138k views
0 votes
An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 95 kPa and 278C. Accounting for the variation of specific heats with temperature, determine (a) the temperature after the heat-addition process, (b) the thermal efficiency, and (c) the mean effective pressure

User Phicon
by
4.0k points

1 Answer

1 vote

Answer:


a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Step-by-step explanation:

a. Internal energy and the relative specific volume at
s_1 are determined from A-17:
u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at
s_2 is calculated from the compression ratio:


\alpha_r_2=(\alpha_r_1)/(r)\\=(621.2)/(16)\\=38.825

#from this, the temperature and enthalpy at state 2,
s_2 can be determined using interpolations
T_2=862K and
h_2=890.9kJ/kg. The specific volume at
s_1 can then be determined as:


\alpha_1=(RT_1)/(P_1)\\\\=(0.287* 300)/(95) m^3/kg\\0.906316m^3/kg

Specific volume,
s_2:


\alpha_2=(\alpha_1)/(r)\\=(0.906316)/(16)m^3/kg\\=0.05664m^3/kg

The pressures at
s_2 \ and\ s_3 is:


P_2=P_3=(RT_2)/(\alpha_2)\\\\=(0.287*862)/(0.05664)\\=4367.06kPa

.The thermal efficiency=> maximum temperature at
s_3 can be obtained from the expansion work at constant pressure during
s_2-s_3


\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+(P\alpha_2)/(R)(r_c-1)\\\\=(862+(4367* 0.05664)/(0.287)(2-1))K\\=1723.84K

b.Relative SV and enthalpy at
s_3 are obtained for the given temperature with interpolation with data from A-17 :
a_r_3=4.553 \ and\ h_3=1909.62kJ/kg

Relative SV at
s_4 is


a_r_4=(r)/(r_c)\alpha _r_3

=
=(16)/(2)*4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;


n=1-(q_o)/(q_i)\\=1-(u_4-u_1)/(h_3-h_2)\\=1-(65903-214.07)/(1909.62-890.9)\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:


MEP=(\omega)/(\alpa_1-\alpa_2)\\=(q_i-q_o)/(\alpha_1(1-1/r))\\=(1909.62-890.9-(65903-214.7))/(0.90632(1-1/16))\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

User Pooya Estakhri
by
4.6k points