Calculate the net work output of the heat engine, in joules, over a single cycle

Question

Calculate the net work output of the heat engine, in joules, over a single cycle of operation

? ? C College Physics-OpenStax × V The Expert TA Human-like G e (a) Calculate The Buoyant For × ? c Secure l https://usq45tx.theexperta.com/Common/TakeTutorialAssignment.aspx l Help Assignment 6 Begin Date: 71/2018 12:01:00 AM - Due Date: 7/6/2018 11:59:00 PM End Date: 7/6/2018 11:59:00 PM (7%) Problem 13: A certain heat engine performs a cycle described by the curve A-B-C-D-A in the p-V diagram shown in the figure. The segments between the p ( 100 Pa) designated points are all straight. The valucs of the pressures and volumes at the designated points are: VA0.65 x 103 m3 VB-33 x 103 m3 PA 2.85x 106 N/m2 PB 24 x 10% N/m2 PC 0.75 x 106 N/m2 PD = 1.32 x 106 N/m2 Assignment Status Click here for detailed view ?theexpert ta.com Problem Status ? Calculate the net work output of the heat engine, in joules, over a single cycle of operation. 2 Completed 3 CompletedW- Grade Summary 0% 100% 5 Partial 6 Completed cos() cotan asi acoso atan acotan sinhO coshO tanh) cotanhO DegreesRadians tan0 Attermpts remaining: 3 (2% per attempt) detailed view 8 Partial 10 Completed 1 Completed I give up 13 14 Completed 15 Completede curve describing the process in the p. V diagram. Hints: 4 fora 0% deduction. Hints rernaining Feedback: 0%, deduction per feedback -Use the relation between the work done during a process and the area under For a cyclic process you need to find the area within the closed curve Be careful with your signs. While in general you would perform integrations, in this case the areas consist only of rectangles and triangles. Submission History All Date times are displayed in Central Standard Time Red subwission date times indicate late work Hints

Explanation / Answer

Net work output = area of the parallelogram = sum of area of two traingles

BC = 1.65 X 106 N/m^2

AD = 1.53 X 106 N/m^2

height (same for both traingles) = V2-V1 = 2.65 X 10-3 m3

thus work = 0.5*(1.65 X 106 N/m2) * (2.65 X 10-3 m3) + 0.5* (1.53 X 106 N/m2) * (2.65 X 10-3 m3) = 4.21 x 103 J

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