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10 months ago

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Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the

1 answer:

DanielleElmas [232]10 months ago

3 0

Option c is correct. Hole filling fasteners (for example, MS20470 rivets) should not be used in composite structures primarily because of the increased possibility of fretting corrosion in the fastener.

Fretting corrosion is the outcome of simultaneous physicochemical and mechanical surface interaction in a tribological contact, and it is the irreversible change of a material. A specific load and cyclic displacement with a limited amplitude must be applied to the contact between the surfaces of two materials for fretting corrosion to take place.

Small gaps develop between the surfaces in contact as a result of cyclic tension. Either material is lost or cracks start to appear as a result. Fretting corrosion's side effect of contact corrosion almost always causes a blockage because the lack of functional movement makes it impossible to remove debris. When debris is still trapped, there is blockage because the oxide takes up more space than the original material.

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2 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

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3 years ago

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2 years ago

The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find

Nataly_w [17]

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

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1 year ago