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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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A furnace wall composed of 200 mm, of fire brick. 120 mm common brick 50mm 80% magnesia and 3mm of steel plate on the outside. I
Liula [17]

Answer:

  • fire brick / common brick : 1218 °C
  • common brick / magnesia : 1019 °C
  • magnesia / steel : 90.06 °C
  • heat loss: 4644 kJ/m^2/h

Explanation:

The thermal resistance (R) of a layer of thickness d given in °C·m²·h/kJ is ...

  R = d/k

so the thermal resistances of the layers of furnace wall are ...

  R₁ = 0.200/4 = 0.05 °C·m²·h/kJ

  R₂ = 0.120 2.8 = 3/70 °C·m²·h/kJ

  R₃ = 0.05/0.25 = 0.2 °C·m²·h/kJ

  R₄ = 0.003/240 = 1.25×10⁻⁵ °C·m²·h/kJ

So, the total thermal resistance is ...

  R₁ +R₂ +R₃ +R₄ = R ≈ 0.29286 °C·m²·h/kJ

__

The rate of heat loss is ΔT/R = (1450 -90)/0.29286 = 4643.70 kJ/(m²·h)

__

The temperature drops across the various layers will be found by multiplying this heat rate by the thermal resistance for the layer:

  fire brick: (4543.79 kJ/(m²·h))(0.05 °C·m²·h/kJ) = 232 °C

so, the fire brick interface temperature at the common brick is ...

  1450 -232 = 1218 °C

For the next layers, the interface temperatures are ...

  common brick to magnesia = 1218 °C - (3/70)(4643.7) = 1019 °C

  magnesia to steel = 1019 °C -0.2(4643.7) = 90.06 °C

_____

<em>Comment on temperatures</em>

Most temperatures are rounded to the nearest degree. We wanted to show the small temperature drop across the steel plate, so we showed the inside boundary temperature to enough digits to give the idea of the magnitude of that.

5 0
2 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
6. During some actual expansion and compression processes in piston–cylinder devices, the gases have been
Katyanochek1 [597]

During some actual expansion and compression processes in piston-cylinder devices, the gases have been are the P1= P2.

<h3>What is the pressure?</h3>

Pressure is something that has the pressure that is physical and that causes the pressure is piston-cylinder devices.

During a few real enlargements and compression procedures in piston-cylinder devices, the gases were located to meet the connection PV n = C, wherein n and C are constants.

Read more about the pressure :

brainly.com/question/25736513

#SPJ1

5 0
1 year ago
The local atmospheric pressure is measured with a water barometer. If the water column is measured to be 30 ft, what is the atmo
ella [17]

Answer:

The atmospheric pressure in atm=0.885 atm

Explanation:

Given that

Local pressure (h)= 30 ft of water height     ( 1 ft= 0.3048 m)

We know that pressure in given by

  P=ρgh

We know that ρ of water is 1000\dfrac{kg}{m^3}

So pressure

P=1000(9.81)(9.144)

P=89.7026 \dfrac{N}{m^2}

We know that 1000 Pa=0.00986 atm

So P=0.885 atm

The atmospheric pressure in atm=0.885 atm

8 0
2 years ago
During January, at a location in Alaska winds at -20°C can be observed, However, several meters below ground the temperature rem
maw [93]

Answer:

a) \eta_{th} = 10.910\,\%, b) Yes.

Explanation:

a) The maximum thermal efficiency is given by the Carnot's Cycle, whose formula is:

\eta_{th} =\left(1-\frac{253.15\,K}{284.15\,K}  \right) \times 100\,\%

\eta_{th} = 10.910\,\%

b) The claim of the inventor is possible since real efficiency is lower than maximum thermal efficiency.

4 0
3 years ago
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