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wlad13 [49]
4 years ago
7

An air-standard cycle with constant specific heats at room temperature is executed in a closed system and is composed of the fol

lowing four processes:
1–2 v = Constant heat addition from 14.7 psia and 80°F in the amount of 300 Btu/lbm
2–3 P = Constant heat addition to 3150 R
3–4 Isentropic expansion to 14.7 psia
4–1 P = Constant heat rejection to initial state
The properties of air at room temperature are cp = 0.240 Btu/lbm·R, cv = 0.171 Btu/lbm·R, and k = 1.4.
a. Show the cycle on P-v and T-s diagrams. (Please upload your response/solution using the controls provided below.) (You must provide an answer before moving on to the next part.)
b. Calculate the total heat input per unit mass
c. Determine the thermal efficiency

Engineering
1 answer:
qwelly [4]4 years ago
5 0

Answer:

a. Please see attachment .

b. 612.377

c. 24.22%

Explanation:

Please see attachment.

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A Carnot power cycle is executed in a piston-cylinder system using 1 kg of water. The cycle consists of an isobaric expansion of
jeka94

Answer:

b) Determine the heat transfer into the cycle and the net work for the cycle, in kJ.

Explanation:

5 0
4 years ago
The criminal and traffic code requires that a driver must have a valid driver's license in his/her
ankoles [38]

Answer:

true

Explanation:

the answer is true because if u don't have a valid license when operating a vehicle and you get pulled over you will get in trouble i know this because my parents got in trouble for it once

5 0
3 years ago
A metal rod is 0.600 m in length at a temperature of 15.0∘C. When you raise its temperature to 37.0∘C, its length increases by 0
Katyanochek1 [597]

Answer:

The coefficient of linear expansion of the metal is ∝ = 2.91 x 10⁻⁵  °C⁻¹.

Explanation:

We know that Linear thermal expansion is represented by the following equation

Δ L = L x ∝ x Δ T ---- (1)

where Δ L is the change in length, L is for length, ∝ is the coefficient of linear expression and  Δ T is the change in temperature.

Given that:

L = 0.6 m

T₁ = 15° C

T₂ = 37° C

Δ L = 0.28 mm

∝ = ?

Solution:

We know that Δ T = T₂ ₋ T₁

Putting the values of T₁  and T₂ in above equation, we get

Δ T = 37 - 15

Δ T =  22 °C

Also Δ L = 0.28 mm

Converting the mm to m

Δ L = 0.00028 m

Putting the values of Δ T, Δ L, L in equation 1, we get

0.00028 = 0.6 x ∝ x 22

Rearranging the equation, we get

∝ = 0.00028 / (0.6 x 16)

∝ = 0.00028 / 13.2

∝ = 2.12 x 10⁻⁵  °C⁻¹

4 0
4 years ago
In the first-order process,a blue dye reacts to form a purple dye. The amount of blue dye at the end of 1 hr is 480 g and the en
Tasya [4]

Answer:

The answer is 960 kg

Explanation:

Solution

Given that:

Assume the initial dye concentration as A₀

We write the expression for the dye concentration for one hour as follows:

ln (C₁) = ln (A₀) -kt

Here

C₁ = is the concentration at 1 hour

t =time

Now

Substitute 480 g for C₁ and 1 hour for t

ln (480) = ln (A₀) -k(1) ------- (1)

6.173786 =  ln (A₀) -k

Now

We write the expression for the dye concentration for three hours as follows:

ln (C₃) = ln (A₀) -k

Here

C₃ = is the concentration at 3 hour

t =time

Thus

Substitute 480 g for C₃ and 3 hour for t

ln (120) = ln (A₀) -k(3) ------- (2)

4.787492 = ln (A₀) -3k

Solve for the equation 1 and 2

k =0.693

Now

Calculate the amount of blue present initially using the expression:

Substitute 0.693 for k in equation (2)

4.787492 = ln (A₀) -3 (0.693)

ln (A₀) =6.866492

A₀ =e^6.866492

= 960 kg

Therefore, the amount of the blue dye present from the beginning is  960 kg

6 0
4 years ago
The activation energy for diffusion in BCC iron is 84 kJ/mol. Which of the following explains why the carbide precipitation acti
k0ka [10]

Answer:

  • B. Precipitation require the diffusional activation energy plus an additional energy to form the precipitate.

Explanation:

Precipitation is the creation of a solid from a solution. When the reaction occurs in a liquid solution, the solid formed is called the precipitate.

The formation of a precipitate indicates the occurrence of a chemical reaction.

Precipitation of carbide requires alot of energy which the diffusion activational energy alone cannot achieve and this was calculated to be 225.6 kJ/mol.

3 0
4 years ago
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