Answer:
Power produced = 90.47 KW
Explanation:
We are given;
R = 0.287 kJ/kg·K
T1 = 15°C = 15 + 273 = 288 K
P1 = 95 KPa
Number of cylinders;N_cyl = 8
Bore;B = 10cm = 0.1m
Stroke;S = 12cm = 0.12m
cp = 1.005 kJ/kg·K
cv = 0.718 kJ/kg·K
k = 1.4
First of all let's find the initial specific volume;
α1 = RT1/P1
α1 = 0.287 * 288/95
α1 = 0.87 m³/kg
Now, let's find the total mass of air from the formula;
m =(V•N_cyl)/α1 =(B²•N_cyl•S•π)/4α1
So, m = (B²•N_cyl•S•π)/4α1
m = (0.1²•8•0.12•π)/(4*0.87)
m = 0.00867 Kg
Now, let's calculate the total mass flow rate;
m' = (m*N_rev)/n'
Where;
N_rev is number of revolutions given as 1500 rpm = 1500/60 rev/s = 25 rev/s
n' is the repetitions per circle = 2.
Thus;
m' = (0.00867*25)/2
m' = 0.108375 kg/s
The temperature at state 2 is gotten from the formula;
T2 = T1*r^(k - 1)
Where r is compression ratio.
We know that formula for compression ratio is;
the ratio of the maximum to minimum volume in the cylinder of an internal combustion engine.
In the question, we are told that minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume.
Thus,
r = 100/5 = 20
So, T2 = 288*20^(1.4 - 1)
T2 = 954.563 K
The cut off ratio is gotten from the formula;
r_c = α3/α2 = T3/T2
T3 = 2000°C = 2000 + 273K = 2273K
Thus; r_c = 2273/954.563
r_c = 2.38
The heat input is gotten from the formula;
q_in = cp(T3 - T2)
q_in = 1.005(2273 - 954.563)
q_in = 1325.03 KJ/Kg
The efficiency is gotten from;
η = 1 - [1/(r^(k - 1)]*[((r_c)^(k) - 1)/(k(r_c - 1))]
Thus;
η = 1 - [1/(20^(1.4 - 1)]*[((2.38)^(1.4) - 1)/(1.4(2.38 - 1))]
η = 0.63
Now, the power output is gotten from the equation;
W' = m'•η•q_in
W' = 0.108375*0.63*1325.03
W' = 90.47 KW
