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Pie
3 years ago
12

An air conditioner using refrigerant R-134a as the working fluid and operating on the ideal vapor-compression refrigeration cycl

e is to maintain a space at 22°C while operating its condenser at 1000 kPa.
a. Determine the COP of the system when a temperature difference of 2°C is allowed for the transfer of heat in the evaporator.
Engineering
1 answer:
Oduvanchick [21]3 years ago
6 0

Answer:

COP=13.37

Explanation:

Considering the allowed temperature difference the temperature in the evaporator will be 20°. The enthalpy and entropy at this temperature are obtained from table for the saturated vapor phase:

h_1=261.64 kJ/kg

s_1=0.92254 kJ/kgK

The enthalpy at state 2 is determined from the given condenser pressure and the condition s_2 = s_1 with data from table using interpolation:  

h_2=273.18 kJ/kg

The enthalpy at states 3 and 4 is determined from the saturated liquid value for the condenser pressure from table:  

h_3=h_4=107.34 kJ/kg

the COP is then:

COP=qL/w

       = h_1- h_4/ h_2- h_1

       =13.37

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Explanation:

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6 0
2 years ago
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
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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)

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

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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

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

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