quiz

I need help with environment quiz 

it is 3 chapters and its tomorrow 3pm KSA time 

it is one hour and a half

I have sample questions of the quiz

it consists of 20 questions only

1

Chapter One

Introduction
HVAC systems are of great importance to architectural design efforts for three main reasons.

1. These systems consume substantial floor space and/or building volume for equipment and

distribution elements that must be accommodated during the design process. According to the

American society of Heating, ventilating and Air-conditioning Engineers (ASHRAE), the required

area is 6-9% of the area served.

2. HVAC systems constitute a single major budget item in building projects, that is not less than 10%

of the total budget.

3. The success of a building depends on the ability to provide thermal comfort with the least

operating costs (maintenance, energy, or replacement). This depends on the HVAC system

design, equipment and controls.

Difference between Passive and Active Systems

Passive design is a system or structure that directly uses natural energy such as sunlight, wind,

temperature differences or gravity to achieve a result without electricity or fuel. Active design is a system

or structure that uses or produces

electricity.

The term passive design is most often used with respect to architecture and infrastructure. For example, a

building may have wide windows that automatically let in more light when the building needs heat and

automatically shade when the building is too hot.

Another common area of passive design is wet infrastructure such as drainage systems that generally

often don’t consume power but use gravity to move water. Most devices and infrastructure have an

active design as they use electricity. The term is typically only used in comparison to passive designs. For

example, solar panels that produce electricity are often referred to as active solar as a comparison to

using solar passively for heat or to grow plants.

Benefits of Passive Systems

Passive designs are often valued for their simplicity and aesthetic appeal. They also tend to have zero

operational costs. As they often contain no moving parts, passive designs potentially last for centuries.

Electrical components are valued for their accuracy and functionality but may need to be regularly

maintained and replaced. They may also have a higher operational cost and environmental impact.

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2

Table: Passive System versus Active System

Passive Design Active Design

Definition

Infrastructure, architecture and devices

that achieve a result by directly using

natural forces without first converting it to

electricity.
Infrastructure, architecture and devices

that use or produce electricity to achieve a

result.

Examples

Passive Heating

Passive Cooling

Green Roofs

Rain Gardens

Solar Panels

Wind Turbines

District Heating

Deep Water Cooling

Operating and

Maintenance Cost
Almost nil Usually high

Economics of Energy Efficiency

Example One

Which car would you buy if both have the same facilities and amenities, same life span, and same gas

consumption, Honda Accord costing BD. 9,000 or Toyota Camry costing BD. 8,000?

Example Two:

Which car would you buy if:

Accord: cost is BD. 9,000, gas consumption is 50 liters/week, Life span is 10 years.

Or,

Camry: cost is BD. 8,000, gas consumption is 60 liters/week, life span is 10 years?

Example Three:

Two refrigerators with same quality and quantity of service:

Option A: costs BD. 600, draws electricity at an average of 150 W, lifespan is 20 years

Option B: costs BD. 700, draws electricity at an average of 100 W, lifespan is also 20 years

If cost of electricity is BD. 0.09/kWh, which one would you choose?

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3

Solution

Example Four

What if option B in example three has a lifespan of 15 years?

Solution

Rule of thumb:

Profitability is Excellent if pbp less than one-third of lifetime of investment;

Profitability is good if pbp less than one-half of lifetime of investment;

Profitability is bad if pbp is more than one-half of lifetime of investment.

ChapterTwo- Handout Four

Radiation Heat Transfer

Reminder: In the first handout of radiation heat transfer, we calculated the radiation exchange between surfaces inside
buildings. This class is the exchange of thermal radiation with outside surfaces. Outside surfaces can be opaque like
walls, roof, and doors, or transparent surfaces like windows. This handout treats radiation between outside opaque
surfaces

Solar Heat Gain Through Opaque Surfaces (walls and Roof)

The sun is the main heat source of the earth, and without the sun, the environment temperature would not be much
higher than the deep space temperature of -270

o
C. The solar energy stored in the atmospheric air, the ground, and the

structures such as buildings during the day is slowly released at night, and thus the variation of the outdoor temperature
is governed by the incident solar radiation and the thermal inertia of the earth. Heat gain from the sun is the primary
reason for installing cooling systems, and thus solar radiation has a major effect on the peak or design cooling load of a
building, which usually occurs early in the afternoon as a result of the solar radiation entering through the glazing
directly and the radiation absorbed by the walls and the roof that is released later in the day. The effect of solar
radiation for glazing such as windows is expressed in terms of the solar heat gain factor (SHGF). For opaque surfaces
such as the walls and the roof, on the other hand, the effect of solar radiation is conveniently accounted for by
considering the outside temperature to be higher by an amount equivalent to the effect of solar radiation. This is done
by replacing the ambient temperature in the heat transfer relation through the walls and the roof by the sol-air
temperature, which is defined as the equivalent outdoor air temperature that gives the same rate of heat transfer to a
surface as would the combination of incident solar radiation, convection with the ambient air, and radiation exchange
with the sky and the surrounding surfaces . Heat flow into an exterior surface of a building subjected to solar radiation
can be expressed as

Figure: The sol-air temperature (Tsol-air) represents the equivalent outdoor air temperature that gives the same rate of heat flow to a
surface as would the combination of incident solar radiation and convection/radiation with the environment

Where qsolar is the solar irradiance falling on the surface, a function of orientation and latitude.

