Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Thermodynamics 1 Sta. Maria Hipolito
Typology: Cheat Sheet
1 / 127
This page cannot be seen from the preview
Don't miss anything!
Basic Principles, Concepts and Definition Thermodynamics is that brach of the physical sciences that treats of various phenomena of energy and the related properties of matter, especially of the laws of transformation of heat into other forms of energy and vice versa.
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.” a= kF m
ma k , k= ma F k is a proportionality constant Systems of units where k in unity but not dimensionless: cgs system: 1 dyne force accelerates 1 g mass at 1 cm/s^2 mks system: 1 newton force accelerates 1 kg mass at 1 m/s^2 fps system: 1 lb force accelerates 1 slug mass at 1 ft/s^2 k = 1 gm ∙ cm dyne ∙ s 2 k^ =^1 k gm ∙ m newton ∙ s 2 k = 1 slug∙ ft lb (^) f ∙ s 2 Systems of units where k is not unity: If the same word is used for both mass and force in a given system, k is neither unity nor dimensionless. 1 lb force accelerates a 1 lb mass at 32.174 ft/s^2 1 g force accelerates a 1 g mass at 980.66 cm/s^2 1 kg force accelerates a 1 kg mass at 9.8066 m/s^2
m k a 1 pound = (1 slug)(1 ft/s^2 ); 1 slug= 1 lb f ∙ s 2 ft F is force in pounds m k is mass in slugs a is acceleration in ft/s^2 Mass and Weight The mass of a body is the absolute quantity of matter in it. The weight of a body means the force of gravity Fg on the body. m k
a
Fg g Where g = acceleration produced by force Fg a = acceleration produced by another force F At or near the surface of the earth, k and g are numerically equal, so are m and Fg Problems:
[ 66 kgm ] [9.^ m s
kgm ∙ m kgf ∙ s 2 = 66 kgf
Fg = 50 lbf g = 32.174 ft/s^2 m= Fg k g
[^50 lbf ] [
lbm ∙ ft lbf ∙ s (^2) ]
ft s 2 = 50 lbm
[ 800 g¿ ¿ f ] [
gm ∙ cm gf ∙ s (^2) ]
cm s 2 =843.91 gm ¿ m 3 k
Fg 3 g
lbm ∙ ft s 2
ft s 2 =[ 0.49 lbm ] [
gm lbm ] =222.26 gm m 4 = Fg 4 k g
[ 3 lbf ] [
lbm ∙ ft lbf ∙ s (^2) ]
ft s 2 [
gm lbm ] =1435.49 gm m 5 k =( 0.10 slug) [
lbm slug ][
gm lbm ] =1459.41 gm Total mass = m 1 + m 2 + m 3 + m 4 + m 5 = 500 + 843.91 + 222.26 +1435.49 + 1459. = 4461.07 gm (b) Total mass = 4461.07 g m
gm lbm
lbm
Specific Volume, Density and Specific Weight The density ρ of any substance is its mass (not weight) per unit volume ρ= m v The specific volume v is the volume of a unit mass. v=
m
ρ The specific weight γ of any substance is the force of gravity on unit volume γ = Fg V Since the specific weight is to the local acceleration of gravity as the density is to the standard acceleration, γ /g = ρ/k, conversion is easily made; ρ= γk g ∨γ= ρg k At or near the surface of the earth, k and g are numerically equal, so are ρ and γ. Problems
g = 9.8066 m/s^2 ρ= 1000 kgm m 3 γ = ρg k
[
kgm m (^3) ][9.^ m s (^2) ]
kgm ∙ m kgf ∙ s 2
kgf m^3
[ 80 kgm ] [9.^ m s
kgm ∙ m kgf ∙ s 2 =78.93 kgf
mechanism is of course enclosed in a case, and a graduated dial, from which the pressure is read, and is placed under the index hand. Gage Pressure Problem A 30-m vertical column of fluid (density 1878 kg/m^3 ) is located where g = 9.65 mps^2. Find the pressure at the base of the column. Solution (p = 0, Pg = Po) (p = po – pg) (pg = 0, p = po) (p = po + pg) p = po + pg pg=γ hg = ρg hg k
