Question

The weight of a diesel engine is directly proportional to the compression ratio (W = kr, k is a constant and r is the compression ratio) because extra metal must be used to strengthen the engine for the higher pressures. Examine the net specific work produced by a diesel engine per unit weight as the pressure ratio is varied and the specific heat input remains fixed. Do this for several heat inputs and proportionality constants k. Are there any optimal combination of k and specific heat inputs?

Support and explain your conclusion using engineering analysis and plots.

Answer 1

The CI cycle is more complex, thus in initial analysis we use a perfect "air-standard" assumption, during which the working fluid may be a fixed mass of air undergoing the entire cycle which is treated throughout as a perfect gas. It is an ideal process, combustion is replaced by heat addition to the air, and exhaust is replaced by a heat rejection process which restores the air to the initial state.
The ideal air-standard diesel undergoes 4 distinct processes, all of which may be separately analysed, as shown within the P-V diagrams below. Two of the four processes of the cycle are adiabatic processes (adiabatic = no transfer of heat), equations for a perfect gas natural process as follows:

The analysis leads to the subsequent three general forms representing an adiabatic process:

k-1 Tv const Tp(-k)/k = const Pok = const

where k is that the ratio of warmth capacities and features a par value of 1.4 at 300K for air.

Process 1-2 is that the adiabatic compression process. Thus the temperature of the air increases during the compression process, and with an outsized compression ratio (usually > 16:1) it'll reach the ignition temperature of the injected fuel. Thus given the conditions at state 1 and therefore the compression ratio of the engine, so as to work out the pressure and temperature at state 2 (at the top of the adiabatic compression process) we have:

=pk I Vi V2 → Compression ratio (A)-4) (6) - Ch k-1 Vi =pk-1

Work W1-2 required to compress the gas is shown because the area under the P-V curve, and is evaluated as follows.

Adiabatic compression process p.vk = Const 3 4 piston W1-2 Wi-2 = SP av = Consefi- W1-2 Pdv = Constv-kav = Const vl-k vl-k) = P. vk 1-k 1-k BV-AV thus: W1-2 1-k (1 - k) since for an ideal gas P.V=m.R.T m.R (T2-T) (1 - k)

An alternative approach using the energy equation takes advantage of the natural process (Q1-2 = 0) leads to a way simpler process.

G- acliabatic W-2 - W1-2 = m. Au = m.Cv.AT W1-2 = m-cy (T- T2)

During process 2-3 the fuel is injected and combusted and this is often represented by a continuing pressure expansion process. At state 3 the expansion process continues adiabatically with the temperature decreasing until the expansion is complete.
Process 3-4 is thus the adiabatic expansion process. the entire expansion work is Wexp = (W2-3 + W3-4) and is shown because the area under the P-V diagram and is analysed as At last the process 4-1 represents the constant volume heat rejection process.

fuel injection 2-3 W2-3: Constant pressure expansion W3-4: Adiabatic expansion = Const P p.vk 3 LW2-3 (Pconst) W3-4 acliabatic W2-3 W3-4 1 3 W2-3 PdV= B.(V3 - V.) ſpav= 44 - W3-4 = m ·Au= m-cy.AT = W3.4 = m-Cy"(T3 - TA) acliabatic Wexp = W2-3 + W3-4

At last the process 4-1 represents the constant volume heat rejection process. In an actual diesel the gas is just exhausted from the cylinder and a fresh charge of air is introduced.The net work Wnet done over the cycle is given by
[Wnet = (Wexp + W1-2)]
as the compression work W1-2 is negative before (work done on the system).
In the Air-Standard Diesel cycle engine the warmth input Qin occurs by combusting the fuel which is injected during a controlled manner, ideally leading to a continuing pressure expansion process 2-3 as shown below. At highest volume (bottom dead center) the burnt gasses are simply exhausted and replaced by a fresh charge of air. this is often represented by the equivalent constant volume heat rejection process Qout = -Q4-1. Both processes are analyzed as follows:

fuel injection 2-3 Qin P Constant pressure expansion 8Q-W=dU = &Q=dU+PDV dH= d(U+PV)= dU+PaV+[vap] 8Q= dH Qin = AH = m.Cp.AT Pconst) in 3 W2-3 Adliabatic processes PVk = Const Qout = Qin = AH = m .Cp.AT = m .Cp (T3-T2) P.V = m.R.T T3_V3 (constant pressure) T2 V2 V3 Qin = m.CpT -1 = m .Cp:T2 (0-1) where: 1c V2 Constant volume heat rejection Qout Qout = -Q4-1 = -AU = -m.Cy.AT = m.cy (T4 - Ti) V3 (cutoff ratio)

At this stage we can conveniently determine the engine efficiency in terms of the heat flow as follows