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RoseWind [281]
3 years ago
12

1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and

300 K, respectively. The heat transfer from the air during the isothermal compression is 80 kJ. At the end of the isothermal compression, the volume is 0.2 m3. Determine the volume at the beginning of the isothermal compression, in m3. Assume the ideal gas model for air and neglect kinetic and potential energy effects.
Physics
1 answer:
liraira [26]3 years ago
3 0

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K  , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

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Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

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r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

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h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

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where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

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