FIGURE CP19.72 shows the Diesel cycle. It is similar to the...

FIGURE CP19.72 shows the Diesel cycle. It is similar to the Otto cycle (see Problem CP19.71), but there are two important differences. First, the fuel is not admitted until the air is fully compressed at point $2 .$ Because of the high temperature at the end of an adiabatic compression, the fuel begins to burn spontaneously. (There are no spark plugs in a diesel enginel) Second, combustion takes place more slowly, with fuel continuing to be injected. This makes the ignition stage a constant-pressure process. The cycle shown, for one cylinder of a diesel engine, has a displacement $V_{\max }-V_{\min }$ of $1000 \mathrm{cm}^{3}$ and a compression ratio $r=$ $V_{\max } / V_{\min }=21 .$ These are typical values for a diesel truck. The engine operates with intake air $(\gamma=1.40)$ at $25^{\circ} \mathrm{C}$ and $1.0 \mathrm{atm}$ pressure. The quantity of fuel injected into the cylinder has a heat of combustion of $1000 \mathrm{J}$ a. Find $p, V,$ and $T$ at cach of the four corners of the cycle. Display your results in a table. b. What is the net work done by the cylinder during one full cycle? c. What is the thermal efficiency of this engine? d. What is the power output in $\mathrm{kW}$ and horsepower $(1 \mathrm{hp}=$ 746 W) of an eight-cylinder diesel engine running at $2400 \mathrm{rpm} ?$

FIGURE CP19.72 shows the Diesel cycle. It is similar to the Otto cycle (see Problem CP19.71), but there are two important differences. First, the fuel is not admitted until the air is fully compressed at point $2 .$ Because of the high temperature at the end of an adiabatic compression, the fuel begins to burn spontaneously. (There are no spark plugs in a diesel enginel) Second, combustion takes place more slowly, with fuel continuing to be injected. This makes the ignition stage a constant-pressure process. The cycle shown, for one cylinder of a diesel engine, has a displacement $V_{\max }-V_{\min }$ of $1000 \mathrm{cm}^{3}$ and a compression ratio $r=$ $V_{\max } / V_{\min }=21 .$ These are typical values for a diesel truck. The engine operates with intake air $(\gamma=1.40)$ at $25^{\circ} \mathrm{C}$ and $1.0 \mathrm{atm}$ pressure. The quantity of fuel injected into the cylinder has a heat of combustion of $1000 \mathrm{J}$
a. Find $p, V,$ and $T$ at cach of the four corners of the cycle. Display your results in a table.
b. What is the net work done by the cylinder during one full cycle?
c. What is the thermal efficiency of this engine?
d. What is the power output in $\mathrm{kW}$ and horsepower $(1 \mathrm{hp}=$ 746 W) of an eight-cylinder diesel engine running at $2400 \mathrm{rpm} ?$

Key Concepts

Diesel Cycle

The Diesel cycle is an idealized thermodynamic cycle that models the operation of a compression-ignition engine. It is characterized by adiabatic compression and expansion processes, and a constant-pressure heat addition phase that represents the gradual injection of fuel and its combustion. This cycle serves as the basis for understanding the performance and efficiency of diesel engines in converting heat into mechanical work.

Adiabatic Process

An adiabatic process is one in which no heat is exchanged with the environment, so any change in the system’s internal energy is solely due to work done on or by the system. In engine cycles such as the Diesel cycle, both the compression and expansion phases are often assumed to be adiabatic, simplifying the analysis and allowing predictions of temperature and pressure changes during these rapid processes.

Constant-Pressure Process

A constant-pressure process occurs when the pressure of a system remains unchanged while its volume changes. In the Diesel cycle, the fuel is injected and combusts during this phase, causing the volume to increase at nearly constant pressure. This characteristic process distinguishes the Diesel cycle from others like the Otto cycle and influences its efficiency and work output.

