Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it is observed

Question

Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it is observed that half of the water in the cooker evaporated in 30 minutes. Assume negligible heat loss from the cooker.

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Calantha 4 years 2021-09-05T02:24:10+00:00 1 Answers 9 views 0

Answers ( )

  1. nguyetanh
    0
    2021-09-05T02:26:05+00:00
    Reply

    Answer:

    The average rate of energy transfer to the cooker is 1.80 kW.

    Explanation:

    Given that,

    Pressure of boiled water = 300 kPa

    Mass of water = 3 kg

    Time = 30 min

    Dryness friction of water = 0.5

    Suppose, what is the average rate of energy transfer to the cooker?

    We know that,

    The specific enthalpy of evaporate at 300 kPa pressure

    h_{f}=561.47\ kJ/kg

    h_{fg}=2163.8\ kJ/kg

    We need to calculate the enthalpy of water at initial state

    h_{1}=h_{f}

    h_{1}=561.47\ kJ/kg

    We need to calculate the enthalpy of water at final state

    Using formula of enthalpy

    h_{2}=h_{f}+xh_{fg}

    Put the value into the formula

    h_{2}=561.47+0.5\times2163.8

    h_{2}=1643.37\ kJ/kg

    We need to calculate the rate of energy transfer to the cooker

    Using formula of rate of energy

    Q=\dfrac{m(h_{2}-h_{1})}{t}

    Put the value into the formula

    Q=\dfrac{3\times(1643.37-561.47)}{30\times60}

    Q=1.80\ kW

    Hence, The average rate of energy transfer to the cooker is 1.80 kW.

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