quiz

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Chapter Three

Human Thermal Comfort

Definition of Comfort: Pursuing an activity without experiencing environmental stress.

Lines of Defense: For Climate –versus –Comfort Challenge

1. Adapting some thermal attributes of the natural environment to assist a person in establishing comfort. For example, setting beneath a tree for shadow in hot day, or warming yourself by sun radiation on a cold day.

2. Modifying the person’s state to reduce environmental stress. For example, putting on a jacket on a cool day.

3. Provide a built enclosure. 4. Employ an active control device to change internal environment (like A.C.).

Thermal Regulation in the Human Body

A human being is a homothermous creature: a creature that maintains a nearly constant

temperature.

Attempt to match between heat productions in the body with the exchange of heat to the

environment.

Heat Production

 Heat is a generated as a by–product of metabolism.

 Metabolism: sum total of all the chemical reactions that occur in all the cells that make up one’s body.

 Metabolic rate: the rate with which heat is produced within the body. Unit is Watt (J/s)

 Heat is produced when:

1. Decomposition of food and formation of ATP (Adenosine tri-phosphate)

2. Energy embodied in the ATP is exchanged to the surrounding tissues.

3. Performance of body functions and activities.

The energy derived from food goes to

20% of the food energy is employed for useful work.

80% exchanged to the surroundings as waste energy.

Human body is 20% efficient

DIRECTION

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How do we get rid of heat?

1. Conduction, Convection, and long-wave radiation (sensible heat loss).

2. Respiration: both sensible heat and latent heat loss.

3. Perspiration: water diffuses through the skin and evaporates at the skin surface.

The heat balance equation between a human being and his/her surrounding.

Heat gain from the environment + metabolic heat production =

Useful work produced + heat loss to the environment

Summary

Human body ingest food

Process it.

Performs useful work.

Sheds heat to surroundings.

At the same time receives heat from the environment.

A heat production, work, and heat exchange occur, the human body tries to maintain a

thermal balance with the environment.

Definition of Thermal Comfort

Givoni: Absence of irritation and discomfort due to heat or cold or a state involving

pleasantness.

Dagostino: Being able to carry on any desired activity without being either chilly or too

hot.

Fanger: the condition of mind which expresses satisfaction with the internal environment.

Conclusion: Thermal comfort is a subjective matter: means is it based on or influenced

by personal feelings, tastes, or opinions.

The most commonly used definition for THERMAL

COMFORT according to the American Society of Heating,

Refrigerating and Air-Conditioning Engineers (ASHRAE) is “That

condition of mind which expresses satisfaction with

the thermal environment and is assessed by subjective evaluation

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As a result, you cannot satisfy all people in the space, but the designer should instead seek

to create a condition that will satisfy the largest number of occupants; 80% according to

ASHRAE.

To shift thermal comfort from subjectivity to objectivity (based on facts) , we rely on eight

parameters.

Eight Parameters that Affect Heat Transfer between the Human Body and

the Environment

Figure: The six Major Environmental and Personal Factors that Influence Thermal Comfort

A. Environment Parameters

1. Dry-bulb temperature (Td) 2. Net mean radiant Temperature (Tmrt) (radiation exchange rate between the body and

the environment). Because walls have different areas, the mean temperature must be a weighted average value according to the following equation:

𝑇𝑚𝑟𝑡 = 𝑇1 × 𝐴1+𝑇2 × 𝐴2 + ⋯ + 𝑇𝑛 × 𝐴𝑛

∑ 𝐴𝑖 𝑛 1

3. Relative humidity (φ).

What is the difference between humidity ratio and relative humidity?

Why do we use relative humidity in weather forecast not humidity ratio?

