THERMODYNAMICS

• INTRODUCTION
• THERMODYNAMIC SYSTEMS
• PROPERTIES
• LAWS OF THERMODYNAMICS
• BASIC CONCEPTS
• ENERGY CONCEPTS
• IDEAL GAS
• PROCESSES OF IDEAL GAS
• GAS CYCLES
 WHAT COMES TO
YOUR MIND WHEN
YOU SAY
THERMODYNAMICS?
HONESTLY.
 THERMO
DYNAMICS
THERMODYNAMICS came from the
word THERMO meaning “heat”
and DYNAMICS meaning
“motion”, in short,
Thermodynamics is the study of
heat in motion.
THERMODYNAMICS
• is a branch of science which deals
with the transformation of energy
from one form to another and the
movement of energy from one
location to another.
transformation of energy

& movement of energy

 Everything in this world is
related to thermodynamics. The
universe is so vast that if we talked
about thermodynamics, it would never
end. So we need to focus on a certain
area of interest. And this certain area
of interest is called the…
…
 So the there needs to be an
indicator that can separate your area
of interest (system) from the wide
universe, and that indicator is this what
we call…
 Lastly, the things outside the
boundary that may or may not affect
your system is the………………
 -a fixed mass or region in a space
chosen for study
- everything external to the system
- separates the system from its
surroundings

*Piston cylinder as an example

of a thermodynamic system.
THREE TYPES OF SYSTEMS
1. OPEN SYSTEM
2. CLOSED SYSTEM
3. ISOLATED SYSTEM
1. OPEN SYSTEM
-is the system where heat, work and
mass all crosses the system boundary.
Examples: jetmatic pumps, compressor, turbines

Energy
out
M in

M out
Energy
in
2. CLOSED SYSTEM
-there’s no mass flow, only heat and
work can cross the system boundary.
Example: refrigeration
Energy
out

NO
MASS
Energy
in
3. ISOLATED SYSTEM
-nothing crosses the system boundary
(no heat, no work, no mass transfer)
Example: insulated gas cylinder

NO
system
MASS

HEAT mass heat

WORK
TWO GENERAL CLASSIFICATION
1. INTENSIVE PROPERTY
- is one that is independent of the mass of the
substance (pressure, density, temperature)

2. EXTENSIVE PROPERTY
- is one that is dependent on magnitude of the mass
(mass, volume, weight, energy)
2.1 Mass (m) – is the quantity of matter in a body

2.2 Weight (W) – the force exerted by gravity on a

given mass

2.3 Volume (V) – is the amount of space occupied by

the matter.

2.4 Energy (E) – is the capacity of a given body to

produce physical effects external to the body.
 I believe you probably heard the
terms Newtons’ Laws of Motion or
probably familiar with this, right?

So how many laws do we have

here and what are they again?

Here in thermodynamics, we also

have laws that we call and they are not
just three but we have four laws. What
are them?
FIRST LAW OF THERMODYNAMICS

Energy in = Work out + Change in internal heat energy

Conservation of Energy – “Energy can neither be
created nor destroyed but only transformed.”
- No engine can do more useful work than the amount of
energy input.
SECOND LAW OF THERMODYNAMICS
- For energy to flow from a colder object to a hotter object, work
must be done. When heat is converted into work, the efficiency, or output
of usable work, will always be less than 100%. The input heat not
converted into work is called waste heat.
THIRD LAW OF THERMODYNAMICS
-states that the absolute zero cannot be reached. It occurs when the
particles in an object stop moving completely and have a temperature of -
273.15 degrees Celsius.
ZEROTH LAW OF THERMODYNAMICS
- If two thermodynamic systems are each in thermal
equilibrium with a third, then they are in thermal equilibrium
with each other.
• Force (F) - Newton's law states that "the acceleration of a
particular body is directly proportional to the resultant force
acting on it and inversely proportional to its mass."

𝐹 𝑚𝑎
a ∝ 𝐹=
𝑚 𝑘
where: F- force (lbf, kgf, N)
a- acceleration (ft/s², m/s²)
m- mass (𝑘𝑔𝑚 , 𝑙𝑏𝑚 )
k- proportionality constant, “k” is often ignored because of
its usual value, 1, but here it’s not always 1.
• System of units where k is unity but not
dimensionless.

𝑚𝑎 𝑘𝑔𝑚 ∙ 𝑚 𝑠𝑙𝑢𝑔 ∙ 𝑓𝑡
k= =1 2
= 1
𝐹 𝑁 ∙ 𝑠 𝑙𝑏𝑓 ∙ 𝑠 2
• If the same word was used for mass and force in a
given system, is neither unity nor dimensionless.
𝑚𝑎 𝑘𝑔𝑚 ∙ 𝑚 𝑙𝑏𝑚 ∙ 𝑓𝑡
k= = 9.81 2
= 32.2
𝐹 𝑘𝑔𝑓 ∙ 𝑠 𝑙𝑏𝑓 ∙ 𝑠 2
Equate the S.I. values of “k” from unity and not unity:
• kgf and N (kilogram force and Newton)

k=k
𝑘𝑔𝑚 ∙ 𝑚 𝑘𝑔𝑚 ∙ 𝑚 NOW, CANCEL THE
1 2
= 9.81 SAME UNITS PRESENT
𝑁 ∙ 𝑠 𝑘𝑔𝑓 ∙ 𝑠 2 ON BOTH EQUATIONS
1 9.81 NEXT, CROSS
= TRANSPOSE THE UNITS
𝑁 𝑘𝑔𝑓 IN THE DENOMINATOR

