Answer to Question #278414 in Mechanics | Relativity for adnan

Question

 

 

1.    An ideal Otto cycle has a compression ratio of 7. At the beginning of the compression

process, air is at 107 kPa and 19°C, and 750 kJ/kg of heat is transferred to air during

the constant-volume heat-addition process. Accounting for the variation of specific

heats of air with temperature, determine (a) the maximum temperature and pressure

that occur during the cycle, (b) the net work output, (c) the thermal efficiency, and

(d) the mean effective pressure for the cycle.

(e) Also, determine the power output from the cycle, in kW, for an engine speed

of 3700 rpm (rev/min). Assume that this cycle is operated on an engine that has four

cylinders with a total displacement volume of 1.9 L.

Answer 1

a)

The maximum temperature and pressure in an Otto cycle occur at the end of the constant-volume heat-addition process.

determine the temperature and pressure of air at the end of the isentropic compression process , using data from table of ideal-gas properties of air:

$T_1=292\ K\to u_1=206.91\ kJ/kg,v_{r1}=676.1$

isentropic compression of an ideal gas:

$v_{r2}/v_{r1}=1/r \to v_{r2}=v_{r1}/r=676.1/7=96.59$

$T_2=620\ K,u_2=450.09\ kJ/kg$

$P_2=P_1\frac{T_2v_1}{T_1v_2}=107\cdot7\cdot\frac{620}{292}=1590\ kPa$

constant-volume heat addition:

$u_3=q_{in}+u_2=750+450.09=1200.09\ kJ/kg$

$T_3=1500\ K,v_{r3}=7.152$

$P_3=P_2\frac{T_3v_2}{T_2v_3}=1590\cdot\frac{1500}{620}=3847\ kPa$

b)

isentropic expansion of an ideal gas:

$v_{r4}=rv_{r3}=7\cdot7.152=50.064$

$T_4=780\ K,u_4=576.12\ kJ/kg$

constant-volume heat rejection:

$q_{out}=u_4-u_1=576.12-206.91=369.21\ kJ/kg$

$w_{net}=q_{in}-q_{out}=750-369.21=380.79\ kJ/kg$

c)

 thermal efficiency of the cycle:

$\eta_{th}=w_{net}/q_{in}=380.79/750=0.51$

Under the cold-air-standard assumptions (constant specific heat values at room temperature):

$\eta_{th,Otto}=1-r^{1-k}=1-7^{1-1.4}=0.54$

d)

mean effective pressure:

$MEP=\frac{w_{net}}{v_{max}-v_{min}}=\frac{w_{net}}{v_{1}(1-1/r)}$

$v_1=RT_1/P_1=0.287\cdot292/107=0.783$ m³/kg

$MEP=\frac{380.79}{0.783(1-1/7)}=567$ kPa

e)

The total air mass taken by all four cylinders when they are charged:

$m=V_d/v_1=0.0019/0.783=0.0024$ kg

net work produced by the cycle:

$W_{net}=mw_{net}=0.0024\cdot380.79=0.924$ kJ

Noting that there are two revolutions per thermodynamic cycle in a fourstroke engine:

$n_{rev}=2$ rev/cycle

power produced by the engine:

$W=W_{net}n/n_{rev}=0.924\cdot3700/(2\cdot60)=28.49$ kW