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a. Humidity Ratio (W)

𝑊 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑟 (𝑚𝑤𝑣)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑑𝑟𝑦 − 𝑎𝑖𝑟 (𝑚𝑑𝑎) 𝑖𝑛 1 𝑘𝑔 𝑜𝑓 𝑚𝑜𝑖𝑠𝑡 − 𝑎𝑖𝑟

Units: dimensionless, or

gwv/gda

gwv/kgda

b. Relative Humidity (φ)

∅ = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑟 (𝑚𝑤𝑣)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑟 𝑎𝑡 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 (𝑚𝑤𝑣,𝑠𝑎𝑡) × 100 𝑎𝑡 𝑠𝑎𝑚𝑒 𝑇, 𝑎𝑛𝑑 𝑃

Units: percentage (%)

Conduct the experiment: bring four different sizes of bakers and record both W and ∅ in the table below:

Take W for the room as 8 gwv/kgda

Humidity Ratio (W) 8 gwv/kgda 8 gwv/kgda 8 gwv/kgda 8 gwv/kgda 8 gwv/kgda

Relative Humidity (∅) 100% (rest

is rain) 100% 50% 30% 10%

Increase temperature

Heating

Cooling Decrease

temperature

Conclusion: Humidity ratio does not give indication about thermal comfort. One single value of humidity ratio can make air either very dry or saturation. That is why, relative humidity is used.

4. Air movement around the body (Vair). Air has to move in order for the body to convect heat to air.

Question 1: Is the ceiling fan in your house a cooling device or a convector?

Question 2: What are the comfort ranges of each of these parameters?

Table: Parameters Comfort Ranges

Parameter Comfort Range

Dry-bulb Temperature (Td) 20oC ≤ Td ≤ 27oC

Mean radiant Temperature (Tmrt) Tmrt = Td

Relative humidity (φ) 35% ≤ φ ≤ 55%

Air Velocity (Vair) 0.25 m/s ≤ Vair ≤ 0.36 m/s

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B. Personal Parameters

1. Activity Level (Metabolic Rate) or heat production rate measured in METS.

1 𝑀𝐸𝑇 = 58.1 𝑊

𝑚2

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The area of the skin can be calculated as:

𝐴𝑠𝑘𝑖𝑛 = 0.202 × 𝑚 0.425 × 𝑙0.725

Where: m = mass of person, kg l = length of person, m The respiratory quotient (RQ) is estimated to be 0.83 for light activity work (MET < 1.5) to 1.0 for extremely heavy work (MET =5.0) Example One Calculate the metabolic rate in W/m2 generated by a student taking notes in a classroom if his skin area is 1.6 m2. How many METS the student is producing? Solution:

The metabolic rate 𝑀 = 21×(0.23×.83+0.77)×8

1.6 = 100.9

𝑊

𝑚2

The number of 𝑀𝐸𝑇𝑆 = 100.9

58.1 = 1.74 𝑀𝐸𝑇𝑆

2. Insulation level of clothing.

Clothing:

a) Restrict the three modes of heat transfer. b) Reduce the ability of the body to exchange heat by perspiration and sweating. c) Clothing has resistance against heat flow OR we say “insulation level”

The clothing insulation level is measured in CLO.

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1 clo = 0.155 (m2. K)

W

 Clo = 0 - corresponds to a naked person.

 Clo = 1 - corresponds to the insulating value of clothing needed to maintain a person in comfort sitting at rest in a room at 21 ℃ with air movement of 0.1 m/s and humidity less than 50% - typically a person wearing a business suit.