1 𝑘𝑔𝑓 = 9.81 𝑁
 Equate the English values of “k” from unity and not unity:
• lbm and slug (pound mass and slug)

k=k

𝑠𝑙𝑢𝑔 ∙ 𝑓𝑡 𝑙𝑏𝑚 ∙ 𝑓𝑡 NOW, CANCEL THE

1 2
= 32.2 SAME UNITS PRESENT
𝑙𝑏𝑓 ∙ 𝑠 𝑙𝑏𝑓 ∙ 𝑠 2 ON BOTH EQUATIONS

1 𝑠𝑙𝑢𝑔 = 32.3𝑙𝑏𝑚

NOTE: Slug is a unit of MASS and not force nor weight.

• WEIGHT (W or Fg) – force due to gravity
𝑚𝑔
W = 𝐹𝑔 =
𝑘

where: Fg- weight (lbf, kgf, N)

g- gravitational acceleration (32.2 ft/s², 9.81 m/s²)
m- mass (𝑘𝑔𝑚 , 𝑙𝑏𝑚 )
• Density – is the mass per unit volume
𝑚 𝑚ሶ
𝜌= =
𝑉 𝑉ሶ
(𝑡ℎ𝑒 𝑑𝑜𝑡 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑖𝑠 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑡𝑖𝑚𝑒)

• Specific Volume – inverse of density

𝑉 1
𝑣= or
𝑚 𝜌
• Specific Weight– weight per unit volume
𝑊 𝑚𝑔 𝜌𝑔
𝛾= = =
𝑉 𝑉𝑘 𝑘

Replaced the value of W with mg/k

By simplifying, we replaced m/V to 𝜌, since it’s the formula for

density
1. What is the weight (N) of a 66- 𝑘𝑔𝑚 man at standard
condition?

𝑚
GIVEN:m = 66 𝑘𝑔𝑚 g = 9.81
𝑠2
REQUIRED: Weight (Fg)
SOLUTION:
𝑚𝑔
𝐹𝑔 =
𝑘
,
since the problem specified the required unit for weight, we
will use not unity for the value of “k”
𝑚
(66𝑘𝑔𝑚)(9.81 2 )
= 𝑠 = 647 𝑁
𝑘𝑔𝑚 ∙ 𝑚
1
𝑁 ∙ 𝑠2
2. The mass of a given airplane at sea level (g = 32.10 fps 2 ) is 10 short tons.
Find its mass in lbm , slugs, and kg m and its (gravitational) weight in lbf when
it is travelling at a 50,000-ft elevation. The acceleration of gravity g decreases
by 3.33 x 10.6 fps 2 for each foot of elevation. (1 short ton = 2000 lbm )
GIVEN: m = 10 short tons
𝑓𝑡 𝑓𝑡
𝑔𝑠𝑒𝑎 = 32.1 g = 32.2
𝑠2 𝑠2
h = 50000 ft a = 3.33 x 10.6 fps 2 / ft
REQUIRED: Mass and Weight
2000 lb
m
SOLUTION: m = 10 short tons x 1 short ton = 20000 lbm
WE JUST USED THE
1 kgm GIVEN CONVERSIONS
= 20000 lbm x = 9090.91 kg m
2.2lbm TO GET THE VALUES
1 𝑠𝑙𝑢𝑔 OF MASS.
= 20000 lbm x = 621.12 𝑠𝑙𝑢𝑔𝑠
32.2lbm
 𝑚𝑔
𝑊=
𝑘
,
we know that this is the formula for weight but given the
situation there will be a difference in gravity because of the word
“decreases” meaning there’s a deceleration that happened with the
given elevation and acceleration.
WE MULTIPLIED ACCELERATION AND
HEIGHT SO THAT WE CAN CANCEL THE
𝑚[𝑔𝑠𝑒𝑎 − 𝑎 ℎ ]
= UNIT FEET.
𝑘
2 fps 2
(20000 lbm )[32.1fps − 3.33 x 10.6 50000𝑓𝑡 ]
𝑓𝑡
=
𝑙𝑏 ∙ 𝑓𝑡
32.2 𝑚 2
𝑙𝑏𝑓 ∙ 𝑠
W = 19834.47 𝑙𝑏𝑓
3. Two liquids of different densities (ρ1 = 1500 kg/m3 and ρ2 = 500 kg/m3 )
are poured together into a 100-liter tank filling it. The resulting density of the
mixture is 800 kg/m3 , find the respective amounts of the liquids used.

