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Svetradugi [14.3K]
4 years ago
13

Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3

00 K, respectively. The magnitude of the heat transfer rejected to the high temperature reservoir is 250 kJ per kg of air. The pressure at the start of the isothermal expansion is 325 kPa. The air can be modeled as an ideal gas with constant specific heat (k=1.4). For the air as a system, determine
a. (5) the coefficient of performance.
b. (5) the net work input to the cycle in kJ/kg.
c. (10) the Pv and Ts diagrams for the cycle with arrows showing the directions of the processes.
You may answer parts (a) and (b) in the response box. Part (c) and your calculations supporting.(a) and (b) must be included in your uploaded response file.

Engineering
1 answer:
Nataly [62]4 years ago
5 0

Answer:

See explaination

Explanation:

for a reverse carnot cycle T-S diagram is a rectangle which i have shown

net work for a complete cycle must be equal to net heat interaction.

Kindly check attachment for the step by step solution of the given problem.

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A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
I need answers for this sheet please.
9966 [12]
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7 0
4 years ago
An electron is traveling with initial kinetic energy K in a uniform electric field. The electron comes to rest momentarily after
V125BC [204]

Answer:

a) E = \frac{K}{e*d}  b) i.

Explanation:

Assuming no other forces acting on the electron, as the field is pushing the electron back till it comes to rest momentarily, applying the work-energy theorem, the loss in kinetic energy, must be equal to the work done by the field on the electron.

The force that does work on the electron is the one due to the electric field E, and by definition of electric field, can be expressed as follows:

W = q*E = -e*E (1)

This work must be equal to the change in Kinetic Energy:

ΔK = Kf -K₀ = 0 - K = -K (2)

From (1) and (2), and solving for E, we have:

E =\frac{K}{e*d}

b) As the electric field (by convention) has the direction that it would have a positive test charge, the electrons (being negative charges),when starting from rest),have the direction opposite to the field.

In this case, as the electron has an initial velocity, and the field is opposing to the electron movement, we conclude that the electric field is in the direction of the electron´s motion, due to the electron slows down, and then comes momentarily to an stop, before changing direction and move opposite to the field as it is its natural behavior.

3 0
3 years ago
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries helium at 200 kPa
Aleks04 [339]

Answer: a 8143.71 kJ/kg

b 393.15 K

Explanation:

This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists

a since no change in pressure =0 the system does work thus

FOR HELIUM  properties in standard thermodynamic chart

cv = 3.1 kJ/kgK

M = Molar mass = 4 kg/kmol

R = Universal gas constant = 8.314 kJ/kg K

cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK  

Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol  = 20.714 kJ/kgkmol

T = 120  °C  to Kelvin = 120 + 273.15k = 393.15 K

W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg

b convert T °C = T K thus 120 + 273.15 K = 393.15 K

P₁/T₁ = P₂/T₂

200 kPa/ 393.15 K = 200 kPa/T₂

T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k

7 0
4 years ago
At the end of the vowel, the soft palate (is lowered) so that the air flows out the (nasal passage). At the same time, the _____
Marat540 [252]

Answer:

At the end of the vowel, the soft palate (is lowered) so that the air flows out the (nasal passage). At the same time, the BACK of the tongue rises to articulate with the VELUM, preventing air from escaping through the MOUTH. The lips remain OPEN. The vocal folds continue to SOUND.

Explanation:

This is the way in which sound is produced in any language through the use of our organs of speech called articulators.

NOTE:

This question should be in English Category not Engineering. I believe that was an oversight from your end.

3 0
3 years ago
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