Engine : Motor Plants and Auxiliary Boilers - 1339/947
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| 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 |
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| Comments |
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| 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. |