TUGAS TERMODINAMIKA KELOMPOK IV
Nama Anggota Kelompok : 1) Arya Dharma 2) Aulia 3) Enda Hutabarat 4) Michael Joy 5) Yos Pawer Ambarita
SOAL DAN PENYELESAIAN PENYELESAIAN g, and what are its units in a system in which the second, 1.1) What is the value of g,
the foot, and the pound mass are defined as in Sec. 1.2, and the unit of force is the poundal, defined as the force required to give l (lbm) an acceleration of l(ft)(s)-2 Penyelesaian
F
1N
lbf
0,225 lbf gc
:
=
=
=
= =
32,2
1.2) Electric current is the fundamental electrical dimension in SI; its unit is the
ampere (A). Determine units for the following quantities, as combinations fundamental SI units. (a) Electric power; (b) Electric charge; (c) Electric potential difference; (d) Electric resistance; (e) Electric capacitance.
Penyelesaian
:
(a) Daya Listrik Daya listrik didefinisikan sebagai laju hantaran energi listrik dalam sirkuit listrik. Satuan SI daya listrik adalah watt yang menyatakan banyaknya tenaga listrik yang mengalir per satuan waktu (Joule/detik). Daya listrik
:
P
=
= I2.R
(b) Muatan Listrik Coulomb, dilambangkan dengan C, adalah satuan SI untuk muatan listrik, dan didefinisikan dalam ampere: 1 coulomb adalah banyaknya muatan listrik yang dibawa oleh arus sebesar 1 ampere mengalir selama 1 detik. Muatan listrik
:
Q = I.t
(c) Beda Potensial Volt (V) adalah satuan turunan di dalam Standar Internasional (SI) untuk mengukur perbedaan tegangan listrik atau beda potensial. 1 Volt berarti beda tegangan yang diperlukan
untuk
membuat
arus
tepat
sebesar
1 ampere di dalam suatu rangkaian dengan resistensi 1 ohm. Beda potensial :
V = I.R
(d) Hambatan Listrik Ohm (lambang: Ω) adalah satuan SI dari impedansi listrik, atau dalam kasus arus
searah, hambatan
listrik. Nama satuan ini berasal dari
ilmuwan Georg Ohm. Satu ohm (yang diukur oleh alat ohm-meter) adalah hambatan listrik pembawa arus yang menghasilkan perbedaan tegangan satuvolt ketika arus satu ampere melewatinya. Hambatan listrik :
R =
(e) Kapasitansi Listrik Kapasitansi atau kapasitans adalah ukuran jumlah muatan listrik yang disimpan (atau dipisahkan) untuk sebuah potensial listrik yang telah ditentukan. Bentuk paling umum dari piranti penyimpanan muatan adalah sebuah kapasitor dua
lempeng/pelat/keping.
Jika
muatan
di
lempeng/pelat/keping adalah +Q dan – Q, dan V adalah tegangan listrik antar lempeng/pelat/keping.
Kapasitansi listrik
:
C =
sat
1.3) Liquidlvapor saturation pressure P
=
is often represented as a function of
temperature by an equation of the form: log10Psat/torr = a -
Here, parameters a, b, and c are substance-specific constants. Suppose it is required torepresent Psat by the equivalent equation: log10Psat/kPa = a -
Show how the parameters in the two equations are related.
1.4) At what absolute temperature do the Celsius and Fahrenheit temperature scales
give the same numerical value? What is the value? Penyelesaian
:
Celcius : Fahrenheit = 5 : 9 (+32) Celcius
= Fahrenheit
5
= 9 (+32)
(kedua ruas dikalikan dengan x)
5x
= 9x + 32
....(1)
-4x
= 32
x
= -8
substitusi nilai x=-8 ke persamaan (1) 5(-8)
= 9(-8) + 32
-40
= -40
Jadi Celcius dan Fahrenheit menunjukkan angka yang sama pada temperatur absolut -40.
1.5) Pressures up to 3000 bar are measured with a dead-weight gauge. The piston
diameter is 4 rnm. What is the approximate mass in kg of the weights required? Penyelesaian Dik
:
: P = 3000 bar
D = 4 mm Dit
: m = .....?
