Transcript-Calculating Total Cooling Requirements

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Calculating Total Cooling Requirements

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Slide 1: Calculating Total Cooling Requirements

Welcome to the Data Center University™ course: Calculating Total Cooling Requirements.

Slide 2: Welcome

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Click the Notes tab to read a transcript of the narration.

Slide 3: Learning Objectives

At the completion of this course, you will be able to:

Estimate the equipment heat output for a data center

Identify data center cooling system capacity measurement components

Employ common capacity measurement conversion tools

Calculate the total cooling requirements for a data center, and you will be able to

Describe the benefits to “right sizing” the cooling solution

Slide 4: Introduction

All electrical equipment produces heat, which must be removed to prevent the equipment temperature from rising to an

unacceptable level. Most Information Technology equipment and other equipment found in a data center or network

room is air-cooled. Sizing a cooling system requires an understanding of the amount of heat produced by the

equipment contained in the enclosed space, along with the heat produced by the other heat sources typically

encountered.

To begin this process, data center managers must first identify how the heat output rate is measured. This is often

handled in different ways, so let’s discuss those ways now.

Slide 5: Benefits of Standardization

In North America, specifications for power and cooling capacity are still often provided in the legacy BTU and Tons



terms. The term “Tons” refers to the cooling capacity of ice and is a relic of the period from 1870-1930 when

refrigeration and air conditioning capacity were provided by the daily delivery of ice blocks. Other parts of the world

have also been known to use the various specifications such as Joules or Calories, in addition to BTUs and Tons.


Heat is energy and is commonly expressed in Joules, BTUs, Tons, or Calories. Common measures of heat output rate

for equipment are BTUs per hour, Tons per day, and Joules per second. Joules represent total energy while Watts

represent energy transferred per unit time, known as power. The SI system, short for 'The International System of Units'

is the modern international standard version of the metric system and there is no compelling reason why confusing

historic measures need to be used to express the same value, yet any and all of them might be used to express power

or cooling capacities.


It’s relatively easy to see why the standards setting organizations are striving for a standard unit of measurement. The

mixed use of these measures causes a great deal of senseless confusion for users and specifiers, and the overarching

benefit being that cooling requirements can be determined by a simple process when all measurements of power and

cooling are expressed in Watts.

As a result, there is a worldwide trend to phase out the archaic terms of BTU and Tons over time, and for that reason,

this course will discuss cooling and power capacities in Watts. The use of the Watt as the common standard is

fortuitous, as data center power, is already measured in Watts, so using Watts as a standard measure simplifies the

work associated with data center design. Now that we’ve discussed the benefits of standardization, let’s move on to

exploring heat conversion factors.

Slide 8: Heat Conversion Factors

The power transmitted by computing or other information technology equipment through the data lines is negligible.

Therefore, the power consumed from the AC power mains is essentially all converted to heat. This fact allows the

thermal heat output of IT equipment in Watts to simply equal its power consumption in Watts. BTUs per hour, as is

sometimes provided in datasheets, is not necessary in determining the thermal output of equipment. The thermal

output is simply the same as the measured power input.



Slide 9: Heat Conversion Factors

The one exception to this rule is the Power over Ethernet (PoE) switches; in these devices up to 30% of the power

consumed by the device may be transmitted to remote terminals, so their heat load may be lower than the electrical

power they consume. Devices powered by PoE switches are often, but not always, located in the same physical space

as the switch in which case they contribute the full 100% of dissipated heat to the space. In this course, we assume

that the entire electrical power is dissipated locally.

Now, let’s take some time and discuss how to best determine the complete system heat output.

Slide 10: Determining Complete System Heat Output

The total heat output of a system is the sum of the heat outputs of the components. The complete system includes the

IT equipment, plus other items such as UPS, Power Distribution, People, Lighting and Air Conditioning Units.

It’s important to note that the air conditioning units in the data center create a significant amount of heat from fans and

compressors. This heat is exhausted to the outside and does not create a thermal load inside the data center. It

does, however, detract from the efficiency of the air conditioning system and is normally accounted for when the air

conditioner is sized.

Slide 11: Determining Complete System Heat Output

A detailed thermal analysis using thermal output data for every item in the data center is possible, but a quick estimate

using simple rules gives results that are within the typical margin of error of the more complicated analysis. The quick

estimate also has the advantage that it can be performed by anyone without specialized knowledge or training.

Let’s begin calculating total cooling requirements by starting with our data center assumptions.

Slide 12: Data Center Assumptions

To demonstrate how to calculate cooling capacities in the data center using the quick estimate method, let’s create a

sample data center.



This pie chart illustrates the components that contribute to the heat output in the data center.

