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3 years ago

Which of the following explains why material properties present challenges for engineers?

Engineering

1 answer:

Maurinko [17]3 years ago

4 0

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

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Air in a piston–cylinder assembly, initially at 3 bar, 142 K, and a volume of 2 m3. The air undergoes a process to a state where

Archy [21]

Answer:

mass of the air = 14.62kg

Workdone = 415.88 kJ

Heat transfer = 415.88 kJ

Explanation:

Please find the attached files for the solution

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3 years ago

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Technician A says that carbon monoxide (CO) and hydrocarbon (HC) levels should increase if the air injection reactor (AIR) hoses

lara31 [8.8K]

Answer:Technician A

Explanation:

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3 years ago

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: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the

Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

$\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}$

$V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s$

$\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}$

taking f = 0.0185

at Z₁ = Z₂

$\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}$

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P= 1.13\ psi$

c) When flow is downhill z₂-z₁ = -2

$\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266$

$\Delta P=-0.601\ psi$

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3 years ago

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at 2308C by rejectin

lord [1]

Answer:

hello your question is incomplete attached below is the missing part and also attached is the solution

answer: a) 0.4801

b) 5.398 kw

c) 2.14

d) 12.72

Explanation:

The quality of the refrigerant at the evaporator inlet

h4 = hf4 + x4(hfx4)

Refrigeration load

Ql = m(h1-h4)

COP of the refrigerator

Ql / m(h2-h1) - Qm

Theoretical maximum refrigeration load

( Ql )max = COPr.rev * [m(h2-h1) - Qin]

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3 years ago

Think about all the things we have investigated in class so far this semester. Identify three examples of climate-related aspect

Free_Kalibri [48]

Answer:

1) the mean surface temperature of earth.

2) precipitation and sunshine

3) air and wind movement.

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3 years ago