THERMODYNAMICS 1

TERM PAPER
MONTES, ELEAZAR JUNOR L.
BSIE-3A
 THERMODYNAMICS
STUDY OF PHYSICS

Newton’s
Law
𝑁1 = 1𝐹 =1 𝐹 𝐹 = 𝑁𝑒𝑤𝑡𝑜𝑛𝑠 /
=0 𝑘𝑔.𝑚 𝑠
𝑚
2

𝑁2 = 𝐹 =
𝑎
𝑁3 = ∑𝐹𝐹 =
−2𝐹𝐹

1. System of Units
𝐹 = 𝑚𝑎 𝐾 =
𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑎 =𝑑⋅1
𝑎 =𝐹
𝑘
𝑡 𝑡
𝑡 = 2𝑠𝑒𝑐 𝑚
𝑚=1
𝑎 = 1 𝑚/𝑠2
4

𝐹 = 𝑚𝑎
1
= 𝑚/𝑠
2
4

𝐹 =4
1 𝑁
𝑘𝑔⋅𝑚
𝑠2

the force is at rest

𝐾 =
𝑚𝑎 /𝑘𝑔⋅𝑚/𝑠2
𝑘 =1
𝐹 𝑁𝐸𝑊𝑇𝑂𝑁

an external force
𝑔 = 9.8 𝑚/𝑠2
˙
𝑘 = 9. 81𝑚/𝑠2
𝑘 = 9.8066 m/𝑠2

proportionality constant
𝐹 = = 1, ≠ 1, ≠ 0
𝑚𝑎
𝑘

UNITY (IDEAL) = 1
NON-UNITY (NON-IDEAL) ≠ 1
𝑔 = 9.8066
𝑚/𝑠2
for accuracy
EXAMPLE 1

UNITY (IDEAL) NON-UNITY (NON-

=1 IDEAL) ≠ 𝟏
𝑔 = 9.8066 𝑚/𝑠2 𝑔 = 9.8066 𝑚/𝑠2
𝑚𝑔
𝐹
𝑚𝑎 𝑘
𝐹
= 𝑚𝑔
𝑘
�

𝐹 = (1)(9.8066 =
�
𝐹
𝑚/𝑠2)
𝑭 = 𝟗. 𝟖𝟎𝟔𝟔 𝑵 (𝑘𝑔𝑚)
𝑘
= 𝑠2
9.8066
𝑘𝑔𝑚
𝑚
𝑘 = ⋅
𝑘𝑔𝑓
𝒌𝒈𝒎
𝒎
𝒌 = 𝟗. 𝟖𝟎𝒄𝒎
⋅ 𝟐 𝒌𝒈𝒇
𝒔
Example
Example 𝑀𝑎𝑛 = 85
𝑀𝑎𝑛 = 85 𝑘𝑔
𝑘𝑔 𝑔 = 9.899
𝑔 = 9.899 𝑚/𝑠
𝐹𝑔
2

𝑚/𝑠
𝐹𝑔
2
= 9.899
𝑚 )
�
=
= (85𝑘𝑔𝑚) 𝑘𝑔𝑚
( 𝑠
� �

= (85𝑘𝑔𝑚)(9.8066 9.8066 ⋅
�
2
𝑚/𝑠2) 𝑚 𝑘𝑔 𝑠
𝑓
𝑭 = 𝟖𝟑𝟑. 2

𝟓𝟔𝟏𝑵 𝑭𝒈 = 𝟖𝟓. 𝟖𝟎𝟏

𝒌𝒈𝒇
 EXAMPLE 2
GIVEN: 𝑔 = 32.688𝑓𝑝𝑠2
𝛥𝑉 = 0.003 𝑓𝑝𝑠2 ℎ = 1000 𝑓𝑡. 𝑎𝑠𝑐𝑒𝑛𝑡

b. weight at 5%
𝑎. 𝑔 𝑎𝑡 30.504
𝑓𝑝𝑠2 𝐹𝑔 = 0.95
ℎ =𝑉𝑆 =−1.584
Δ𝑔
30.504−32.088 𝑎 = 0.95 𝑔
𝑓𝑡/𝑠2
𝑎 = 0.95 (32.088
𝑓𝑡/𝑠2)
−1.584
= 𝑓𝑡/𝑠2
− 0.00
3
𝑓
𝑡

