Engineering, 07.07.2020 17:01 gracie0818

The pressure and temperature at the beginning of compression of an air-standard Diesel cycle are 95 kPa and 300 K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2150 K. Determine: a. the compression ratio, r.
b. the cutoff ratio, rc.
c. the thermal efficiency of the cycle, ?th (%).
d. the mean effective pressure, mep (kPa).

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Answer from: fvmousdiana

a) 23.19

b) 2.19

c) 0.597 ( 59.7%)

d) 975 KPa

Explanation:

Solution:-

- We will determine the properties of the working fluid (Air) at each state value.

- We will seek help from air standard property tables ( A-22 ) and write down the following:

State 1:

P1 = 95 KPa , T1 = 300 K

u1 = 214.07 KJ/kg , vr1 = 621.2 , Pr1 = 1.386

- Using relations for isentropic compression we determine the reduced pressure value at state 2:

State 2:

P2 = P3 = 7200 KPa

$P_r_2 = P_r_1*(\frac{P_2}{P_1} )\\\\P_r_2 = 1.386*(\frac{7200}{95} ) = 105.04\\$

T2 = 979.6 K , vr2 = 26.793 , h2 = 1022.82 KJ/kg

State 3:

P3 = 7200 KPa , T3 = 2150 K

h3 = 2440.3 KJ/kg , vr3 = 2.175 , Pr1 = 1.386

- Using relations for isentropic expansion we determine the reduced volume value at state 4:

State 4:

$\frac{v_4}{v_3} = \frac{v_1}{v_2}*\frac{v_2}{v_3} = \frac{v_r_1}{v_r_2}*\frac{T_2}{T_3} \\\\\frac{v_4}{v_3} = \frac{621.2}{26.793}*\frac{979.6}{2150} = 10.56$

$v_r_4 = \frac{v_4}{v_3}*v_r_3 = 10.56*2.175\\\\v_r_4 = 22.98$

T4 = 1031 K , u4 = 785.75 KJ/kg

a) the compression ratio ( r )

- It is the ratio of volume decreased in the compression stroke ( State 1 -> state 2 ):

$r = \frac{v_1}{v_2} = \frac{v_r_1}{v_r_2} =\frac{621.2}{26.793}\\\\r = 23.19$

b) the cutoff ratio ( rc )

- It is the ratio of volume increased in the heat-addition process ( State 2 -> state 3 ):

$r_c = \frac{v_3}{v_2} = \frac{T_3}{T_2}\\\\r_c = \frac{2150}{979.6} = 2.19$

c) the thermal efficiency of the cycle ( nth )

- It is the ratio of total work-output from the cycle and the heat addition (Q23). The net work of the cycle is comprised of both heat addition(St2 - St3) and exhaust (St4 - St1 ). Hence, the work done by the cycle is:

$w_n_e_t = w_e - w_c\\\\w_n_e_t = (h_3 - h_2) - (u_4 - u_1 )\\\\w_n_e_t = (2440.3 - 1022.82) - (785.75 - 214.07 )\\\\w_n_e_t = 845.8 KJ/kg$

- The thermal efficiency of the cycle would be:

$n_t_h = \frac{w_n_e_t}{h_3 - h_2} = \frac{845.8}{2440.3-1022.82} \\\\n_t_h = \frac{845.8}{1417.48}\\\\n_t_h = 0.597$

d) the mean effective pressure, mep

- It is the ratio of net work done by the cycle and the displaced volume of the compression stroke ( V1 - V2 ).

$mep = \frac{w_n_e_t}{V_1 - V_2} = \frac{w_n_e_t}{v_1( 1 - \frac{v_2}{v_1} )} = \frac{w_n_e_t}{v_1( 1 - \frac{1}{r} )} \\$

- We need to determine the specific volume of air at state 1. We will use ideal gas equation to determine.

$v_1 = \frac{RT_1}{P_1} = \frac{0.287*300}{95} \\\\v_1 = 0.9063 \frac{m^3}{kg}$

- Now we can use the calculated specific volume ( v1 ), compression ratio ( r ) and the net cycle work ( wnet ) to determine the ( mep ):

$mep = \frac{845.8}{0.9063*(1-\frac{1}{29.13}) } \\\\mep= 975 KPa$

Answer from: Quest

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Answer from: Quest

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