For the purposes of our example, let’s assume a 5,000 square foot, (465 square meter) data center that is rated at 250

kW. It holds 150 racks and a maximum staff of 20 persons. The data center is assumed to be loaded at 30% of capacity,

which is typical. The total IT load of the data center in this case would be 30% of 250 kW, or 75 kW. Additionally, for the

purposes of the example, it is assumed that the UPS/Batteries and PDUs are located in the same physical space as the

IT load.

The thermal load is 180 kW, which is 50% more than the IT load.

Note that the contributions to the thermal output of the UPS and the Power Distribution are amplified by the fact that the

system is operating at only 30% of capacity. If the system was operating at 100% of capacity, the efficiency of the

power systems would increase and their relative contributions to the thermal output of the system would decrease.

The significant loss of efficiency is a real cost to oversizing a system.

Let’s take each of our heat producing components one by one so that we can understand how to work through our heat

IT Load 70%

People 2%

UPS 13%

Lighting 9%

Power Distribution 6%



output calculation worksheet. Our first component will be the IT equipment.

Slide 13: IT Equipment—Determining Heat Output

To calculate the heat output of the IT equipment, we will need to identify our total IT load in kilowatts. In our original

assumption, we stated that our sample data center is rated at 250 kW (a kW is equivalent to 1000 Watts). We also are

assuming that our data center is loaded to 30% of capacity. Therefore, our heat output on the IT load side is 75 kW (250

kW x .30).

For more information on how to calculate present and future IT power loads, please consider participating in the DCU

course “Calculating Total Power Requirements”.

Slide 14: UPS and Battery—Determining Heat Output

Our next heat producing component is the UPS with battery. The heat output of the UPS consists of a fixed loss and a

loss proportional to operating power. These losses are sufficiently consistent across equipment brands and models

and so they can be estimated without significant error. Therefore they are determined via simple and standardized

rules.

To calculate this loss, we simply take the power rating of our sample data center in this case, 250 kW and multiply by

.04 this calculates to the number 10. We then add this result (10 kW) to the product of our total IT load (75 kW)

multiplied by the loss proportional to the operating factor (.06) to get 4.5 kW. The sum of these two results, 10 kW +

4.5 kW = 14.5 kW.

Now that we’ve calculated the heat output we are generating from our UPS and batteries, let’s move on to the power

distribution system.

Slide 15: Power Distribution—Determining Heat Output

Our next heat producing component is the power distribution system. Like the UPS, the power distribution system

consists of a fixed loss and a loss proportional to operating power. These losses can also be estimated without

significant error based on standard rules.

To calculate this loss, we simply take the power rating of our sample data center (250 kW) and multiply by the fixed loss



factor of .02. We then add this result (5 kW) to the product of our total IT load (75 kW) and multiply by the .02 loss

proportional to operating power, which gives us a figure of 1.5 kW. The sum of these two results, 5 kW + 1.5 kW = 6.5

kW. This is the heat output we are generating from our power distribution system.

Lighting and People will be the next component we need to calculate.

Slide 16: Lighting and People-Determining Heat Output

Lighting and People can also be readily figured using standard values.

To calculate the lighting heat loss, we take the flooring area of our sample data center. If measuring in square feet, we

take our 5,000 square feet of space and multiply it by 2.0. This will give us a measurement of 10000 Watts. If

measuring in meters we take our 465 square meter data center and multiply it by 21.53. This gives us a measurement

of 10000 Watts. We also need to factor in the heat that humans give off while in the data center. For our people

calculation, we take our assumption of 20 people and multiply by 100. This gives us a calculation of 2,000 Watts.

Now, we take each of the measurements and convert them to kilowatts by dividing by 1000, so the Lighting Heat Output

is 10 kW of heat output, and the People Heat Output is 2 kW.

Now let’s take a look at a summation of our calculations.

Slide 17: Heat Output Calculation Summary

This worksheet is a summary of our previous calculations, and provides us with the rapid calculation of the heat load.

Using the worksheet, it is possible to determine the total heat output of a data center quickly and reliably.

The data center will also experience additional heat sources that are encountered in the data center. So let’s explore

those next.

Slide 18: Factoring Other Heat Sources

The prior analysis ignores sources of environmental heat such as sunlight through windows and heat conducted in from

outside walls. Many small data centers and network rooms do not have walls or windows facing the outside, so there

is no error resulting from this assumption. However, for large data centers with walls or a roof exposed to the



outdoors, additional heat enters the data center which must be removed by the air conditioning system. If the data

room is located within the confines of an air-conditioned facility, these other heat sources (sunlight through windows

and heat conducted in from outside walls) may be ignored. However, if the data center has significant wall or ceiling

exposure to the outside, then an HVAC consultant will need to assess the maximum thermal load. These are extra heat

sources and it must be accounted for in the calculation of the thermal output of the complete system.

