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3 years ago

Difference between theory and practice?

Engineering

1 answer:

Ronch [10]3 years ago

5 0

Answer:

There is a huge difference between theory vs. practice. Theory assumes an outcome, while practice allows you to test the theory and see if it is accurate.

Theory and Practice Explained

Practice is the observation of disparate concepts (or a phenomenon) that needs explanation. A theory is a proposed explanation of the relationship between two or more concepts, or an explanation for how/why a phenomenon occurs.

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An automotive fuel has a molar composition of 85% ethanol (C2H5OH) and 15% octane (C8H18). For complete combustion in air, deter

slava [35]

Answer:

a) 1

b) 1813.96 MJ/kmol

c) 32.43 MJ/kg , 1980.39 MJ/Kmol

Explanation:

molar mass of ethanol (C2H5OH) = 46 g/mol

molar mass of octane (C8H18) = 114 g/mol

therefore the moles of ethanol and octane

ethanol = 0.85 / 46

octane = 0.15 / 114

a) determine the molar air-fuel ratio and air-fuel ratio by mass

attached below

mass of air / mass of fuel = 12.17 / 1 = 12.17

b ) Determine the lower heating value

LHV of ( C2H5OH) = 26.8 * 46 = 1232.8 MJ/kmol

LHV of (C8H18). = 44.8 mj/kg * 114 kg/kmol = 5107.2 MJ/Kmol

LHV ( MJ/kmol) for fuel mixture = 0.85 * 1232.8 + 0.15 * 5107.2 = 1813.96 MJ/kmol

c) Determine higher heating value ( HHV )

HHV of (C2H5OH) = 29.7 * 46 = 1366.2 MJ/kmol

HHV of C8H18 = 47.9 MJ/kg * 114 = 5460.6 MJ/kmol

HHV in MJ/kg = 0.85 * 29.7 + 0.15 * 47.9 = 32.43 MJ/kg

HHV in MJ /kmol = 0.85 * 1366.2 + 0.15 * 5460.8 = 1980.39 MJ/Kmol

4 0

2 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

Label the following statements as either T (true) or F (false).

IgorLugansk [536]

Answer:

1 its 1 bc you have to do it step by step

Explanation:

step by step

6 0

2 years ago

A 18000 Kg truck is driving along a level road with a velocity of 32 m/s. If the driver suddenly applies the brakes for 12 secon

evablogger [386]

Answer:

See it in the pic

Explanation:

See it in the pic

7 0

3 years ago

Q1. A truck traveling at 40 mph is approaching a stop sign. At time ????0 and at a distance of 80ft, the truck begins to slow do

densk [106]

Answer:

The truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

Explanation:

The distance that the truck starts slowing down = 80 ft from the stop sign

Using equations of motion, we can calculate the distance it will take the truck to stop, then check of it is less than or more than 80 ft.

u = initial velocity of the truck = 40 mph = 58.667 ft/s

v = final velocity of the truck = 0 ft/s (since it comes to a stop eventually)

x = horizontal distance covered during the deceleration

a = Deceleration = -12 ft/s² (it'll have a negative sign, since it is negative acceleration

v² = u² + 2ax

0² = 58.667² + 2(-12)(x)

24x = 3441.816889

x = 143.41 ft

143.41 ft > 80 ft; hence, the truck will not stop in time. The truck passes the stop sign by about 63.41 ft before it stops.

7 0

3 years ago

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Difference between theory and practice?​

Difference between theory and practice?