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Svetradugi [14.3K]
3 years ago
13

Air within a piston cylinder assembly executes a Carnot refrigeration cycle between hot and cold reservoirs at TH=600 K and TC=3

00 K, respectively. The magnitude of the heat transfer rejected to the high temperature reservoir is 250 kJ per kg of air. The pressure at the start of the isothermal expansion is 325 kPa. The air can be modeled as an ideal gas with constant specific heat (k=1.4). For the air as a system, determine
a. (5) the coefficient of performance.
b. (5) the net work input to the cycle in kJ/kg.
c. (10) the Pv and Ts diagrams for the cycle with arrows showing the directions of the processes.
You may answer parts (a) and (b) in the response box. Part (c) and your calculations supporting.(a) and (b) must be included in your uploaded response file.

Engineering
1 answer:
Nataly [62]3 years ago
5 0

Answer:

See explaination

Explanation:

for a reverse carnot cycle T-S diagram is a rectangle which i have shown

net work for a complete cycle must be equal to net heat interaction.

Kindly check attachment for the step by step solution of the given problem.

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A steady stream (1000 kg/hr) of air flows through a compressor, entering at (300 K, 0.1 MPa) and leaving at (425 K, 1 MPa). The
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Answer:

The work furnished by the compressor is 69.77kJ/s

The minimum work required for the state to change is 55.26kW

Explanation:

The explanation to these solution is on the first, second , third and fourth uploaded image respectively

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Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

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If the surface energy of a magnesium oxide - nickel oxide (MgO-NiO) solid solution is 1.05 J/m2 and its elastic modulus is 198 G
Nastasia [14]

Answer:

The maximum length is 3.897×10^-5 mm

Explanation:

Extension = surface energy/elastic modulus

surface energy = 1.05 J/m^2

elastic modulus = 198 GPa = 198×10^9 Pa

Extension = 1.05/198×10^9 = 5.3×10^-12 m

Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4

Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm

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