Engine : Motor Plants and Auxiliary Boilers - 1339/947
|
|---|
| « Previous Question |
| A four cylinder, four-stroke cycle, single acting diesel engine has a 740 mm bore and a 1500 mm stroke. What indicated power will be developed if the average mean effective pressure is 18 kg/cm2 at a speed of 90 RPM? |
| A) 3416 kW |
| B) 4644 kW |
| C) 7296 kW |
| D) 9290 kW |
loading answer...
| Comments |
|---|
| PuzzleAddict - 2025-09-12 13:56:41 Registered (41) |
| Salut ! Le soir, fatigué après le travail, j'ai lancé le plinko casino. Au début, je pensais juste y jeter un œil, mais je me suis laissé prendre au jeu ! Les boules sont trop mignonnes, le son est apaisant et les petites victoires font plaisir. C'est un jeu génial pour se détendre et passer un bon moment. |
| gurkururze - 2024-10-09 01:39:57 Registered (194) |
| I just discovered www.signnow.com/ and it's a game changer for businesses looking to simplify signing and managing documents online. Highly recommend giving it a try! |
| Artemis17 - 2017-05-30 06:22:35 Member (2) |
| There isn't really a conversion factor for kW, only HP. IHP=PLAN will yield W so you simply divide by 1000 to convert to kW. Also, I apologize, I forgot to write (1/60) to convert RPM to seconds above, but that will get you the correct answer "3416". |
| gigem_das05 - 2017-05-30 03:04:21 Member (4) |
| I used IHP = (PLAN*n/2)/4500, where P = pressure in kg/cm^2, L = length in meter, A is area in cm^2, N = RPM, n = # of cylinders. 4500 conversion factor for kW and HP. I appreciate the feedback but I am not following your math and when I used your calculation I got a huge number. Could you offer a little more detail. Thanks |
| Artemis17 - 2017-05-29 20:56:21 Member (2) |
| This answer is correct. IHP=(4 CYL)(18kg/cm2)(9.81m/s2)(10,000cm2/m2)(1.5m)(pi*.74^2/4)(90/2) |
| gigem_das05 - 2017-05-25 21:36:00 Member (4) |
| I don't think this question is correct. Every time I calculate I come up with answer B 4644KW. |