Clothing Insulation

Clo m2K/W

Nude

0 0

Underwear - pants

Pantyhose 0.02 0.003

Panties 0.03 0.005

Briefs 0.04 0.006

Pants 1/2 long legs made of wool

0.06 0.009

Pants long legs 0.1 0.016

Underwear - shirts

Bra 0.01 0.002

Shirt sleeveless 0.06 0.009

T-shirt 0.09 0.014

Shirt with long sleeves 0.12 0.019

Half-slip in nylon 0.14 0.022

Shirts

Tube top 0.06 0.009

Short sleeve 0.09 0.029

Light blouse with long sleeves 0.15 0.023

Light shirt with long sleeves 0.20 0.031

Normal with long sleeves 0.25 0.039

Flannel shirt with long sleeves 0.30 0.047

Long sleeves with turtleneck blouse

0.34 0.053

Trousers

Shorts 0.06 0.009

Walking shorts 0.11 0.017

Light trousers 0.20 0.031

Normal trousers 0.25 0.039

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Clothing Insulation

Clo m2K/W

Flannel trousers 0.28 0.043

Overalls 0.28 0.043

Coveralls Daily wear, belted 0.49 0.076

Work 0.50 0.078

Highly-insulating coveralls Multi-component with filling 1.03 0.160

Fiber-pelt 1.13 0.175

Sweaters

Sleeveless vest 0.12 0.019

Thin sweater 0.20 0.031

Long thin sleeves with turtleneck

0.26 0.040

Thick sweater 0.35 0.054

Long thick sleeves with turtleneck

0.37 0.057

Jacket

Vest 0.13 0.020

Light summer jacket 0.25 0.039

Smock 0.30 0.047

Jacket 0.35 0.054

Coats and over-jackets and over-trousers

Overalls multi-component 0.52 0.081

Down jacket 0.55 0.085

Coat 0.60 0.093

Parka 0.70 0.109

Sundries

Socks 0.02 0.003

Thin soled shoes 0.02 0.003

Quilted fleece slippers 0.03 0.005

Thick soled shoes 0.04 0.006

Thick ankle socks 0.05 0.008

Boots 0.05 0.008

Thick long socks 0.10 0.016

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Clothing Insulation

Clo m2K/W

Skirts, dresses

Light skirt 15 cm. above knee 0.01 0.016

Light skirt 15 cm. below knee 0.18 0.028

Heavy skirt knee-length 0.25 0.039

Light dress sleeveless 0.25 0.039

Winter dress long sleeves 0.40 0.062

Sleepwear

Under shorts 0.10 0.016

Short gown thin strap 0.15 0.023

Long gown long sleeve 0.30 0.047

Hospital gown 0.31 0.048

long pajamas with long sleeve 0.50 0.078

Body sleep with feet 0.72 0.112

Robes Long sleeve, wrap, short 0.41 0.064

Long sleeve, wrap, long 0.53 0.082

An overall insulation - or Clo - value can be calculated by simply taking the Clo value for each individual garment worn by a person and adding them together. The

mean surface area of the human body is approximately 1.8 m 2

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Insulation Values According to ASHRAE Fundamentals Handbook (Table 8)

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3. Moisture permeability of the clothing ensemble (how moisture passes). 4. Compressibility of the clothing ensemble (clothing assembly).

When clothing are compressed, they will be less insulative. Why?

Both permeability and compressibility are difficult to measure and they are not accounted for.

Then, in thermal comfort, we use only the first six parameters

Measuring Mean radiant temperature in a Space:

To describe the net radiant exchange heat rate, we use the mean radiant temperature (MRT)

which can be measured by a Globe thermometer.

The globe thermometer consists of temperature sensing device placed inside and at the center of

6 inch (150 mm) diameter hollow copper ball and painted flat black.

𝑇𝑚𝑟𝑡 = [(𝑇𝑔𝑙𝑜𝑏𝑒 + 273) 4

+ 1.1 × 108 × 𝑉𝑎𝑖𝑟

0.6

𝜀 × 𝐷0.4 (𝑇𝑔𝑙𝑜𝑏𝑒 − 𝑇𝑑)]

0.25

− 273

Where:

TMRT = Mean radiant temperature, oC

Tgloble = Globe temperature, oC

Vair = Air velocity, m/s

ε = Emissivity of the globe thermometer (if black= 0.95)

D = Globe diameter, m

Tair = Dry-bulb temperature, oC

Or, you can calculate the Tmrt by the below equation, where A is the surface area = 4 π r2 of the

sphere of the globe thermometer, and hconv is the convective heat transfer coefficient from table

next page.

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𝑇𝑚𝑟𝑡 4 = 𝑇𝑔𝑙𝑜𝑏𝑒

4 + ℎ𝑐𝑜𝑛𝑣 × 𝐴 × (𝑇𝑔𝑙𝑜𝑏𝑒 − 𝑇𝑑)

𝜀 × 𝜎

Tmrt, Tg, and Td are in Kelvin.

