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Bezzdna [24]
3 years ago
8

There are two piston-cylinder systems that each contain 1 kg of an idea gas at a pressure of 300 kPa and temperature of 350 K. T

he two systems then undergo two different processes:
System 1 undergoes process 1 as follows:
Isobaric heating to 500 K, then compressed to a pressure of 500 kPa and a temperature of 600 K.

System 2 undergoes process 2 as follows:
Directly compressed in one step to a pressure 500 kPa and a temperature of 600 K

We "measure" the entropy of each system before and after the process. Which system had the biggest entropy change?

a. System 1
b. System 2
c. Entropy change is the same for both systems
Engineering
1 answer:
Nataly_w [17]3 years ago
8 0

Answer:

Entropy change is same for both systems because entropy is a state function i.e., it doesn't depend on the path by which the system arrived at it's present state (500 kPa, 600 K) from initial state (300 kPa, 350 K) .

Explanation:

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3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

F = 23.33 kN

Applying the impulse momentum principle on the three cars;

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F(75-0) = 1.125 × 10⁶

The force T developed is then;

T =  1.125 × 10⁶ / 75

T = 15000 N

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7 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

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