Question 1: Ideal Otto Cycle Consider an idea Otto cycle...

Question 1: Ideal Otto Cycle Consider an idea Otto cycle with intake pressure of 120 KPa and 300 K. 850 KJ/Kg of heat added to the air-fuel mixture during the constant-volume heat-addition process. Consider compression ratio of 7 in the I.C cycle. Referring to figure 1, determine the following: A) Find the internal energy at all states $u_1$, $u_2$, $u_3$ and $u_4$ B) Cycle maximum temperature. C) Cycle maximum Pressure. D) Net output work. E) Thermal efficiency. $P$, kPa 3 Isentropic $q_{in}$ 2 4 $q_{out}$ Isentropic 120 1 $V_2 = V_3 = \frac{1}{7}V_1$ $V$ $V_1 = V_4$ Figure 1: P-V diagram of Ideal Otto Cycle

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00:01 Hello students, it given that pi is equal to 25 kilopascal and it is given that ti is equal to 27 degree centigrade which is equal to ti is equal to 27 plus 273 equals to 300 kelvin and heat q is given as 750 kilojoule per kg.

00:26 We need to calculate pressure at the end of the process for 1.

00:33 It is given that p3 by p2 equals to t3 by t2.

00:41 We need to calculate t3, t2 and p2.

00:45 Here it can be given that t2 by t1 is equal to v1 by v2 gamma power minus 1.

00:56 Here gamma is referred to be 1 .4.

01:00 Therefore, gamma equal to 1 .4 and r equal to v1 by v2 equal to 8.

01:09 It is given that these values.

01:12 Now t2 is equal to t1, 8 power 1 .4 minus 1 into t1 is equal to the value of 300 kelvin.

01:25 Therefore, the obtained value of t2 is equal to 689 .2 kelvin.

01:33 Now we will calculate the value of t3 is equal to from the relation q is equal to mcv delta t.

01:48 Here m is equal to 1 and cv is given as 0 .718 into delta t equal to t3 minus t2.

02:00 Here t2 is 689 .2 kelvin.

02:04 From this on solving here q is given that 750.

02:09 Therefore, the obtained value of t3 here is equal to 1733 .99 kelvin.

02:18 Now we need to calculate another value here to calculate the p3.

02:26 Here it is the value of p2.

02:29 For p2, p2 by p1 is equal to r power gamma.

02:35 Here p2 is equal to p1 is equal to 95 into 8 power 1 .4.

02:46 Therefore, p2 is equal to 1746 kilopascal.

02:55 Therefore, we will calculate now the value of p3...

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