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topjm [15]
3 years ago
15

Yes I’m very cool I promise.

Engineering
1 answer:
Alex3 years ago
8 0
Okay I believe you I swear
You might be interested in
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
Bad White [126]

Answer:

the net work per cycle \mathbf{W_{net} = 0.777593696}  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 V_2 = 0.16 V_1

the crankshaft N rotates at a speed of  2400 RPM.

At the beginning of the compression , temperature T_1 = 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:

V_1-V_2 = \dfrac{\pi}{4}D^2L

V_1-0.16V_1= \dfrac{\pi}{4}(3.7)^2(3.4)

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

the mass in air ( lb) can be determined by using the formula:

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

Now; the net work per cycle can now be calculated as by using the following formula:

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

W = 88.0144746 hp

8 0
3 years ago
The underground cafe has an operating cash flow of $187,000 and a cash flow to creditors of $71,400 for the past year. During th
Serggg [28]

Answer:

cash flow to stockholders = $39,700

Explanation:

Operating cash flow = $187,000

cash flow to creditors = $71,400

Net working capital = $28,000

Net capital spending =  $47,900

Cash flow to stockholders = ?

CFF = operating cash flow - net working capital - net capital spending

CFF = $187,000 - $28,000 - $47,900 = $111,100

CFF = cash flow to creditors + cash flow to stockholders

cash flow to stockholders = CFF - cash flow to creditors

cash flow to stockholders = $111,100 - $71,400 = $39,700

Hence $39,700 is the amount of the cash flow to stockholders for the last year.

3 0
3 years ago
Vehicles begin to arrive at a parking lot at 8:10 am at a constant rate of 6 veh/min until 8:25 am. There is no arrival from 8:2
tresset_1 [31]

1.i am superman

2.175 175 175

3.em dilisues

4.bye classmate magbabalik pa ako

8 0
3 years ago
The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large
SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

C(d) = 220 / (1.221 * 625)

C(d) = 220 / 763.125

C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0
3 years ago
"Design a sequential circuit with two T flip-flops A and B, and one input x. When x = 0, the circuit remains in the same state.
Harlamova29_29 [7]

Answer:

See attached images for the diagrams and tables

3 0
3 years ago
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