Heat Engines Thermal Efficiency Energy Flow Diagrams Thermodynamics Physics Problems

in this video we're going to focus on heat engines so what exactly is heat engine heat engine is device that converts heat energy into mechanical energy now heat naturally flows from high temperature to cold temperature and as it flows down some of that energy can be used to perform mechanical work so let's work on this problem we have heat engine that absorbs 4 500 joules of heat energy from hot reservoir and discards 2500 joules into the environment how much work is performed by this heat engine so let's draw an energy flow diagram so this device represents the heat engine basically this circle in the middle now the amount of heat energy that's being absorbed is 4 500 joules this is the hot reservoir and this is the cold reservoir so heat naturally flows from hot to cold and so this is the engine the circle in the middle so 4 500 joules of heat energy is absorbed from the hot reservoir and 2500 joules of heat energy is expelled into the environment how much work is performed by this heat engine well we need to know what the variables are in this problem qh is the heat energy absorbed by the engine per cycle so that's positive 4500 joules it's positive because the system the engine is absorbing heat energy qc is negative 2500 joules it's negative because the system releases heat energy into the environment and the mechanical work is the sum of these two values now if you're dealing with just positive numbers and you don't want to be bonded with negative sign you can also use this equation it's the absolute value difference of these two numbers it's qh minus qc if you're just dealing with positive numbers so that's going to be 4500 joules minus 2500 joules and so that's 2000 joules so in this section that's how much heat energy is being converted to mechanical work it's 2000 joules of heat energy this makes sense the engine absorbs total of 4 500 joules of heat energy and then it releases 2500 joules and 2 000 joules if you add 2500 and 2 000 that's 4 500 joules so if 4500 joules of energy goes into the system in order for the system to maintain its temperature it has to release 4 500 joules of energy so the amount of energy that comes into the engine must equal the total amount of energy that leaves the engine so that it can maintain its original state now let's move on to part calculate the thermal efficiency of this engine the thermal efficiency is the mechanical work divided by qh multiplied by 100 so in this example the mechanical work is 2 000 joules the amount of heat energy that was absorbed by the hot reservoir per cycle is 4500 joules and then we need to multiply this by 100 and so the efficiency in this example is 44.4 percent and so that's the answer to this problem now let's move on to number two steam engine performs six thousand joules of work while discarding 14 000 joules into cold reservoir per cycle how much heat energy is absorbed by the hot reservoir per cycle so let's draw picture so here we have our energy flow diagram and so this is going to be the hot reservoir just like before and this is going to be the cold reservoir so 6 000 joules of mechanical work is being performed so that's so we have that and the steam engine discards 14 000 joules into the cold reservoir so that's the absolute value of qc qc is negative 14 000. how much heat energy is absorbed by the hot reservoir per cycle so we're looking for qh so we know that is the absolute value of qh minus qc so qh is going to be plus the absolute value of qc and qh is going to be positive so in this example is 6 000 joules and qc is 14 000 joules so qh is going to be 20 000 joules and so 20 000 joules of thermal energy is taken from the hot reservoir per cycle part calculate the thermal efficiency of the steam engine the thermal efficiency just like before it's going to be the work divided by qh times 100 so the work output is 6 000 joules qh is 20 000. let's multiply that by hundred so the efficiency in this example is thirty percent so that's the answer for part now let's move on to part what is the power rating of this engine in watts and also in in horsepower if it takes point 25 seconds to complete cycle so how can we calculate the power power is work divided by time and so we have the work it's 6 000 joules per cycle however we have the time it takes to complete cycle watts mean work is in joules and time is in seconds so one joule per second is equal to one watt which is the unit for power so keep that in mind now what i'm going to do is start with this value the work output is 6 000 joules per cycle so in two cycles the machine or the engine will do twelve thousand joules of work in three cycles eighteen thousand in four cycles twenty four thousand so every cycle that the machine goes through six thousand joules of mechanical work will be performed now it takes point 25 seconds for one cycle to be completed so if we set up this way notice that the unit cycles will cancel and this is going to give us the unit joules per second which is equivalent to watt so it's 6 000 joules divided by 0.25 seconds so the power rating is 24 000 watts so that's the first answer to part now we also want the power rating in horsepower and you need to know that one horsepower is equivalent to 746 watts so 24 000 divided by 746 and so this is equal to 32.2 horsepower so that's the power rating of this particular engine or steam engine now how much mechanical work is performed by this engine in one hour so how can we get that answer what we need to do is start with the power rating but first let me get rid of some stuff we said that the power rating is 24 000 watts which is equivalent to 24 000 joules per second but actually i'm going to start with hours so the time is one hour now we know that power is work divided by time so if you rearrange the equation work is power multiplied by time so i'm going to convert hours into seconds one hour is 60 minutes and in each minute there are 60 seconds so now have the time in seconds i'm going to multiply by the power rating which is 24 000 watts or 24 000 joules per second so the unit hours will cancel