Section 4.6 Internal Combustion Engines 159
Example 4.8 Thermal efficiency of the Otto engine (Continued)
Solution: = − Basis: model as ideal gas, QH CV(T3 T2)(*ig) = − QC CV (T1 T4)(*ig) = + = − + − WS,net QH QC CV(T3 T2 T1 T4)(*ig) η = − + − − + − − − = − − CV(T3 T2 T1 T4)/[CV(T3 T2)] 1 (T3 T2)/(T3 T2) 1 + (T1 T4)/(T3 T2)(*ig)
RC/ V RC/ V RC/ V T2 = V1 T4 = V3 = V2 ; (*ig) T1 V2 T3 V4 V1
= –R/C = –R/C ⇒ T4 T3 r V; T1 T2 r V (*ig)
−R/C (1−γ) η = 1 − r V = 1 − r (*ig) 4.8
The Diesel Engine The Diesel Engine is similar to the Otto engine except that the fuel is injected after the compression and the combustion occurs relatively slowly. This necessitates “fuel-injectors,” but has the advan- tage that higher compression ratios can be obtained without concern of pre-ignition (ignition at the wrong time in the cycle). Pre-ignition occurs when the temperature rise due to compression goes past the spontaneous ignition temperature of the fuel, creating an annoying pinging and knocking sound, and reducing the efficiency of the cycle. Pre-ignition does not occur in a diesel engine because there is no fuel during compression. Some spontaneous ignition temperatures are given below.
Spontaneous ignition temperatures (°C) of sample hydrocarbons
Isooctane 447 Benzene 592 Toluene 568 n-Octane 240 n-Decane 232 n-Hexadecane 230
Methanol 470 Ethanol 392
Diesel fuel is much like decane and hexadecane and burns without a spark. Auto fuel is much like isooctane and benzene and, ideally, will not ignite until the spark goes off. But all fuels are mixtures, and if the gasoline contains enough n-octane instead of isooctane, then undesirable pre-ignition will occur.
The thermodynamics of the air-standard Diesel engine can be analyzed like the Otto Cycle. But, instead of rapid combustion at constant volume, the Diesel engine has relatively slow combus- tion at constant pressure. In the air-standard Diesel cycle shown in Figure 4.18, step 1-2 is an adia- 160 Unit I First and Second Laws
2 3
P 4
1
V
Figure 4.18 Schematic of air-standard Diesel cycle. batic compression, step 2-3 represents the combustion where heat is added, step 3-4 is adiabatic, step 4-1 is an isochoric cooling.
Example 4.9 Thermal efficiency of a Diesel engine
Develop an expression for the thermal efficiency of the air-standard diesel cycle as a function of = = the compression ratio rc V1/V2 and the expansion ratio re V4 /V3. Assume the working fluid is an ideal gas, and the volume effect of moles of gas generated is small relative to the effect of heating from combustion.
Solution: This process is a little more complicated than the Otto cycle because heat addition and work occurs during constant pressure combustion. The energy balance for the combus- tion step is: = ∆ − = ∆ − = ∆ = ∆ dU = QH + WS,combustion ⇒ QH U WS,combustion U + P2 (V3 V2) (U + PV) H = ∆ = − ⇒ QH H CP(T3 T2) (*ig) For isochoric cooling = − QC CV (T1 T4) (*ig)
For the cycle, the energy balance is 0 = QH + QC + WS,net, giving the thermal efficiency
CT()− T T T η = −W /Q = (Q + Q )/Q = 1 +(*ig)V 14=+1 1 − 4 S,net H H C H − 1 γ − − CTP ()32 T TT32TT32 where
TT- T T PV 32=-=3 2 33 -rcgg--11 = rc/ re - rc g (*ig) T1 T1 T1 PV11
where for T2/T1 we have used Eqn 2.53, and for T3/T1 we have used: the ideal gas law; P3 = P2; Eqn. 2.55; and V1 = V4, as Section 4.7 Fluid Flow 161
Example 4.9 Thermal efficiency of a Diesel engine (Continued)
T P V P V V γ V 3 3 3 2 3 1 3 γ ⁄ ----- ==------------------=------------=rc re (*ig) T1 P1V1 P1 V1 V2 V4 For the last term in brackets from the formula for efficiency,
TT- T T PV PV 32=-=-=-=-3 2 regggg---1 22 re 1 32 re1 re / rrc (*ig) T4 T4 T4 PV44 PV44
where for T3/T4 we have used Eqn 2.53, and for T2/T4 we have used: the ideal gas law; P2 = P3; Eqn. 2.55; and V4 = V1, as
γ T P V P V V V γ -----2 ==------2 2 -----3- -----2- =-----4- -----2- =re ⁄ rc (*ig) T4 P4V4 P4 V4 V3 V1 Substuting the intermediate results and rearranging the formula for efficiency,
11Ï 1¸ h =+1 Ì - ˝ g rcgggg// re- rc--11 re- re rc Ó ˛ (*ig) 4.9 1 Ï re rc ¸ =+1 Ì - ˝ g Órcgg-◊ re rc -1 rc◊- regg-1 re ˛
4.7 FLUID FLOW
Consider the general flow system of Fig. 4.19a in which work and heat are transferred and the fluid undergoes changes in kinetic and potential energy. Recognize that the compressor or pump in the schematic could be replaced with an expander or turbine. Rather than deriving an integral equation between points 1 and 4 in the schematic, let us consider a balance over a differential element at steady-state as shown in Fig. 4.19b where the possibility of heat and work transfer are permitted. The steady-state balance for a single stream becomes:1
2 in 2 out u gz u gz 0 = lim H ++------– H ++------++dQ dWS dL → 02gc gc 2gc gc
For a differential element, Hout = Hin + dH, and
2 out 2 in 2 u u u gz out gz in gz ------= ------+ d ------, ----- = ----- + d----- 2gc 2gc 2gc gc gc gc
The differential balance becomes
2 u gz 0 = – dH – d ------– d----- ++dQ dW S 2gc gc
1. Note that the velocity in this equation is a mean velocity.