ENGINE DESCRIPTION: Wankel Engine: The Wankel engine (Figure 1.1) is a type of i
ID: 1826946 • Letter: E
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ENGINE DESCRIPTION: Wankel Engine: The Wankel engine (Figure 1.1) is a type of internal combustion engine using an eccentric rotary design to convert pressure into a rotating motion instant of using reciprocating pistons. However, the three cavities provide the four actions of a piston engine. INTAKE; The intake is at its minimum volume and it increases volume drawing in the air-fuel mixture COMPRESSION: The following chamber is reduced in volume, compressing the air-fuel mixture COMBUSTION: At the point of minimum volume the mixture is ignited using two sets of spark-plugs to assure that the combustion is complete and high pressure rotates the rotor increasing volume. EXHAUST: The high pressure combustion gases pass the exhaust port, reducing the volume and forcing the combustion gas to exit the chamber. Tables 1.1 and 1.2, lists the assumed properties of the air-fuel mixture at cycle states 1, 2, 3 and 4. The air is assumed to be an ideal gas, with an Ideal Gas Constant, R= .287 [kNm/kg degree K] Table 1.1: Assumed Properties of the Air-Fuel mixture Table 1.2: Assumed Paths between States Note that paths 1-2 and 3-4 are at constant pressure, while path 4-1 is polytrophic and the return path, to complete the cycle, is polytropic; P(v)k where k=1.4 for air. Calculate the temperatures at states l, 2, 3 and 4. Calculate the net work done in a cycle (Approximated below)Explanation / Answer
Gas constant for air, R = 287 J/kg-K Problem 1: 1) P1*v1 = RT1 T1 = P1*v1/R = (100*10^3)*2 / 287 = 697 K 2) T2 = P2*v2/R = (100*10^3)*1/287 = 348.4 K 3) T3 = P3*v3/R = (200*10^3)*1 / 287 = 697 K 4) T4 = P4*v4 / R = (200*10^3)*1.22 / 287 = 850.2 K 5) T1 = P1*v1 / R = 100.11*10^3 *2 / 287 = 697.6 K Problem 2: Work done in process 1-2: = p1*(V2 - V1) = (100*10^3)*(1 - 2) = -100 kJ/kg Work done in process 2-3 = 0 (since constant volume process) Work done in process 3-4 = P3*(v4 - v3) = (200*10^3)*(1.2 - 1) = 40 kJ/kg For process 4-1:T4 / T1 = (v1/v4)^(k - 1) So, 850.2 / 697 = (2 / 1.22)^(k - 1) Solving this, k = 1.4 Work done in process 4-1 = R(T4 - T1) / (k-1) = 287*(850.2 - 697)/(1.4 - 1) = 109.9 kJ/kg Net work done in cycle = -100 + 0 + 40 + 109.9 = 49.9 kJ/kg
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