Jawab :
P = 3000 bar x D = 4 mm x
A =
=
= 3000 x 105 N/m2
= 4 x 10-3 m
2
πD
(3,14) (4 x 10-3 m)2
= 12,56 x 10-6 m2
P =
F
↔ F =P.A
= (3000 x 105 N/m2) (12,56 x 10-6 m2) = 3768 N F = m.g (g = 9,8 m/s2) m=
F
=
= 384,49 kg
1.6) Pressures up to 3000 atm are measured with a dead-weight gauge. The piston
diameter is 0.17 (in). What is the approximate mass in (lb,) of the weights required? Penyelesaian : Dik
: P
= 3000 atm
D = 0,17 inch Dit
: m (dalam lbm) = ....?
Jawab :
1 inch = 2,54 cm D = 0,17 inch x
x
= 0,4318 x 10-2 m
P = 3000 atm x
A =
=
x
= 3039,75 x 105 Pa
2
πD
(3,14) (0,4318 x 10-2 m)2
= 0,146 x 10-4 m2
P =
F
↔ F =P.A
= (3039,75 x 105 N/m2) (0,146 x 10-4 m2) = 4438,035 N F = m.g (g = 9,8 m/s2) m=
F
=
= 452,86 kg
m = 452,86 kg x
= 998,38 lbm
1.7) The reading on a mercury manometer at 298.15 K (25°C) (open to the
atmosphere at one end) is 56.38 cm. The local acceleration of gravity is 9.832 m.s-2. Atmospheric pressure is 101,78 kPa. What is the absolute pressure in kPa being measured? The density of mercury at 298.15 K (25°C) is 13.534 g.cm-3. Penyelesaian : Dik
: h = 56,38 cm
g = 9,832 m/s2 Patm = 101,78 kPa 3
ρ = 13,543 gr/cm Dit
: Pabs (dalam kPa) = ....?
Jawab :
3
ρ = 13,543 gr/cm x
h = 56,38 cm x
x
= 13,543 x 103 kg/m3
= 56,38 x 10-2 m
Pgauge = ρ = (13,543 x 103 kg/m3) (9,832 m/s2) (56,38 x 10-2 m) = 75072,66 Pa
Pgauge = 75072,66 Pa x
= 75,07266 kPa
Pabs = Pgauge + Patm = 75,07266 kPa + 101,78 kPa = 176,85266 kPa o
1.8) The reading on a mercury manometer at 70 ( F) (open to the atmosphere at one
end) is 25,62 (in). The local acceleration of gravity is 32,243 ft.s-2. Atmospheric pressure is 29,86 (inHg). What is the absolute pressure in Psia being measured? The density of mercury at 70 (°F) is 13.534 g.cm-3. Penyelesaian : Dik
: h = 25,62 in
g = 32,243 ft/s2 Patm = 28,86 inHg 3
ρ = 13,543 g/cm Dit
: Pabs (dalam Psia) = ....?
Jawab : 2
g = 32,243 ft/s x h = 25,62 in x
x
3 ρ = 13,543 g/cm x
Pgauge
= 9,802 m/s2
x
= 65,07 x 10-2 m
= 13,543 x 103 kg/m3
= ρ
= (13,543 x 103 kg/m3) (9,802 m/s2) (65,07 x 10-2 m)
= 86379,4 Pa Patm
= 29,86 inHg x
Pgauge
= 86361,8 Pa x
Pabs
= Patm + Pgauge
= 14,666 Psia
= 12,5257 Psia
= (14,666 + 12,5257) Psia = 27,19 Psia
1.9) Liquids that boil at relatively low temperatures are often stored as liquids under
their vapor pressures, which at ambient temperature can be quite large. Thus, n butane stored as a liquid/vapor system is at a pressure of 2.581 bar for a temperature of 300 K. Largescale storage (>50 m3) of this kind is sometimes done in spherical tanks. Suggest two reasons why. Penyelesaian :
Tangki bola (spherical tanks) banyak digunakan untuk menyimpan cairan yang mudah mendidih pada suhu yang relatif rendah, karena : 1) Tekanan dalam tangki bola didesain berada di bawah tekanan uap cairan pengisi. 2) Tangki bola menawarkan volume maksimum untuk luas permukaan yang kecil dan memiliki ketebalan 1½ kali ketebalan tangki silinder dengan diameter yang sama
1.10) The first accurate measurements of the properties of high-pressure gases were
made by E. H. Amagat in France between 1869 and 1893. Before developing the dead-weight gauge, he worked in a mine shaft, and used a mercury manometer for measurements of pressure to more than 400 bar. Estimate the height of manometer required. Dik
: P = 400 bar
g = 9,8 m/s2 3
ρ = 13,5 gr/cm Dit
: h = ....?