Then, the heat transfer through a wall (or roof) can be expressed as:

accounting for conduction, convection, and radiation

Note the temperature rise due to solar radiation is

Then, the rate of additional heat gain from solar through the wall becomes:

accounting for radiation only

The total solar radiation incident on the entire wall is Qsolar=A× qsolar. Therefore, the fraction of the incident solar heat
transferred to the interior of the house is

Example:

The west masonry wall of a house located in Bahrain (latitude 26

o
N) has an area of 19.5 m

2
and is made of 200-mm thick

concrete blocks with overall U-value of 1.65 W/(m
2
.K) including both outside and inside convection coefficients. The

reflectivity of the wall surface is 55%. If the interior of the house is to be maintained at 24
o
C, answer the following

questions:

a. The sol-air temperature on July 21 which the design date for cooling load. The solar irradiance is 772 W/m
2
on

west.

αs=1-ρ =1-0.55 =0.45

Tsol-air = To + αs×q/ho = 39.2 +0.45×772/22.7 = 54.5

o
C

b. The total heat gain through the wall (conduction, convection, and radiation).

Qwall = UA (Tsol-air – Ti) = 1.65×19.5 (54.5-24) = 981 W

c. What is the temperature rise and heat gain due to radiation only.

Temperature rise due to radiation only (∆Tsol) = 54.5 – 39.2 =15.3

o
C

The heat gain due to radiation only Qrad =UA ∆Tsol = 1.65×19.5×15.3 =492 W

d. Calculate the fraction of radiation heat gain from the total heat gain.

Fraction = (492/981)×100 =50%

e. Calculate the fraction of radiation transferred to that of total incident on the wall. Comment on the answer.

Solar fraction transferred =

=1.65×0.45/22.7 = 0.033 (3.3%)

University of Bahrain

College of Engineering

Department of Mechanical Engineering

Second Semester 2017-1

8

Mechanical Installations in Buildings (MEG 435)
Environmental Control Systems II (ARCG 226)

First Examination

March 21, 2018

Student ID#:_________________
(Please DO NOT write your name)

(circle your section)

ARCG 226 1 2 3

MEG 435 1

Selection √ Question Highest Score Student Score

1 25

2 25

3 25

4 25

Total 75*

*Based on answering three questions only.

Instructions

1. This exam is closed book, closed notes.
2. Make sure that you have four questions, in six pages numbered from 1 to 6, plus

four additional pages of graphs and tables and formulae sheets.
3. You are required to answer three questions only. Tick the three selected

questions in the table above.
4. You have seventy-five minutes to complete the exam.
5. Read each question with care and be sure to answer all parts of each question.
6. All answers must be placed in the question sheets directly below the question

where space is provided.
7. You should show units in your solutions. If you do not indicate the units or the

units indicated are wrong, you will lose one point for each mistake even though
the numerical answer is correct.

8. As a suggestion, work the easiest question first; question number one is not
necessarily the easiest.

Good Luck…

1

Question One A (Payback Period)

Suppose that the annual bill for heating water for a 4-star hotel is BD.1000/year using an
electrical centralized heating system. The hotel engineer suggested to the hotel’s
management investing BD. 5,000 for a solar heating system that can make the annual bill
BD. 600/year only.

1. How much savings the hotel would acquire if this solar heating system was installed.

2. Calculate the pay-back period (pbp) for this new system.

Question One B (Shape Factors)

The figure below shows a triangular sunroom for a house. The sunroom has five surfaces,
namely: floor (f), Wall (w), Roof (r), and two triangular sides (s). Use the relationship
below to answer the questions:

Ff→f + Ff→w + Ff→r + Ff→s + Ff→s = 1

1. What is the value of the shape factor between the floor and itself (Ff→f)?

2. Calculate the shape factor between the floor and the wall (Ff→w).

3. Calculate the shape factor Ff→r if Ff→s =0.1

5

5 m

5 m
10 m

2

Question Two (Conduction/Convection Heat Transfer)

The Figure shown below is a schematic diagram of a wall constructed of layers arranged
in series as in the Figure below. The wall has a length of 20 m and a height of 3 m. The
inside air temperature (Ti) = 23oC, and the outside air temperature (To) = 39oC. Use the
information listed in the table to answer the following questions.

