practice problem 1
The electric cycle, 1,825 volts at approximately 7.5 amps for 30 seconds, then 240 volts at approximately 1.5 amps for 60 seconds… a 5 second pause intervenes, and the cycle is repeated, was designed to render the condemned brain dead within the first few moments. The function of the remainder of the cycle is to stop the body's organs so that a physician can certify that death has occurred…
- the total electric energy delivered to Mr. Stewart on the day of his execution.
- his change in temperature assuming assuming a standard 70 kg mass and a body made entirely of water.
Fairly straightforward. Energy is power times time. Electric power is voltage times current. Energy is a scalar, so just add up the parts of the cycle and double each to get the total.
E = ΣPt = ΣVIt E = 2[(1825 V)(7.5 A)(30 s) + (240 V)(1.5 A)(60 s)] E = 864,000 J
The electric energy applied to Stewart's body was transformed into heat. Use the sensible heat equation and assume that the human body is basically a bag of water.
E = Q = mcΔT ΔT = Q mc ΔT = (864,000 J) (70 kg)(4,190 J/kg℃) ∆T ≈ 3 ℃
practice problem 2
Now answer these questions…
Most of the energy used to fuel the CCAT house comes from the sun, but a battery bank is also charged with 24 V permanent magnet DC generators run by exercise bikes in the basement. The flywheel on each bike is furnished with a groove. Through this groove is a belt that is attached to the generator's central shaft. The average person pedaling 70 rpm spins the shaft creating an electrical current of 4 A…. This allows someone to get a comfortable workout while at the same time charging the batteries.
- Determine the power output of person pedaling a stationary bicycle at the CCAT house.
- How much work does the person do in one minute?
- Over what total distance have the pedals moved in one minute? (Assume a crank arm length of 18 cm.)
- How fast are the person's feet moving while generating power at the CCAT house?
- With what average force does a typical person in the CCAT house push down on the pedals?
Pick the appropriate electric power equation.
P = VI P = (24 V)(4 A) P = 96 W
Work is power times time.
W = Pt W = (96 W)(60 s) W = 5,760 J
This is a geometry problem. In one minute, the pedaler's foot would crank through 70 circumferences.
Δs = 70 × 2πr Δs = 140π(0.18 m) Δs = 79 m
This is a basic kinematics problem.
v = Δs Δt v = 79 m 60 s v = 1.3 m/s
Use the force–velocity form of the equation for power (P̅ = F̅v cos θ). Solve it for force. (We need to assume that the force applied by the foot to the crank arm is always parallel to the pedal's velocity.) Insert numbers. Compute solution.
F = P v F = 96 W 1.3 m/s F = 73 N
practice problem 3
- charge in…
- milliamp hours [mAh]
- coulombs [C]
- potential energy in…
- milliwatt hours [mWh]
- joues [J]
Compile your findings in a table like this one.
There is no correct sequence for completing this table. Here's one I just thought of. Start with the definition of potential difference or voltage…
Solve it two ways.
|W = qV||⇐||
Compute the charge of the NiZn battery first.
|qNiZn =||2,500 mWh|
|qNiZn = 1,563 mAh|
Compute the energy of the NiMH battery (or the work it can do) using similar logic.
|WNiMH = qV|
|WNiMH = (2,500 mAh)(1.2 V)|
|WNiMH = 3,000 mWh|
Convert the charges to coulombs using the definition of current.
|q = I∆t||⇐||
Shift the decimal 3 places when converting from milliamps to amps. Replace an hour with 3,600 s.
|qNiZn = I∆t|
|qNiZn = (1.563 A)(3,600 s)|
|qNiZn = 5,625 C|
|qNiMH = I∆t|
|qNiMH = (2.500 A)(3,600 s)|
|qNiMH = 9,000 C|
Convert the energies to joules using the defintion of power.
|W = P∆t||⇐||
Shift the decimal 3 places when converting from milliwatts to watts. Replace an hour with 3,600 s.
|WNiZn = P∆t|
|WNiZn = (2.500 W)(3,600 s)|
|WNiZn = 9,000 J|
|WNiMH = P∆t|
|WNiMH = (3.000 W)(3,600 s)|
|WNiMH = 10,800 J|
Compile the results in a table.
The nickel metal hydride battery can hold more energy than the nickel zinc battery, which translates to longer periods between rechargings. That makes the NiMH the "better" battery. Higher voltage might be useful for some applications, but I don't know of any. AA, AAA, C, and D cells all should be around 1.5 V. Electrical engineers who design battery powered devices count on this.
practice problem 4