Question: What is the value of Tmrt if Tgloble = Td?

Table 3.2: Correlation for calculating convective Heat Transfer Coefficient

Type of Activity Correlation of hconv (W/m2.K)

Applicable Range of velocity (m/s) or met

Seated Person 3.1 0≤V≤0.2

8.3×V0.6 0.2≤V≤4.0

Active Person 5.7×(M-0.85)0.39 1.1≤M≤3.0

Example Two

If the dry-bulb temperature of air (Td) is 20oC, the Globe temperature (Tglobe) is 25oC, its emissivity

(ε ) is 0.95, and air velocity (Vair) in the space is 0.25 m/s, what is the mean radiant temperature?

Applying the equation:

𝑇𝑚𝑟𝑡 = [(𝑇𝑔𝑙𝑜𝑏𝑒 + 273) 4

+ 1.1 × 108 × 𝑉𝑎𝑖𝑟

0.6

𝜀 × 𝐷0.4 (𝑇𝑔𝑙𝑜𝑏𝑒 − 𝑇𝑑)]

0.25

− 273

The answer is Tmrt = 30oC.

Combining Mean Radiant and dry-bulb temperatures

The heat produced by the body must be dissipated to the environment, otherwise the body

would overheat.

If the rate of heat transfer is higher than the rate of heat production, the body cools down and

we feel cold. If the rate is lower, we feel hot.

Complex problem: since it involves radiation, convection, evaporation, and many other

variables.

Question: All heat produced by the body is dissipated. Why?

Answer: The temperature of the body is constant 36-37oC.

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Total energy production rate by the body = production rate of heat (Q) + production rate of

work (W).

Q + W = M * Askin

Where M = rate of energy production per surface area of skin, expressed in units of Mets.

W is ignored because it is included in Q.

The value of met varies with the activity level. Refer to values of metabolic rates.

Total Q = Qconv + Qrad + Qevap + Qresp,sensible + Qresp,latent

Skin Respiration

1. Convection heat Transfer

Qconv = Acl hconv (Tcl – Td)

Acl = area of clothing, m2

Tcl = temperature of clothing, K

Td = dry-bulb temperature, K

Hconv = convective heat transfer coefficient, W/(m2.K)

2. Radiative Heat Transfer

Definition: The mean radiant temperature (Tmrt) is the temperature of the environment with

which a human body would exchange the same radiation with the actual environment.

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𝑇𝑚𝑟𝑡= ∑ 𝐹𝑐𝑙−𝑛𝑛 × 𝑇𝑛

Where Fcl-n =Area i/Total Area = radiation factor

Tn = surface temperature

Then, 𝑄𝑟𝑎𝑑 = 𝐴𝑐𝑙 ℎ𝑟𝑎𝑑 (𝑇𝑐𝑙 − 𝑇𝑚𝑟𝑡)

Combining Equations

Let us sum both the convective and the Radiative coefficients.

hc+r = hconv + hrad

and create an operative temperature (Top), where:

𝑇𝑜𝑝 = ℎ𝑐𝑜𝑛𝑣 × 𝑇𝑑 + ℎ𝑟𝑎𝑑 × 𝑇𝑚𝑟𝑡

ℎ𝑐𝑜𝑛𝑣 + ℎ𝑟𝑎𝑑

Usually hconv and hrad are close at indoor conditions (hconv = hrad), then:

𝑇𝑜𝑝 = 𝑇𝑑 + 𝑇𝑚𝑟𝑡

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Then;

𝑄𝑐𝑜𝑛𝑣 + 𝑄𝑟𝑎𝑑 = 𝐴𝑐𝑙 ℎ𝑐𝑜𝑛𝑣+𝑟𝑎𝑑 (𝑇𝑐𝑙 − 𝑇𝑜𝑝)