Layer ∆x
(cm)

k
(W/(m.K)

Rth
(m2.K/W)

Outside Convection (ho) – –

Red brick veneer 10 0.20

Air space 5 0.08

Rigid insulation 5 0.04

Moisture/vapor barrier 0.1 0.21

Limestone aggregate concrete Blocks filled with perlite 20 (1)

Inside convection (hi) (ρ=10%) – –

Rth,total =

1. From the attached ASHRAE Fundamentals Handbook table 10, find the values of both the
outside and inside convection coefficients.

Outside convection coefficient (ho) =

Inside convection coefficient (hi) =

2. Fill in cell (1) in the table with the value of the thermal resistance (Rth) of limestone
aggregate concrete blocks from the attached ASHRAE fundamentals Handbook table 1.

3

3. Calculate the total thermal resistance (Rth,total) of the wall.

4. Calculate the heat transfer rate (Q) across the wall.

5. As an architect, suggest four ways from the course materials you learned to reduce the
heat transfer through the wall.

a. ______________________________________________________________________________________________

b. ______________________________________________________________________________________________

c. ______________________________________________________________________________________________

d. ______________________________________________________________________________________________

Question Three (Radiation Heat Transfer)

4

The South wall of a building in Bahrain is made with the layer arrangement shown in the
figure below. The design outdoor temperature for Bahrain is 39oC. The building is
maintained at 24oC. The sky temperature is 10oC. The incident solar radiation (qsol) falling
on the wall in July is 420 W/m2. If the area of the wall is 150 m2, its U-value is 0.35
W/(m2.K) and its outside reflectivity (ρ) is 85%, answer the following questions:

1. Draw the thermal resistance network through the wall accounting for both conduction
and convection.

2. Calculate the sol-air temperature (Tsol-air) for the wall in July. Take ho =22.7 W/(m2.K).
ignore infra-red radiation portion.

3. Calculate the total heat gain (Qtotal) through the wall by conduction, convection, and
radiation.

A

B

C D
To
=

Ti

Tsky = 10oC

qsol= 420 W/m
2

5

4. Calculate the fraction of the above heat gain due to solar only.

5. If the outside surface of the wall is at 38oC, calculate the radiation exchange (Qrad) between
the wall and the sky. The wall is totally exposed to sky.

6

Question Four (Ideal Gas Law and First Law of Thermodynamics)

A single storey building located in Manama, Bahrain (Patm=101.325 kPa, ρ=1.2 kg/m3)
with a floor area of 1000 m2 and a height of 3 m has air change per hour (ACH) = 0.3. The
outdoor air temperature (To) is 36oC. Take the specific heat of air (cp) = 1006 J/(kg.K) and
Rair = 287 J/(kg.K) and answer the following questions:

1. Use the ideal gas law to calculate the room air temperature (T).

2. Calculate the infiltration volumetric flow rate (Vinf) through the building.

3. How much heat (Qinf) must be removed by the building HVAC system to cool this

infiltrating outside air.

4. Convert the value above to refrigeration tons.

5. If the same building is located in Boulder, Colorado (1,500 m above sea level), will

the cooling requirements be higher or lower than the value found for Bahrain?
Explain. No need to do calculations.

7

8

Formulae Sheets

Chapter 1

Chapter 2

𝑅𝑡ℎ =
∆𝑥

𝑘

𝑅 =
𝑅𝑡ℎ
𝐴

𝑈𝐴 =

1

𝑅𝑡𝑜𝑡𝑎𝑙

𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑈 × 𝐴 × (𝑇𝑠𝑜𝑙−𝑎𝑖𝑟 − 𝑇𝑖)

𝑄𝑠𝑜𝑙 = 𝑈 × 𝐴 × ∆𝑇𝑠𝑜𝑙

𝑄𝑟𝑎𝑑 = 𝐴𝜀𝜎(𝑇𝑠

4 − 𝑇𝑠𝑘𝑦
4 )

𝑃𝑣 = 𝑅𝑎𝑖𝑟𝑇 where T is Kelvin

𝑃𝑉 = 𝑚𝑅𝑎𝑖𝑟𝑇

Savings

Investment
pbp



TAUQ cond 

.

R

T

R

T

A

Q

th







.

)( sfconvconv TTAhQ 

 

2

2

12

1

21

1

4

2
4

11
.

1
A
A

F

TTA
Qrad















1
11

4
21

3







 avg
rad

T
h

TcmQ p
..

V

V
ACH

.


