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AnnyKZ [126]
2 years ago
10

After combustion has taken place, the _______________ open, releasing unwanted gases from the cylinder's combustion chamber.

Engineering
1 answer:
Likurg_2 [28]2 years ago
7 0

Answer: Do not know

Explanation: Sorry

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Helium gas expands in a piston-cylinder in a polytropic process with n=1.67. Is the work positive, negative or zero?
IRINA_888 [86]

Answer:

work will be positive when it is under polytropic expansion process

Explanation:

It states a polytropic  process with n equal to 1.67. there is a polytropic expansion that mean work is positive and if it was polytropic compression then it would   be negative

PV^n = const

P_1V_1 = P_2V_2

Also work during the process of polytropic is given as

W_{1-2} =\frac{P_1V_1 -P_2V_2}{n-1}

the work will be positive when it is under the polytropic expansion process

3 0
3 years ago
Suppose that the time (in hours) required to repair a machine is an exponentially distributed random variable with parameter ???
Andrej [43]

We need to define the variables,

So,

F_x (x) = 1-e^{-\lambda x}\\F_x (x) = 1-e^{-0.5x}

Therefore, the probability that the repair time is more than 4 horus can be calculate as,

P(x>4)=1-P(x4)= 1-F_x(4)\\P(x>4) = 1-e^{-0.5*4}\\P(x>4) = 1-0.98\\P(x>4) = 0.018

The probability that the repair time is more than 4 hours is 0.136

b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,

P(x\geq 12|x>7)=P(X\geq7+5|x>7)\\P(x\geq12|x>7)=P(X\geq5)\\P(x\geq12|x>7)=1-P(x\leq 5)\\P(x\geq12|x>7)=1-e^{-0.5(2)}

P(x\geq 12|x>7)=0.6321

The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63

3 0
3 years ago
A piston–cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine t
drek231 [11]

Answer:

See explanation

Explanation:

Given:

Initial pressure,

p

1

=

15

psia

Initial temperature,

T

1

=

80

∘

F

Final temperature,

T

2

=

200

∘

F

Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

R

=

0.04513

Btu/lbm.R

C

v

=

0.158

Btu/lbm.R

Find the work done during the isobaric process.

w

1

−

2

=

p

(

v

2

−

v

1

)

=

R

(

T

2

−

T

1

)

=

0.04513

(

200

−

80

)

w

1

−

2

=

5.4156

Btu/lbm

Find the change in internal energy during process.

Δ

u

1

−

2

=

C

v

(

T

2

−

T

1

)

=

0.158

(

200

−

80

)

=

18.96

Btu/lbm

Find the heat transfer during the process using the first law of thermodynamics.

q

1

−

2

=

w

1

−

2

+

Δ

u

1

−

2

=

5.4156

+

18.96

q

1

−

2

=

24.38

Btu/lbm

7 0
3 years ago
The rolling process is governed by the frictional force between the rollers and the workpiece. The frictional force at the entra
adell [148]

Answer:

b)false

Explanation:

Rolling is a process in which work piece passes through rolls to produce desired out put of the work piece.Rolling  is a metal forming process.

We know that friction force is responsible for motion of work piece between rolls.If friction force is so small at the entrance side then work piece will not enter in the forming zone and forming process will not occurs.So the friction force should be high at the entrance side and low at the exit side.

So given statement is wrong.

3 0
3 years ago
The coolant heat storage system:
Monica [59]

Answer:

c

Explanation:

This is because many things, such as pcs, over heat

8 0
3 years ago
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