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Travka [436]
3 years ago
9

We have considered heat and work to be path-dependent. However, if all heat transfer with surroundings is performed using a reve

rsible heat transfer device (some type of reversible Carnot-type device), work can be performed by the heat transfer device during heat transfer to the surroundings. The net heat transferred to the surroundings and the net work done will be independent of the path. Demonstrate this by calculating the work and heat interactions for the system, the heat transfer device, and the sum for each of the following paths where the surroundings are at Tsurr = 273 K. The state change is from state 1, P1 = 0.1 MPa, T1 = 298 K and state 2, P2 = 0.5 MPa and T2 which will be found in part (a). CP = 7R/2. a. Consider a state change for an ideal gas in a piston/cylinder. Find T2 by an adiabatic reversible path. Find the heat and work such that no entropy is generated in the universe. This is path a. Sketch path a qualitatively on a P-V diagram. b. Now consider a path consisting of step b, an isothermal step at T1, and step c, an isobaric step at P2. Sketch and label the step on the same P-V diagram created in (a). To avoid generation of entropy in the universe, use heat engines/pumps to transfer heat during the steps. Calculate the WEC and WS as well as the heat transfer with the surroundings for each of the steps and overall. Compare to part (a) the total heat and work interactions with the surroundings. c. Now consider a path consisting of step d, an isobaric step at P1, and step e, an isothermal step at T2. Calculate the WEC and WS as well as the heat transfer with the surroundings for each of the steps and overall. Compare to part (a) using this pathway and provide the same documentation as in (b).
Engineering
1 answer:
NeX [460]3 years ago
3 0

Answer:

path-dependent. However, if all heat transfer with surroundings is performed using a reversible heat transfer device (some type of reversible Carnot-type device), work can be performed by the heat transfer device during heat transfer to the surroundings. The net heat transferred to the surroundings and the net work done will be independent of the path. Demonstrate this by calculating the work and heat interactions for the system, the heat transfer device, and the sum for each of the following paths where the surroundings are at Tsurr = 273 K. The state change is from state 1, P1 = 0.1 MPa, T1 = 298 K and state 2, P2 = 0.5 MPa and T2 which will be found in part (a). CP = 7R/2. a. Consider a state change for an ideal gas in a piston/cylinder. Find T2 by an adiabatic reversible path. Find the heat and work such that no entropy is generated in the universe. This is path a. Sketch path a qualitatively on a P-V diagram. b. Now consider a path

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If welding is being done in the vertical position, the torch should have a travel angle of?
siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
2 years ago
What is the simplified meaning of the word "engineering"​
juin [17]

Answer:

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Explanation:

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3 years ago
The major resisting force in gravity dam? ​
inna [77]

The self weight of dam.

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3 years ago
Read 2 more answers
.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves
vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 m^3/kg

Specific volume of saturated liquid=v_f=0.00102 m^3/kg

Now the relation  of total specific volume for a specific dryness value is given as

                                  v=v_f+x(v_g-v_f)

Substituting the values give

v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg

Now the volume of vessel is given as

v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3

So the volume of vessel is 4.7680m^3.

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680m^3 and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ

So the quality of the mixture is 90.3%  or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0
3 years ago
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