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Air is compress from 100 kPa and 50°C to 2000 kPa and 500°C. Assuming constant specific heats, determine the change in the specific entropy of air.Group of answer choices0.0478 kJ/kg.K1.767 kJ/kg.K-1.767 kJ/kg.K-0.0478 kJ/kg.K

Question

Air is compress from 100 kPa and 50°C to 2000 kPa and 500°C. Assuming constant specific heats, determine the change in the specific entropy of air.Group of answer choices0.0478 kJ/kg.K1.767 kJ/kg.K-1.767 kJ/kg.K-0.0478 kJ/kg.K

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Solution 1

To solve this problem, we need to use the formula for change in entropy for an ideal gas with constant specific heats:

Δs = cp * ln(T2/T1) - R * ln(P2/P1)

where:

  • cp is the specific heat at constant pressure,
  • R is the specific gas constant,
  • T1 and T2 are the initial and final temperatures in Kelvin,
  • P1 and P2 are the initial and final pressures.

For air, cp = 1.005 kJ/kg.K and R = 0.287 kJ/kg.K.

First, convert the temperatures from Celsius to Kelvin: T1 = 50°C + 273.15 = 323.15 K, T2 = 500°C + 273.15 = 773.15 K.

Then, substitute these values into the formula:

Δs = 1.005 * ln(773.15/323.15) - 0.287 * ln(2000/100)

Calculate the natural logarithms and the products:

Δs = 1.005 * 0.857 - 0.287 * 2.303 = 0.862 - 0.661 = 0.201 kJ/kg.K

So, the change in the specific entropy of air is 0.201 kJ/kg.K. This option is not in the given answer choices, so there might be a mistake in the problem or the answer choices.

This problem has been solved

Solution 2

To solve this problem, we need to use the formula for change in entropy for an ideal gas with constant specific heats:

Δs = cp * ln(T2/T1) - R * ln(P2/P1)

where:

  • cp is the specific heat at constant pressure,
  • R is the specific gas constant,
  • T1 and T2 are the initial and final temperatures in Kelvin,
  • P1 and P2 are the initial and final pressures.

For air, cp = 1.005 kJ/kg.K and R = 0.287 kJ/kg.K.

First, convert the temperatures from Celsius to Kelvin: T1 = 50°C + 273.15 = 323.15 K, T2 = 500°C + 273.15 = 773.15 K.

Then, substitute these values into the formula:

Δs = 1.005 * ln(773.15/323.15) - 0.287 * ln(2000/100)

Calculate the natural logarithms and the products:

Δs = 1.005 * 0.857 - 0.287 * 2.303 = 0.862 - 0.661 = 0.201 kJ/kg.K

So, the change in the specific entropy of air is 0.201 kJ/kg.K. This option is not in the given answer choices, so there might be a mistake in the problem or the answer choices.

This problem has been solved

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