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A Carnot engine receives 250 kJ·s−1 of heat from a heat-source reservoir at 525°C and rejects heat to a heat-sink reservoir at 50°C. What are the power developed and the heat rejected?

Answer :

Hashirriaz830

Answer:

W = -148.8 kJ/s

Qc= -101.2 kJ/s

Explanation:

note:

Solution is attached in word form due to error in mathematical equation.  please find the attachment

${teks-lihat-gambar} Hashirriaz830
Lanuel

a. The quantity of heat rejected by the Carnot engine is equal to -101.2 kJ/s.

b. The power developed by the Carnot engine is equal to 148.8 kJ/s.

Given the following data:

  • Quantity of heat received = 250 kJ/s
  • Temperature of heat-source = 525°C
  • Temperature of heat rejected = 50°C

Conversion:

Temperature of heat-source = 525°C to Kelvin = 525 + 273 = 798K

Temperature of heat rejected = 50°C to Kelvin = 50 + 273 = 323K

To find the power developed and the heat rejected, we would use Carnot's equation:

[tex]-\frac{Q_R}{T_R} = \frac{Q_S}{T_S}[/tex]

Where:

  • [tex]Q_R[/tex] is the quantity of heat rejected.
  • [tex]T_R[/tex] is the heat-sink temperature.
  • [tex]T_S[/tex] is the heat-source temperature.
  • [tex]Q_S[/tex] is the quantity of heat received.

Making [tex]Q_R[/tex] the subject of formula, we have:

[tex]Q_R = -( \frac{Q_S}{T_S})T_R[/tex]

Substituting the given parameters into the formula, we have;

[tex]Q_R = -( \frac{250}{798}) \times 323\\\\Q_R = -( \frac{80750}{798})[/tex]

Quantity of heat rejected = -101.2 kJ/s

Now, we can determine the power developed by the Carnot engine:

[tex]P = -Q_S - Q_R\\\\P = -250 - (-101.2)\\\\P = -250 + 101.2[/tex]

Power, P = 148.8 kJ/s.

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