1000 ℎ = 𝑣𝑠2−32.088
30.484−Δ𝑎
= −1000
0.003
𝑎 𝑓𝑡

ℎ = 528,000
ft ℎ = 534666.667
1𝑚𝑖𝑙
528000 𝑓𝑡 = 𝟏𝟎𝟎 𝑓𝑡
× 5.280 𝐦𝐢𝐥𝐞𝐬 1
𝑚𝑖𝑙𝑒
Convert in 534666.667 𝑓𝑡 𝑥
kilometers 5,280
𝑓𝑡
100 𝑚𝑖𝑙𝑒𝑠 1.609𝑘 = 𝟏𝟔𝟎. 𝟗 𝒌𝒎
× 1
𝑚𝑖𝑙𝑒

c. Force @ 180 lbm, height of

29,131 ft
𝐹 𝑚𝑔
= �
�
(180 𝑙𝑏𝑚)( 30.088
= 𝑓𝑡/𝑠2)
32.174
𝑓𝑡 ⋅ 𝑙𝑏𝑚
𝑠
𝑭 = 𝟏𝟕𝟗. 𝟓𝟏𝟗
2

𝒍𝒃𝒇
SPECIFIC VOLUME
𝜌= ,𝛾 =
𝑚 𝐹𝑔 DENSITY
𝑚
𝜌
𝑉 𝑉

= �

𝜌=𝛾
𝑘
�

or 𝛾
𝑃𝑔
𝑔
=
𝑘

𝑉 = 𝑉𝑜𝑙𝑢𝑚𝑒
m = mass
𝜌 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐷𝑒𝑛𝑠𝑖𝑡𝑦
SPEDIFIC VOLUME
� 1
𝛾 = �=
𝑚
Problem 1
g= 9.8066 𝑚/𝑠2
𝑘𝑔𝑚
𝜌 = 1000 SPECIFIC WEIGHT
𝑚3
� 𝐹
𝜌 𝛾 −� = 𝛾 =
𝛾 =𝑔 𝜌𝑔 𝑉
( 𝑘
𝑚
)( 3)
=
1000𝑘𝑔𝑚 𝑚
9.8066 2)
𝑘𝑔𝑚 𝑚 𝑠
9.8066 ⋅
𝑘𝑔𝑓 𝑠
2

𝛾= 𝐷𝑒𝑛𝑠𝑖𝑡𝑦
𝟏𝟎𝟎𝟎𝒌𝒈𝒇/𝒎𝟑 𝑆𝑔 =
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓
𝑊𝑎𝑡𝑒𝑟
SOLUTION
𝑃𝑥
1. Given: 𝑃𝑞𝑖𝑟 = =
𝑃𝐻2
1.3𝑘𝑔/𝑚3
𝑂
V= 10𝑚 × 10 ×
3
Formula: 𝜌𝑎𝑖𝑟
=
�
�

𝑉
2. Given: 𝑉 = 2𝐿
𝜌=
800𝑘𝑔/𝑚3

= 2𝐿 ×1000
3

=2×
10−3𝑚3
800𝑘
𝑚 = (2 × 𝑔 )
𝑚
m = 1.60 3

kgm

3. 𝐺𝑖𝑣𝑒𝑛: 𝑃𝐵𝑙𝑜𝑜𝐷 =
1060
𝑔 =𝑃1𝐻2
𝑃𝑥

𝑃𝐻 =
1000𝑘𝑔/𝑚3
2
𝑂

𝑆𝑔 =(1000𝑘𝑔/
(1000𝑘𝑔/𝑚 )
𝑚3)

Problem 2
Given:
𝜌1 = 1500𝑘 𝑔/𝑚3
𝜌2 = 500𝑘 𝑔/𝑚3
𝜌𝑇 = 800𝑘 𝑔/𝑚3
𝑉𝑇 = 100 L

𝑔 = 9.675𝑚/𝑠2
𝑀𝑇 =?
SOLUTION: 𝑉1 + 𝑉2 = 𝑉1
𝜌=
𝑚

𝑀𝑇 = 𝜌 + (𝑉)

𝑀𝑇 = 𝑀1 + 𝑀2

( 0 )
= (800
10
)1000
= 80𝑘𝑔𝑚

𝐹 =
(80)(9.675)
(9.8066)
= 78.926 𝑘𝑔𝑓

𝑚𝑇 = 80𝑘𝑔 = 𝑉1𝑃1 + 𝑉2𝑃2 eq.1

100𝐿 = 𝑈1 + 𝑉2

100𝐿 = 𝑉1 + 𝑉2
0.1𝑚3 = 𝑉1 + 𝑉2 eq. 2
0.1𝑚3 − 𝑉1 = 𝑉2

80 = (𝑉1)(1500) + (0.1𝑚3 − 𝑉1)(500)

80 = 1500𝑉1 + 50 − 500𝑉2
80 = 1000 𝑉1 + 50
𝑉1
= 1000
80−50

𝑉 = 0.03𝑚3
𝑚 = (0.03𝑚3)(1500𝑘𝑔/𝑚3)

𝒎=
𝟒𝟓𝒌𝒈𝒎
𝐹𝑚
=
(45 𝑘𝑔)(9.675)
9.8066

𝑭 = 𝟒𝟒. 𝟑𝟗𝟔
𝒌𝒈𝒇
 PRESSURE
Force
/F
Area
P=
A

EXAMPLE
m=50 kg
2
a=4.25 m
P=?
F=mg
m
F=50 kg (9.8061 2
)
s
F=490.33 N
F
P=
A
490.33 N
P= 2
4.25 m