Jawab :
P = 400 bar x
= 4 x 107 Pa
3
ρ = 13,5 gr/cm x
x
= 13,5 x 103 kg/m3
P = ρ h = =
= 0,03023 x 104 m = 302,3 m
1.11) An instrument to measure the acceleration of gravity on Mars is constructed of
a spring from which is suspended a mass of 0.40 kg. At a place on earth where the local acceleration of gravity is 9.81 m.s-2, the spring extends 1.08 cm. When the instrument package is landed on Mars, it radios the information that the spring is extended 0.40 cm. What is the Martian acceleration of gravity? Penyelesaian : Dik
: m
= 0,4 kg
g bumi = 9,8 m/s2 x bumi = 1,08 cm xmars = 0,4 cm Dit
: gmars = ....?
Jawab :
d Bu →
x bumi = 1,08 cm x
= 1,08 x 10-2 m
F = m.g bumi = K. x bumi (0,4 kg)( 9,8 m/s2) = K. 1,08 x 10-2 m K = 3,6296 x 102 kg/s2 d M
→
x bumi = 0,4 cm x
= 0,4 x 10-2 m
F = m. gmars = K. xmars (0,4 kg) gmars = (3,6296 x 102 kg/s2) (0,4 x 10-2 m) gmars = 3,629 m/s2
1.12) The variation of fluid pressure with height is described by the differential
equation: Here, ρ is specific density and g is the local acceleration of gravity. For an ideal gas, ρ = MP/RT , where M is molar mass and R is the universal gas constant. Modeling the atmosphere as an isothermal column of ideal gas at 283.15 K (l0 °C), estimate the ambient pressure in Denver, where z = l (mile) relative to sea level. For air, take M = 29 g mol-1; values of R are given in App. A. Penyelesaian : : z = 1 mile
Dik
T = 10 oC M = 29 g/mol R = 82,06
g = 9,8 m/s2 : PDenver = ....?
Dit
Jawab :
T = 10 oC + 273,15 = 283,15 K z = 1 mile x
g = 9,8 m/s2 x d d
x
.....(1)
M
.....(2)
substitusi pers (2) ke pers (1) : d d
=-
∫
M
d = -
ln = -
g
M M
∫ d z
= 980 cm/s2
= - ρ
ρ =
x
= 1609,36 x 102 cm
ln
=-
ln P = -
M M
=
P
z z
M
-
..... (1)
M
z= = - 0,196 .....(2)
Substitusi pers (2) ke pers (1) : P
=
P
=
M
-
= 0,822 atm
1.13) A group of engineer has landed on the moon, and they wish to determine the
mass of some rocks. They have a spring scale calibrated to read pounds mass of at a location where the acceleration of gravity is 32,186 (ft)(s)-2. One of the moon rocks gives a reading of 18,76 on this scale. What is its mass? What is its weight on the moon? Take g (moon) = 5,32 (ft)(s)-2. Penyelesaian : Dik
Dit
: g bumi
= 32,186 (ft)(s)-2
g bulan
= 5,32 (ft)(s)-2
Δ bulan
= 18,76
: m
w bulan
= ....? = ....?