The sensible heat transfer is

𝑄𝑐𝑜𝑛𝑣+𝑟𝑎𝑑 = 𝐴𝑐𝑙𝑜𝑡ℎ𝑖𝑛𝑔 × (𝑇𝑠𝑘𝑖𝑛 − 𝑇𝑜𝑝)

𝑅𝑐𝑙𝑜𝑡ℎ𝑖𝑛𝑔 + 1

ℎ𝑐𝑜𝑛𝑣+𝑟𝑎𝑑

The latent heat transfer is

𝑄𝑙𝑎𝑡 = 𝑚𝑤𝑣 × ℎ𝑓𝑔

Where:

mwv =The rate of evaporation from the body, kg/s (approximately =0.012 kgwv/s)

The enthalpy of vaporization of water =2430 kJ/kg at 30oC.

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ASHRAE Comfort Chart

ASHRAE has developed an industry consensus standard to describe comfort requirements in buildings. The standard is known as ASHRAE Standard 55-2004 Thermal Environmental Conditions for Human Occupancy. The purpose of this standard is to specify the combinations of indoor thermal environmental factors and personal factors that will produce thermal environmental conditions acceptable to a majority of the occupants within the space. One of the most recognizable features of Standard 55 is the ASHRAE Comfort Zone as portrayed on a modified psychrometric chart given in Figure shown in the next page. The Standard allows the comfort charts to be applied to spaces where the occupants have activity levels that result in metabolic rates between 1.0 met and 1.3 met and where clothing is worn that provides between 0.5 clo and 1.0 clo of thermal insulation. The comfort zone is based on the PMV values between -0.5 and +0.5.

If the moisture leaving an average resting

person’s body in one day were collected

and condensed it would fill a 1-L container

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Optimal Air Temperature for Comfort

𝑇𝑑,𝑜𝑝𝑡𝑖𝑚𝑎𝑙 = 27.2 − 5.9 × 𝑅𝑐 − 3.0 × (1 + 𝑅𝑐)(𝑀 − 1.2)

Where:

Rc = clothing resistance, clo

M = Metabolic rate, MET

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Example Three

A room 3 m × 3 m × 3 m with five surfaces at Ts =20oC except one surface which is a window at

11oC. The dry-bulb temperature is (Td) 21oC, and the relative humidity (ø) is 40%. Calculate the

mean radiant temperature (Tmrt) and the operative temperature (Top), and use ASHRAE comfort

chart to specify whether the conditions are comfortable or not.

Solution

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Example Four It is well established that a clothed or unclothed person feels comfortable when the skin temperature is about 33oC. Consider an average man (mass = 64 kg, length = 1.68 m) wearing summer clothes whose thermal resistance is 0.6 clo. The man feels very comfortable while standing in a room maintained at Td= 22oC. The air motion in the room is small and can be neglected, and the interior surface temperature of the room (Tmrt) is about the same as the air temperature. If this man were to stand in that room unclothed, determine the temperature at which the room must be maintained for him to feel thermally comfortable. Assume that the latent heat transfer rate from the person remains constant. Solution: Hconv = 4.0 W/(m2.K) Hrad = 4.7 W/(m2.K) Hconv+rad = 8.7 W/(m2.K) Askin = 0.202 * 640.425 * 1.680.725 = 1.72 m2 Rclothing = 0.6 * 0.155 = 0.093 (m2.K)/W Top = 22oC Then; Qsensible, clothed = 91 W When the person is unclothed, Rclothing = 0 Then; In the same equation, Qsensible is known but Td is unknown. Td = 33- 91/(8.7*1.72) = 26.9oC.

The person will feel comfortable when nude at 26.9oC.

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Example Five

Use the new ASHRAE comfort chart to determine whether the conditions listed below are

expected to be comfortable for light office work (MET ≤ 1.1).

a. Summer: Td = Tmrt = 23oC, ø =60%, hconv=hrad

b. Summer: Td = 22oC, Tmrt= 28oC, hconv=hrad, ø =30%

c. Winter: Td=25oC, Tmrt =20oC, ø =30%, hconv=hrad

d. Summer: Td = 22oC, Vair = 1 m/s, Tglobe =25oC, hrad= 4.5 W/(m2.K), hconv= 3.5 W/(m2.K),

W = 11 gwv/kgda.

