Question

An ideal diesel cycle has a compression ratio is 16 to 1. The maximum cycle temperature is 1700 C and minimum cycle temperature is 15 centigrade calculate
the specific heat transfer to the cycle
the specific work of the cycle
the thermal efficiency of the cycle​

Answer 1

Answer:

The Air Standard Diesel cycle is the ideal cycle for Compression-Ignition (CI) reciprocating engines, first proposed by Rudolph Diesel over 100 years ago. The following link by the Kruse Technology Partnership describes the four-stroke diesel cycle operation including a short history of Rudolf Diesel. The four-stroke diesel engine is usually used in motor vehicle systems, whereas larger marine systems usually use the two-stroke diesel cycle. Once again we have an excellent animation produced by Matt Keveney presenting the operation of the four-stroke diesel cycle.The actual CI cycle is extremely complex, thus in initial analysis we use an ideal "air-standard" assumption, in which the working fluid is a fixed mass of air undergoing the complete cycle which is treated throughout as an ideal gas. All processes are ideal, 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 engine undergoes 4 distinct processes, each one of which can be separately analysed, as shown in the P-V diagrams below. Two of the four processes of the cycle are adiabatic processes (adiabatic = no transfer of heat), thus before we can continue we need to develop equations for an ideal gas adiabatic process as follows:

The Adiabatic Process of an Ideal Gas (Q = 0)

The analysis results in the following three general forms representing an adiabatic process:



where k is the ratio of heat capacities and has a nominal value of 1.4 at 300K for air.

Process 1-2 is the adiabatic compression process. Thus the temperature of the air increases during the compression process, and with a large compression ratio (usually > 16:1) it will reach the ignition temperature of the injected fuel. Thus given the conditions at state 1 and the compression ratio of the engine, in order to determine the pressure and temperature at state 2 (at the end of the adiabatic compression process)

Answer 2

Answer:

given, (V1/V2)=16:1

Max temp. T3=1700°c = 1973K

Min temp. T1=15°c = 288K

for process 1-2

(T2/T1) = (V1/V2)^(1.4-1)

T2= 288(16)^0.4

T2= 873K

1). specific heat transfer,

Qs = Cp(T3-T2)

=1.005(1973-873) = 1105.5KJ/Kg

for process 3-4

(T3/T4) = (V4/V3)^(1.4-1)

since(V4=V1, V3=V2)

(1973/T4) = (16)^0.4

T4 = 650.8K

Qr= Cv(T4-T1) = 0.718(650.8-288)

= 260.5KJ/Kg

2). net specific work,

Wnet = Qs - Qr

= 1105.5 - 260.5 = 845KJ/Kg