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Julli [10]
3 years ago
5

using the following data for july, calculate the cost of goods manufactured: beginning finished goods inventory 150,475. Ending

finished goods inventory 145,750. sales 400,000. Gross Margin 120,000. The cost of goods manufactured was
Engineering
1 answer:
Alla [95]3 years ago
5 0

Answer:

The correct response is "$275,275".

Explanation:

The given values are:

Sales,

= 400,000

Gross margin,

= 120,000

Beginning Inventory goods,

= 150,475

Finished inventory goods,

= 145,750

Now,

The cost of goods sold will be:

= Sales-Gross \ margin

On substituting the values, we get

= 400,000-120,000

= 280,000

As we know,

⇒ Cost \ of \ goods \ sold=Beginning \ inventory \ goods+ cost \ of \ goods \  manufactured-Ending \ inventory \ goods

⇒ 280,000=150,475+ cost \ of \ goods \ manufactured-145750

⇒ 280,000=cost \ of \ goods \ manufactured+4,725

⇒ Cost \ of \ goods \ manufactured=280,000-4,725

⇒                                                 =275,275 ($)

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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
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Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  Btu per cycle

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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 V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

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and;

Otto cycle with a pressure =  14.5 lbf/in² = (14.5 × 144 ) lb/ft²

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The maximum temperature in the cycle is 5200 R

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V_1-0.16V_1= 36.55714291

0.84 V_1 =36.55714291

V_1 =\dfrac{36.55714291}{0.84 }

V_1 =43.52040823 \ in^3 \\ \\  V_1 = 43.52 \ in^3

V_1 = 0.02518 \ ft^3

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m = \dfrac{P_1V_1}{RT}

where;

R = 53.3533 ft.lbf/lb.R°

m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R  \times 519 .67 ^0 R}

m = 0.0018962 lb

From the tables  of ideal gas properties at Temperature 519.67 R

v_{r1} =158.58

u_1 = 88.62 Btu/lb

At state of volume 2; the relative volume can be determined as:

v_{r2} = v_{r1}  \times \dfrac{V_2}{V_1}

v_{r2} = 158.58 \times 0.16

v_{r2} = 25.3728

The specific energy u_2 at v_{r2} = 25.3728 is 184.7 Btu/lb

From the tables of ideal gas properties at maximum Temperature T = 5200 R

v_{r3} = 0.1828

u_3 = 1098 \ Btu/lb

To determine the relative volume at state 4; we have:

v_{r4} = v_{r3} \times \dfrac{V_1}{V_2}

v_{r4} =0.1828 \times \dfrac{1}{0.16}

v_{r4} =1.1425

The specific energy u_4 at v_{r4} =1.1425 is 591.84 Btu/lb

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W_{net} = Heat  \ supplied - Heat  \ rejected

W_{net} = m(u_3-u_2)-m(u_4 - u_1)

W_{net} = m(u_3-u_2- u_4 + u_1)

W_{net} = m(1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (1098-184.7- 591.84 + 88.62)

W_{net} = 0.0018962 \times (410.08)

\mathbf{W_{net} = 0.777593696}  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:

W = 4 \times N'  \times W_{net

where ;

N' = \dfrac{2400}{2}

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

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