A four-cylinder, four-stroke internal combustion engine has a bore of 66 mm and a stroke of 54

A four-cylinder, four-stroke internal combustion engine has...

Key Concepts

Engine Power Calculation

Engine power calculation involves converting the net work per cycle into a power output by considering the engine speed (RPM) and the number of working cycles per unit time. In multi-cylinder engines, the total engine power is computed by summing the work contributions of individual cylinders, taking into account the mechanical configuration (e.g., four strokes per cycle) and firing order.

Otto Cycle

The Otto cycle is a thermodynamic cycle that models the operation of a spark-ignition internal combustion engine. It consists of two adiabatic (isentropic) processes interspersed with two constant-volume processes, representing the idealized compression and expansion strokes along with the heat addition and rejection. This cycle provides a basis for analyzing engine efficiency and net work output under the air?standard assumptions.

Air?Standard Assumption

The air?standard assumption simplifies the analysis by treating the working fluid as air with constant specific heats, and by assuming that the combustion process is represented by heat addition at constant volume. This idealization allows the use of ideal gas relationships to compute state properties and cycle performance, even though real combustion processes involve complex chemical reactions.

Engine Geometry

Engine geometry refers to the physical dimensions of the engine components, such as bore (diameter of the cylinder), stroke (distance traveled by the piston), and clearance volume (the minimum volume in the cylinder when the piston is at the top dead center). These parameters directly influence the displacement volume and compression ratio, which are critical in determining the thermodynamic efficiency and performance of an engine.

Compression Ratio

The compression ratio is the ratio of the cylinder volume at bottom dead center (when the piston is at its maximum displacement) to the clearance volume at top dead center. It plays a central role in the thermodynamic efficiency of the Otto cycle; higher compression ratios generally lead to higher thermal efficiencies, although they are limited by engine knocking and material constraints.

Net Work per Cycle

Net work per cycle is the difference between the work output during expansion and the work input during compression in the engine cycle. This concept is fundamental in thermodynamics, as it represents the useful mechanical work extracted per cycle after accounting for energy losses, and it provides a measure for engine efficiency and mechanical performance.

Transcript

00:01 We're looking at power plants for this particular air standard auto cycle, costro, four cylinder engine.

00:09 We want to determine the network, the work net in kilojou.

00:21 We also want to determine the power in kilowatt, right? so we just applied the standard auto cycle problem equations for the analysis.

00:35 All right so we have tea temperature in kelvin have a specific volume so this is the ts diagram kilo joe kilogram kelvin so we have this process one to two two two two two two three constant volume heat addition three to four eccentric expansion and four to one constant volume hit rejection this is the pressure volume diagram project and this is the volume to put it in cubic meter all right so this gives us this so let's so let's reduce be one two three four okay so we have we can just go ahead now the lot of information is actually given and we can just look for the square volume that's called the square volume b so to s is equal to pi d square about four times the length okay and then stroke so if we plug in we solve this now, we know pi, we know d.

02:48 What is this? d is 0 .066 meter, okay? which is, we are giving us this m and value to meter and l is the stroke.

03:07 So you are going to have this to be equal to 0 .001, 8, 4 ,000.

03:30 Seven four four four four four all right so we know that uh hundred percent is uh it's seven hundred percent uh is actually a hundred minus thirteen percent because we are given that the clearance volume is thirty percent of the cylinder two volume uh base of the t here okay so to do this that we are going to have this one to be four to which is seven percent okay that's let's remove it 18 percent we're not we're not the clearance volume okay so that means what we have here is 87 percent so this is 87 all right now that is in cubic meter so let me put it cubic meter all right it's in cubic meter let me fix it well cubic meter all right then how do you get it what would be 13 % now so 13 % if you solve this out 13 % i'm gonna give you 0 .0 0 0 027 -67 5 65 5 cubic meter so this 13 % but the clarang volume all right so it does do your use of mathematics or soft dance all right so the total volume now is everything.

05:41 The total volume now, the total volume now is swept volume.

05:52 Okay? so this 13 % is clearance volume.

05:56 So we can just clear all the v or c.

05:59 Okay.

06:00 This is a swap volume.

06:02 So then total volume now is equal to sweep volume plus clearance volume.

06:12 If we give us to be equal to a v1, which of course is equal to a.

06:19 Everything that is equal to 0 .000000 cubic meter all right now we can at t1 at t1 at t1 we are given equal to 300 300 kelvin so if you if you look for that at using your table for i'm going to give 3 to 114 .0 .07 kilo joe, that kilo gram.

07:19 Alright, so we can just get the relative volume there.

07:23 From the table there, it's going to give you 621 .2.

07:28 And of course, if at centrifugal compression, process 1 to 2, this are point 2, the relative volume there is going to give you the r1 over what you have at sorry over the compression ratio all right so if you just plug in value there of course we need to know the compression ratio compression ratio compression ratio is v1 over v2 which is r okay so if you divide the volume at point one okay the volume at point two okay and what's the volume at point two uh this is the volume at point two this is the volume at point two this is the volume at point two this one i see same thing at volume at point uh so if you divide them you'll get r if you give you equal to 7 .69 or approximately 8 so if we work with 8 you are going to have this to be equal to 77 .65 all right so if we interpolate you are going to have the temperature at point 2 to be equal to 673 .1 kelvin and we will have the internal energy at 2 to be equal to 673 to 4991 .2 .3 kilojou 1 .5.

09:48 All right so you have that now...

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