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At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJ∙kg-1∙K-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJ∙kg-1∙K-1, respectively, the work output from the turbine is ________ MW (in integer).
    Correct answer is '450'. Can you explain this answer?
    Most Upvoted Answer
    At steady state, 500 kg/s of steam enters a turbine with specific enth...

    Applying steady flow energy equation,

    We know, Power

    Since, reversible heat transfer,

    By putting in equation (ii),
    P = 500 (3500 - 2500) -100]
    P = 450000 kW
    P = 450 MW
    Hence, the work output from the turbine is 450 MW.
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    Community Answer
    At steady state, 500 kg/s of steam enters a turbine with specific enth...
    In order to calculate the work output from the turbine, we need to use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added minus the work done by the system:

    ΔU = Q - W

    Given that the process is steady state, we can assume that there is no change in internal energy (ΔU = 0), and therefore the equation simplifies to:

    Q = W

    We can calculate the heat added (Q) by using the specific enthalpy and specific entropy values at the turbine inlet and outlet. The heat added can be calculated as:

    Q = mass flow rate * (specific enthalpy at outlet - specific enthalpy at inlet)

    Q = 500 kg/s * (2500 kJ/kg - 3500 kJ/kg)

    Q = -5 * 10^5 kJ/s

    Since the heat loss occurs at a temperature of 500 K, we can assume it is a reversible process and use the Carnot efficiency formula to calculate the work output. The Carnot efficiency is given by:

    η = 1 - (Tc/Th)

    Where η is the efficiency, Tc is the temperature of the cold reservoir (500 K), and Th is the temperature of the hot reservoir (specific enthalpy at inlet).

    Using the specific enthalpy at inlet (3500 kJ/kg) and the specific entropy at inlet (6.5 kJ/kg-K), we can find the temperature at the inlet using the steam tables.

    From the steam tables, we find that the temperature at the inlet is approximately 641.6 K.

    Now we can calculate the work output (W) using the Carnot efficiency formula:

    W = η * Q

    W = (1 - (Tc/Th)) * Q

    W = (1 - (500 K / 641.6 K)) * (-5 * 10^5 kJ/s)

    W = 4.5 * 10^5 kJ/s

    Converting the work output to MW:

    W = 4.5 * 10^5 kJ/s * (1 MW / 1000 kJ/s)

    W = 450 MW

    Therefore, the work output from the turbine is 450 MW.
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    At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJkg-1K-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJkg-1K-1, respectively, the work output from the turbine is ________ MW (in integer).Correct answer is '450'. Can you explain this answer?
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    At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJkg-1K-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJkg-1K-1, respectively, the work output from the turbine is ________ MW (in integer).Correct answer is '450'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJkg-1K-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJkg-1K-1, respectively, the work output from the turbine is ________ MW (in integer).Correct answer is '450'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At steady state, 500 kg/s of steam enters a turbine with specific enthalpy equal to 3500 kJ/kg and specific entropy equal to 6.5 kJkg-1K-1. It expands reversibly in the turbine to the condenser pressure. Heat loss occurs reversibly in the turbine at a temperature of 500 K. If the exit specific enthalpy and specific entropy are 2500 kJ/kg and 6.3 kJkg-1K-1, respectively, the work output from the turbine is ________ MW (in integer).Correct answer is '450'. Can you explain this answer?.
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