I ve got the power! Power and Percent Efficiency Wednesday April 16, 2014.
Power? WATT s that? Power is defined as energy per unit of time Electrical power describes the amount of electrical energy that is converted into heat, light, sound or motion every second Power = Energy/time or P = E/t! The unit for energy = joules (J) The unit for time = seconds (s) A joule/second (Power) = watt (W)
Another way to calculate power: Power = Current x Potential Difference P = I x V All these equations are really a PET PIV of mine! P = E/t P = I x V
Power Problem #1: A battery uses 810 J of energy to run a portable radio for 30 minutes. What is the power of the radio? (Remember: a watt = joules/second) In order to solve any mathematical problem you can use this acronym to help you with the steps: G: Givens U: Unknowns E: Equation S: Substitute S: Solve
Power Problem #1: Solution G: Givens We know: energy = 810 J, time = 30 minutes = 1800 sec. U: Unknowns We want to know: power (in watts) E: Equation The equation we want to use is: Power = Energy/time! or! P = E/t! S: Substitute Power = Energy/time! or! P = E/t! Power = 810 J/1800 sec. S: Solve = 0.45 W (watts) Therefore the radio produces 0.45 watts of power.
Power Problem #2: A current of 13.6A passes through an electric baseboard heater when it is connected to a 110 V wall outlet. What is the power of the heater? G: Givens We know: current (I) = 13.6A, voltage (V) = 110 V U: Unknowns We want to know: power (in watts) E: Equation Which equation do we use THIS time?
Whoa, whoa, whoa...watt?! S: Substitute Power = current x voltage = I x V Power = 13.6 A x 110 V S: Solve = 1496 W Therefore the baseboard heater produces 1496 watts of power.
Watt is the difference between a 60W and 100W lightbulb? P = I x V 60W = I x 120V I = 0.5A Less current = less light P = I x V 100W = I x 120V I = 0.83A More current = more light
Percent Efficiency Definition: Electrical devices are not perfect so the total amount of energy that they receive is not the total amount of energy they provide. It is the ratio of useful power output to power input. For example: the purpose of a light bulb is to produce light energy HOWEVER some of the electrical energy that the light bulb receives is converted into heat energy output (not ALL converted to light) input
% Efficiency Equation % efficiency = Useful energy output x 100%!!!! Total electrical energy input Example #1: An oven requires an input of 130000J of energy to run in order to make cookies, which need 125000J to bake. Calculate the percent efficiency of this oven.
Solution G, Givens: input of 130000J, output of 125000J. U, Unknowns: percent efficiency E, Equation: % efficiency = Useful energy output x 100%!! Total electrical energy input S, Substitute: % efficiency = 125000J x 100%!!!! 130000J S, Solve: = 96% efficient
Problem #2 An electric kettle has a power rating of 1000 W. It takes the kettle 4.00 min to heat 600 ml of water from 22.0 C to 100 C. If it takes 196000J of energy to heat the water, what is the efficiency of the kettle? STEP 1: Solve for useful energy output (i.e. the energy needed to do the job). Already done! It s given to you in the question. = 196000J
Take it one step at a time! STEP 2: Solve for electrical energy input (i.e. the energy the kettle is receiving). Givens: We know that the kettle needs 1000 W and heated the water in 4 min. Unknowns: We want to know how much energy was provided to the kettle to do this.
Take it one step at a time! Equation: Which equation do we use? P = E/t rearranged... E = t x P Substitute: convert 4 min to seconds! E = t x P E = 240 sec. x 1000 W = 240000 J This is the energy input. Now what? You re not done!
And solve. STEP 3: Solve for percent efficiency. % efficiency = Useful energy output x 100%!!!! Total electrical energy input % efficiency = 196000 J x 100%!!!! 240000J = 82% Therefore the kettle is 82% efficient.
Energy at Home Since we pay for the power, it is important to know how much is being used and how much is being wasted. Energy can be wasted as, heat, light or sound. It is now law that the energy efficiency of an appliance must be listed on any new appliance that is sold. There is a meter outside of your house that measures energy in kilowatts. The official reading is in kilowatt hours. This stands for 1000 watts being used up every hour (1 kwh) and you are billed based on a dollar amount per kilowatt hour (for example, $0.02 for every kilowatt hour used).
Please complete the practice problems! ALL THE POWER TO YOU!