10

�̇�𝑠𝑒𝑛 = 𝑚 × 𝑐𝑝̇ × ∆𝑇

𝑄𝑖𝑛𝑓 = 𝑚 × 𝑐𝑝̇ × ∆𝑇

𝐶𝑂𝑃 =
𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡

𝜂 =
𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡

Constants

Density of air (ρair) =1.2 kg/m3
Density of water (ρwater) =1000 kg/m3
Universal gas constant = 8314.41 J/(kg.mol.K)
Gas constant for air = 287 J/(kg.K)
Stefan-Boltzmann constant = 5.67 ×10-8 W/(m2.K4)

Unit conversions

1 m = 3.281 ft
1Ibm= 0.45356 kg
1 Btu = 1.055 kJ
oC =(oF-32)*(5/9)
K= oC+ 273.15
1 W =3.412 Btu/h
1 ton refrigeration = 3.517 kW

Profitability is excellent if pbp less than one-third of lifetime of investment;
Profitability is good if pbp less than one-half of lifetime of investment;
Profitability is bad if pbp is more than one-half of lifetime of investment.

)./(006.1, KkgkJc ap 

)./(186.4, KkgkJc wp 

1

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

2

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

3

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?

4

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

5

B. Personal Parameters

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

1 𝑀𝐸𝑇 = 58.1

𝑊

𝑚2

6

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.

7

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

8

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

9

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

.

Insulation Values According to ASHRAE Fundamentals Handbook (Table 8)

10

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.

11

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.

12

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.

13

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

2

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.

14

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

15

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

16

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

17

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.

18

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.

19

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

https://en.wikipedia.org/wiki/P._Ole_Fanger

https://en.wikipedia.org/wiki/Skin_temperature

https://en.wikipedia.org/wiki/Dry-bulb_temperature

https://en.wikipedia.org/wiki/Mean_radiant_temperature

https://en.wikipedia.org/wiki/Relative_humidity

https://en.wikipedia.org/wiki/Clothing_insulation

20

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.

21

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?

22

23

Chapter

Two

Handout Five

Thermodynamic Processes in Buildings
There are eight properties of air needed in buildings and in HVAC systems:

1. Pressure (P) force per unit area measured in kPa

The pressures of air and water are very important.

– Absolute pressure (Pabs): pressure exerted by fluid above zero pressure (vacuum)

– Gage pressure (Patm): pressure exerted by fluid above atmospheric pressure patm = 101 kPa

at sea level

– Vacuum pressure (Pvac): pressure exerted by fluid below atmospheric pressure patm

Pressure Effect on Properties

Pressure Effect on Properties

Atmospheric Layers

Note: Height of Mount Everest is 8.848 km

Unlike the Ocean, there is no definite boundary. The atmosphere gradually thins out as you get higher, at
a rate of roughly 1% per hundred meters for the first few thousand meters, more slowly after that. About
half the mass of the atmosphere is within the first 5.5 km, about three quarters within 11 km.

The atmosphere is breathable for people in normal health up to about 4 km, about the elevation of the
highest cities towns, for example Lhasa in Tibet at 3,656 m, after which it becomes difficult without
acclimatization. No doubt by coincidence, the Summit of Earth’s highest Mountain, Everest, at 8.8 km,
seems to be pretty close to the boundary where a very fit person can survive without oxygen equipment,
at least for a short time.

Nearly all weather occurs in the lowest layer of the atmosphere, the troposphere, extending to about 12
km (higher at the equator, lower at the poles). Above that, there are several layers with their own distinct
properties.

The Kármán line, located at 100 km, considered to be the boundary of outer space, beyond which a
traveller is considered to be an astronaut. This isn’t entirely arbitrary. It approximates the height
beyond which no aircraft flight would be possible because the atmosphere is too thin to provide
aerodynamic lift at a speed below orbital velocity.

ratio of height of atmosphere to diameter of Earth

Pabs = Patm + Pg

Pabs = Patm – Pvac

Components in dry air
Volume ratio = Molar ratio,

compared to dry air
Molar Mass Molar mass in air

Name Formula [mol/molair] [vol %]
[g/mol]

[kg/kmol]
[g/molair]

[kg/kmolair]
[wt %]

Nitrogen N2 0.78084 78.084 28.013 21.873983 75.52

Oxygen O2 0.20946 20.946 31.999 6.702469 23.14

Argon Ar 0.00934 0.934 39.948 0.373114 1.29

Carbon dioxide CO2 0.00033 0.033 44.010 0.014677 0.051

Neon Ne 0.00001818 0.001818 20.180 0.000367 0.0013

Helium He 0.00000524 0.000524 4.003 0.000021 0.00007

Components in dry air
Volume ratio = Molar ratio,
compared to dry air
Molar Mass Molar mass in air
Name Formula [mol/molair] [vol %]
[g/mol]
[kg/kmol]
[g/molair]
[kg/kmolair]
[wt %]

Methane CH4 0.00000179 0.000179 16.042 0.000029 0.00010

Krypton Kr 0.0000010 0.0001 83.798 0.000084 0.00029

Hydrogen H2 0.0000005 0.00005 2.016 0.000001 0.000003

Xenon Xe 0.00000009 0.000009 131.293 0.000012 0.00004

Average molar mass of air 28.9647

Gibbs Dalton law:

Mixture Pressure = Sum of the partial pressures of the constituents

For moist-air:

All dry gases can be considered as one gas, then:

2. Temperature (T): A measure of the thermal activity in a body measured in

• Thermal activity depends on the velocity of the molecules and other particles of which a

matter is composed.

