This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: 824W % 7,0734 Template Unit_02
Energy- capability of a system to do work (symbol: E or W) The result of the work may have a great variety of forms:
1. Displacement of mass in gravitational ﬁeld,
AW = mg(Ah), m - mass in kilograms [kg], g = 9.81[m/sz], h - displacement in meters [in]. 2. Overcoming friction at a body movement,
AW= F, Ax, F, - force in newtons [N] along x, x - displacement in meters [m]. 3. Change of a body velocity,
AW = vav , m - mass in kilograms [kg], v — velocity in [m/s]. 4. Change of rotation speed,
AW = In) An) , I - moment of inertia [kg m2], (2) - angular speed [rd/s]. 5. Change of capacitor voltage,
AW = Cu Au, C - capacitance in farads [F], u - voltage in volts [V] . 6. Change of inductor current,
AW = Li Az', L- inductance in henrs [H], i - current in arnpers [A] . 7. Change of a body temperature,
AW= c m AT, c — speciﬁc heat [kcal/kg/OC], m — mass in [kg], AT — temperature in [0C]. Water c = 1 [kcal/kg/0C]
Ice c = O.49[kcal/kg/°C]
Aluminum c = 0.215[kcal/kg/°C]
Copper c = 0.092[kcal/kg/°C]
8. Change of body structure or micro-structure (phase), (melting, vaporization),
AW= LF m, LF — latent heat of fusion [kcal/kg], for water: 79.7 [kcaJ/kg], m - mass [kg].
AW= Lv m, Lv — latent heat of vaporization [kcal/kg], for water: 540 [kcal/kg]. 9. Radiation of energy
AW= e o AT“(At), e — emmisitivity of the area: black: e = 1, white: e = 0,
o — Stefan’s Constant or= 5.67 10'8 [W/m2 OK], T — temp. [0K], t = - time [s]
A — Radiation area [ml] 10. Change of a gas volume, pressure or temperature
dW=ﬂdP,dV,dT) P — pressure, V— volume, T — temperature : =>> Thermodmamics
PV Since, according to Boyle-Gay-Lussac Law: 7.— = Const., the effect of energy
change depends on which of three parameters, P, Vand T is kept constant.
11. Change of chemical bindings. =>> Electrochemisz
W = ac U ac — acumulation capacity [Ah], U — voltage. 12. Change of nucleus structure of sub-atomic particles that affects the particle mass, =>> Nuclear Phy_sics
E= m C], m — maSS in [kg], c — speed oflight: c = 3x108 [m/s]. imam
One joule [J] is the work of force of l newton [N] at the distance of 1 meter [In]. One newton [N] is a force that
accelerates an object of the mass of 1 kg by a = 1 m/sz. When an object of mass m is elevated by h in the gravitational
Earth ﬁeld, with acceleration equal to a = g = 9.81m/sz, then its energy increases by W], = mgh.
Illustration 1: A person of the weight m = 150 lb = 150 x 0.4536 kg/lb = 68 kg, that go upstairs from the ﬁrst to the
second ﬂoor, by h = 10 ft = 0.3048 m/ft = 3.048 m, increases his/her potential energy by Wp=mgh = 68 kgx 9.81 m/szx 3.048 m=2034 J= 2.034kJ 1 kilowatthour IkWhI One kilowatthour [kWh] is energy transmitted with the power of 1000 W = 1 [kW] over one hour time. Since 1 [J] is
energy transmitted with power of 1 [W] in 1 second, i.e., 1 [J] = 1 [W5], hence 1 [kWh] = 3600 [kWs] = 3600 [kJ] = 3.6 [MJ]
It means that joule [J] is a very small unit of energy as compared to kilowatthour [kWh].
Illustration 2: A bulb of 100 W switched on by 24 hours takes energy:
W= P t = 0.1 [kW] x 24 [hour] = 2.4 [kWh] = 2.4 [kWh] x 3.6 MJ/kWh = 8.64 [MJ] Illustration 3: Assuming that average power of a residential home P = 1500 W, energy delivered to such a home per one
month is W=Pt= 1.5 [kW] x24 [hour] x30 [days] = = 1080 [kWh] = 1080 [kWh] x3.6 MJ/kWh = 3.88 [G1]. 1 kilocalorie kcal
Energy needed to increase temperature of 1 kg of water (1 liter) by ldegree in Celsius scale.
1 [kcal] = ke [kJ]
[CC = 4.19 [kJ/kcal] — Joule constant One kilocalorie [kcal], similarly like joule [J] or kilojoule [U], is a very small unit of energy. Nutrition value of
industrially processed food products are speciﬁed, erroneously, in calories [cal]. Indeed, the numerical value provided on
a processed food speciﬁes not calories, but kilocalories [kcal]. For example, a cup of yogurt provides energy of 170 [cal].
As a matter of fact, it provides energy of 170 [kcal]. Illustration 4: A human daily need of nutrition calories is of the order of W= 2500 [kcal] = 4.19 [kJ/kcal] x 2500 [kcal] = 10.5 [MJ].
Assuming that a person has mass m = 70 [kg], this energy is sufﬁcient for climbing to the height such that
mgh = W= 10.5 106 [J] _1_m_ a =
_mg_ 70x9_51_15.05><10[m] 15.3 [km]. Illustration 5: A dead person cools from body temperature of 38°C to ambient temperature of 18°C in 24 hours. The
body looses heat energy independently; a person is alive or dead. To balance the loss of heat energy, nutrition has to provide energy
W = c m AT, where c is speciﬁc heat of human body. Since it contains about 85% of water, we could assume that c ~ 1 [kcal/kg/OC].
Thus, for a person of m = 80 kg, ' W = c m AT = l [l<cal/kg/°C] 80 [kg] 20 [0C] = 1600 [koal] 1 British thermal unit [Btu]
Energy needed for heating 1 pound of water by l Farenhait degree. 1 [Btu] = 1 [lb F]. Since 10 F = 0.55560 C, thus heating of 1 [kg] by 1° [F] of water requires 0.5556 [kcal]. Since 1 [lb] = 0.4536 [kg], hence
1 [Btu] = 1 [1b F] = 1 [lb F] 0.4536 [kg/lb] 0.5556 [C/F] = 0.252 [kg 0C] = 0.252 [kcal].
1 [Btu] = 0.252 [kcal] = 0.252 [kcal] 4.19 [kJ/kcal] = 1.06 [1d]
1 [kBtu] = 1.06 [MJ] 1[kBtu] =1.06 [MJ]=1.06[MJ]><%%(% 1 [MBtu] = 0.293 [MWh] = 0.293 [kWh] 1 |ton TNT|=1.167 [MWh] .
Illustration 6: One day of operation of 100 MW generator produces electric energy W= 100 [MW] 24 [hour] = 2400 [MWh] = 2056 ton TNT = 2.056 kiloton TNT
1 [Quad] = 1015 [Btu] ~ 45 million ton of coal: ~ 170 million barrels (1 barrel = 138.87 liters) of oil ~ One day of world energy use
Illustration 7: Calculate the coal equivalent to one kg of mass converted to energy E= mc2 =1[kg]><(3><108[m/s])2 = 9x1016[J] = 9x 101°[MJ] = 9 x 101°ﬁu<3m1 = 8.5 x1013[Btu] = 0.085 [Quad] 1 [kg] ~ 0.085 [Quad] ~ 45 x 0.085 = 3.8 [millon ton] coal ~ 170 x 0.085 = 14.4 [million barrel] oil ...
View Full Document
This note was uploaded on 02/06/2012 for the course EE 3410 taught by Professor Staff during the Fall '08 term at LSU.
- Fall '08