Jawab :
m bulan = m bumi Δ u u
=
Δ bumi =
=
u Δ u u
g bumi
-
= 113,498
32,186 (ft)(s)-2
m = Δ bumi lbm = 113,498 lbm Wdibulan
= m . g bulan = 113,498 lbm . 5,32 ft s-2 = 603,809 lbf
1.14) A 70 W outdoor security light burns, on average, 10 hours a day. A new bulb
costs $5.00, and the lifetime is about 1000 hours. If electricity costs $0.10 per kWh, what is the yearly price of "security," per light? Penyelesaian : Dik
: P = 70 W
t = 10 jam/hari
Dit
cost bulb
= 5 dollar/1000 jam
costelec
= 0,1 dollar/kWh
: costtotal per tahun
= ....?
Jawab :
cost bulb per tahun
= cost bulb . t . 365 hari = (5 dollar/1000 jam)(10 jam/hari)(365 hari)
= 18,25 dollar costelec per tahun
= costelec . P . t . 365 hari = (0,1 dollar/kW.jam)(0,07 kW)(10 jam/hari)(365 hari) = 25,55 dollar
costtotal per tahun
= cost bulb per tahun + costelec per tahun = (18,25 + 25,55) dollar = 43,8 dollar
1.15) A gas is confined in a 1,25(ft)-diameter cylinder by a piston, on which rests a
weight. The mass of the piston and weight together is 250 (lbm). The local acceleration of gravity is 32,169 (ft)(s-2), and atmospheric pressure is 30,12 (in Hg).
(a) What is the force in lbf exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 1,7 ft, what is the work done by the gas in (ft lbf)? What is the change in potential energy of the piston and weight? Penyelesaian
:
: D = 1,25 ft
Dik
m = 250 lbm = 32,169 ft.s-2
g
Patm
= 30,12 in Hg
: a) F = ..?
Dit
b) P = ..? dalam kPa w = ? d Ep = ? j ∆ = Jawab
:
A =
=
2
πD
(3,14) (1,25 ft)2
= 1,227 m2 a) F = patm . A + m.g
patm = 30,12 inHg x
p
d- p
-
- d -
= 2130,35 lbf ft-2 F = 2130,35 lbf ft-2 . 1,226 ft2 + 250 lbm . 32,169 ft s-2 F = 10,1726 x 103 lbf
b) P =
F
=
= 8,294. 103 lbf ft-2
P = 8,294. 103 lbf ft-2
- - -
-
P = 57,595 psia c) w = F ∆ = 10,1726 x 103 lbf . 1,7 ft = 17,2934 lbf ft Ep = ∆
= 250 lbm . 32,169 ft.s-2 . 1,7 ft = 13,671. 103 lbf ft
1.16) A gas is confined in a 0.47-m-diameter cylinder by a piston, on which rests a
weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m sP2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight? Dik :
D = 0,47 m A = 0,1734 m2 m = 150 kg g = 9,813 ms-2 patm = 101,57 kPa
Dit
:
a) F = ..?
b) P = ..? dalam kPa w = ? d Ep = ? j ∆ = Jawab : A =
=
2
πD
(3,14) (0,47 m)2
= 0,1734 m2 a) F = P atm . A + m.g F = 101,57. 103 Pa . 0,1734 m2 + 150 kg . 9,813 ms-2 F = 19,08395. 103 N
b) P =
F
=
= 110,057 kPa
w = F ∆
= 19,08395. 103 N. 0,83 m = 15,8396 kJ Ep = ∆ = -2 . 0,83 m = 1,2217 kJ
1.17) Verify that the SI unit of kinetic and potential energy is the joule. Penyelesaian :
Ek =
EP
½ mv2 = m.g.h ½ (kg)(m/s)2 = ½ kg.m2/s2 Joule
kg.m/s2.m
= kg.m2/s2 =
Joule -1
1.18) An automobile having a mass of 1250 kg is traveling at 40 m.s . What is its
kinetic energy in kJ? How much work must be done to bring it to a stop? Penyelesaian :
Dik
:
m = 1250 kg v = 40 m.s-1
Dit
:
Ek = ....? W = ....?
Jawab :
Ek = ½ mv2 = ½ (1250 kg)( 40 m/s)2 = 1000000 J = 1000 kJ W = ΔEk = 1000 kJ
