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3 years ago

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Ammonia enters the expansion valve of a refrigeration system at a pressure of 10 bar and a temperature of 22oC and exits at 2.0

bar. The refrigerant undergoes a throttling process. Determine the temperature, in oC, and the quality of the refrigerant at the exit of the expansion valve.

1 answer:

KonstantinChe [14]3 years ago

6 0

Answer:

The the quality of the refrigerant at the exit of the expansion valve is 0.179.

Explanation:

Given that,

Initial pressure = 10 bar

Temperature = 22°C

Final pressure = 2.0 bar

We using the value of h

$h = 293.4\ kJ/kg$

The refrigerant during expansion undergoes a throttling process

Therefore, $h_{1}=h_{2}$

We need to calculate the quality of the refrigerant at the exit of the expansion valve

At 2.0 bar,

The property of ammonia

$h_{f}=47.8 kJ/kg$

$h_{g}=1417.7 kJ/kg$

Using formula

$h_{2}=h_{f}+x(h_{g}-h_{f})$

Put the value into the formula

$293.4 =47.8+x(1417.7-47.8)$

$x=\dfrac{293.4-47.8}{1417.7-47.8}$

$x=0.179$

Hence, The the quality of the refrigerant at the exit of the expansion valve is 0.179.

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Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

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4 years ago

A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and

frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

$m=\sqrt{\dfrac{hP}{KA}}$

For circular fin

$m=\sqrt{\dfrac{4h}{KD}}$

$m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}$

m = 37.08

$\eta_f=\dfrac{tanhmL}{mL}$

$0.65=\dfrac{tanh37.08L}{37.08L}$

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

$\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}$

$\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}$

ε = 23.52

5 0

3 years ago

Read 2 more answers

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.

fgiga [73]

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776 N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F = ∆p/∆t = I/∆t

let the upward velocity be the positive

Δp = mv2 - m(-v1)

Δp = mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t = 0.00125

F = ∆p/∆t

F = 2.22/0.00125

F = 1776 N

4 0

3 years ago

A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

<u>v_f = 10.85 m/s</u>

4 0

3 years ago

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

Ymorist [56]

Answer:

Explanation:

Given

mass of tree stump is $M$

mass bullet is $m$

velocity of bullet is $v$

Conserving momentum for bullet and tree stump

Initial Momentum $P_i=mv$

Suppose $v_0$ is the velocity of the system

Final Momentum $P_f=(M+m)v_0$

Initial momentum =Final Momentum

$mv=(m+M)v_0$

$v_0=\frac{mv}{m+M}$

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3 years ago