GIVEN: ρ1 = 1500 kg/m3 ρT = 800 kg/m3

ρ2 = 500 kg/m3 VT = 100 L (1 m3 = 1000 L)

REQUIRED: Masses
SOLUTION: m •
1 m2 FIRST CONVERT VOLUME
ρ1 ρ2 1m3
VT = 100L × = 0.1m3
1000 𝐿
• GET THE TOTAL MASS USING THE
ρT = 800 FORMULA FOR DENSITY
VT = 100 L mT = ρT ∙ VT = 800 0.1 = 80kg m
 ρ2 = 500 kg/m3 • WE KNOW THAT:
ρ1 = 1500 kg/m3
m2 mT = m1 + m2
m1
80kg m = ρ1 V1 + ρ2 V2 EQ.1

ρT = 800 VT = V1 + V2
0.1m3 = V1 + V2
VT = 100 L V1 = 0.1m3 − V2 EQ.2

• WE HAVE TWO VARIABLES: 𝑉1 &𝑉2 NOW SUBSTITUTE

EQ.2 TO EQ.1 TO ELIMINATE ONE VARIABLE

80kg m = (1500 kg/m3 )(0.1m3 − V2 ) + (500 kg/m3 )V2

V2 = 0.07 m3
• SUBSTITUTE 𝑉2 TO EQ. 2 TO GET 𝑉1
V1 = 0.1m3 − 0.07 m3 = 0.03m3
• NOW WE CAN GET THE MASSES
m1 = ρ1 V1 = (1500 kg/m3 )(0.03m3 ) = 45 kg m
m2 = ρ2 V2 = (500 kg/m3 )(0.07 m3 ) = 35kg m
PRESSURE – is the normal force exerted by the fluid per
unit area of the surface.
𝑭
✓ P = , (Pa, psi)
𝑨

where: F- normal force (lbf, kgf, N)

A- area normal to the force (in², ft², m²)
ENGLISH UNIT S.I. UNIT
𝒍𝒃𝒇 𝑵
𝟐 (psi) (Pa, Pascal)
𝒊𝒏 𝒎𝟐
Patm = atmospheric pressure (barometric)
1atm = 14.7 psi (lbf/in²)
= 101.325 kPa (kN/m²)
= 1.0332 kgf/cm²
= 760 mmHg = 760 torr
= 29.92 inHg
= 1.01325 bar
= 10.33 mH20 (meter of water)
 ABSOLUTE PRESSURE AND GAGE PRESSURE
✓Pabs = Patm ± Pg
where: Pabs – absolute pressure
Patm – atmospheric or
barometric pressure
Pg – gage pressure

Note: For gage pressure value:

use (-) if it’s indicated that
the pressure is in vacuum, if not
stated use (+) otherwise
 𝑭
✓P=
𝑨

𝒎𝒈 • WE CHANGED THE VALUE OF FORCE TO MASS

𝒌 TIMES GRAVITY, WHILE WE SUBSTITUTED
= 𝑽 AREA FROM THE FORMULA FOR VOLUME
WHICH IS AREA TIMES HEIGHT.
𝒉

• WE PUT HEIGHT ON THE NUMERATOR AND

𝑚𝑔ℎ
= THE PROPORTIONALITY CONSTANT IN THE
𝑉𝑘 DENOMINATOR.

• WE KNEW THAT “mg/Vk” IS EQUAL TO THE

P =𝛾ℎ FORMULA OF GAMMA OR THE SPECIFIC
WEIGHT.
PROBLEM:

A pressure gage register 40 psig in a region where the

barometer is 14.5 psia. Find the absolute pressure in
psia and kPa.
SOLUTION:

A pressure gage register 40 psig in a region where the

barometer is 14.5 psia. Find the absolute pressure in
psia and kPa.

ANSWER: 54.5 psia & 375.66 kPa

Heat - is the total energy of molecular motion
in a substance
Temperature - is a measure of the average
energy of molecular motion in a substance.
WATER °C °F

BOILING 100 212

tc tf

ICE 0 32
 Linear interpolation:
100 −0 212 −32
= CROSS MULTIPLY
𝑡𝑐 −0 𝑡𝑓 −32

100 − 0 𝑡𝑓 − 32 = 212 − 32 (𝑡𝑐 − 0) SOLVE THE OPERATIONS

LET’S ARRANGE THE TERMS IN

100)(𝑡𝑓 − 32 = (180)(𝑡𝑐) TERMS OF “tc”

100(𝑡𝑓 − 32)
= 𝑡𝑐 SIMPLIFY
180
Fundamental formulas:

𝟓
𝒕𝒄 = (𝒕𝒇 − 𝟑𝟐) NOW IN TERMS OF “tf”
𝟗
𝟗
𝒕𝒇 = 𝒕𝒄 + 𝟑𝟐
𝟓

ABSOLUTE TEMPERATURES:
KELVIN (K) = tc + 273 RANKINE (˚R) = tf + 460
The change in temperature in degree Celsius is just the same
with the change in temperature in Kelvin.
∆𝑡𝑐 = ∆𝑇𝐾
All is true with degree Fahrenheit and degree Rankine.
∆𝑡𝑓 = ∆𝑇°𝑅

IN CONVERTING CHANGE IN TEMPERATURE FROM ONE UNIT TO ANOTHER WE USE

THE FOLLOWING RELATIONS:

