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There is a heat engine whose cycle is described in the figure. P (10° N/m²) A 2.6 2.0 В 1.0 0.6 1.0 4.0 V (10° m³) ©theexpertta.com Part (a) What is the net work output of the heat engine that follows path ABDA in the figure, with a straight line from B to D? Give your answer in J. The net work output of a heat engine is the area contained within its path going in a clockwise direction on a P-V diagram. For this problem, the triangular area contained by the straight lines connecting A to B, B to D, and D to A. We can divide this triangle into two right angle triangles created by the horizontal dashed line at 2*10€ N/m². The area of a right angle triangle is A = b.hJ where A is the area in units of J, b is the base of the triangle in N/m², and h is the height of the triangle in m-3. The net work output is therefore the sum of the two triangular areas, W= A1 + Ag= 아어시 + bzh2) = (2.6 – 2) · 10° N/m² · (4 – 1) - 10-3 m³ + (2 – 1) · 10° N/m² - (4 – 1) · 10-3 W = 2.4 · 103 J

There is a heat engine whose cycle is described in the figure. P (10° N/m²) A 2.6 2.0 В 1.0 0.6 1.0 4.0 V (10° m³) ©theexpertta.com Part (a) What is the net work output of the heat engine that follows path ABDA in the figure, with a straight line from B to D? Give your answer in J. The net work output of a heat engine is the area contained within its path going in a clockwise direction on a P-V diagram. For this problem, the triangular area contained by the straight lines connecting A to B, B to D, and D to A. We can divide this triangle into two right angle triangles created by the horizontal dashed line at 2*10€ N/m². The area of a right angle triangle is A = b.hJ where A is the area in units of J, b is the base of the triangle in N/m², and h is the height of the triangle in m-3. The net work output is therefore the sum of the two triangular areas, W= A1 + Ag= 아어시 + bzh2) = (2.6 – 2) · 10° N/m² · (4 – 1) - 10-3 m³ + (2 – 1) · 10° N/m² - (4 – 1) · 10-3 W = 2.4 · 103 J

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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There is a heat engine whose cycle is described in the figure. P (10° N/m?) A 2.6 2.0 D 1.0 0.6 1.0 4.0 V (10° m³) ©theexpertta.com Part (a) What is the net work output of the heat engine that follows path ABDA in the figure, with a straight line from B to D? Give your answer in J. The net work output of a heat engine is the area contained within its path going in a clockwise direction on a P-V diagram. For this problem, the triangular area contained by the straight lines connecting A to B, B to D, and D to A. We can divide this triangle into two right angle triangles created by the horizontal dashed line at 2*10° N/m². The area of a right angle triangle is 1 A = -b .hJ where A is the area in units of J, b is the base of the triangle in N/m2, and h is the height of the triangle in m3. The net work output is therefore the sum of the two triangular areas, W = A1 + A2 =(b,h1 + byh2) = -3 m³ + (2 – 1) · 10° N/m² · (4 – 1) · 10-3 m³) ((2.6 – 2) - 10° N/m² - (4 – 1) - 10 W = 2.4 · 10° J B.
Transcribed Image Text:There is a heat engine whose cycle is described in the figure. P (10° N/m?) A 2.6 2.0 D 1.0 0.6 1.0 4.0 V (10° m³) ©theexpertta.com Part (a) What is the net work output of the heat engine that follows path ABDA in the figure, with a straight line from B to D? Give your answer in J. The net work output of a heat engine is the area contained within its path going in a clockwise direction on a P-V diagram. For this problem, the triangular area contained by the straight lines connecting A to B, B to D, and D to A. We can divide this triangle into two right angle triangles created by the horizontal dashed line at 2*10° N/m². The area of a right angle triangle is 1 A = -b .hJ where A is the area in units of J, b is the base of the triangle in N/m2, and h is the height of the triangle in m3. The net work output is therefore the sum of the two triangular areas, W = A1 + A2 =(b,h1 + byh2) = -3 m³ + (2 – 1) · 10° N/m² · (4 – 1) · 10-3 m³) ((2.6 – 2) - 10° N/m² - (4 – 1) - 10 W = 2.4 · 10° J B.
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