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ycow [4]
3 years ago
15

A 0.40-m3 insulated piston-cylinder device initially contains 1.3 kg of air at 30°C. At this state, the piston is free to move.

Now air at 500 kPa and 70°C is allowed to enter the cylinder from a supply line until the volume increases by 50%. Using constant specific heats at room temperature, determine (a) the final temperature, (b) the amount of mass that has entered, and (d) the work done; and (d) entropy generation

Engineering
1 answer:
Setler79 [48]3 years ago
6 0

Answer:

(a) The Final Temperature is 315.25 K.

(b) The amount of mass that has entered  0.5742 Kg.

(c) The work done is 56.52 kJ.

(d) The entrophy generation is 0.0398 kJ/kgK.

Explanation:

Explanation is in the following attachments.

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g An analog voice signal, sampled at the rate of 8 kHz (8000 samples/second), is to be transmitted by using binary frequency shi
slamgirl [31]

Answer:

The module is why it’s goin to work

Explanation:

4 0
3 years ago
A cylindrical metal specimen having an original diameter of 10.55 mm and gauge length of 54.5 mm is pulled in tension until frac
kumpel [21]

Answer:

(a) 53.94%

(b) 26.61%

Explanation:

Change in area will be given by

\triangle A=\frac {\pi(R_o^{2}-r_n^{2})}{\pi R_o^{2}} where \triangle A represent change in area R is radius and subscripts O and n represent original and new respectively.

Substituting 10.55/2 for original radius and 7.16/2 for new radius then

\triangle A=\frac {\pi(5.275^{2}-3.58^{2})}{\pi 5.275^{2}}\times 100\approx 53.94

(b)

Similarly, percentage elongation will be found by dividing the change in length by the the original length. In this case, rhe original length was 54.5mm and goes to 69 mm so the change in length is given by subtracting the final length from the original length

Percentage elongation is \frac {69-54.5}{54.5}\times 100\approx 26.61

6 0
4 years ago
A parallel circuit with two branches and an 18 volt battery. Resistor #1 on the first branch has a value of 220 ohms and resisto
likoan [24]

Answer:

  2.455 W

Explanation:

The power dissipated in each branch is ...

  P = V^2/R

So, the branch powers are ...

  branch 1: 18^2/220 ≈ 1.473 W

  branch 2: 18^2/330 ≈ 0.982 W

Total power is ...

  1.473 W + 0.982 W = 2.455 W

8 0
3 years ago
Help please all of the numbers b4 the equal sign are wrong
Nostrana [21]

Answer:

  • 3/5" = 12'
  • 1 3/4" = 35'
  • 1 1/4" = 25'
  • 9/10" = 18'
  • 2 13/20" = 53'

Explanation:

One number is wrong; they all lack units.

The basic ratio is 1" = 20', so you can divide feet by 20 to find inches.

  • 3/5" = 12'
  • 1 3/4" = 35'
  • 1 1/4" = 25'
  • 9/10" = 18'
  • 2 13/20" = 53'

Perhaps you want decimal inches:

  • 0.60" = 12'
  • 1.75" = 35'
  • 1.25" = 25'
  • 0.90" = 18'
  • 2.65" = 53'
7 0
3 years ago
A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 Mpa root m is exposed to a stress of 1030 MPa. W
cupoosta [38]

Answer:

It will not  experience fracture when it is exposed to a stress of 1030 MPa.

Explanation:

Given

Klc = 54.8 MPa √m

a = 0.5 mm = 0.5*10⁻³m

Y = 1.0

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:

<em>σc = KIc / (Y*√(π*a))</em>

Thus

σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))

⇒ σc = 1382.67 MPa > 1030 MPa

Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.

3 0
3 years ago
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