• Thermometer is used to measure temperature

– rely on the fact that most liquids expand and contract when their temperature is raised or

lowered

• Temperature scale: Fahrenheit (˚F) and Celsius (˚C), Rankine (˚R) and Kelvin

– Fahrenheit
• 0°F as the stabilized

temperature when equal
amount of ice, water, and salt
are mixed

– Celsius
• 0°C as melting point of ice

(water) and 100°C as boiling
point of water

• ˚ F = 1.8 ˚ C + 32
– Kelvin

• 0 K as absolute zero
• K = ˚ C + 273.15

– Rankine
• ˚ R = ˚ F + 459.67

a. Dry-bulb temperature (Td): The dry-bulb temperature (DBT) is the temperature of air measured

by a thermometer freely exposed to the air but shielded from radiation and moisture. DBT is the

temperature that is usually thought of as air temperature, and it is the true thermodynamic

temperature. It indicates the amount of heat in the air and is directly proportional to the mean

kinetic energy of the air molecules. Temperature is usually measured in degree Celsius (°C),

Kelvin (K), or Fahrenheit (°F) in IP units.

Unlike wet bulb temperature, dry bulb temperature does not indicate the amount of moisture in

the air. In construction, it is an important consideration when designing a building for a certain

climate. Niall called it one of “the most important climate variables for human comfort and

building energy efficiency.

b. Wet-bulb temperature (Tw): The thermodynamic wet-bulb temperature is the lowest temperature

which may be achieved by evaporative cooling of a water-wetted (or even ice-covered),

ventilated surface.

c. Dew-point temperature (Tdew): the dew point is the temperature to which the ambient air must

be cooled to reach 100% relative humidity assuming there is no evaporation into the air; it is the

point where condensate (dew) and rain would form.

https://en.wikipedia.org/wiki/Temperature

https://en.wikipedia.org/wiki/Air

https://en.wikipedia.org/wiki/Thermometer

https://en.wikipedia.org/wiki/Radiation

https://en.wikipedia.org/wiki/Moisture

https://en.wikipedia.org/wiki/Celsius

https://en.wikipedia.org/wiki/Kelvin

https://en.wikipedia.org/wiki/Fahrenheit

https://en.wikipedia.org/wiki/Wet_bulb_temperature

https://en.wikipedia.org/wiki/Construction

https://en.wikipedia.org/wiki/Design

https://en.wikipedia.org/wiki/Building

https://en.wikipedia.org/wiki/Climate

https://en.wikipedia.org/wiki/Evaporative_cooling

https://en.wikipedia.org/wiki/Dew_point

https://en.wikipedia.org/wiki/Relative_humidity

Dry-bulb and Wet-bulb Temperatures

Water absorbs heat from the probe (bulb)-in fact from the air- as it evaporates. If the weather is dry, more

water evaporates and thus absorbs more heat from the bulb- then the temperature difference between

dry and wet becomes large. If the weather is wet, less water will evaporate and thus absorbs less heat

from the bulb- then the temperature difference between dry and wet becomes little. At 100% relative

humidity, the wet-bulb temperature equals the dry-bulb temperature.

In a weather station, the thermometer is usually housed in a Stevenson Screen. A Stevenson Screen is a
white boxed shelter that contains temperature and relative humidity equipment. It shields the
instruments from sunshine and precipitations and has louvered sides to permit the free movement of air.
Ideally the shelter is placed over grass, mounted at 1 meter above the ground and as far from any
buildings as circumstances permit.

https://en.wikipedia.org/wiki/Relative_humidity

https://en.wikipedia.org/wiki/Relative_humidity

3. Humidity

a. Humidity Ratio (W)

Units: dimensionless

gwv/gda

gwv/kgda

b. Relative Humidity (φ)

Units: percentage

What is the difference between humidity ratio and relative humidity and why do we use relative humidity

in weather forecast not humidity ratio?

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%

Heating

Cooling

Relationship between Temperature and Relative Humidity During the Day

Relative humidity ( ) is inversely proportional to the air temperature .i.e. if temperature increases, the

Relative humidity decreases and vice versa.

Effect of Relative Humidity

Room humidity can have a major impact on the quality of the living environment. A relative humidity ( )

of 40-60% is generally considered to be optimal for a comfortable and healthy home. Too much moisture

can lead to mold and overheating. Too little causes dry eyes, chapped lips and an environment in which

bacteria and viruses can thrive.

