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3 years ago

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A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

ength of steel is 0.58oy. What is the maximum load before yielding? A. 2313 psi B. 1025 psi C. 3207 psi D. 1654 psi

1 answer:

just olya [345]3 years ago

3 0

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is $\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}$

The shear strength= $0.58\sigma_y$ and in this case $\sigma_y=36 000 psi$

Shear strength= $\frac {Load}{Shear area}$ hence making load the subject then

Load=Shear area X Shear strength

Load= $\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi$

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A heat pump designer claims to have an air-source heat pump whose coefficient of performance is 1.8 when heating a building whos

Anit [1.1K]

Answer:

The claim is valid.

Explanation:

Let assume that heat pump is reversible. The coefficient of performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{300\,K}{300\,K-260\,K}$

$COP_{HP} = 7.5$

The claim is valid as real heat pumps have lower coefficients of performance.

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3 years ago

A thermoelectric refrigerator is powered by a 16-V power supply that draws 2.9 A of current. If the refrigerator cools down 3.1

Viktor [21]

Answer:

COP = 0.090

Explanation:

The general formula for COP is:

COP = Desired Output/Required Input

Here,

Desired Output = Heat removed from water while cooling

Desired Output = (Specific Heat of Water)(Mass of Water)(Change in Temperature)/Time

Desired Output = [(4180 J/kg.k)(3.1 kg)(25 - 11)k]/[(12 hr)(3600 sec/hr)]

Desired Output = 4.199 W

And the required input can be given as electrical power:

Required Input = Electrical Power = (Current)(Voltage)

Required Input = (2.9 A)(16 V) = 46.4 W

Therefore:

COP = 4.199 W/46.4 W

<u>COP = 0.090</u>

8 0

3 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

7 0

3 years ago

BlackZzzverrR [31]

Answer:

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2 years ago

B)<br>State the essential difference between a plain carbon steel<br>and an alloy steel

choli [55]

Answer:

Plain carbon steel has no or trace external elements while alloy steel has high amount of other elements.

Explanation:

Plain carbon steel has no or trace amount of other elements while alloy steel has high amount of other elements in their composition.

The presence of other elements in alloy steel improvise several physical properties of the steel while plain carbon steel has the basic properties.

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3 years ago