Calculate Tmrt for the equation

Tmrt = 31.8oC

𝑇𝑜𝑝 = 3.5×22+4.5×31.8

3.5+4.5 = 27.5 oC

From the comfort chart at Top =27.5oC and W= 11gwv/kgda, the answer is NO.

Question: What do you do to bring thermal comfort conditions back to the room?

Referring to chart, the maximum Top to enter the comfort zone is = 27oC, then repeat the

same equation by substituting Top=27oC and find Td again.

𝑇𝑜𝑝 = 3.5 × 𝑇𝑑 + 4.5 × 31.8

3.5 + 4.5 = 27℃

Then, Td =20.8oC.

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Fanger’s Thermal Comfort Model When discussing thermal comfort, there are two main different models that can be used: the static model (PMV/PPD) and the adaptive model. The PMV/PPD model was developed by P.O. Fanger using heat-balance equations and empirical studies about skin temperature to define comfort. Standard thermal comfort surveys ask subjects about their thermal sensation on a seven-point scale from cold (-3) to hot (+3).

Table: ASHRAE Thermal Sensation Scale

Value Sensation

+3 Hot

+2 Warm

+1 Slightly warm

0 Neutral

-1 Slightly cool

-2 Cool

-3 Cold

Fanger's equations are used to calculate the Predicted Mean Vote (PMV) of a group of subjects for a particular combination of air temperature, mean radiant temperature, relative humidity, air speed, metabolic rate, and clothing insulation. PMV equal to zero is representing thermal neutrality, and the comfort zone is defined by the combinations of the six parameters for which the PMV is within the recommended limits (-0.5<PMV<+0.5). Although predicting the thermal sensation of a population is an important step in determining what conditions are comfortable, it is more useful to consider whether or not people will be satisfied. Fanger developed another equation to relate the PMV to the Predicted Percentage of Dissatisfied (PPD). This relation was based on studies that surveyed subjects in a chamber where the indoor conditions could be precisely controlled. The PMV/PPD model is applied globally but does not directly take into account the adaptation mechanisms and outdoor thermal conditions. ASHRAE Standard 55-2017 uses the PMV model to set the requirements for indoor thermal conditions. It requires that at least 80% of the occupants be satisfied.

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Example Six (Fanger Thermal Comfort Chart)

A room 10 m  8 m  4.0 m at sea level (P= 101,325 kPa) is maintained at a dry temperature (Td) of 22oC.

The 20 persons occupying this room must perform a medium activity work and wear light clothing. If the

average velocity of the air in the room is 0.5 m/s, and the globe temperature as measured by a globe

thermometer is 22oC, answer the following questions:

1. Calculate the mean radiant temperature (Tmrt) in this room. The globe diameter is 150 mm, and its emissivity is 0.95.

2. Use Fanger Comfort charts and determine the relative humidity and the wet-bulb temperature in the room.

3. If the relative humidity is considered low and people decided to raise it to 65%, what must be the new air velocity in the room.

4. What must be the air temperature in winter when people wear more clothing (clo = 1)? (based on above conditions).

5. How much moisture is released into the air if latent heat generated by each person is 55 W and enthalpy of water vapor is 2450 kJ/kgwv.

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𝑚𝑜𝑖𝑠𝑡𝑢𝑟𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 =

55 𝐽 𝑠. 𝑝𝑒𝑟𝑠𝑜𝑛

𝑠. 2450 × 1000 𝐽 𝑘𝑔𝑤𝑣

× 20 𝑝𝑒𝑟𝑠𝑜𝑛 = 4.5 × 10−4 𝑘𝑔𝑤𝑣

𝑠 = 1.62

𝑘𝑔𝑤𝑣 ℎ𝑟

Question: Which device throws this excess moisture to outside?

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