g hg kv pg= Fg A
γV A
γA hg A
pg= gρhg k
[
m s (^2) ][^1878 kgm m (^3) ] 1 kgm ∙ m N ∙ s 2 ( 30 m) ¿ 543,
m 2 ∨543.68^ kPa^ (^ gage^ ) Atmospheric Pressure A barometer is used to measure atmospheric pressure. Where ho = the height of column of liquid supported by atmospheric pressure Po Problems
[
lb ft (^3) ]¿^ ¿ where ho = column of mercury in inches then, pg=0.491^ hg lb ¿. 2 and, p=0.491h^ lb ¿. 2 Problems
lbm slug =0.06853 slug ¿
m s 2 =[^1 m s (^2) ][3.^ ft m ]
ft s 2 F= ma k =( 0.06863 slug) [
ft s (^2) ]=0.2248^ lbf 1 newton = 0.2248 lbf 1 lbf = 4.4484 newtons 1 lb ¿
( 1 lb) [
lb ][
m ] ¿ 2 1 lb ¿
m 2 p=[54. lb ¿ (^2) ] [
m 2 lb ¿. 2 ] =375,780 Pa∨375.78 kPa
t ° F− 32 212 − 32
t ° C− 0 100 − 0 t ° F=
t ° C+ 32 t ° C=
t ° F− 32 Absolute temperature is the temperature measured from absolute zero. Absolute zero temperature is the temperature at which all molecular motion ceases. Absolute temperature will be denoted by T, this T ° R=t ° F+ 460 , degreed Rankine T K =t ° C+ 273 , Kelvin Degrees Fahrenheit (ºF) and degrees Centigrade (ºC) indicate temperature reading (t). Fahrenheit degrees (Fº) and Centigrade degrees (Cº) indicate temperature change or difference (∆t). 180 Fº = 100 Cº 1 F °=
It follows that, 1 F °= 1 R ° and 1 C °= 1 K °
Btu ( lb) ( F °)
[
cal Btu ] [ lb] (^) [ 454 g lb ] [ F ° ][
9 F ° ] 1 Btu ( lb) ( F °)
cal ( g ) ( C ° ) Conservation of Mass The law of conservation of mass states that mass is indestructible. The quantity of fluid passong through a given section is given by the formula ⩒ = Av ṁ =
v
Av v = Avρ Where ⩒ = volume flow rate A = cross sectional area of the stream v = average speed ṁ = mass flow rate Applying the law of conservation of mass, ṁ = A 1 v 1 v 1
A 2 v 2 v 2 A 1 v 1 ρ 1 = A 2 v 2 ρ 2 Problems
Conservation of Energy Conservation of Energy Gravitational Potential Energy (P) The gravitational potential energy of a body is its energy due to its position or elevation. Kinetic Energy (K) The energy or stored capacity for performing work possessed by a moving body, by virtue of its momentum is called kinetic energy. Internal Energy (U,u)
Internal energy is energy stored within a body or substance by virtue of the activity and configuration of its molecules and of the vibration of the atoms within the molecules. u = specific internal energy (unit mass) Δ u = u 2 – u 1 U = mu = total internal energy ( m mass) Δ U = U 2 -U 1 Work (W) Work is the product of the displacement of the body and the component of the force in the direction of the displacement. Work is energy in transition; that is is exists only when a force is “moving through a distance”. Work of a nonflow System The area under the curve of the process on the pV plane represents the work done during a nonflow reversible process. Work done by the system is positive ( outflow of energy) Work done on the system is negative (inflow of energy) Flow Work (Wf) Flow work or flow energy is work done in pushing a fluid across a boundary, usually into or out of a system.
