Question
Download Solution PDFAn engine at full load delivers 200 kW brake power. It requires 25 kW to rotate it without fuel at the same speed. The mechanical efficiency at half load is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)
Calculation:
Given:
Brake power (BP) = 200 kW, Half load = 100 kW Friction Power (FP) = 25 kW
Mechanical efficiency at half load \( = \frac{{BP}}{{BP+ FP}}\)
Mechanical efficiency at half load \( = \frac{{100}}{{125 }}\)
Mechanical efficiency at half load = 0.8 ⇒ 80 %
Last updated on May 28, 2025
-> SSC JE ME Notification 2025 will be released on June 30.
-> The SSC JE Mechanical engineering application form will be available from June 30 to July 21.
-> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 2 to 31.
-> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government.
-> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination.
-> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-.
-> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination.