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. ​

in progress 0
Orla Orla 4 years 2021-09-05T05:25:46+00:00 1 Answers 19 views 0

Answers ( )

  1. Eirian
    0
    2021-09-05T05:27:37+00:00
    Reply

    Answer:

    The time taken is  t = 11.23 \ sec  

    Explanation:

    From the question we are told

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

          The the voltage  of the heater is  is  V = 240 V

           The volume of the heater  Z = 0.2085\ m^3

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

             The initial temperature is T_1 = 20^oC

            The final temperature is T_2 = 80^oC

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

    The current of the heater is mathematically represented as

                I = \frac{P}{V }

    substituting values we have

               I = \frac{5.5 *10^{3}}{240}

               I = 22.91 \ A

    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

           Q = m c_w \Delta T

    Where m is the mass which is mathematically evaluated as

             m = \rho_w * Z

             m =1000 * 0.2085

             m =208.5 \ kg

    Therefore  

                     Q = 208.5 * 4.200 * (80 - 20)

                    Q = 52542 J

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

              Q_h = \frac{52542}{0.85}

              Q_h =61814.1 J

    Now Power is mathematically  represented as

             P = \frac{Q}{t}

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

    =>      t = \frac{Q}{P}

    Substituting values  

              t = \frac{61814.1}{5.5*10^{3}}

              t = 11.23 \ sec          

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