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fenix001 [56]
4 years ago
9

An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFar

ad, and R = 89.4 Ohms, find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, and (c) the charge on the capacitor at a time equal to one time constant after the battery is connected
Engineering
1 answer:
Musya8 [376]4 years ago
6 0

Answer:

a) RC = 1.03 mseg.

b) Qmax = CV = 85.2 μC

c) Q = 53.9 μC

Explanation:

a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.

This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.

In this case , it can be calculated as follows:

ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.

b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:

Q = C V = 11.5 μF . 7.41 V = 85.2 μC

c) During the charging process, the charge increases following this equation:

Q = CV (1 - e⁻t/RC)

When t = RC, the expression for Q is as follows:

Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC

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b) QL = 5.85 KW

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Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

              P1 = 60 KPa, Tsat = -36.5°C  

              T1 = -34°C , h1 = hg = 230.03 KJ/kg

              Qin = 450 W - surrounding heat  

- Condenser Inlet, ( Super-heated R-134a vapor ):

              P2 = 1.2 MPa , Tsat = 46.32°C  

              T2 = 65°C   , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

             P3 = P2 = 1.2 MPa , Tsat = 46.32°C

             T3 = 42°C   , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

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             P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

             h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

             x4 = ( h4 - hf ) / hfg

             x4 = ( 111.23 - 3.9 ) / 223.9

             x4 = 0.4795      Answer ( a )        

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

             Ti = 18°C , h = hf = 75.47 KJ/kg  

- Condenser Outlet, ( Saturated liquid water ):

             To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

             Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

             flow ( mr ) = Qh / [ h2 - h3 ]

             flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

             flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

           

            Win = flow ( mr )*[ h2 - h1 ] - Qin

            Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

            Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

            QL = Qh - Win

            QL = 8.3675 - 2.513

            QL= 5.85 KW  .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

           

             COP = QL / Win

             COP = 5.85 / 2.513

             COP =  2.33      Answer ( c )            

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

           

             COP_max = [ TL ] / [ Th - TL ]

Where,

            TL : The absolute temperature of heat sink, refrigerated space

            TH : The absolute temperature of heat source, water inlet

                 

            COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]          

            COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

     

           COP_max = QL_max / Win

           QL_max = Win*COP_max

           QL_max = 2.513*5.063

           QL_max = 12.72 KW     Answer ( d )

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