∆𝑡𝑓 = 1.8 ∆𝑡𝑐

∆𝑇°𝑅 = 1.8 ∆𝑇𝐾

-mass is indestructible
-mass in = mass out

𝑚ሶ 1 𝜌1 𝑚ሶ 2 𝜌2
𝐴1 SYSTEM 𝐴2

vത1 vത 2
Where: mሶ – mass flow rate
𝜌 − density
vത − velocity
A – area
Recall: The formula for volume flow rate is Area x Velocity Vሶ = A vത
 𝑚ሶ 1 𝜌1 𝑚ሶ 2 𝜌2
𝐴1 SYSTEM 𝐴2

vത1 vത 2

𝑚ሶ 1 = 𝑚ሶ 2 • REPLACE MASS WITH THE

DENSITY FORMULA
•
𝜌1 Vሶ 1 = 𝜌2 Vሶ 2
SUBSTITUTE FORMULA OF
VOLUME

•
𝜌1 𝐴1 vത1 = 𝜌2 𝐴2 vത 2 USING THEIR RELATION
WE CAN ALSO CHANGE
DENSITY INTO SPECIFIC
𝐴1 vത1 𝐴1 vത1 VOLUME
=
𝑣1 𝑣2
1. Given the barometric pressure of 14.7 psia (29.92 in. Hg),
make these conversions: (a) 80 psig to psia and to atm (b) 20
in. Hg vacuum to in. Hg abs and psia

GIVEN: Patm = 14.7 psia

REQUIRED: (a) & (b)
SOLUTION:
✓ 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 ± 𝑃𝑔
80psig is in gage, we need to convert it into psi absolute.
(a) 𝑃𝑎𝑏𝑠 = 14.7 𝑝𝑠𝑖𝑎 + 80𝑝𝑠𝑖𝑔 = 94.7 𝑝𝑠𝑖𝑎
To convert the answer to atm, we will use the following
conversions we learned above.
1 𝑎𝑡𝑚
𝑃𝑎𝑏𝑠 = 94.7psia × = 6.44 𝑎𝑡𝑚
14.7 𝑝𝑠𝑖𝑎
 We can use 29.92 inHg for the Patm so that we won’t
convert other units, then gage pressure will be negative since it’s
in vacuum as stated.
(b) 𝑃𝑎𝑏𝑠 = 29.92𝑖𝑛𝐻𝑔 − 20𝑖𝑛𝐻𝑔 = 9.92 𝑖𝑛 𝐻𝑔
Now convert the absolute pressure we got into psia
unit.
14.7 𝑝𝑠𝑖𝑎
𝑃𝑎𝑏𝑠 = 9.92inHg × = 4.87𝑝𝑠𝑖𝑎
29.92𝑖𝑛𝐻𝑔
2. A Fahrenheit and a Celsius thermometer are both immersed in
a fluid. (a) If the two numerical readings are identical, what is the
fluid temperature expressed in K and in deg R? (b) What is the
fluid temperature if the Fahrenheit reading is numerically twice
that of Celsius reading?
GIVEN: (a) it says that tf is equal to tc so, t˚f = t˚c
(b) tf is twice tc so, t˚f = 2 (t˚c)
REQUIRED: (a) & (b)
SOLUTION:
5
✓ t˚c = 9
(t˚f − 32)
(a) t˚f = t˚c substitute it to the above equation
5
t˚f = (t˚f − 32)
9 t˚f = t˚c
9t˚f= 5t˚f −160 t˚f = −40℉ t˚c = −40℃
 5
✓ t˚c = 9
(t˚f − 32)
(b) t˚f = 2 (t˚c)
substitute the value of t˚f to the above equation
5
t˚c = (2t˚c
− 32)
9
9t˚c= 10t˚c −160
t˚c = 160 ℃, but t˚f = 2 (t˚c)
t˚f = 2(160) = 320 ℉
convert the answers into ABSOLUTE TEMPERATURE
VALUES
𝑇𝐾 = t˚c +273 = 160 +273 = 433K
𝑇°𝑅 = t˚f +460 = 320 +460 = 780°R
3. A fluid moves in a steady flow manner between two sections
in a flow line. At section 1: A1=10 ft2, vത1 = 100 fpm, 𝑣1 =4ft3/𝑙𝑏𝑚 .
At section 2: A2 = 2 ft2, 𝜌2 = 0.20 lb/ft3. Calculate (a) the mass
flow rate in 𝑙𝑏𝑚 /hr and (b) the speed at section 2 in fps.
GIVEN:
A1=10 ft2 A2 = 2 ft2
vത1 = 100 fpm 𝜌2 = 0.20 lb/ft3
𝑣1 =4ft3/𝑙𝑏𝑚

REQUIRED: (a) 𝑚ሶ (b) vത 2

SOLUTION:
A1vത1 (10 ft2)(100 ft/min) 𝑙𝑏𝑚 60𝑚𝑖𝑛 𝑙𝑏𝑚
𝑚ሶ 1 = = = 250 × = 1500
𝑣1 4ft3/𝑙𝑏𝑚 𝑚𝑖𝑛 1ℎ𝑟 ℎ𝑟
A1=10 ft2 A2 = 2 ft2
vത1 = 100 fpm 𝜌2 = 0.20 lb/ft3
𝑣1 =4ft3/𝑙𝑏𝑚