4. Specific volume (ʋ) (which is the reciprocal of density) measured in m
3
/kg

Density – mass/volume (used for solids and liquids)

air =1.2 kg/m
3
, water = 1000 kg/m

3
 Ratio= 833

Changes slightly with temperature, why? Because of volume change

5. Specific Heat (c)

 Amount of heat that is required to change the temperature of 1 kg of the substance

1 °C.

 Units (J/kg-°C)

 Without phase change !!

 Property of material which changes slightly with temperature

Specific heat for water is 4.186 kJ/(kg-K)

air is 1.00 kJ/(kg-K)

6. Specific internal energy (μ) measured in kJ/kg

• Internal energy (U): microscopic energy possessed by a system caused by the

motion/vibration of the molecules and/or intermolecular forces.

– the motion/vibration increases with temperature

• Internal energy is thus often measured by the body’s temperature (this is not

true when the body is a liquid or a solid (such as ice) which is changing phase!)-

this leads to sensible and latent heat discussed later

Important: a body does not contain heat; it contains thermal energy

7. Specific Enthalpy (h) measured in kJ/kg

Can be defined as the energy per unit mass a fluid transports across a system boundary.

– A property of a body that measures its heat content

– Enthalpy includes: (i) Internal energy U and

(ii) pv or energy due to flow work

Which is internal energy + work needed to move a mass of fluid across a system boundary.

– Enthalpy is a combined property which is widely used in thermal analysis

– When T, p or V changes, H changes

specific enthalpy h = u + p.v in kJ/kg (v is specific volume=1/density)

Instead of sensible or latent heat equations, enthalpy equation is widely used since one does not have to

worry about state of fluid.

Enthalpy is the total energy (sensible + latent)

8. Specific entropy (s). not need in this course.

Two other properties needed:

Specific heat at constant volume (cv)

Specific heat at constant pressure (cp)

There are two sourced to obtain properties of air:

1. Simple equations such as the ideal gas law (for air, or moist air).

2. Tables computed from complex equations (for water and refrigerants).

Sensible and Latent Heats

Note to consider from the chart

1. Since the air is a mixture of dry gases and water vapor, the water vapor carries the latent heat,

while the dry gases carries the sensible heat.

2. Sensible heat is a function of temperature f(T), while latent heat is a function of humidity ratio

f(W).

3. The total enthalpy of the moist-air (hT) = enthalpy of dry-air (hda)+ enthalpy of water vapor (hwv).

4. To convert 1 pound of water from solid to liquid you need 144 Btu. But to convert the same

amount from liquid to gas you need 970 Btu. That is 6.7 times (970/144). What does this mean?

Water vapor carries a huge amount of energy

http://www.google.com/url?sa=i&rct=j&q=&esrc=s&frm=1&source=images&cd=&cad=rja&uact=8&ved=0ahUKEwjQiurC5q3JAhUCuBQKHZdECIQQjRwIBw&url=http://www.tpub.com/fireman/25.htm&psig=AFQjCNEYBqh_clQ7XP3xHNrZJVqq5CB-Wg&ust=1448617402744803

Ideal Gas Law

The ratio of pressure P times molar volume ʋ divided by the absolute temperature T is observed to

approach a constant value, when the pressure is allowed to approach zero.

Where R is the universal gas constant = 8314.41 J/(kg.mol.K)

The gas constant R = R/Molecular weight of gas

For example: Rair= 8314.41/28.97 = 287 J/(kg.K)

Mole: the standard unit to measure the amount of a substance.

Avogadro’s principle: Equal volumes of all gases at the same temperature and pressure contain the same

number of molecules (6.023 *10
23

molecules). That is, one mole of a substance contains 6.023 *10
23

molecules.

P=Rair*T

P=Rair*T

P=(m/V) Rair*T

PV=m×Rair×T

Since mass of substance (m) = number of moles in kmol (n) * molecular weight of the substance (M)

Then, PV= n*M* Rair *T

Also Rair =R/M, then M=R/Rair where R is the universal gas constant =8314.41 J/(kg.mol.K)

For example Rair = 8314.41/28.97= 287 J/(kg.K)

Then, PV =n*(R/Rair)*Rair*T =n*R*T

Forms of the Ideal Gas Law:

TRP air

TmRPV air

nRTPV  (not used in HVAC)

Where;

P= pressure, Pa

ʋ= specific volume, m
3
/kg

Rair= gas constant of air = 287 J/(kg.K)

T = absolute temperature, K

V = volume, m
3

M = mass, kg

Relationship between Humidity ratio and Pressure:

From above, since TmRPV air

Since Rgas= universal gas constant (R)/ molecular weight (Mgas), and Patm= Pda + Pwv

Then,

Example:

The dimensions of your classroom is 12 m × 6 m × 3 m. If the temperature (Td) is 22
o
C, and the humidity

ratio is 10 gwv/kgda, answer the following questions.

