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2 years ago

8

g ideal air conditioner maintains a house at 20 by dumping heat into the atmosphere at 40 this refrigerator has a coefficient of

performance of

1 answer:

masya89 [10]2 years ago

4 0

AnsweCalculate a heat pump's coefficient of performance. ... Heat pumps, air conditioners, and refrigerators utilize heat transfer from cold to hot.Explanation:

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Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

A certain piece of property is assessed at $150,000. If the tax rate is $2.50 per $100, what is the tax on this property?

stiks02 [169]

Answer:

The tax on this property is $3750$ dollars

Explanation:

Given

Tax on per $100 is $2.50

Tax on every $1 is $\frac{2.5}{100} = 0.025$ dollars

Tax on property of value $150,000 is

$150,000 * 0.025 = 3750$ dollars

The tax on this property is $3750$ dollars

7 0

3 years ago

A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.

Alex73 [517]

Answer: you can watch a video on how to solve this question on you tube

6 0

3 years ago

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

3 years ago

Ignoring any losses, estimate how much energy (in units of Btu) is required to raise the temperature of water in a 90-gallon hot

Rudik [331]

Answer:

Q=36444.11 Btu

Explanation:

Given that

Initial temperature = 60° F

Final temperature = 110° F

Specific heat of water = 0.999 Btu/lbm.R

Volume of water = 90 gallon

Mass = Volume x density

$1\ gallon = 0.13ft^3$

Mass ,m= 90 x 0.13 x 62.36 lbm

m=729.62 lbm

We know that sensible heat given as

Q= m Cp ΔT

Now by putting the values

Q= 729.62 x 0.999 x (110-60) Btu

Q=36444.11 Btu

5 0

3 years ago