𝑚ሶ 1 = 𝑚ሶ 2 WE KNOW THAT MASS1 IS THE

𝑚ሶ 2 = 𝜌2 𝐴2 vത 2 JUST SAME MASS AS MASS 2

𝑙𝑏
𝑚ሶ 2 1500 𝑚
vത 2 = = ℎ𝑟
𝜌2 𝐴2 (0.20 𝑙𝑏𝑚 )(2 ft2)
ft3
𝑓𝑡 1ℎ𝑟 CONVERT HOUR INTO
vത 2 = 37500 ×
ℎ𝑟 3600 𝑠𝑒𝑐 SECONDS

𝑓𝑡
vത 2 = 10.42
𝑠𝑒𝑐
ENERGY – capacity of the system to perform work of produced
heat
FORMS OF ENERGY – energy possessed by a body which is to be
considered when analyzing thermodynamic system.
A. STORED ENERGY – B. TRANSITION ENERGY - not
dependent upon the flow dependent on the flow of
of mass mass
1. POTENTIAL ENERGY 5. HEAT
2. KINETIC ENERGY 6. WORK (MECHANICAL)
3. INTERNAL ENERGY
4. FLOW WORK
1. POTENTIAL ENERGY – stored energy due to its elevation
✓∆PE = mg ∆z (Joule, BTU, ft-lbf)
PE -> f (m,z)
1 m
z1
2 z2
where: m – mass
∆ z
g – gravity
z – elevation / height
2. KINETIC ENERGY – stored energy of a body by virtue of its
motion (velocity)
✓∆KE = m∆ vത ² / 2g (Joule, BTU, ft-lbf)
KE -> f (m,തv)
q where: vത − velocity
m m
dx

1 2
As the coaster goes up the potential energy increases,
and when it goes down the kinetic energy increases
while the potential energy decreases.
– it is the energy required to move the fluid across the boundary of
the system
Wf -> f (P , V)

The unit for work would

be Joule (J) or ft-lbf

Derivation:
Wf= PV
𝑁
= ( 2 ∙ 𝑚3 )
𝑚
=N∙m
N∙m = J (Joule)
 P1
𝑚1
V1
SYSTEM P2
𝑚2
V2

∆ Wf = ‫𝑉𝑑𝑃 ׬‬

∆Wf = 𝑊𝑓2 − 𝑊𝑓1

∆Wf = P2 V2 − P1 V1
NOTE: V can also be
∆Wf = ∆PV expressed as 𝑉ሶ
1. A 600𝑘𝑔𝑚 hammer of a pile driver is lifted 2m above a pilling head. What
is the change of potential energy? If the hammer is released, what will be
its velocity and the instant if it sticks the pilling? Local g= 9.65 m/s2
GIVEN: m= 600𝑘𝑔𝑚
g = 9.65 m/s2 vത =?
𝑧2 = 2𝑚
𝑧1 = 0 (since it is the datum line)
𝑧2

REQUIRED: ∆PE & vത

SOLUTION:
𝑚𝑔∆𝑧 𝑚𝑔(𝑧2 −𝑧1 )
✓ ∆PE = =
𝑘 𝑘
(600𝑘𝑔𝑚 )(9.65 m/s2)(2𝑚 − 0)
= = 11580 𝐽 𝑜𝑟 11.58 𝑘𝐽
𝑘𝑔𝑚 ∙ 𝑚
1
𝑁 ∙ 𝑠2
The potential energy at 2m elevation is also the energy that will
be used when the hammer is released, that means
∆PE = ∆KE
vത =?
𝑚തv 2
∆KE =
2𝑘
600𝑘𝑔𝑚 (തv 2 ) 𝑧2
11580 J =
𝑘𝑔𝑚 ∙ 𝑚
2(1 )
𝑁 ∙ 𝑠2
𝑘𝑔𝑚 ∙ 𝑚
2 1 11580𝑁𝑚 𝑚2
𝑁 ∙ 𝑠2 SQUARE ROOT BOTH SIDES
vത 2 = = 38.6 2 OF THE EQUATION
600𝑘𝑔𝑚 𝑠
𝑚2
vത = 38.6 2 = 6.21 𝑚/𝑠
𝑠
2. The flow energy of 142 L/min of a fluid passing a boundary to a system is
108.5 kJ/min. Determine the pressure at this point.

GIVEN: 𝑉ሶ = 142 L/min Wf = 108.5 kJ/min

REQUIRED: 𝑃 =?
SOLUTION:
✓ 𝑊𝑓 = 𝑝Vሶ
𝑘𝐽 𝑘𝑁 ∙ 𝑚
𝑊𝑓 108.5
𝑚𝑖𝑛 =
108.5
𝑚𝑖𝑛
CANCEL AND CONVERT
𝑃= = NECESSARY UNITS
𝑉ሶ 𝐿 𝐿 1𝑚 3
142
𝑚𝑖𝑛 142 ×
𝑚𝑖𝑛 1000𝐿

𝑘𝑁
𝑃 = 764.08 𝑜𝑟 764.08𝑘𝑃𝑎
𝑚2
– energy crossing a system boundary because of a temperature
difference between the system and surroundings.
✓Q = mc∆T (Joule, ft-lbf)

where: m – mass (lbm, kgm)

c – specific heat (BTU/lb°R, kJ/kgm-K)
∆T – change in temp. (°R, K)

This is the total heat flow of the gas

• When Q is negative (-), heat is being removed from the system
• When Q is positive (+), heat is being added to the system
BTU or British Thermal Unit is another unit that we
can use in English system.

