Assignment 4 - Solutions - Solution Two reversible heat...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solution Two reversible heat engines operate in series between a source at 527 °C and a sink at 17°C. If the engines have equal efficiencies and the first rejects 400 U to the second, calculate: (i) the temperature at which heat is supplied to the second engine; (ii) the heat taken from the source; (iii) the work done by each engine. Assume that each engine operates on the Carnot cycle. (1) Let the temperature at which heat is supplied to the second engine be T, then 1 - (T/800) : 1 - (29o/T) ie T = [(290x800) = 481.7'K (ii) 208.7 °C Let the heat from the source be 0. then 1 * (290/481.7) = 1 ‘ (400/0) ie 0 = 400x482/29O = 664.4 RJ (iii) For the first engine: W i 400 - 664.4 3 * 264.4 kJ ie work output 264.4 kJ For the second engine: - W/400 = 1 - (290/481.7) w = ‘ 159.2 kJ ie work output = 159.2 kJ , SE”! 3% Asst/Viki 94 W W! 5.6 A four-cylinder petrol engine has a swept volume of 2000 cm 3, and the clearance volume in each cylinder is 60 cm". Calculate the air standard cycle efficiency. If the introduction conditions are 1 bar and 24 °C, and the maximum cycle temperature is 1400 °C, calculate the mean effective pressure based on the air standard cycle. Sohfibn Swept volume per cylinder : 2000/4 = 500 cm3 Total volume = 500 + 60 = 560 cm3 560/60 = 9.333 ie compression ratio Then, cycle efficiency = 1 — (1/9.333)°-4 = 0.591 or 59.1% \/ 53 Now, T2 = 297X(9.333)°-4 = 725.7 K \ heat supplied = CV(T3 ~ T2) 2 0.718(1673 — 725.7) 2 680.2 kJ/kg Then, net work output = 680-2x0.591 = 402.0 kJ/kg The mean effective pressure is given by the area on the p~v diagram divided by the length of the diagram. ie mean effective pressure, pm = Q§§_work,bgtgut v1 — V2 Then, V1 = RTi/px = 287x297/1x105 = 0.852 m3/kg and, V2 1 v1/9.333 = 0.852/9.333 0.091 m3/kg Therefore, pm = 402x1000 = 5.28 bar W 2 “Cr” (We—'6 -> + Cram) : L005” K (ma—m + }‘0(*4‘02 > + #935“ % -_—_ lo} 313’ xo 07}(1y;(0”6kj IV) I 1 I- \~ 1 5,! (PU/r a 7 no “731’6' If: y; 1: [2 2f 0 623 VOL/JUN: O‘HE W062“ (7 < ; , .. 3 7 _ " l . ' >4? ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern