In case of a heat pump, second law efficiency is given as

Related MCQs

• Both first law efficiency and second law efficiency indicate how effectively the input has been converted into the product.
• A heat engine is supplied with 250 kJ/s of heat at a constant fixed temperature of 227°C. The heat is rejected at 27°C. The cycle is reversible, if the amount of heat rejected is
• Consider the following statements: The definition of 1. temperature is due to zeroth law of thermodynamics 2. entropy is due to first law of thermodynamics 3. internal energy is due to second law of thermodynamics 4. reversibility is due to Kelvin-Planck's ststement of these statements
• Second law efficiency can also be given as
• For a power plant, second law efficiency can be given by desired output(W) / available energy(Wmax).
• A Rankine cycle has efficiency of 19.4%. The steam enters the turbine with enthalpy of 32.85 kJ/Kg while heat of water entering the boiler is 423.6 kJ/Kg.the work of the turbine is 560 kJ/ Kg. the pump work is
• A steam power plant working on Rankine cycle produces steam in the boiler at a rate of 100 kg/s with enthalpy of 3280 kJ/kg from water whose total heat content is 420 kJ/kg. The plant has Rankine efficiency of 24.65 while pump requires 5 kJ/kg of work. The turbine generates power of
• Correlate the following statements i) the volume of water into which steam is condensed in condenser is very much smaller than the volume of steam before condensation ii) the net work of a Rankine cycle is heat supplied in boiler - heat rejected to condenser - pump work
• A simple Rankine cycle produces 40 MW of power, 50 MW of process heat and rejects 60 MW of heat to the surroundings. What is the utilization factor of this co generation cycle neglecting the pump work?
• a heat engine receives heat at the rate of 1500 kJ/min.& gives an output of 8.2 kW. Its thermal efficiency is equal to