I know this may sound a little dumb but what is the result of using a 50 HP electric motor to drive a compressor thats rated as a 100 HP
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I know this may sound a little dumb but what is the result of using a 50 HP electric motor to drive a compressor thats rated as a 100 HP
Not dumb question at all. A compressor may be "rated" for a 100HP motor, but at only certain conditions. This conditions include Refrigerant used, Discharge Pressure, Suction Pressure, and Compressor Speed.
For example, a Compressor will have a much larger HP requirement at Medium Temp than at Low Temp.
Also a two stage system will use a much smaller motor on the Low Stage than would be required on a High Stage (assuming that the High Stage is doing most of the work).
This does require care in selection as you have to get to the operating conditions where you selected the motor. Some compressor are set up with an unloading pressure control for startup of the system.
A Blast Freezer is a excellent example of this.
NH3LVR is absolutely correct. The compressor may be rated for a higher Hp at some operational conditions. The same compressor when being operated for a lower evaporative temperature, will consume less power.
A motor that is correctly loaded will run much more efficiently as well. Lightly loaded motors are inefficient as the 'standing' losses make up the majority of the electrical load, whereas when loaded, they are only a small proportion of the electrical load.
A refrigeration compressor capacity can never be assessed on the basis of HP. A compressor will have diffrent power consumption for different conditions. In case the requirement is for a 50 HP and the Compressor is rated for 100 HP, it does not mean that only 100 HP motor shall be required. It is quite possible that the required rating is quite different.
I feel it is wrong to rate a compressor by HP . I have seen a number of people rate it at HP. A compressor capacity will depend on factors such as the Refrigerant, Evaporation Temperature and Condensing Temperature.
You should also never equate compressor HP equal to any given capacity. In air conditioning you may sometimes hear a 5 Ton compressor is the same thing as a 5 HP compressor. That may be true for air conditioning applications only. This is a bad habit to get into.
Power input is determined by the capacity produced by the compressor.
In our country I have heard so many times even from an engineer that he wants a say 5 Tons compressor. I tell them what is a 5 Ton compressor. The TR will depend on three factors.
1. The Refrigerant being used.
2. The Evaporative Temperature.
3. The Condensing Temperature.
I fail to understand that how could a refrigeration engineer can say things like that.
I've been starting to set the conditions then say the BTU rating, ie: 6,000btu @ 35C condencing, 7.4C evap, just like it says on its paper work. Now they know the conditions and can rufely extrapulate from there.
If you want to correctly state the compressor performance you would list:
The operating suction pressure and temperature will tell you the suction superheat at that operting condition.
- Capacity at the operating suction pressure & temperature
- Power input at the discharge pressure & liquid feed temperature to the expansion device
- RPM
- Refrigerant used
The discharge pressure and liquid feed temperature help you to find the amount of subcooling provided.
RPM & Refrigerant are usual requirements.
The compressor capacity is determined by the suction pressure and discharge pressure and liquid feed temperature.
Using the evaporating and condensing temperatures are only useful when you are describing the performance of those components.
Hi there,
I am with you on this.:confused:Quote:
Originally Posted by Samarjit Sen
Here they say "I need 10HP condenser and evaporator"???????:confused:
I am tired repeating "HP is an electrical unit not thermal".
What can we do? Just keep correcting and tolerate.
and get mad:eek:.
