A multicylinder gasoline engine in an airplane, operating at 2.50 ×10^3 rev/min, takes in energy

step 1 Okay, so it's going to be a long question, but we can do this. We have this, you know, airplane

A multicylinder gasoline engine in an airplane, operating at...

A multicylinder gasoline engine in an airplane, operating at $2.50 \times 10^{3}$ rev/min, takes in energy $7.89 \times 10^{3} \mathrm{J}$ and exhausts $4.58 \times 10^{3} \mathrm{J}$ for each revolution of the crankshaft. (a) How many liters of fuel does it consume in 1.00 $\mathrm{h}$ of operation if the heat of combustion of the fuel is equal to $4.03 \times 10^{7} \mathrm{J} / \mathrm{L}^{2}$ (b) What is the mechanical power output of the engine? Ignore friction and express the answer in horsepower. (c) What is the torque exerted by the crank-shaft on the load? (d) What power must the exhaust and cooling system transfer out of the engine?

Key Concepts

Horsepower Conversion

Horsepower is a unit of power commonly used to quantify the output of engines and motors. Converting mechanical power from SI units (like watts) to horsepower is important for comparing engine performance using different measurement systems.

Torque

Torque is a measure of the rotational force applied to an object, causing it to rotate about an axis. In engine systems, torque is directly related to the work done per cycle and is calculated by dividing the work done per revolution by the angular displacement in radians.

Angular Speed Conversion

Angular speed conversion is the process of translating rotational speeds (e.g., revolutions per minute) into angular units (e.g., radians per second). This conversion is necessary for accurately linking rotational motion with mechanical quantities such as torque and work.

Mechanical Power Output

Mechanical power output is the rate at which work is performed by the engine's moving parts. It represents the useful power available to drive loads and is derived from the net work produced per cycle multiplied by the number of cycles per unit time.

Energy Balance in Engine Cycles

Engine energy balance involves accounting for all energy flows within the system. In a heat engine, the total energy input from fuel combustion is partitioned into work output, energy lost as heat in the exhaust, and other inefficiencies. This concept is critical for evaluating engine performance and efficiency.

Fuel Consumption Calculation

Fuel consumption refers to the determination of the volume or mass of fuel required to deliver a predetermined energy output. This involves relating the energy provided per unit of fuel (via the heat of combustion) to the total energy input needed by the engine over a period of operation.

Heat of Combustion

This is the amount of energy released when a given volume or mass of fuel is burned completely. It serves as a measure of the fuel’s energy content and is used to calculate how much fuel is needed to supply a specified amount of energy to an engine.

Transcript

00:01 Okay, so it's going to be a long question, but we can do this.

00:03 We have this, you know, airplane engine, and we have the amount of energy that gets pulled in from the gasoline combustion per revolution, and we also have the amount of exhaust per revolution, and then we have the specific heat of combustion of the fuel, and we also have the fact that we're going to have 2 .5 times sensitive three revolutions per minute, okay? 2 ,500 revolutions per minute.

00:28 So, first we got to do is figure out how.

00:30 How much fuel this thing is going to use.

00:33 So let's calculate the amount of energy that it's going to use per hour.

00:40 Okay? in order to do per hour, first we need to figure out the amount per minute.

00:45 And in order to do that, we need to multiply.

00:47 So we'll call q per minute.

00:50 We need to multiply the amount of key in that's being generated by the engine times the frequency or the, you know, rotational speed of the, the engine so our rpm and that's just going to be equal to 7 .89 times 10 to the 3 joules per revolution times 2 .5 times 10 to the 3 revolutions per minute and that'll give us a energy per minute of 1 .97 times 10 to the 7 joules per minute.

01:26 Now we need to turn this energy per hour so that's just going to be q min times 60 minutes per hour and that's going to give us i'm not going to write it down you just won't this number by 60 it'll give us 1 .18 times 10 to the 9 joules per hour okay and now in order to find the volume we'll call it v fuel all we need to do is take our q per hour and divided by 4 .0 or our h fuel right our combustion the heat of combustion of the fuel and so all we got to do is we have 1 .18 times 10 to the 9 joules in the one hour that we're going to think about this.

02:16 Divided by 4 .03 times 10 to the 7 joules per liter.

02:21 It'll give us a volume of 29 .4 liters.

02:25 So we need 29 .4 liters every hour to run this engine.

02:31 Cool.

02:32 Next we have to figure out the power output of the engine.

02:36 So that's done pretty simply.

02:38 Now, power is going to be equal to work divided by time.

02:45 And so, oops.

02:49 And so let's figure out the work...