A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16% of the cylinder volume at bottom dead center and the crankshaft rotates at 2400 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.5 lbf/in.2 and a temperature of 60F at the beginning of compression. The maximum temperature in the cycle is 5200R. Based on this model, calculate the net work per cycle, in Btu, and the power developed by the engine, in horsepower.

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:

Note: The cylindrical volume is equal to [tex]0.16V_1[/tex]

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.

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

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