the unit minutes will cancel and also the unit seconds will cancel as well so this is going to give me the mechanical work performed by this steam engine in one hour so it's 60 times 60 times 24 000. so that's going to be very big number so it's 86 million 400 000 joules or you can write it as 8.64 times 10 to the 7 joules so that's how much mechanical work can be performed by this steam engine in just one hour number three jet engine has thermal efficiency of if it performs 1500 joules of mechanical work per cycle how much heat energy is absorbed from the hot reservoir per cycle so we need to calculate qh we have the mechanical work and the efficiency so we need to use this formula efficiency is equal to mechanical work divided by qh multiplied by hundred percent so the efficiency is 18 the mechanical work is 1500 joules so the first thing we should do is divide both sides by one hundred percent so eighteen percent divided by hundred percent is point eighteen and that's equal to fifteen hundred divided by qh so at this point we can cross multiply so this is going to be 1 times 1500 which is 1500 and then that's going to be 0.18 times qh so to calculate qh we need to divide both sides by point 18. so qh in this example is 1500 divided by 0.18 which works out to be thirty 8333 point three joules i'm just going to round into nearest whole number and so this is the answer for part now let's move on to the next part how much heat energy is discarded to the environment per cycle that was not supposed to happen so we need to calculate qc so we know that is the difference between qh and qc the mechanical work is 1500 qh is 8 33 and so to calculate qc what we need to do is basically subtract both sides by 83 33. so 1500 minus 83 33 that's negative 6833 so we can see that the absolute value of qc is 6833 joules so that's how much heat energy is discarded to the environment now if you want the actual value of qc keep in mind that it's really negative because the heat engine or the jet engine releases that heat that's an exothermic process and anytime the system releases heat energy that process is negative for but if you're not worried about any negative signs you could say that 6833 joules of heat energy was discarded now let's move on to part how much mechanical work does it perform over 200 cycles now keep in mind that it's 1500 joules of mechanical work per cycle so all we need to do is take this value and multiply by 200 cycles so 1500 times 200 that is equal to 300 000 joules so that's how much mechanical work the jet engine performs over the course of 200 cycles and so that's it for this problem number four gasoline engine is burning 1.5 kilograms of fuel per hour the energy of combustion of this fuel is 450 000 joules per gram if the engine goes through 100 cycles per minute how much thermal energy is absorbed per cycle now our goal is to calculate the energy per cycle so we need to perform unit conversion process and our goal is to get joules on top and the number of cycles on the bottom so i'm going to start with the rate at which the fuel is burned so 1.5 kilograms of fuel is burned per hour so we need to ride it like this now the combustion is joules per gram so that indicates that need to convert kilograms to grams so one kilogram is equal to thousand grams and so we could cancel the unit kilograms so now can use this value to convert to joules so the energy of combustion is 450 000 joules per gram now still have unit of time that have to get rid of and this value connects cycles with time so have 100 cycles per minute which means need to convert hours into minutes so one hour is equivalent to 60 minutes so the last thing need to do is multiply by the number of cycles per minute so in one minute there are 100 cycles so we can cancel these units so now have the energy in joules per cycle so it's going to be 1.5 times thousand times 450 000 divided by 60 and then divide that by so this comes out to be 112 500 joules per cycle now because the thermal energy is absorbed by the engine this has to be qh and so that's it for part now let's move on to part if 30 000 joules of mechanical work is performed per cycle how much heat energy is discarded per cycle so we know that is equal to qh minus qc so our goal is to calculate qc so let's rearrange the equation i'm going to take this term and move it to this side where it's going to become positive qc and i'm going to take move it to that side where it's going to be negative so this is how we can calculate qc so qh is hundred twelve thousand five hundred joules per cycle and the mechanical work is thirty thousand joules per cycle so if we subtract these two numbers this will give us 82 500 joules and need the extra zero that's supposed to be 30 000 as opposed to 3 so this is the answer for part now let's move on to part what is the power radian of this engine in kilowatts and horsepower so how can we get that answer we know that power is work divided by time now we have the mechanical work it's 30 000 joules per cycle and we have the time it takes to complete one cycle which is currently 100 cycles per minute so let's multiply by that now we need it in second stone because one watt is one joule per second so let's convert minutes to seconds one minute is equal to 60 seconds so it's going to be 30 000 times 100 divided by 60. and so the power is 50 000 watts now to convert that to kilowatts we need to divide it by thousand one kilowatt is equal to thousand watts and so 50 000 divided by 1000 will give us power rating of 50 kilowatts so that's the first answer to part now let's move on to the second part of part so we need to convert watts to horsepower so let's start with 50 000 watts and one horsepower is 746 watts so let's take 50 000 and divide it by 746 and so this works out to be 67 horsepower and so that's it for part now the last thing we need to do is calculate the thermal efficiency of this engine so the efficiency is going to be divided by multiplied by 100 so the work is 30 000 joules we have qh which is 112 500 and then let's multiply the ratio by hundred percent so the efficiency is 26.7 percent and so that's it for this problem you