A industrial (large) pressure cooker operates at 275 kPa. Initially there is 10 kg of water at 20°C, the cooker is operated until 1 kg of s

Question

A industrial (large) pressure cooker operates at 275 kPa. Initially there is 10 kg of water at 20°C, the cooker is operated until 1 kg of saturated vapor is remaining in the cooker with the remaining mass leaving as vapor. The cooker is not insulated and10 MJ of heat is lost to the surroundings. How much heat was transferred to the cooker (use information below)?

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RobertKer 7 months 2021-07-29T20:13:41+00:00 1 Answers 3 views 0

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    2021-07-29T20:15:37+00:00

    Answer:

    Q_{in} = 25349.92\,kJ

    Explanation:

    Let establish a control volume in the industrial pressure cooker, which is a transient state system. From the First Law of Thermodynamics, the heating process is modelled:

    Q_{in} + m_{1}\cdot h_{1} - m_{2}\cdot h_{2} = (m_{1}-m_{2})\cdot u_{2} - m_{1}\cdot u_{1}

    The heat transfered to the cooker is:

    Q_{in} = m_{2}\cdot h_{2} - m_{1}\cdot h_{1} + (m_{1}-m_{2})\cdot u_{2}-m_{1}\cdot u_{1}

    Properties at each state are described below:

    State 1

    u_{1} = 83.913\,\frac{kJ}{kg}

    h_{1} = 83.915\,\frac{kJ}{kg}

    State 2

    u_{2} = 2540.1\,\frac{kJ}{kg}

    h_{2} = 2720.9\,\frac{kJ}{kg}

    The heat transfered to the cooker is:

    Q_{in} = (9\,kg)\cdot (2720.9\,\frac{kJ}{kg} ) - (10\,kg)\cdot (83.915\,\frac{kJ}{kg} ) + (1\,kg) \cdot (2540.1\,\frac{kJ}{kg} )-(10\,kg)\cdot (83.913\,\frac{kJ}{kg} )

    Q_{in} = 25349.92\,kJ

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