1. Use the ideal gas law to calculate the density (ρ) of air in the classroom.

2. Calculate the mass of water vapor and the mass of dry air in the classroom.

First find Pwv and Pda and then use the ideal gas law equation:

Note that volume and temperature of both dry air and water vapor is the same.

From above relationship:

Pda= Patm -Pwv=101.3-1.6=99.7 kPa

Note that Rair =287.05 J/(kg.K)

And Rwv= 461.52 J/(kg.K)

Zi

W

Q

mi

Vi

Pi

Ui

Hi

Mo

Vo

Po

Uo

Ho

System

Zo

Open-system First Law of Thermodynamics:

.2..2.

22
Wh

V
zgmQh

V
zgm o

o
ooi

i
ii 





























Q = heat transfer rate, W

M = mass flow rate, kg/s

Ho= final enthalpy, kJ/kg

Hi = initial enthalpy, kJ/kg

To = final temperature, K

Ti = initial temperature, K

Cp = specific heat of air =1006 J/(kg.K)

Important Note

Is heat gain to air

Is heat removal from air

UA is equivalent to m-dot cp

Example 1:

Air enters an auditorium through the diffusers of an HVAC system at a temperature of 12
o
C at a

volumetric flow rate of 0.5 m
3
/s. The air collects the heat and leaves through the air inlets at the set-point

temperature of the space at T=24
o
C. Calculate the following:

1. the air mass flow rate entering the auditorium.

2. the heat removed from the auditorium?

Air enters at 12
o
C, carries the heat and leaves at 24

o
C, then the amount 0f heat removed by air as it

leaves:

Example 2

In a windy day in Bahrain (sea level, Patm=101.325 kPa, ρ=1.2 kg/m
3
), air leaks to a building at a rate of 1

air change per hour. All air in the building is exchanged with fresh air every one hour “once each hour”. It

occurs through cracks, around windows, through loose fittings building materials, open doors, etc… This is

called infiltration. Take the outdoor temperature (To) is 39
o
C and the area of the building is 150 m

2
, with

an average height of 7 m and answer the following questions:

1. Calculate the volumetric flow rate (Vinf) of this infiltrating air.

Air change per hour (ACH) =
(V) Volume

V)( Rate Flow Volumetric
.

2. Calculate the mass flow rate (minf) of this infiltrating air

3. How much heat must the cooling system of the building remove from this infiltrating air to keep

the building at its set-point indoor temperature (Ti) of 22
o
C.

4. How many refrigeration tons does this heat equal to? Comment on the answer and draw a

conclusion.

1 ton refrigeration is = 3.517 kW (refer to conversion factors sheet)

Then,

Comment: a small quantity of air leaking through windows causes a lot of heat

gain that the HVAC systems has to remove. In hot climates, buildings must be air-

tight. Windows has two basic functions: daylight and view.

Chapter Two

Basic Thermal Science

• The science behind heat flow is called heat transfer

• There are three ways by which heat can be transferred: conduction, convection, and
radiation

• Heat flow depends on the temperature difference (sensible heat transfer)

98.6oF= 37oC is the body temperature

Conduction Heat Transfer

Conduction – The form of heat transfer through a body that occurs without any movement of the body.

It is due to molecular or electron vibration or movement.

Fourier’s Law of Heat Conduction:

The rate of heat transfer by conduction (Qcond) is proportional to the temperature difference (∆T) and the
heat flow area (A) and is inversely proportional to the distance (∆x) through which conduction occurs.

Where:

Qcond = conductive heat transfer rate, W
K = thermal conductivity, W/(m.K)
T1-T2 = temperature difference, K
A = area through which conduction occurs, m

2

= thickness, m

Q: What are the other forms of the equation?
Q: How the total thermal resistance of a composite structures is calculated?

24

o
C

10

o
C 38

o
C

Q:What is meant by steady-state?

A: Temperatures T1 and T2 do not
change in time

Fig 2.13. Thermal conductivity varies with temperature.

However, for most HVAC
applications, the range of
operation in temperature

is so narrow that
conductivity can be
assumed constant.

Definitions:

Thermal conductivity (k) is the heat transfer rate of a material per meter thickness when the temperature
difference (∆T) across is equal to 1

o
C. Units (W/m.K)

Resistance (R) is the temperature difference (∆T) needed to transfer 1 watt across the material. Units
(K/W).

Thermal resistance (Rth) is the temperature difference (∆T) needed to transfer 1 watt across the material
“holding the area equal to 1 m

2
. Or, the area (A) needed to transfer 1 watt across the material “holding

the temperature difference equal to 1 m
2
. Units (m

2
.K)/W.

Thermal conductance (U) is the heat transfer rate of a material per meter squared of its area when the
temperature difference (∆T) across is equal to 1

o
C. Units (W/m

2
.K).