Cheers
Hi claude11Quote:
Originally Posted by claude11
It should be firstly noted that there is an inverse relationship between the torque and rotational speed for the compressor for a fixed value of input power "HP". AT zero RPM "starting mode or braking mode" the torque has a maximum sometimes reaches from 5 to 6 times the rated one and it is called brake or initial torque, this is typically occurred in case of starting electrical motor the strating current reaches from 5 to 6 times the rate one and it named inrush current. Therefore, u may see there is two values of the current rated/inrush in motor catalouges and similarly in the ICE "internal combustion engine" there is something named brake HP and u can also take example of the pump operation when u closed the discharge valve in centrifugal pump (not positive displacement because will blow off" the head pressure raise up stoutly . Alternatively, At maximum RPM of the compressor or motor the torque will be zero which is so-called "unloading condition or free rotation. After establish this fundamental we back to our question "what is the result of using a 50 HP electric motor to drive a compressor thats rated as a 100 "?. Firstly when the motor of 50 HP tries to start compressor of 100 HP which will have an initial torque of doubled value of that for the motor so the motor will squeeze hardly itself to whirl the hefty compressor on expense of raising up the inrush current and eventually the motor may be tripped off by thermal overload (as a result of Imax^2*Relect) or overload switch itself. If the motor hashes itself and overcomes the initial torque of compressor it causes the compressor to rotate clumsy with almost no difference in pressure. Let us assume somebody says ok i will make a prime mover to just start the compressor to rotate and convey the duty to this motor, unfortunately this unwise solution because the imposed rated torque of the motor will be half the required torque of the compressor to sustain in the same RPM. I wish it could help
Best regards
Hi Samarjit SenQuote:
Originally Posted by Samarjit Sen
Yes but there is not a general rule or to say that u should put a constraint which is "The same compressor when being operated for a lower evaporative temperature, will consume less power for the same system pressure ratioirrespective of the cooling capacity. Viz, if two compressors, one works with -8°C as evaporative temperature and the other works with +8°C, if u ask me which will consume power. I will reply ..tell me what is the pressure ratio or in another word the condensing temperature to judge. By the way, the explaintion why at a lower evaporative temperature the compressor consumes less power is the density of refrigerant entering the compressor is higher and therefore the compressor volumetric efficiency is enhanced therefore the system is charged by less amount of refrigerant Kg or Ib to give the required capacity and this is answer of why the refrigeration system is more efficient "higher COP" than other peers in AC systems. I wish it could clear
Best regards:)
Dear Mohamed Khamis,
What I wanted to say that you can not judge the capacity based on the HP. The compressor which may be rated for 100 HP must be based on the TE, Tc, refrigerant and the rpm. If the conditions are changed and made suitable for 50 HP then this very compressor shall operate with 50 HP.
What I feel proper is to use the term for capacity is kw at ... Te, ... Tc,...rpm and .... Refrigerant. I am sure you will agree to this.
I would like to know more about this from you.
At lower evaporating temperatures the gas density decreases (while the specific gas volume increases).Quote:
Originally Posted by mohamed khamis
Since the compressor is a constant volume device, if the gas density decreases, the mass flow is reduced.
Hi Us IcemanQuote:
Originally Posted by US Iceman
Absoultely, the decrease in temperture is followed by a increase in Density for the same pressure, I will expalin to you from two sides "thortical and practiacl":
Thoritcal ....gas general law, density = pressure /(gas constant*temperture)
As temperature decreases the density is increased and vice versa however as mentioned in constant pressure which this is the constraint for that.
Practical..if u open ur doemstic fridge door u find the saturated air (from the frezeer) goes up or down? i am very sure u will say it will goes down because the incoming ambinet air is cooled as a result of the cold air and cold surface meeting, thus the cooled air goes down responding to the increase in its density.
Back to the explainstion why in refrigeration application the compressor consumes less power compared with that of AC system, my prevoius explaintion unfortuntey was incorrect due to i miss one paramter ( the pressure in both cases) in account and i will correct here.
Suppose we have two typical compressors of cooling capaicty 5 kW and the same pressure ratio of 4.0 and one will work in refrigeration application of -10°C (compressor A), R22 and the second will service in Ac system of +10°C (compressor B), R22. the question which compressor will be efficent or another word which system in which the both compressors will be installed has the higher COP.
