 = 873 K Engine exhaust pressure p3 = 1.8 bar Turbine exhaust pressure p4 = 1.05 bar Isentropic efficiency of the compressor ηc = 70 % Isentropic efficiency of the turbine ηt = 80 % Air properties: cpair = 1.01 kJ/kg.K, γair = 1.4 Exhaust properties: cpex = 1.15kJ/kg K, γex = 1.33 Solution: Referring to figure 2, the real and ideal temperatures can be evaluated along with the work expressions. (i) If the compression were isentropic, T2s = T1(p2/p1)(γ-1)/γ T2s = 288(2.0/0.95)(1.4-1)/1.4 = 356 K. or 83°C

From the definition of compressor isentropic efficiency, ηc = (T2s - T1)/(T2 - T1) T2 = (T2s - T1)/ηc + T1 = (356-288)/0.7 + 288 = 385 or 133°C (ii) If the turbine were isentropic, T4s = T3(p4/p3)(γ-1)/γ T4s = 873(1.05/1.8)(1.33-1)/1.33= 762.9 K. or 490°C

Figure 2. Temperature/entropy diagram for a turbocharger. From the definition of turbine isentropic efficiency, ηt = (T3-T4)/(T3 - T4s) T4 = T3 - ηt(T3 - T4s) = 600 - 0.8(600 - 490) = 512°C (iii) Compressor power Wc = mair cpair(T2 - T1) = mair 1.01(113 - 15) kW = mair 98.98 kW from the air/fuel ratio mex = mair ( 1+ 1/18) = mair (1.056). and turbine power Wt = mex cpex(T3 - T4) = mair (1.056) x 1.15(600 - 512) = mair 106.82 kW Thus, the mechanical power loss as a percentage of the power generated in the turbine is (Wt - Wc)/Wt x100 = (106.82 - 98.98)/106.82 x 100 = 7.34 per cent Soal 4: Compare the cooling effect of fuel evaporation on charge temperature in a turbocharged spark ignition engine for the following two cases: (a) the carburettor placed before the compressor (b) the carburettor placed after the compressor. The specific heat capacity of the air and the latent heat of evaporation of the fuel are both constant. For the air/fuel ratio of 12.5 : 1, the evaporation of the fuel causes a 25 Κ drop in mixture temperature. The compressor efficiency is 70 per cent for the pressure ratio of 1.5, and the ambient air is at 15°C. Assume the following property values: - for air cp = 1.01 kJ/kg K, γ = 1.4 - for air/fuel mixture cp = 1.05 kJ/kg K, γ = 1.34 Finally, compare the compressor work in both cases.

Data: Air/fuel ratio = 12.5 : 1 Mixture temperature drop ΔT = 25 Κ Compressor efficiency ηc = 70 % Pressure ratio rp = 1.5 Ambient air temperature T1 = 15°C Air properties cp = 1.01 kJ/kg K, γ = 1.4 Air/fuel mixture properties cp = 1.05 kJ/kg K, γ = 1.34 Solution: Both arrangements are shown in figure 3. (a) T1 = 15°C = 288 Κ Τ2 = T1 - ΔT = 288 - 25 = 263 Κ If the compressor were isentropic, T3s = Τ2 (rp)( γ -1)/γ = 263 (1.5)(1.34 -1)/1.34 = 291.5 Κ From the definition of compressor isentropic efficiency T3 = (T3s + T2)/ηc + T2 = 291.5 - 263)/ 0.7 + 263 = 303.7 Κ (b) T4 = 288 Κ From isentropic compression T5s = Τ4 (rp)( γ -1)/γ = 288 (1.5)(1.4-1)/1.4 = 323.4 Κ

Figure 3. Possible arrangement for the carburettor and compressor in a spark ignition engine: (a) carburettor placed before the compressor; (b) carburettor placed after the compressor. From the definition of compressor isentropic efficiency T5 = (T5s - T4)/ηc + T4 = (323.4 - 288)/ 0.7 + 288 = 338.5 Κ T6 = T5 - 25 = 338.5 - 25 = 313.5 Κ Since T6 > T3, it is advantageous to place the carburettor before the compressor. Comparing the compressor power for the two cases: (Wc)a = mmix cpmix (T3 - T2) = 1.08 mair 1.05(303.7 - 263) = 46.15 mair kW (Wc)b = mair cpair (T5 - T4) = mair 1.01(338.5 - 288) = 51.01 mair kW

Thus placing the carburettor before the compressor offers a further advantage in reduced compressor work.