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2 years ago

What's the difference between motor, engine, and machine?.

Engineering

1 answer:

harina [27]2 years ago

7 0

Answer:

A machine consumes energy whereas an engine produces energy. Engine converts heat energy to Mechanical Energy but machine converts all forms of energy to mechanical energy except heat energy. A engine runs a machine,without a engine a machine can't run.

Explanation:

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An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.

Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

$density=\frac{P}{RT}$

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

$density=\frac{120}{(0.287)(293)}=1.43kg/m^3$

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0

3 years ago

Suppose that the cost of electrical energy is $0.15 per kilowatt hour and that your electrical bill for 30 days is $80. Assume t

mr_godi [17]

Answer:

a) 740 W b) 6.2 A c) 8.1%

Explanation:

We need first to get the total energy spent during the 30 days, that can be calculated as follows:

1 month = 30 days = 720 hr

If the total cost amounts $80 (for 720 hr), and the cost per kwh is 0.15, we have:

80 $/mo = 0.15 $/Kwh*x Kwh/mo

Solving for the total energy spent in the month:

$x (kwh/mo) = \frac{80}{0.15} = 533.3 kwh$

Assuming that the power delivered is constant over the entire 30 days, as power is the rate of change of energy, we can find the power as follows:

$P = \frac{E}{t} = \frac{533.3 kWh}{720 h} = 0.74 kW = 740 W$

b) If the power is supplied by a voltage of 120 V, we can find the current I as follows:

$I =\frac{P}{V} =\frac{740W}{120V} = 6.2A$

c) If part of the electrical load is a 60-W light, we can substract this power from the one we have just found, as follows:

P = 740 W - 60 W = 680 W

The new value of the energy spent during the entire month will be as follows:

E = 0.68 kW*24(hr/day)*30(days/mo) = 490 kWh/mo

The reduction in percentage regarding the total energy spent can be calculated as follows:

ΔE = $\frac{533.3-490}{533.3} = 0.081*100= 8.1%$

⇒ ΔE(%) = 8.1%

$%(E) =\frac{533.3-490}{533.3} = 0.081 * 100 = 8.1%$

4 0

4 years ago

A ship tows a submerged cylinder, 1.5 m in diameter and 22 m long, at U = 5 m/s in fresh water at 20°C. Estimate the towing powe

shusha [124]

Answer:

(a) 120 kW

(b) 800 kW

Explanation:

Given:

diameter: 1.5 m

length = 22 m

U = 5 m/s

temperature = 20°C

For water at 20°C, take ρ = 998 kg/m³ and µ = 0.001 kg/m⋅s

To find:

power in kW

(a) if the cylinder is parallel

Length / Diameter = L / D = 22 / 1.5 = 14.6 = 15

Re(L) = ρ*U*L / µ = 998 * 5 * 22 / 0.001 = 109780000 = 1.1E8

C(D.Frontal) ≈ 1.1

Force = F = 1.1 * ρ/2 * U² * π / 4 * D

= 1.1 (998 / 2) (5)²(π / 4)(1.5)²

= 1.1 * 499 * 25 * 0.785 * 2.25

= 24000 N

Power = Force * Displacement / time

= F * U

= 24000 * 5

= 120000

Power = 120 kW

b) if the cylinder is normal to the tow direction.

Re(L) = ρ*U*D / µ = 998 * 5 * 1.5 / 0.001 = 7485000 = 7.5E6

C(D.Frontal) ≈ 0.4

Force = F = 0.4 * ρ/2 * U² * D * L

= 0.4 (998 / 2) (5)²(1.5)(22)

= 164670 ≈ 165000

Power = Force * Displacement / time

= F * U

= 165000 * 5 = 825000

≈ 800 kW

Power = 800 kW

7 0

4 years ago

For many clients, a decision is based primarily on time and money, which are the two most abundant

saw5 [17]

True is what I would pick

8 0

3 years ago

Read 2 more answers

A steel plate of width 120mm and thickness of 20mm is bent into a circular arc radius of 10. You are required to calculate the m

notka56 [123]

Answer:

Hence the magnitude of the pure moment m will be $2\times 10^5.$

Explanation:

Width of steel fleet = 120 mm
The thickness of steel fleet = 10 mm
Let the circle of radius = 10 m

Now,

We know that,

$\frac{M}{I} = \frac{E}{R}$

Thus, $M =\frac{EI}{R}$

Here

R = 10000 mm

$I=\frac{1}{12}\times 120\times 10^{3}\\= 10^{4} mm^{4}$

$E=2\times 10^{5}n/mm^{2}\\\\E=2\times 10^{5}n/mm^{2}\\\\M={(2\times 10^{5}\times 10^{4})/{10000}}\\\\M=2\times 10^{5}$

Hence, the magnitude of the pure moment m will be $2\times 10^5.$

6 0

3 years ago