-it is defined as the amount of heat required to raise

the temperature of one pound of water by one
Fahrenheit degree.
 -is defined as the quantity of heat required to change the temperature
of unit mass to one (1) degree.
USE: c = cp @ constant pressure process
where Q = mcp∆T = ∆H (change in
enthalpy)

c = cv @ constant volume process

where Q = mcv∆T = ∆U (change in internal
energy)

Unit Derivation:
𝑄 J 𝑄 𝑓𝑡−𝑙𝑏𝑓
(SI) c = =
m∆T kgm∙K (English) c = =
m∆T lbm∙˚R
– stored energy due to its motion of molecules and forces
of attraction between them.

CHANGE OF
STATE ✓∆ U = mcv ∆ T

✓U = m∙u
Q
✓∆U= m ∆u
= m(𝑢2 − 𝑢1 )
Where: cv – specific heat
u – specific internal
energy
ENTHALPY
-combination of useful energy, the amount of
energy absorbed, or heat expelled by the system to
cause a change in the system. ✓ ∆ H = mcp ∆ T

✓ H = m∙ h
✓ ∆H= m ∆h
= m(ℎ2 − ℎ1 )
✓ ∆ H = ∆ U + ∆ Wf
Where: cp – specific heat
h – specific enthalpy
Wf – work flow
 Gas constant:

R = cp – cv

Specific heat ratio: Some values can be found on

the Table next slides.

k = cp / cv

NOTE: That cp is always, if not equal, to cv.

 - energy in transit
- a process done by or on the system, but a
system contains no work
- the action of force on the object through a
distance (W = F x d)

Standard formula for work: 𝑊 = ‫𝑉𝑑𝑃 ׬‬

This is the total work done on or by the gas

• When W is negative (-), work is being
done by the system
• When W is positive (+), work is being
done on system
-the measure of a system's thermal energy per unit
temperature that is unavailable for doing useful work.

-often interpreted as the degree of disorder or

randomness in the system.

𝑄 𝐽 𝑓𝑡 − 𝑙𝑏𝑓 𝐵𝑇𝑈
𝑆= 𝑈𝑁𝐼𝑇𝑆: , 𝑜𝑟
𝑇 𝐾 °𝑅 °𝑅
1. Provide 4 𝑘𝑔𝑚 of a gaseous substance with 300kJ of
heat at constant volume so that it undergoes a
temperature change of 80K. (a) Find the average
specific heat cv of substance during this process. (b)
if k=1.55 for this gas, find cp and R.
1. Provide 4 𝑘𝑔𝑚 of a gaseous substance with 300kJ of heat at
constant volume so that it undergoes a temperature change
of 80K. (a) Find the average specific heat cv of substance
during this process. (b) if k=1.55 for this gas, find cp and R.
GIVEN: m= 4 𝑘𝑔𝑚 ∆𝑇 = 80𝐾
Q = 300kJ
REQUIRED: (a) cv (b) cp & R if k=1.55
SOLUTION:
(a)
✓ 𝑄 = 𝑚𝑐𝑣 ∆𝑇 (we will use cv since V = constant)
𝑄 300𝑘𝐽 𝑘𝐽
cv = = = 0.9375
𝑚∆𝑇 (4𝑘𝑔𝑚 )(80𝐾) 𝑘𝑔𝑚 ∙ 𝐾
(b) NOW USING THE FORMULAS FOR SPECIFIC HEAT CONSTANT AND GAS CONSTANT
𝑐𝑝
✓ 𝑘=
𝑐𝑣
𝑘𝐽
cp = kcv = (1.55)(0.9375 )
𝑘𝑔𝑚 ∙𝐾
𝑘𝐽
cp = 1.4531
𝑘𝑔𝑚 ∙𝐾

