A 5.5-kW water heater operates at 240 V. i. Should the heater circuit have a 20A or a 30A circuit breaker? [2 marks]

Question

A 5.5-kW water heater operates at 240 V.

i. Should the heater circuit have a 20A or a 30A circuit breaker? [2 marks]

ii. Assuming 85% efficiency, how long will the heater take to heat the water in

a 0.2085-m3 tank from 20°C to 80°C? The density of water is 1,000 kg/m3

and the specific heat capacity is 4,200 J/kg/°C. ​

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Orla Orla 1 year 2021-09-05T05:25:46+00:00 1 Answers 5 views 0

Answers ( )

    0
    2021-09-05T05:27:37+00:00

    Answer:

    The time taken is  [tex]t = 11.23 \ sec[/tex]  

    Explanation:

    From the question we are told

          The power the water is [tex]P = 5.5KW = 5.5 *10^ {3} W[/tex]

          The the voltage  of the heater is  is  [tex]V = 240 V[/tex]

           The volume of the heater  [tex]Z = 0.2085\ m^3[/tex]

            The specific heat of water is  [tex]c_w = 4200 J /kg/^oC[/tex]

             The initial temperature is [tex]T_1 = 20^oC[/tex]

            The final temperature is [tex]T_2 = 80^oC[/tex]

             The density of water is  [tex]\rho_w = 1000 \ kg/m^3[/tex]

    The current of the heater is mathematically represented as

                [tex]I = \frac{P}{V }[/tex]

    substituting values we have

               [tex]I = \frac{5.5 *10^{3}}{240}[/tex]

               [tex]I = 22.91 \ A[/tex]

    So since the current produced is greater than 20 A hence the heater current rating would be 30A

          The quantity of heat required to heat the water is mathematically represented as

           [tex]Q = m c_w \Delta T[/tex]

    Where m is the mass which is mathematically evaluated as

             [tex]m = \rho_w * Z[/tex]

             [tex]m =1000 * 0.2085[/tex]

             [tex]m =208.5 \ kg[/tex]

    Therefore  

                     [tex]Q = 208.5 * 4.200 * (80 – 20)[/tex]

                    [tex]Q = 52542 J[/tex]

    Since the heater has an  efficiency then the heat generate by the heater is  

              [tex]Q_h = \frac{52542}{0.85}[/tex]

              [tex]Q_h =61814.1 J[/tex]

    Now Power is mathematically  represented as

             [tex]P = \frac{Q}{t}[/tex]

    Where t is the time taken for heater to heat the water

    =>      [tex]t = \frac{Q}{P}[/tex]

    Substituting values  

              [tex]t = \frac{61814.1}{5.5*10^{3}}[/tex]

              [tex]t = 11.23 \ sec[/tex]          

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