### Machine room heat release calculation

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I need calculate the heat release in machine room. It is a hydro elevator. It has an oil cooler equiped in the hydro system. Normally, we just use 1HP = 2544 BTU/Hr, then consider the effeciency, we use H = 0.3 X HP X 2544 = 382 HP. So for a 40 HP motor, the BTU/Hr is 15280. I am not sure with an oil cooler, is it gonna be a greater than 15280? Because oil cooler will release most of the heat in the oil to the air per a certain time, that is , release the heat much faster. And BTU/Hr will be greater? Am I correct? If so, then how can I calculate the BTU/Hr with an oil cooler? Thanks a lot.

-- M Ts (cronept@msn.com), December 08, 2004

Help

-- MTS (cronept@msn.com), December 09, 2004.

In the case of an oil cooler I agree with you that the heat will be released faster. However, take a scenario where the motor is running continuously for two hours: the fact that the oil cooler is releasing the heat faster makes no difference once the system has arrived at a steady state (it will only prevent the oil temperature rising to dangerous limits). The oil cooler would only make a difference in cases where the lift runs continuously for say 10 minutes and then stays idle for 50 minutes. During the heavy 10 minutes, the presence of an oil cooler would ensure that most of the heat is released during those 10 minutes. without an oil cooler, it will take longer than 10 minutes to release the heat from the oil, and thus the same amount of heat would be released over a longer period of time. But the case of 10 heavy minutes/50 idle minutes is not your worst case. So I think it will not make a difference.

However, this all depends on the rule of thumb formula you show above (i.e., 1 hp = 2544 btu/hr, and the 0.3 in the second formula). If you can explain the reasoning behind those, I might be able to help you further. For example, 1 btu is approximately equal to 1055 Joules. A motor of 1 hp running continuously for an hour would generate 746 W (J/s) x3600 s = 2685600 J. 2544 btu/hr=2683920. So this assumes that the motor is running full time at full load. I assume then the 0.3 is assuming that the efficiency of the system is 30% (i.e, 30% of the energy is dissipated as heat, while 70% is used as useful energy). I think this is excessive for the following reasons: 1. the motor when it runs is does not always run at full load even at peak times. 2. The motor certainly does not run continuously, as it only works during the up direction for a hydraulic lift. Then there are the door opening/closing times....etc. Generally a hydraulic lift will only do around 120 starts per hour. 3. The efficiency figure you quote is probably slightly pessimistic: I think 0.25 to 0.2 is more representative of modern systems.

You also need to remember that some modern systems employ regenerative VVVf drives to the pump motor allowing some of the energy to be sent back to the mains.

If you would like to discuss further, please e-mail (al_sharifvtc@compuserve.com)

-- Lutfi Al-Sharif (al_sharifvtc@compuserve.com), December 10, 2004.

You are right, I use 0.3 as a factor by assuming the heat generated by a motor with 70% efficiency. And 1HP = 2544 BTU/hr comes from 1 HP = 0.7068 BTU/sec based on conservation of energy.

I made a mistake in my previous formulas. It should be H = 0.3 X HP X 2544 = 763 HP. And the hydraulic motor only works 50% of the time, then H=382 HP. Sorry for the confusion. Yes, as you can see, all the calculation is based on rough estimates, I think it is really hard to calculate the effect of the oil cooler on heat release.

By the way, what formulas or methods are you using to calculate the MR heat release?

-- MTS (cronept@msn.com), December 13, 2004.

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