Compression Ratio

The compression ratio is defined as the ratio of the maximum volume in the cylinder to the minimum volume reached during the compression stroke. It is a crucial parameter in engine performance because it affects the temperature and pressure attained during compression, which in turn impacts the efficiency and combustion characteristics of the engine cycle.

Thermal Efficiency

Thermal efficiency is a measure of how well an engine converts the heat from fuel combustion into useful work. It is defined as the ratio of the net work output of the cycle to the heat input. In the context of engine cycles, a higher thermal efficiency indicates a more effective conversion of heat into mechanical energy, which is crucial for engine performance evaluation.

Net Work

Net work in a thermodynamic cycle is the difference between the work produced during the expansion phase and the work required during the compression phase. This net work is the useful mechanical energy output of the engine and is fundamental in assessing the engine’s overall performance and efficiency.

Engine Power

Engine power quantifies the rate at which an engine performs work, typically expressed in kilowatts or horsepower. It is calculated by considering the net work per cycle and the number of cycles executed per unit time. Understanding engine power is essential in rating the performance of engines in practical applications, such as determining the output of multi-cylinder engines operating at a given speed.

Transcript

00:01 So now we're talking about a diesel engine, which has a little different cycle than the compression rate and then the gasoline engine.

00:14 And so we can see here we have heat coming in during a constant pressure process and not during a constant volume process.

00:22 So basically we're just kind of shifting things around.

00:26 And they talk about it a little bit in the problem.

00:30 So the fuel isn't brought into the cylinder until the full air is fully compressed.

00:36 So we fully compress it here and then we inject fuel.

00:41 And again, because of the high temperature of the air here because of the compression, we don't need spark plugs.

00:49 So just injecting the fuel at a high enough temperature that it will just spontaneously combust.

00:56 So the combustion takes place more slowly with the fuel and continuing to be injected.

01:05 So again, we can keep, as we spray this fuel and it's obviously combusting immediately, so we kind of keep spraying it in.

01:12 So it occurs basically roughly at a constant pressure process of the ignition stages here.

01:18 Again, we have adiabatic expansion, adabatic compression, and then we have an isochoric process where the exhaust is let out, is dumped into the atmosphere.

01:34 So in this problem, we're told you have this a pretty high compression ratio here.

01:40 Again, this is about 21.

01:43 And again, you need that to get the temperature way up, so you don't need the spark plug.

01:49 But obviously, it comes with some other problems too that you have to overcome.

01:54 Or everything would be a diesel engine.

01:58 So let's see here.

02:01 We're given t .e .v .p.

02:04 So we can find a number of moles because we know everything at one here.

02:07 So we have 0 .043 moles of air, you know, per coming in per stroke of the cylinder.

02:17 We know v2.

02:19 So we can find p2 from knowing the adibatic compression.

02:25 And we wind up with 7 .19 megapascal.

02:30 So it's very highly compressed gas right there.

02:34 And then we can get the temperature now, because we know v2 and p2, we can get temperature is roughly 1 ,000 kelvin.

02:42 So pretty high temperature, and that's why we can get this fuel to combust spontaneously.

02:50 To figure out what t3 is, we can use the fact that we're told that we have one megajoule or one kilojoule.

02:59 Yeah, one kilojoule of fuel coming in or burning.

03:07 So we're getting one kilojoule of heat being added as the fuel burns during the process.

03:13 So we find out that t3 is then 1800 kelvin.

03:16 So again, the fuel combusts and we obviously get a much hotter, hotter gas.

03:23 And then, so now we know, let's see here, we know t3, and we know we had a constant pressure process so we can get v3, and that's about, about 90 cubic centimeters.

03:36 So, you know, it's not really much different from here on this scale anyway.

03:41 This is much exaggerated here.

03:43 So you don't have a whole lot of change in pressure there or change in volume.

03:50 Now, we know v4 is v3 or v1.

03:54 So we can find p4, again because of the adiabatic expansion here...

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