The answer is the compressor in refigeration and the reason is as follows:
The density in refigerntant entering compressor A is smaller becuase both of the pressure and temperture will decresaed togther. However, the decrease in pressure is more signifcantly that the decrease in temperture and accordingly the refrigerant density is decreased and these some figures to emphsize what i say:
at Te = -10°C and assume saturated refrigernt vapor entering the compressor the evporating pressure is Pe = 0.3543 mPa while at Te= +10°C the Pe = 0.6807 mPa, suppose gas constat in 0.287 (as the air the density for sucked refrigerant will be :
For Compressor A = 4.694 kg/m3 : For Compressor B = 8.38 kg/m3
mass flow rate = Cylinder volume *RPM/60*volumteric efficeny * sucked refigaernt density
We assume two typical compressors therefore the cylinder volume, clearance ratio and RPM are the same
volumteric efficeny = 1 + cleance ratio - clearnce ration *(Pressure ration)^(1/n)
The same pressure ratio and the same refrigernat type thus the volumteric effciey is the same , eventually the compressor A (refrigeration appliction) will handle a less a refigerant mass to give the required specifications responding to the decrease in the inlet density and thus the power consumption will be samller.
Regarding to the compressor is a constant volume device, I regret to say i totally disagree with u US iceman becuase it is only constant volume device in one case if the two service valves are closed :cool:and on the same time the compressor has four processes, no one of these process is accomplished under a constant volume
the suction process is under constnat pressure apromxetly regardless the drop in sutiobn valve
88888the compression process is polytropic , the re-expansion of residual gas is also polytropic and discharge process is isobaric process. The only cycle which has the constant volume process is the SIC "spark ignition cycle" which is called otto cycle or car engine cycle and this hypothesized thoery. In addition in majority of compressors'type being used in HVAC systems are positive displacement volume i.e. the boosting in discharge pressure is generated on expense of the change of volume, am i right or wrong?, if u mean that the suction volume flow rate equals the dischage volume rate also i am sorry to say No the mass flow rate only is fixed constant while both are differnt due to the differnt densities in both sides. Sorry for long post
Regards:)
Mohamed khamsi
Dear Samarjit SenQuote:
Originally Posted by Samarjit Sen
Certainly i agree totally with u about this saying the compressor should be identified by the abovementioned sepecifcations
Best regards:)
Hi every body
If the compressor is rated for all conditions using 50 Hp Motor on 100 Hp is simple. For air-conditioning you may need 100 Hp motor, but not for low temperature application where the suction & discharge pressures are very much lower.
Suny
Well that's OK. It is, but it is not a constant volume process. I think we are getting some of out terms confused.Quote:
Originally Posted by mohamed khamis
A compressor operating at a fixed speed has a specific theoretical volume flow rate (cubic meters per second or cubic feet per minute). If the speed remains fixed, the only possible change to this volume flow is when the volumetric efficiency changes as the pressure ratio changes.
For all practical and theoretical considerations, the compressor is a constant volume device. It pumps the same relative volume flow (again only changed by VE affects or speed).
I did not mean the process occurs at constant volume, that would be quite a feat!
You said the same thing I said earlier. At lower temperatures the gas density does decrease because the pressure is lower at the lower temperatures. Therefore the total power input goes down, but the specific power input increases for higher pressure ratios.Quote:
Originally Posted by mohamed khamis
We are not comparing a constant pressure process.
We are looking at the differences between the effects of low temperature operation and higher temperature operation.
Two different evaporating temperature will result in two different mass flows for the same refrigerant and compressor.
Lower saturated temperature/pressures result in lower density and higher specific gas volume, and...
Higher saturated temperature/pressures result in an increase in density and lower specific gas volumes.
You can see this from any refrigerant data.
I think you are trying to explain too many details.;)
The other part I agree with. This explains why air density increases as the temperature of air is lowered. Because it does occur at a relatively constant pressure.
This is different from a refrigeration system. In this case, the compressor is changing the pressures.
Ok US iceman, i got what u meant and thanks for this clarificationQuote:
Originally Posted by US Iceman
Regards:)
Hello !
Interesting problem .
What did you end up doing with the motor and compressor ?
Victorman