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3 years ago

Air is compressed in a piston-cylinder device. List three examples of irreversibilities that could occur

Engineering

1 answer:

KiRa [710]3 years ago

4 0

Answer:

While air is compressed in a piston cylinder there are following types of irreversibilities

1.Due to finite temperature difference heat transfer take place between cylinder and surrounding.

2.Due friction force between cylinder and piston .

3.Compression process is so fast due to this ,it leads in the irreversibility of system.

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Six forces act on a beam that forms part of a building's

IrinaVladis [17]

Answer:

Explanation:

write each force in terms of magnitude and directions

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let

FA = FA sin (110) + FA cos (110)

FB = 20 sin (270) + 20 cos (270)

FC = 16 sin (140) + 16 cos (140)

FD = 9 sin (40) + 9 cos (40)

FE = 20 sin (270) + 20 cos (270)

FG = FG sin (50) + FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110) + 0 + 16 sin (140) + 9 sin (40) + 0 + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110) + 0 + 10.285 + 5.785 + 0 + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110) + 16.070 + FG sin (50) = 0

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0

3 years ago

What is the weight density of a 2.24 in diameter titanium sphere that weights 0.82 lb?

nasty-shy [4]

Answer:

$0.14\ lb/in^{3}$

Explanation:

Density is defined as mass ler unit volume, expressed as

$\rho=\frac {m}{v}$

Where m is mass, $\rho$ is density and v is the volume. For a sphere, volume is given as

$v=\frac {4\pi r^{3}}{3}$

Replacing this into the formula of density then

$\rho=\frac {m}{\frac {4\pi r^{3}}{3}}$

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

$\rho=\frac {0.82}{\frac {4\pi 1.12^{3}}{3}}=0.13932044952427\approx 0.14 lb/in^{3}$

4 0

4 years ago

It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

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2 years ago

Which of the following does the electrical technician typically use?

kkurt [141]

Answer:

What are the choices?

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3 years ago

Show the ERD with relational notation with crowfoot. Your ERD must show PK, FKs, min and max cardinality, and correct line types

zhenek [66]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.

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4 years ago