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Pie
4 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]4 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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3 years ago
Write equations used to calculate the diode reverse saturation current, the voltage at which diode goes into resistive behavior,
laiz [17]

Answer:

Diode equation for reverse saturation current

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

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Voltage at which avalanche multiplication occurs:V=5volts

Explanation:

we take here forward and reverse 0.7 volt and -5 volt  

As Diode current equation is express as

I_D = I_o \times (e^{\frac{V_D}{\eta V_T}} -1 )   ....................1

here I_D is total current through the diode and I_o is reverse saturated current and V_D is  voltage drop across diode and \eta is idealized factor and V_T is thermal voltage

so here we know that when Bios is forward than

V_D  = V_T     .................2

ans Bios is Reverse than  

V_D  = V_R      ..................3

so here

1.  diode reverse saturation current is express as

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

and

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6 0
3 years ago
A freezer is maintained at 20°F by removing heat from it at a rate of 75 Btu/min. The power input to the freezer is 0.70 hp, and
Igoryamba

Answer:

Explanation:

Cop of reversible refrigerator = TL / ( TH - TL)

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TH = temperature of air around = 75 °F

Heat removal rate QL = 75 Btu/min

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conversion on F to kelvin = (T (°F) + 460 ) × 5 / 9

COP ( coefficient of performance) reversible = (20 + 460) × 5/9 / (5/9 ( ( 75 +460) - (20 + 460) ))

COP reversible = 480 / 55 = 8.73

irreversibility expression, I = W actual - W rev

COP r = QL / Wrev

W rev = QL /  COP r  where 75 Btu/min = 1.76856651 hp  where W actual = 0.70 hp

a) W rev =  1.76856651 hp  /  8.73  = 0.20258 hp is reversible power

I = W actual - W rev

b) I = 0.7 hp - 0.20258 hp = 0.4974 hp

c) the second-law efficiency of this freezer = W rev / W actual =  0.20258 hp / 0.7 hp = 0.2894 × 100 = 28.94 %

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4 years ago
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Answer:

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Explanation:

as long as the engine and evrything is running it should be good

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