2.) Joule’s law: P= izr, where P is the rate at whichPower, Energy, Work, And Efficiency
Why This Family Matters
Section titled “Why This Family Matters”Routes equations where electrical quantities become work, heat, output, loss, or system performance.
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Candidate Formula Cards
Section titled “Candidate Formula Cards”3.) The power equation : P0 = ei, where P0 is the2. Joule’s law: P = ^^r, where P is the power, or the rate at2.) Joule’s law : P= i^r, where P is the rate at whichThe e.m.f. consumed by resistance r is EI = 7r, thus theThe kinetic energy of 1 kg. weight of matter, Eq = mc^,Thus the alternating current i = IQ since 0 consumes in a resist-E = V2 *fn$,E = 7 \/2the current / = omn = 44.2 amperes effective.r = - - -rT^ - ” = 0.71 ohms per circuit.(Id = 0.82 cm.), 45 cm. apart, transmitting the output of thisE = A/2 */ $T = e3 = ^Te1 = 0.2 E + 2 Xe2 = 2 ei - 0-3 irE,5. In Joule’s law, P = i^r, r is not the true ohmic resistance,5. In Joule’s law, P = /^r, r is not the true ohmicmechanical energy t^o =Fl by (1), and therefrom the mechanical3. The power equation: Po = ei, where Po is the power3.) The power equation : P^ = ei, where P^ is theThis energy, for y = 0, or the mass at rest, becomes:power consumed by resistance r is P = 72r.Since E0 = 2 irfn$, it follows thator, since 0 = 2 irft = 210 t we have,ei = 0.8 e;ez = | ei = 0.4 ire,5. In Joule’s law, P = i2r, r is not the true ohmicgive an annual loss of 33 = $7, or 7 perenergy of mass,” is £“00 = mc”-.gives 4 X ; = jg or altogether ^ + ^ = jg of theor glof 4 = 24 » ^^’^ so gives a total copper economy ofThe value / = -7= is called the effective value of the alter-effective gives 44.2\/2 = 62.5 amp. maximum.3, as 2 = 0.094 X 106/ for each wire.and since 1 joule = 10^ absolute units, the mechanical work is16 x .79 = 12.6 c. p., and at an efficiency of 3.1 watts per= 4.44 fn& ;= 4.44 X 0.333 X 360 X 6.4= 3400 volts per circuit.= 5900 volts effective.X 0.71 = 1400 watts per circuit, or a total of 3 X 1400 = 4200$ = 22.3 X 62.5 X 0.094 X 106 = 131 megalines.= 4.44 X 0.333 X 131= 1.006;= 0.6 E.power of 5 X 10* K. W. Estimating the energy of the discharge,thus must be such that e volts times 1 amp.-sec, or e watt-sec. orenergy produced, + increase of the stored magnetic energy. (1)FIG. 8. - E.m.f. of a single conductor, direct-current machine1.8 X 10~6, what is the e.m.f. consumed in the machine by theoutput, or the efficiency of the electromagnet, thus is, by (22)Fife. 13. Single-Phase Llfehtlnfe and Three-Phnse Power.life of 500 hours; since obviously any efficiency can be pro-32.4 gram-cm. pull per volt-ampere suppUed to its terminals.cost, to $33, as the loss thereby incurred, of7 per cent., is less than the loss in standing idle.LOAD FACTOR AND COST OF POWER 53For instance, Fig. 14 shows an approximate load curveLOAD FACTOR AND COST OF POWER 55shape shown in Fig. 16: fairly constant from the openingLOAD FACTOR AND COST OF POWER 57LOAD FACTOR AND COST OF POWER 59Average energy of the discharge: 10,000 K. W. sec., orsq. cm. and a mean length of 240 cm. per turn. At a resistivityresistance, and what the power consumed at 450 kw. output?The resistance of 360 turns of 240 cm. length, 0.22 sq. cm.section and 1.8 X 10~6 resistivity, is360 X 240 X 1.8 X 10~6apart and 0.82 cm. diameter was found in paragraph 1, exampleC, as shown in Fig. 6, to B, E, constant voltage E exists betweenrising on a parabolic curve from 0 to E2 during 40 per cent, of