A four cylinder two stroke petrol engine with stroke to bore ratio 1.2 develops 35
kW brake power at 2200 rpm. The mean effective pressure in each cylinder is 9
bar and mechanical efficiency is 78%. Determine (1) diameter and stroke of each
cylinder, (2) Brake thermal efficiency, (3) Indicated thermal efficiency. If fuel
consumption is 8 kg/hr having C.V. =43000 KJ/kg
Answers
Answer:
four cylinder two stroke petrol engine with stroke to bore ratio 1.2 develops 35
kW brake power at 2200 rpm. The mean effective pressure in each cylinder is 9
bar and mechanical efficiency is 78%. Determine (1) diameter and stroke of each
cylinder, (2) Brake thermal efficiency, (3) Indicated thermal efficiency. If fuel
consumption is 8 kg/hr having C.V. =43000
Explanation:
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Answer:
BTE = BP / (FC * CV) = 35,000 / (8 * 43,000) = 0.1028 or 10.
Explanation:
To solve this problem, we'll use the following formulas:
Brake power (BP) = 2πNT/60, where N is the engine speed in rpm and T is the torque developed by the engine in Nm.
Torque (T) = MEP * A * L, where MEP is the mean effective pressure, A is the area of the piston head, and L is the stroke length.
Piston displacement (PD) = (π/4) * D^2 * L, where D is the diameter of the cylinder.
Brake thermal efficiency (BTE) = BP / (FC * CV), where FC is the fuel consumption rate in kg/hr and CV is the calorific value of the fuel in kJ/kg.
Indicated thermal efficiency (ITE) = (IMEP * PD * N * 2) / (FC * CV * 60), where IMEP is the indicated mean effective pressure.
We're given:
Brake power (BP) = 35 kW
Engine speed (N) = 2200 rpm
MEP = 9 bar
Mechanical efficiency (ME) = 78%
Fuel consumption rate (FC) = 8 kg/hr
CV = 43000 kJ/kg
Stroke-to-bore ratio = 1.2
We need to find:
Cylinder diameter (D) and stroke length (L)
Brake thermal efficiency (BTE)
Indicated thermal efficiency (ITE)
First, we can find the torque developed by the engine:
T = MEP * A * L
We know that MEP = 9 bar and we can assume that the area of each piston head is the same. Let's call this area A_piston. Then:
T = 9 * A_piston * L
Next, we can find the piston displacement:
PD = (π/4) * D^2 * L
Since we know the stroke-to-bore ratio, we can write:
L = 1.2 * D
Substituting this into the piston displacement formula:
PD = (π/4) * D^2 * (1.2 * D) = 0.3 * π * D^3
Now we can find the diameter of each cylinder. We'll use the brake power formula to find the torque and then use the torque formula to find the area of each piston head:
BP = 2πNT/60
35,000 = (2π * 2200 * T)/60
T = 246.95 Nm
From T = MEP * A_piston * L, we have:
A_piston = T / (MEP * L) = 246.95 / (9 * 1.2 * D) = 18.072/D
Substituting this into the piston displacement formula:
PD = 0.3 * π * D^3 = 18.072 * L * D^2
Solving for D:
D = (PD / (0.3 * π * L))^(1/2) = (8.025/L)^(1/2)
Substituting L = 1.2D, we get:
D = (8.025/(1.2D))^(1/2) = 1.97 cm
So the diameter of each cylinder is approximately 1.97 cm, and the stroke length is 1.2 times this, or approximately 2.36 cm.
Next, we can find the brake thermal efficiency:
BTE = BP / (FC * CV) = 35,000 / (8 * 43,000) = 0.1028 or 10.
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