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Tju [1.3M]
3 years ago
11

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 21 m3/min and exits at 12 b

ar, 400 K. Heat transfer occurs at a rate of 3.5 kW from the compressor to its surroundings.
Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
Engineering
1 answer:
omeli [17]3 years ago
3 0

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

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A cylindrical metal specimen having an original diameter of 12.8 mm and gauge length of 50.80 mm is pulled in tension until frac
Sedaia [141]

Answer:

%Reduction in area = 73.41%

%Reduction in elongation = 42.20%

Explanation:

Given

Original diameter = 12.8 mm

Gauge length = 50.80mm

Diameter at the point of fracture = 6.60 mm (0.260 in.)

Fractured gauge length = 72.14 mm.

%Reduction in Area is given as:

((do/2)² - (d1/2)²)/(do/2)²

Calculating percent reduction in area

do = 12.8mm, d1 = 6.6mm

So,

%RA = ((12.8/2)² - 6.6/2)²)/(12.8/2)²

%RA = 0.734130859375

%RA = 73.41%

Calculating percent reduction in elongation

%Reduction in elongation is given as:

((do) - (d1))/(d1)

do = 72.14mm, d1 = 50.80mm

So,

%RA = ((72.24) - (50.80))/(50.80)

%RA = 0.422047244094488

%RA = 42.20%

3 0
3 years ago
If you are driving a 30-foot
Ratling [72]

Answer:

3 sec

If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.

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5 0
3 years ago
A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =
oee [108]

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

5 0
2 years ago
(Specific weight) A 1-ft-diameter cylindrical tank that is 5 ft long weighs 125 lb and is filled with a liquid having a specific
lina2011 [118]

Answer:

The answer to the question is 514.17 lbf

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Weight of tank = 125 lb

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Mass of content =  66.4×3.92699 = 260.752 lb

Total mass = 260.752 + 125 = 385.75 lb = 174.97 kg

=Weight = mass * acceleration = 174.97 *9.81 = 1716.497 N

To have an acceleration of 10.7 ft/s² = 3.261 m/s²

we have F = m*a = 174.97*(9.81+3.261) = 2287.15 N = 514.17 lbf

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