Tutor profile: Azhie W.
Subject: Mechanical Engineering
What is the maximum possible cycle efficiency of a heat engine operating between a heat source at 478 K and a heat sink at 273 K?
Maximum possible cycle efficiency is the Carnot efficiency (ƞc) of that cycle. The equation for Carnot efficiency is given below η_c=1-T_C/T_H where T_H is the temperature of the heat source and T_C is the temperature of the heat sink Thus η_c=1-273/478 =0.4288 ≅0.43
Evaluate ∫▒〖(x+1)/(x^2 (x-1)) dx〗
We start by using partial fractions:(x+1)/(x^2 (x-1))=A/x+ B/x^2 +C/(x-1) which gives x + 1 = Ax(x - 1) + B(x- 1) + Cx^2 = (A + C) x^2 + (B - A)x - B x + 1 from which we deduce A + C = 0, B − A = 1, and −B = 1. Therefore, B = −1, A = −2, and C = 2. Thus ∫▒〖(x+1)/(x^2 (x-1)) dx〗=∫▒〖-2/x-∫▒〖1/x^2 +∫▒〖2/(x-1) 〗〗〗 = -2In(x)+1/x+2In(x-1)+C
(x-1)^2=(x-1)(x-1) Solving using foil (x-1)(x-1)=x^2-x-x+1 =x^2-2x+1 Answer is x^2-2x+1 Alternatively: You can identify that (x-1)^2 can be written as (a-b)^2;where a=x and b=1 From algebra standard identities, (a-b)^2=a^2-2ab+b^2 This implies (x-1)^2=x^2-2(x)(1)+1^2 =x^2-2x+1 Answer: x^2-2x+1
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