Respuesta :
Answer:
the net work per cycle [tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, W = 88.0144746 hp
Explanation:
the information given includes;
diameter of the four-cylinder bore = 3.7 in
length of the stroke = 3.4 in
The clearance volume = 16% = 0.16
The cylindrical volume [tex]V_2 = 0.16 V_1[/tex]
the crankshaft N rotates at a speed of 2400 RPM.
At the beginning of the compression , temperature [tex]T_1[/tex] = 60 F = 519.67 R
and;
Otto cycle with a pressure = 14.5 lbf/in² = (14.5 × 144 ) lb/ft²
= 2088 lb/ft²
The maximum temperature in the cycle is 5200 R
From the given information; the change in volume is:
[tex]V_1-V_2 = \dfrac{\pi}{4}D^2L[/tex]
[tex]V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)[/tex]
[tex]V_1-0.16V_1= 36.55714291[/tex]
[tex]0.84 V_1 =36.55714291[/tex]
[tex]V_1 =\dfrac{36.55714291}{0.84 }[/tex]
[tex]V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3[/tex]
[tex]V_1 = 0.02518 \ ft^3[/tex]
the mass in air ( lb) can be determined by using the formula:
[tex]m = \dfrac{P_1V_1}{RT}[/tex]
where;
R = 53.3533 ft.lbf/lb.R°
[tex]m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}[/tex]
m = 0.0018962 lb
From the tables of ideal gas properties at Temperature 519.67 R
[tex]v_{r1} =158.58[/tex]
[tex]u_1 = 88.62 Btu/lb[/tex]
At state of volume 2; the relative volume can be determined as:
[tex]v_{r2} = v_{r1} \times \dfrac{V_2}{V_1}[/tex]
[tex]v_{r2} = 158.58 \times 0.16[/tex]
[tex]v_{r2} = 25.3728[/tex]
The specific energy [tex]u_2[/tex] at [tex]v_{r2} = 25.3728[/tex] is 184.7 Btu/lb
From the tables of ideal gas properties at maximum Temperature T = 5200 R
[tex]v_{r3} = 0.1828[/tex]
[tex]u_3 = 1098 \ Btu/lb[/tex]
To determine the relative volume at state 4; we have:
[tex]v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}[/tex]
[tex]v_{r4} =0.1828 \times \dfrac{1}{0.16}[/tex]
[tex]v_{r4} =1.1425[/tex]
The specific energy [tex]u_4[/tex] at [tex]v_{r4} =1.1425[/tex] is 591.84 Btu/lb
Now; the net work per cycle can now be calculated as by using the following formula:
[tex]W_{net} = Heat \ supplied - Heat \ rejected[/tex]
[tex]W_{net} = m(u_3-u_2)-m(u_4 - u_1)[/tex]
[tex]W_{net} = m(u_3-u_2- u_4 + u_1)[/tex]
[tex]W_{net} = m(1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)[/tex]
[tex]W_{net} = 0.0018962 \times (410.08)[/tex]
[tex]\mathbf{W_{net} = 0.777593696}[/tex] Btu per cycle
the power developed by the engine, in horsepower. can be calculated as follows;
In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:
[tex]W = 4 \times N' \times W_{net[/tex]
where ;
[tex]N' = \dfrac{2400}{2}[/tex]
N' = 1200 cycles/min
N' = 1200 cycles/60 seconds
N' = 20 cycles/sec
W = 4 × 20 cycles/sec × 0.777593696
W = 62.20749568 Btu/s
W = 88.0144746 hp
The net work per cycle and the power developed by this combustion engine are 0.7792 Btu and 88.20 hp.
Given the following data:
- Diameter of bore = 3.7 in
- Stroke length = 3.4 in
- Clearance volume = 16% = 0.16
- Speed of 2400 RPM.
- Initial temperature = 60 F to R = 519.67 R.
- Initial pressure = 14.5 [tex]lbf/in^2[/tex] to [tex]lbf/ft^2[/tex] = 2088 [tex]lbf/ft^2[/tex]
- Maximum temperature = 5200 R.
Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]
How to calculate the net work per cycle.
First of all, we would determine the volume, mass and specific energy as follows:
[tex]V_1-V_2=\frac{\pi D^2L}{4} \\\\V_1-0.16V_1=\frac{3.142 \times 3.7^2 \times 3.4}{4}\\\\0.84V_1=36.56\\\\V_1=\frac{36.56}{0.84} \\\\V_1=43.52\;in^3 \;to \;ft^3 = 0.0252\;ft^3[/tex]
For the mass:
[tex]M=\frac{PV}{RT} \\\\M=\frac{2088 \times 0.0252}{53.3533 \times 519.67} \\\\M=\frac{52.6176}{27726.109411}[/tex]
M = 0.0019 lb.
At a temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r1}=158.58\\\\u_1 = 88.62\;Btu/lb[/tex]
For the relative volume at the second state, we have:
[tex]v_{r2}=v_{r1}\times \frac{V_2}{V_1} \\\\v_{r2}=158.58\times 0.16\\\\v_{r2}=25.3728[/tex]
Note: At 25.3728, specific energy ([tex]u_2[/tex]) is 184.7 Btu/lb.
At a maximum temperature of 519.67 R, the relative volume and specific energy are:
[tex]v_{r3}=0.1828\\\\u_3 = 1098\;Btu/lb[/tex]
For the relative volume at state 4, we have:
[tex]v_{r4}=v_{r3}\times \frac{V_1}{V_3} \\\\v_{r4}=0.1828\times \frac{1}{0.16}\\\\v_{r4}=1.1425[/tex]
Note: At 1.1425, specific energy ([tex]u_4[/tex]) is 591.84 Btu/lb.
Now, we can calculate the net work per cycle by using this following formula:
[tex]W=Heat\;supplied -Heat\rejected\\\\W=m(u_3-u_2)-m(u_4-u_1)\\\\W=0.0019(1098-184.7)-0.0019(591.84-88.62)\\\\W=1.73527-0.956118[/tex]
W = 0.7792 Btu.
How to calculate the power developed.
In a four-cylinder, four-stroke internal combustion engine, power is given by this formula:
[tex]W=4N'W_{net}[/tex]
But;
[tex]N'=\frac{N}{2 \times 60} \\\\N'=\frac{2400}{120} \\\\N'=20\;cycle/sec[/tex]
Substituting the given parameters into the formula, we have;
[tex]W=4 \times 20 \times 0.7792[/tex]
W = 62.336 Btu/sec.
In horsepower:
W = 88.20 hp.
Read more on net work here: https://brainly.com/question/10119215