𝑘𝐽
✓ R = cp − cv = (1.4531 − 0.9375)
𝑘𝑔𝑚 ∙𝐾
𝑘𝐽
R= 0.5156
𝑘𝑔𝑚 ∙𝐾
2. If 6 𝑙𝑏𝑚 of argon undergo a constant pressure heating
process from 80℉ to 230℉, determine (a) the heat (b)
the change in internal energy (c) change in enthalpy
2. If 6 𝑙𝑏𝑚 of argon undergo a constant pressure heating
process from 80℉ to 230℉, determine (a) the heat (b) the
change in internal energy (c) change in enthalpy
GIVEN: m= 6 𝑙𝑏𝑚
𝑇1 = 80℉ + 460 = 540°𝑅 we already converted the
given temperatures into
𝑇2 = 230℉ + 460 = 690°𝑅 absolute values
REQUIRED: (a) Q (b) ∆𝑈 (c) ∆𝐻
SOLUTION:
(a)
✓ 𝑄 = 𝑚𝑐𝑝∆𝑇 (we will use cp since P = constant)
𝐿𝑜𝑜𝑘𝑖𝑛𝑔 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 𝑤𝑒 𝑘𝑛𝑒𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑐𝑝 𝑜𝑓 𝑎𝑟𝑔𝑜𝑛 𝑖𝑛 𝐸𝑛𝑔𝑙𝑖𝑠ℎ 𝑆𝑦𝑠𝑡𝑒𝑚
𝐵𝑇𝑈
𝑖𝑠 0.12
𝑙𝑏𝑚 ∙ °𝑅
 𝐵𝑇𝑈
(a) 𝑄 = 𝑚𝑐𝑝∆𝑇 = (6𝑙𝑏𝑚 )(0.12 )(690 − 540)°𝑅
𝑙𝑏𝑚 ∙ °𝑅
𝑄 = 108 BTU

(b) ∆𝑈 = 𝑚𝑐𝑝∆𝑇
𝐴𝑔𝑎𝑖𝑛 𝑙𝑜𝑜𝑘 𝑜𝑛 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 𝑏𝑢𝑡 𝑡ℎ𝑖𝑠 𝑡𝑖𝑚𝑒 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑐𝑝 𝑜𝑓 𝐴𝑟𝑔𝑜𝑛.

𝐵𝑇𝑈
∆𝑈 = 6𝑙𝑏𝑚 0.07 690 − 540 °𝑅
𝑙𝑏𝑚 ∙ °𝑅
∆𝑈 = 63 𝐵𝑇𝑈

(c) ∆H =? 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑖𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 𝑖𝑠 𝑢𝑛𝑑𝑒𝑟 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑡ℎ𝑒𝑛

∆H = Q = 108 BTU
3. The ratio of specific heats is k=cp/cv and, for an ideal gas,
their difference is cp – cv = R, a constant. Combine these
two expressions and arrive at a new formula for cp and cv.
𝑐𝑝
GIVEN: k = EQ. 1 R = cp − cv EQ. 2
𝑐𝑣

REQUIRED: cp & cv formulas

SOLUTION:
Let, 𝑐𝑝 = 𝑘𝑐𝑣 SUBSTITUTE THE VALUE OF cp TO EQUATION 2
R = kcv − cv
R = cv k − 1
𝑅
cv =
k−1
Let,
𝑐𝑝
cv = SUBSTITUTE THE VALUE OF cv TO EQUATION 2
𝑘

R = cp − cv
𝑐𝑝
R = cp −
𝑘
𝑐𝑝𝑘 − 𝑐𝑝
R= TRANSPOSE “k” TO THE LEFT SIDE
𝑘

kR = cp(k − 1) 𝑘𝑅 𝑅
cp = & cv =
k−1 k−1
𝑘𝑅
cp = These formulas can be used when finding
k−1
the values for cp and cv.
THERMODYN ANSWERS to QUIZ 3

Test I:

1. Energy quantity that is a function of both mass and elevation

Potential Energy

2. Energy quantity that is a function of both mass and velocity.

Kinetic energy

3. It is the energy required to move the fluid across the boundary of the system.
Flow work

4. A Joule unit is the same as

Newton-meter

5. 1 Btu is equal to
778 ft-lbf

6. When the heat is being removed from a system, which statement is true?
Heat is negative

7. It is defined as the amount of heat required to raise the temperature of one pound of water by
one Fahrenheit degree.
British thermal unit

8. It is defined as the quantity of heat required to change the temperature of unit mass to one
degree.
Specific heat

9. A stored energy due to its motion of molecules and forces of attraction between them.
Internal energy

10. It is called the combination of useful energy, the amount of energy absorbed, or heat expelled
by the system to cause a change in the system.
Enthalpy

11. If the system undergoes a constant pressure process, which is true?

The heat is equal to the change of enthalpy

12. If the system undergoes a constant volume process, which is true?

The heat is equal to the change of internal energy

13. Which of the following gas has the greatest specific heat ratio?
Argon
14. What is the specific heat constant of Oxygen if the volume is constant in the system?
0.6595 kJ/kgm-K

15. What is the gas constant value of Methane?

0.1238 Btu/lbm-oR

16. If a certain gas has the following specific heat values; cp = 1.25, and Cv = 0.70 Btu/lbm-oR, what is
the gas constant, R?
Solution:
R = cp – cv = 1.25 – 0.70
R = 0.55 Btu/lbm-oR

17. If a certain gas has the following specific heat values; cp = 1.25, and Cv = 0.70 Btu/lb m-oR, what is
the specific heat ratio, k?
Solution:
k = Cp/Cv = 1.25/0.70
k = 1.7857

18. If the given system undergoes a pV = C curve, which of the following relations is true?
V1/V2 = p2/p1

19. For a 5 kgm in a system, what is the total enthalpy if the specific enthalpy is 80 kJ/kg m?
Solution:
H = m x h = 5 kgm x 80 kJ/kgm = 400 kJ

20. For a 3 lbm system, what is the specific internal energy if the total internal energy is 300 Btu?
Solution:
u = U / m = 300 Btu / 3 lbm = 100 Btu/lbm

