An ideal Diesel cycle has a maximum cycle temperature of 2000°C. The state of the air at the beginning of the compression is P1 = 95 kPa and T1 = 15°C. This cycle is executed in a four-stroke, eight-cylinder engine with a cylinder bore of 10 cm and a piston stroke of 12 cm. The minimum volume enclosed in the cylinder is 5 percent of the maximum cylinder volume. Determine the power produced by this engine when it is operated at 1500 rpm. Use constant specific heats at room temperature. The properties of air at room temperature are cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, R = 0.287 kJ/kg·K, and k = 1.4.

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