Cooling capacity

[jaison]

Hai sirs,
I am working in many types of refrigeration and ac equipments,my question is how we can find compressor capacity by the given measurements, i mean condenser length, evaporator length, diameter,fan speed, compressor LRA, aplication and temperature.we can not get these all but we try it to get more, can any body give detailed reply.

[US Iceman]

Everything you want to know would take several college size text books to complete. These are all intensive engineering questions.

However, you can estimate some basic capacities if you make some assumptions. Assume you know the fan delivery volume (cubic meters per second). You can find this from the original manufacturers specifications.

Second measure the inlet and outlet temperature of the air across a coil.

If you know the estimated air volume flow and the temperature difference of the air you can estimate how much heat is being removed or rejected.

The same applies to water flow.

These are what I refer to as "reality checks". The heat being rejected should be larger than the quantity of heat being removed as a general note.

This also tells you the general cooling capacity of the system and also helps to verify the basic equipment capacity requirements.

[mohamed khamis]

Hai sirs,
I am working in many types of refrigeration and ac equipments,my question is how we can find compressor capacity by the given measurements, i mean condenser length, evaporator length, diameter,fan speed, compressor LRA, aplication and temperature.we can not get these all but we try it to get more, can any body give detailed reply.

Hi Jaison

there are two ways (one is short run and second is long run)to get the system cooling capacity:

The short run


1. measure the evaporator length and width and measure frontal air velocity at evaporator face area otherwise it is normally taken as of 2.5 m/s for evaporator with plain fins and 3 m/s for evaporator with corrugated fins. Thus:

Evaporator air mass flow rate = air density (1.2 kg/m3)*frontal area (length *width)*frontal air velocity

2. measure the average air temperature before and after the evaporator

The system cooling capacity = Evaporator air mass flow rate*wet specific heat (1.18)*(inlet evaporator air temperature - outlet evaporator air temperature)/SHF

SHF = coil sensible heat factor and it is ususally from 0.6 to 0.65

If u want the derivation of this equation i can post it to u

The long run

1. measure the condenser length and width and measure frontal air velocity at condenser face area otherwise it is normally taken as of 2.5 m/s for condenser with plain fins and 3 m/s for condenser with corrugated fins. Thus:

air mass flow rate = air density (1.2 kg/m3)*frontal area (length *width)*frontal air velocity

2. measure the average air temperature before and after the condenser

3. condenser heat rejection = air mass flow rate
2. measure the compressor electrical current and voltage , thus

compressor input power= V*I* power factor
(measured or usually taken of 0.8 at a rated compressor speed) then

cooling capacity = condenser heat rejection - compressor input power /(efficiency of motor or mechanical efficiency and normally taken of 0.8)

For water chillers

cooling capacity = water flow rate (kg/s) *4.187*(inlet water temperature - outlet water temperature)

I wish it could help

Cheeers

[jaison]

Thank you Iceman & Mohamed khamis,
About refrigerators and water coolers , with condenser pipe length and ***** type and aplication temperature how we can calculate compressor capacity.some systems evaporator length we can not calculate some times we can calculate evaporator length,we ca not assume compressor data.

[Brian_UK]

Jaison, are you trying to design new equipment by disassembling old equipment?

It seems that you need to undertake some college work to enable you to progress further.

Reverse engineering does not always work successfully ;)

[Samarjit Sen]

Hello Mohamed Khamis,

Could you please provide further details about assessing the cooling capacities. Further what about a Shell & Tube Condenser. I am interested to know about this also.

[Samarjit Sen]

There are times which happens quite frequently, a client who may be the actual user or a refrigeration contractor wants me to check a system and give them my opinion as the plant is not working properly or is not maintaining the desired conditions. In such cases it is very difficult to make an assessment as most cases, the equipments installed are either of local made or are second hand used units from a ship cutting yard.