Part II:

1. A 100-lb mass has a potential energy of – 4 Btu with respect to a given datum within the
earth’s standard gravitational field. Find its height relative to the datum.
Solution:
z = PE x k / m x g
= (-4 Btu)(32.2 ft-lbm/lbf-sec2) / (100 lbm)(32.2 ft/sec2) x 778 ft-lbf/1 Btu
z = - 31.12 feet

2. A 64,400-lbm airplane is travelling at 1,000 feet/sec, how much is its kinetic energy?
Solution:
KE = m v2 / 2k
= (64,400 lbm)(1,000 ft/sec)2 / 2 (32.2 ft-lbm/lbf-sec2) x 1 Btu/778 ft-lbf
KE = 1,285,347 Btu
3. For a constant pressure system whose mass is 50 lb, 1 hp-min is required to raise its
temperature 1oF. Determine the specific heat for the system.
Solution:
Cp = Q / m ΔT = 1 hp-min / 50 lbm (1 oF) x 42.4 Btu/hp-min
Cp = 0.848 Btu/lbm-oF

4. If a certain gas has k = 1.30 and R = 310 Joules/kg-K, what is the specific heat at constant
pressure?
Solution:
Cp = k R / (k-1) = 1.30 (310 J/kgm-K) / (1.30 – 1) x (1 kJ/1,000 J)
Cp = 1.3433 kJ/kgm-K

5. If 20 kgm of acetylene (C2H2) undergo a constant volume process from 32oC to 85oC,
determine the heat flow within the system.
Solution: Cv = 1.37 kJ/kgm-K
Q = m Cv ΔT = 20 kgm (1.37 kJ/kgm-K) [(85 + 273) – (32 + 273)] K
Q = 1,452.20 kJ

6. A 1-kg gaseous system enclosed in a piston-and-cylinder arrangement receives heat while its
pressure remains constant at 345 kPaa. The enthalpy increases 420 kJ and the temperature
increases 70 K. What is the amount of heat received by the system?
Solution:
For any constant pressure system, Q = ΔH, hence,
Q = 420 kJ

7. Determine the moving boundary work done by a 1-lbm fluid system with an initial pressure
and volume of 80 psia and 1 ft3, respectively, to a pressure of 400 psia, in accordance with
the relation V = C.
Solution:
For when V = C, the differential dV = 0, hence,
Wmb = ꭍ pdV = 0 Btu

8. A very hot substance at 950 psia, 600oF has a specific volume of 0.5485 ft3/lbm. Calculate the
flow energy to move this substance.
Solution:
Wf = p V = 950 lbf/in2 (0.5485 ft3/lbm) x (144 in2/1 ft2) x (1 Btu/778 ft-lbf)
Wf = 96.45 Btu/lbm

9. During a thermodynamic process executed by a system, the pressure increases from 100
kPaa to 400 kPaa in accordance with pV = C; the initial volume is 50 liters. Determine the
work.
Solution: 1 m3 = 1,000 liters
Wmb = p1V1 ln(p1/p2) = 100 kN/m2 (0.05 m3) ln (100/400) = - 6.93 kN-m or
 Wmb = -6.93 kJ

10. Given the relations pV1.4 = C, find the final volume in a system if the pressure decreases from
240 psia to 14.7 psia, and the initial volume was 1 cubic foot.
Solution:
P1V11.4 = p2V21.4 or V2 = V1 (p1/p2)1/1.4 = 1 ft3 (240/14.7)1/1.4
V2 = 7.35 cu. ft.

11. During a thermodynamic process executed by a system, the pressure increases from 100
kPaa to 400 kPaa in accordance with pV = C; the initial volume is 50 liters. Find the change of
flow work.
Solution:
For when pV = C, p1V1 = p2V2, hence, ΔpV = p2V2 – p1V1 = 0. Therefore,
ΔWf = ΔpV = 0 kJ

12. A 1-kg gaseous system enclosed in a piston-and-cylinder arrangement receives heat while its
pressure remains constant at 345 kPaa and the temperature increases 70 oC. If the gas is air,
find the change of internal energy.
Solution: Cv = 0.7186 kJ/kgm-K; and ΔT = 70oC = 70 K
ΔU = m Cv ΔT = 1 kgm (0.718 kJ/kgm-K) (70 K)
ΔU = 50.30 kJ

13. A 1-kg gaseous system enclosed in a piston-and-cylinder arrangement receives heat while its
pressure remains constant at 345 kPaa and the temperature increases 70 oC. If the gas is air,
find the change of enthalpy.
Solution: Cp = 1.0062 kJ/kgm-K
ΔH = m Cp ΔT = 1 kgm (1.0062 kJ/kgm-K) (70 K)
ΔH = 70.43 kJ

14. A 1-kg gaseous system enclosed in a piston-and-cylinder arrangement receives heat while its
pressure remains constant at 345 kPaa and the temperature increases 70 oK. If the gas is air,
find the heat.
Solution:
When pressure is constant, Q = m Cp ΔT.
Q = 1 kgm (1.0062 kJ/kgm-K) (70 K)
Q = 70.43 kJ