Engineering Thermodynamics The condition for the reversibility of a cycle is The pressure and temperature of the working substance must not differ, appreciably, from those of the surroundings at any stage in the process The working parts of the engine must be friction free All the processes, taking place in the cycle of operation, must be extremely slow All of these The pressure and temperature of the working substance must not differ, appreciably, from those of the surroundings at any stage in the process The working parts of the engine must be friction free All the processes, taking place in the cycle of operation, must be extremely slow All of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics Which of the following process can be made reversible with the help of a regenerator? Constant volume process Constant pvn process Constant pressure process All of these Constant volume process Constant pvn process Constant pressure process All of these ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics In an isothermal process There is no change in internal energy All of the listed here There is no change in enthalpy There is no change in temperature There is no change in internal energy All of the listed here There is no change in enthalpy There is no change in temperature ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics The more effective way of increasing efficiency of Carnot engine is to Decrease lower temperature Increase lower temperature Increase higher temperature Decrease higher temperature Decrease lower temperature Increase lower temperature Increase higher temperature Decrease higher temperature ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics The increase in entropy of a system represents Decrease in pressure Degradation of energy Increase in temperature Increase in availability of energy Decrease in pressure Degradation of energy Increase in temperature Increase in availability of energy ANSWER DOWNLOAD EXAMIANS APP
Engineering Thermodynamics The polytropic index (n) is given by log [(p1v1)/(p2v2)] log (p2/ p1)/log (v1/ v2) log (p1p2)/log (v1v2) log (v1/ v2)/ log (p1/p2) log [(p1v1)/(p2v2)] log (p2/ p1)/log (v1/ v2) log (p1p2)/log (v1v2) log (v1/ v2)/ log (p1/p2) ANSWER DOWNLOAD EXAMIANS APP
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