I would like to know as to how ( Short run) assess the capacity of the Evaporator Coils, Air Cooled Condensers and the Shell & Tube Condensers. Is ther any thumb rule for this. Specially the Shell & Tube condensers, which has inside tubes of various diameters.

[mohamed khamis]

Hello Mohamed Khamis,

Could you please provide further details about assessing the cooling capacities. Further what about a Shell & Tube Condenser. I am interested to know about this also.

Hi Samarjit Sen

Generally speaking, the sensible heat is quite easy to be determined comparing to the latent heat. So, in the process of combined sensible and latent we try to shun off the calculation of the latent heat by knowing the SHF for the coil.

Generally, the sensible heat = mair *cpa*(tai - tao) = SHF * cooling capacity

Cp : is the specific heat, kJ/kg.K

therefore, cooling capacity = {mair *cpair*(tai - tao)}/SHF

And generally SHF is 0.6 to 0.65 for frontal air velocity of 2.5m/s (500fpm) and 0.65 to 0.7 for frontal air velocity of 3.0 m/s (600fpm) as a result of enhancing in air heat transfer coefficient and in turn the sensible removal heat. Thus u can know roughly the range of ur rated "design" capacity.

For the coil in which the heat transfer is only sensible such as liquid chiller "shell-and-tube evaporator" or water -cooled condenser "shell-and-tube, double pipe,ect" or air-cooled condenser, the rule is

Heat or cooling capacity = coolant mass flow rate *CPa or CPw*air or water temperature difference

Me personally i tried this rule of thumb and it gives interesting results...I will give a real evaporator

The coil specifications

design dbti = 80F and wbti = 67 F, dbto = 55F and wbti = 53 F, face velocity = 500 fpm (2.5 m/s), R-12, saturation temperatures are of 35F/125F, Length = 24 in (0.61 m) and width = 18 in (0.46 m), 10 fpi , 1500.0 CFM, 0.625 in outside diamter,tubing rows = 4, tubes spaces =1.5*1.3 inch and 6 tubing circuitry the system cooling capacity = 61403.7 BTU/HR (17.98 kW)

Now we will apply the rule,

the cooling capacity = {2.5 * (0.61*0.46)*1.2*1.005*(80-55)*(5/9)}/0.6 = 19.58 kW
(66877.8 BTUh)

and when SHF = 0.65 ....cooling capacity = 18.07 kW (61722.2 BTU)

Therefore, cooling capacity in the range of 18.07 kW to 19.58 kW, is it marvelous rule. I wish it could help and I'd like to be in ur service.

Best regards:)

[mohamed khamis]

Hi Samarjit Sen

Generally speaking, the sensible heat is quite easy to be determined comparing to the latent heat. So, in the process of combined sensible and latent we try to shun off the calculation of the latent heat by knowing the SHF for the coil.

Generally, the sensible heat = mair *cpa*(tai - tao) = SHF * cooling capacity

Cp : is the specific heat, kJ/kg.K

therefore, cooling capacity = {mair *cpair*(tai - tao)}/SHF

And generally SHF is 0.6 to 0.65 for frontal air velocity of 2.5m/s (500fpm) and 0.65 to 0.7 for frontal air velocity of 3.0 m/s (600fpm) as a result of enhancing in air heat transfer coefficient and in turn the sensible removal heat. Thus u can know roughly the range of ur rated "design" capacity.

For the coil in which the heat transfer is only sensible such as liquid chiller "shell-and-tube evaporator" or water -cooled condenser "shell-and-tube, double pipe,ect" or air-cooled condenser, the rule is

Heat or cooling capacity = coolant mass flow rate *CPa or CPw*air or water temperature difference

Me personally i tried this rule of thumb and it gives interesting results...I will give a real evaporator