Example (1):

Calculate the conductive heat transfer rate through a 200-mm think concrete block wall of area 10 m
by 3 m if its k =0.35 W/(m.K) and the outside temperature is 39oC, and the inside air temperature is
24oC.

Example (2):

Draw the thermal resistance network (TRN) for the wall structure shown and write its total
resistance (Rtot) equation.

Example (3):

Draw the thermal resistance network (TRN) for the wall structure shown and write its total
resistance (Rtot) equation.

ules when drawing TRN’s

1. when you split paths, you should split from the beginning not when the split occurs.

2. when layers have different areas in a structure, you should use

R

not Rth, because R is
independent of area.

R

Example (4):

A wall with an area of (14 × 3) m

2
is constructed from the layers listed in the table below and shown

graphically in the drawing. If the outside temperature (To) is 39
o
C and the inside temperature (Ti) is 24

o
C,

answer the following questions:

Layer
Thickness
(∆x) (cm)

Thermal
Conductivity (k)

(W/(m.K)

Resistance (Rth)
(m

2
.K)/W

1 Cement plaster- Sand Aggregate 1 – 0.013

2
Normal-weight concrete blocks filled
with perlite

20 – 0.35

3 Air space 5 0.024 =0.05/0.024=2.08

4
Molded beads expanded polystyrene
insulation 24 kg/m

3

8 0.037 =0.08/0.037=2.16

5 Cement board 1150 kg/m
3
1.1 0.25 =0.011/0.25=0.044

Total thermal resistance = 4.65

Heat flow (Qcond)

To =39oC Ti=24oC

1. Draw the thermal resistance network (TRN) for the wall.

2. Calculate the total thermal resistance of the wall.

Resistances are in series, then:

Rth, total = Rth1 + Rth2 + Rth3 + Rth4 + Rth5 = 4.65 (m

2
.K)/W

3. Calculate the conductive heat transfer rate (Qcond)

Qcond = A×∆T/Rth,total = 42 m
2
× (39 – 24)K /4.65 (m

2
.K)/W = 135 W

4. What is the temperature of the insulation( Tins)?

To =39oC Ti=24oC

Cavities & Air Spaces

In principle, use of cavities is similar to use of a insulating material. If an air space is left between two
layers making a wall or roof in any building, the air trapped between two layers being poor conductor of
heat acts as a barrier to heat transfer.

Heat is transferred across an air space by a combination of conduction, convection and radiation. Heat
transfer by conduction is inversely proportional to depth of the air space. Convection is mainly dependant
on the height of the air space and its depth. Heat transfer by radiation is relatively independent of both
thickness and height, but is greatly dependent on the reflectivity of the internal surfaces. All three
mechanisms are dependent on the surface temperatures. The mathematical treatment of air cavity would
be similar to that of insulation if natural convection in air is neglected. The thickness of air cavity is a very
important design parameter that governs its effectiveness by controlling the heat transfer coefficient as in
case of insulation.

It has been found that with gaps broader than 50 mm, movement of trapped air due to temperature
gradient starts that in turn increases the coefficient of heat transfer. This increase in heat transfer takes
place due to convective heat transfer taking place in addition to conductive heat transfer. Therefore,
cavities broader than 50 mm are normally not preferred. However, if more thickness of air cavity is
required for getting heavy insulation, by putting partitions in the main broad cavity multiple cavities can
be used as an alternative.

Some typical values of thermal resistance for air cavities are given below:

Air cavity Placement Thickness of air layer (mm) Thermal resistance (m
2
K/W)

Vertical
10- 20 0.14

20- 50 0.17

Horizontal- heat flow from bottom to top 10- 50 0.17

Horizontal- heat flow from top to bottom 10- 50 0.21

If it is possible to ventilate the air gap between the roof and the ceiling, then we could expect a reduction
of heat transfer especially by convection. If the ventilation is effective then the air in the void will remain
close to the ambient temperature, thus reducing the convective heat transfer to zero. Ventilated air,
however, does not reduce the radiative heat transfer from the roof to the ceiling. The radiative
component of the heat transfer may be reduced by using low emissivity or high reflective coating (e.g.
aluminum foil) on either surface facing the cavity.

In addition to application on walls and roofs, the concept of air cavities also finds very important place in
development of insulating windows using double and triple glazing details.

Temperature Profile in a Construction

In a steady-state situation, because there is no heat-storage or heat-production in the construction, the
heat flow through every layer in the building construction must be the same. The temperature change
across each layer is linear and the rate of change depends on the thermal resistance. When the thermal
resistance is small, the temperature difference across the layer is small, but when the thermal resistance
is large, the temperature difference across the layer is also large.

The same calculation method is followed as in electrical theory. Ohm’s Law says that the voltage
differences over resistances, connected in series, are proportional to the measure of the resistances