Humidification will also have an impact on the data center, so let’s discuss that next.

Slide 19: Humidification

In addition to removing heat, a cooling system for a data center is designed to control humidity. Ideally, when the

desired humidity is attained, the system would operate with a constant amount of water in the air and there would be no

need for ongoing humidification. Unfortunately, in most cooling systems the air-cooling function causes significant

condensation of water vapor and consequent humidity loss. Therefore, supplemental humidification is required to

maintain the desired humidity level.

Slide 20: Humidification

Supplemental humidification creates an additional heat load on the CRAC, or cooling unit, effectively decreasing the

cooling capacity of the unit and consequently requiring oversizing.

Slide 21: Humidification—Small Data Centers

For small data rooms or large wiring closets, an air conditioning system which isolates the bulk return air from the bulk

supply air by using ducting can result in a situation where no condensation occurs and therefore no continuous

supplemental humidification is needed. This allows 100% of the rated air conditioning capacity to be utilized and

maximizes efficiency. The required oversizing for a CRAC unit therefore ranges from 0% for a small system with ducted

exhaust air return, to 30% for a system with high levels of mixing within the room.

Slide 22: Humidification—Large Data Centers

For large data centers with high amounts of air mixing, the CRAC unit must deliver air at low temperatures to overcome

the recirculation effects of the higher temperature equipment exhaust air. This results in substantial dehumidification

of the air and creates the need for supplemental humidification. This causes a significant decrease in the performance

and capacity of the air conditioning system. The CRAC system rating must be 1.3 times the anticipated IT load rating



plus any capacity added for redundancy. (This approach works well with smaller network rooms of under 4,000 square

feet / 372 square meters.) This results in the CRAC system being oversized up to 30%.

For more information on humidity in the data center, please consider participating in the DCU course “Fundamentals of

Cooling II Humidity in the Data Center”.

Slide 23: Sizing the Air Conditioning

Once the cooling requirements are determined, it is possible to size an air conditioning system. The following factors,

must be considered:

The size of the cooling load of the equipment (including power equipment)

The size of the cooling load of the building

Oversizing to account for humidification effects

Oversizing to create redundancy

Oversizing for future requirements

The Watt loads of each of these factors can be summed to determine the total thermal load.

Now let’s discuss future trends, along with the benefits to right-sizing the cooling solution.

Slide 24: Future Trends

The air conditioning system in a data center has two key functions: to provide bulk cooling capacity, and distribute cool

air to the IT loads. Room cooling solutions can address low to medium density data center cooling in a traditional open

supply and return air environment. Higher density loads can be handled with containment strategies designed to

provide higher return temperature air back to room cooling units. This increases both room cooling capacity and

efficiency while updating a traditional cooling design. Additionally, high density and variable density IT equipment has

led to the development of newer rack-oriented and row-oriented cooling architectures to address these

challenges. Look for row-oriented cooling to emerge as the preferred solution for state-of-the-art, next generation data

centers. With these new modular cooling technologies, oversizing of cooling systems can become a thing of the past.

For more information on the advantages of row and rack oriented cooling architectures, please consider participating in

the DCU course “Advantages of Row and Rack–Oriented Cooling Architectures I & II”.



Slide 25: Summary

Let’s wrap up with a quick review of the course material we have covered.

A proper assessment of cooling requirements can cut costs significantly. The challenge is to identify the balance

medium between insufficient cooling (and the threat of constant downtime) and over sizing (which is a common

problem resulting in unnecessary investments in unneeded capacity).

A simplified calculation technique with measurements involving IT load, power distribution, UPS, lighting and people

can help to provide a quick cooling estimate.

A first step to simplification is to standardize on a measurement methodology. The industry is looking to replace tons

and BTU measurements with kW measurements.

For larger data centers, the cooling requirements alone are typically not sufficient to select an air conditioner. The

effects of other heat sources such as walls and roof, along with recirculation, are significant and must be examined for

a particular installation.

The design of the air handling ductwork or raised floor has a significant effect on the overall system performance, and

also greatly affects the uniformity of temperature within the data center.

The adoption of a simple, standardized, and modular air distribution system architecture, combined with the simple

heat load estimation method described, could significantly reduce the engineering requirements for data center design.

Containment (HACS & CACS) should be considered as another key approach towards improving data center cooling

performance and efficiency.

Slide 26: Thank You!

Thank you for participating in this course.

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