The coil specifications

design dbti = 80F and wbti = 67 F, dbto = 55F and wbti = 53 F, face velocity = 500 fpm (2.5 m/s), R-12, saturation temperatures are of 35F/125F, Length = 24 in (0.61 m) and width = 18 in (0.46 m), 10 fpi , 1500.0 CFM, 0.625 in outside diamter,tubing rows = 4, tubes spaces =1.5*1.3 inch and 6 tubing circuitry the system cooling capacity = 61403.7 BTU/HR (17.98 kW)

Now we will apply the rule,

the cooling capacity = {2.5 * (0.61*0.46)*1.2*1.005*(80-55)*(5/9)}/0.6 = 19.58 kW
(66877.8 BTUh)

and when SHF = 0.65 ....cooling capacity = 18.07 kW (61722.2 BTU)

Therefore, cooling capacity in the range of 18.07 kW to 19.58 kW, is it marvelous rule. I wish it could help and I'd like to be in ur service.

Best regards:)

Since we are talking about the rule of thumb and in order to make a comprehensive rule. The condenser heat rejection also can be obtained by ;

condenser heat rejection = cooling capacity *{1 + (1/COP)}

Compressor input power = cooling capacity *{(1/COP)}

and COP is usually 3 to 4 and it is of 2 to 3 in harsh conditions and i don not know Lana will quarrel again with me about these values or not.

thus the condenser heat rejection is around 22.5 kW to 24 kW and therefore the compressor input power is 4.5 kW to 6 kW.
Cheers

[Samarjit Sen]

Hi Mohamed Khamis,

Thank you for such detail information. Today I posted a seperate thread seeking these details. What I want to know that at site when I am called to check a plant for inefficient performance, how can I find out the capacities on the basis of the physical dimensions.

I am sure Lana would not quarell with any of the dimensions.

[mohamed khamis]

Hi Mohamed Khamis,

Thank you for such detail information. Today I posted a seperate thread seeking these details. What I want to know that at site when I am called to check a plant for inefficient performance, how can I find out the capacities on the basis of the physical dimensions.

I am sure Lana would not quarell with any of the dimensions.

U welcome at any time Samarjit Sen;

Cheers

[anwarzaw]

hi...

I just register today and I am a junior HVAC engineer. I am work,ing on a conderser design for A/C. I have selected the compressor according to cooling capacity and power input from a compressor manufacturer company named TECUMSEH.

my question is that can I design the condenser as condenser capacity equal to cooling capacity of compressor + power input to the compressor? Do I need to consider any motor efficiency ?

thanks...

anwar


Hi Jaison

there are two ways (one is short run and second is long run)to get the system cooling capacity:

The short run


1. measure the evaporator length and width and measure frontal air velocity at evaporator face area otherwise it is normally taken as of 2.5 m/s for evaporator with plain fins and 3 m/s for evaporator with corrugated fins. Thus:

Evaporator air mass flow rate = air density (1.2 kg/m3)*frontal area (length *width)*frontal air velocity

2. measure the average air temperature before and after the evaporator

The system cooling capacity = Evaporator air mass flow rate*wet specific heat (1.18)*(inlet evaporator air temperature - outlet evaporator air temperature)/SHF

SHF = coil sensible heat factor and it is ususally from 0.6 to 0.65

If u want the derivation of this equation i can post it to u

The long run

1. measure the condenser length and width and measure frontal air velocity at condenser face area otherwise it is normally taken as of 2.5 m/s for condenser with plain fins and 3 m/s for condenser with corrugated fins. Thus:

air mass flow rate = air density (1.2 kg/m3)*frontal area (length *width)*frontal air velocity

2. measure the average air temperature before and after the condenser

3. condenser heat rejection = air mass flow rate
2. measure the compressor electrical current and voltage , thus

compressor input power= V*I* power factor
(measured or usually taken of 0.8 at a rated compressor speed) then

cooling capacity = condenser heat rejection - compressor input power /(efficiency of motor or mechanical efficiency and normally taken of 0.8)

For water chillers

cooling capacity = water flow rate (kg/s) *4.187*(inlet water temperature - outlet water temperature)

I wish it could help

Cheeers