N
P=115.372 2
m

P FLUID=Pg h
Pex = ATM
P=P ex + P FLUID

EXAMPLE
h=20 m
P=?
kg
ρ=1030 3
m
kg m
P FLUID=(1030 3 )(9.8066 2 )(20M)
m s

N
P FLUID=202,015.96 2
m

N
Pex =101,325 2
m
PT =Pex + PFLUID
N N
PT =101,325 2 + 202,015.96 2
m m

N
PT =303,340.96 2
m

P gage=P|¿|− P ¿
ATM

PVAC =P ATM −P|¿|¿

PTABS =Pvac + P atm - Close System
PTABS =P ATS + Pg - Open System

PROBLEM
Convert 100 PSIgauge to PSIABS
1 atm = 101.325 Kpa
=14.7 psi
= 760 mmHg
= 760 torr
= 29.9 inHg
= 1.013 bar
= 2,116.2 psf
PT =P ATM + P g
PT =14.7 psi+100 psi

PT =114.7 psi

TEMPERATURE
9
℉= ℃+3 0
5
5
℃= [T (℉ )−32]
9
K=T ( ℃ ) +273.1 5
R=T ( ℉ ) +459.6 7
CONVERT
540 ℉ ¿℃
5
℃= (540 ℉ −32)
9
℃=282.222

120 ° R ¿ K
120 ° R=T ( ℉ ) + 459.6 7
℉=−339.67
5
℃= (−339.67 ℉−32)
9
℃=−206.48 3
K=−206.483+273.1 5
K=66.66 7

−69 ℃ ¿ ℉
9
℉= (−69 ℃)+3 0
5
℉=−92.2

1025 K ¿℉
1025 K=T ( ℃ ) +273.1 5
℃=1025−273.1 5
℃=751.8 5
 9
℉= ( 751.85 ) +3 0
5
℉=1385.3 3

−256 ℉ ¿ K
9
−256 ℉= ℃+3 0
5
9
−256 ℉−30= ℃
5
(−256−30)(5)
=℃
9
℃=−16 0
K=−160℃ +273.15
K=113.15

THERMAL EXPANSION
∆ L=∝ L0 (T 1−T 0)

Example
L=3 mL
T 0=20 ℃ ; T 1=100 ℃
∆ L=(∝)(L)(∆ T )
−5
2.4 ×10
∆ L=( )(3 m)(100℃−20 ℃)
℃
−3
∆ L=5.76 ×10 m
∆ V =β V 0 (T F−T 0 )

EXAMPLE
−5
18 ×10
Hg=
℃
V =50 mL
T i=25 ℃ ; T f =100 ℃

( )
−5
18 ×10
∆V = (50 mL)(100 ℃−25 ℃)
℃
∆ V =0.675 m
HEAT =Q=mC ∆ T
WORK =W =F H × D
EXAMPLE 1
F=20 N ; D=5 m
W =(20 N )(5 m)
W =100 joules

EXAMPLE 2
500 N
F A=
tan30 °
F A=866.025 N - Horizontal Force
W =F H × D
W =( 866.025 N ) (10 m )
W =8660.25 joules
 POWER
W
P=
t

EXAMPLE 1
m=500 Kg
d=1km
t=15 min
P=?
Find Power in Watts and Horsepower
F=mg
m
F=(500 Kg)(9.8066 2
)
s
F=4903.3 N
W =Fcosθd
W =( 4903.3 )( 1 ) ( 1000 )
W =4,903,300 Nm
W
P=
t
4,903,300
P=
( 15 )( 60 )
P=5448.111 W ≈ 5.45 KW
1 HP
¿ 5448.111W ×
746 W
¿ 7.303 HP

EXAMPLE 2
P=25 KW
m=1500 Kg
h=30 m
t=?
m
P=(1500 Kg)(9.8066 2
)
s
F=14,709.9 N
W =F × D
W =( 14,709.9 N )( 30 m )
W =441,297 J
W
P=
t
441,297 J
25,000 W =
t
441,297 J
t=
25,000 W
t=17.652 s
 IDEAL GAS LAW
P1V 1 P2V 2
= =mRT
T1 T2

3
V 1=0.6 m ; V 2=?
P1=760 mmHg ; P2=720 mmHg
T 1=25 ℃ ; T 2=20℃
(760 mmHg)(0.6 m ) (720 mmHg)(V 2 )
2
=
25 ℃ 20 ℃
2
(760 mmHg)(0.6 m )(20 ℃)
V 2=
(25 ℃)(720 mmHg)
3
V 2=0.507 m

CHARLES LAW
V1 V2
=
T 1 T2

Find V 2
V 1=2000 mL
T 1=32℃ ; T 2=20 ℃
2000 mL V2
=
32℃ 20 ℃
V 2=1250 mL

BOYLES LAW
P1 V 1=P2 V 2

P1=2 atm ; P2=?

V 1=100 L ; V 2=50 L
( 2 atm ) ( 100 L )=( P2 ) ( 50 L )
( 2 atm ) ( 100 L )
P 2=
50 L

P2=4 atm