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Note: this is a multi-part question. once an answer is submitted, you will be unable to return to this part. a 0.4-m3 rigid tank contains refrigerant-134a initially at 160 kpa and 40 percent quality. heat is now transferred to the refrigerant until the pressure reaches 600 kpa. note: this is a multi-part question. once an answer is submitted, you will be unable to return to this part. determine the amount of heat transferred. (please provide an answer before moving on to the next part.)

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Based on the calculations, the amount of heat transferred is equal to 2,317 kJ.

Given the following data:

Volume of tank = 0.4 m³.

Initial pressure = 160 kPa.

Final pressure = 600 kPa.

Initial quality, x = 40% = 0.4

How to calculate the amount of heat transferred?

From the superheat table A-12, the specific volumes of the saturated liquid and vapor at 160 kPa are:

  • Vf(160 kPa) = 0.0007437 m³/kg.
  • Vg(160 kPa) = 0.12348 m³/kg.

Next, we would determine the initial specific volume by using this formula:

V₁ = (1 - x)Vf + xVg

V₁ = (1 - 0.4)0.0007437 + (0.4)0.12348

V₁ = 0.049838 m³/kg.

Also, we would find the mass of this refrigerant at the initial state by using this equation:

m = V/V₁

m = 0.4/0.049838

Mass, m = 8.03 kg.

From the superheat table A-12, the internal energy of the saturated liquid and vapor at 160 kPa are:

  • Uf(160 kPa) = 31.09 kJ/kg.
  • Ug(160 kPa) = 221.35 kJ/kg.

At the initial state, we would find U₁ from the quality:

U₁ = (1 - x)Uf + xUg

U₁ = (1 - 0.4)31.09 + (0.4)221.35

U₁ = 107.19 kJ/kg.

From the superheat table A-13, the final specific volume and internal energy of the saturated liquid and vapor at 600 kPa are:

  • Vf = 0.057006 m³/kg.
  • Uf = 367.81 kJ/kg.

At the final state, we would find U₂ by extrapolating with the last two points in the table:

[tex]U_2 = 357.96 + (\frac{367.81 \; - \; 357.96}{0.057006\;-\;0.055522}) \times (0.049838\;-\;0.055522)\\\\U_2 = 357.96 + (\frac{9.85}{0.001484}) \times (-0.005684)\\\\U_2 = 357.96 + 37.73\\\\[/tex]

U₂ = 395.69 kJ/kg.

Now, we can calculate the amount of heat transferred:

Q = m(U₂ - U₁)

Q = 8.03(395.69 - 107.19)

Q = 8.03(288.5)

Q = 2,316.66 ≈ 2,317 kJ.

Read more on heat transfer here: https://brainly.com/question/14814826

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