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icang [17]
3 years ago
13

For each of the cases below, determine if the heat engine satisfies the first law (energy equation) and if it violates the secon

d law.
1. ˙QH=6 kW,˙QL=4 kW,˙W=2 kW
2. ˙QH=6 kW,˙QL=0 kW,˙W=6 kW
3. ˙QH=6 kW,˙QL=2 kW,˙W=5 kW
4. ˙QH=6 kW,˙QL=6 kW,˙W=0 kW
Engineering
1 answer:
stepan [7]3 years ago
6 0

Answer:

From first law of thermodynamics(energy conservation)

Qa= Qr+W

Qa=Heat added to the engine

Qr=heat rejected from the engine

W=work output from the engine

Second law:

It is impossible to construct a heat engine that will deliver the work with out rejecting heat.

In other word ,if engine take heat then it will reject some amount heat and will deliver some amount of work.

1.

QH=6 kW,

QL=4 kW,

W=2 kW

6 KW= 4 + 2  KW

It satisfy the first law.

Here heat is also rejected from the engine that is why it satisfy second law.

2.

QH=6 kW, QL=0 kW, W=6 kW

This satisfy first law but does not satisfy second law because heat rejection is zero.

3.

QH=6 kW   ,   QL=2 kW,      W=5 kW

This does not satisfy first as well as second law.Because summation of heat rejection and work can not be greater than heat addition or we can say that energy is not conserve.

4.

QH=6 kW,   QL=6 kW,   W=0 kW

This satisfy first law only and does not satisfy second law.

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A laboratory in the Y building keep a vacuum pressure of 0.1 kPa abs. What is the net force acting on the door considering the a
seropon [69]

Answer:

net force acting on the floor is 100 kN

Explanation:

Given data:

P_{vaccum} = 0.1 kPa

P_{atm} = 101.325 kPa

dimension of floor = 2 m \times 0.5 m

we know that

Net force can be calculated as follow

f_{net} = P_{vaccum} \times area

f_{net} = 0.1\times 10^3 \times 2\times 0.5

f_{net} = 0.1\times 10^3 \times 1

f_{net} = 100 kN

Therefore net force acting on the floor is 100 kN

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4 years ago
1. When working on charging systems, all of the fol-
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Answer:

c

Explanation:

You never want short system terminals

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2 years ago
A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was
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Answer:

Time of submersion in years = 7.71 years

Explanation:

Area of plate (A)= 16in²

Mass corroded away = Weight Loss (W) = 3.2 kg = 3.2 x 106

Corrosion Penetration Rate (CPR) = 200mpy

Density of steel (D) = 7.9g/cm³

Constant = 534

The expression for the corrosion penetration rate is

Corrosion Penetration Rate = Constant x Total Weight Loss/Time taken for Weight Loss x Exposed Surface Area x Density of the Metal

Re- arrange the equation for time taken

T = k x W/ A x CPR x D

T = (534 x 3.2 x 106)/(16 x 7.9 x 200)

T = 67594.93 hours

Convert hours into years by

T = 67594.93 x (1year/365 days x 24 hours x 1 day)

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4 years ago
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