Homework Help is Here – Start Your Trial Now!
Science
Advanced Physics
R-134a enters a compressor operating at steady state at -4°C. At this state, 7% of the mass is in the liquid phase. R-134a then exits at a pressure of 1600 kPa. If the actual exit temperature is 70°C, determine the (a) isentropic compressor efficiency in %, (b) the work input, in kJ/kg of refrigerant flowing and the (c) change in the entropy of the refrigerant, in kJ/kg-K. Heat transfer between the compressor and its surroundings as well as the kinetic and potential energy effects can be ignored. Answers: Energy Balance Equation: Entropy Balance Equation: DO NOT insert extra space in between variable, e.g. Q-W+minhin- mouthout=mfhf-mihi. Use i and f to denote the initial and final states, in and out for the entry and exit points. State at the Inlet: State at the Outlet (isentropic): State at the Outlet (actual): Isentropic efficiency = Win kJ/kg %3D As = kJ/kg-K

R-134a enters a compressor operating at steady state at -4°C. At this state, 7% of the mass is in the liquid phase. R-134a then exits at a pressure of 1600 kPa. If the actual exit temperature is 70°C, determine the (a) isentropic compressor efficiency in %, (b) the work input, in kJ/kg of refrigerant flowing and the (c) change in the entropy of the refrigerant, in kJ/kg-K. Heat transfer between the compressor and its surroundings as well as the kinetic and potential energy effects can be ignored. Answers: Energy Balance Equation: Entropy Balance Equation: DO NOT insert extra space in between variable, e.g. Q-W+minhin- mouthout=mfhf-mihi. Use i and f to denote the initial and final states, in and out for the entry and exit points. State at the Inlet: State at the Outlet (isentropic): State at the Outlet (actual): Isentropic efficiency = Win kJ/kg %3D As = kJ/kg-K

icon
Related questions
Question
PLEASE ANSWER IT ASAP FOR AN UPVOTE.
R-134a enters a compressor operating at steady state at -4°C. At this state, 7% of the mass is in the liquid phase. R-134a then exits at a pressure of 1600 kPa. If the actual exit temperature is 70°C, determine the (a) isentropic compressor efficiency in %, (b) the work input, in kJ/kg of refrigerant flowing and the (c) change in the entropy of the refrigerant, in kJ/kg-K. Heat transfer between the compressor and its surroundings as well as the kinetic and potential energy effects can be ignored. Answers: Energy Balance Equation: Entropy Balance Equation: DO NOT insert extra space in between variable, e.g. Q-W+minhin- mouthout=mfhf-mihi. Use i and f to denote the initial and final states, in and out for the entry and exit points. State at the Inlet: State at the Outlet (isentropic): State at the Outlet (actual): Isentropic efficiency = Win kJ/kg %3D As = kJ/kg-K
Transcribed Image Text:R-134a enters a compressor operating at steady state at -4°C. At this state, 7% of the mass is in the liquid phase. R-134a then exits at a pressure of 1600 kPa. If the actual exit temperature is 70°C, determine the (a) isentropic compressor efficiency in %, (b) the work input, in kJ/kg of refrigerant flowing and the (c) change in the entropy of the refrigerant, in kJ/kg-K. Heat transfer between the compressor and its surroundings as well as the kinetic and potential energy effects can be ignored. Answers: Energy Balance Equation: Entropy Balance Equation: DO NOT insert extra space in between variable, e.g. Q-W+minhin- mouthout=mfhf-mihi. Use i and f to denote the initial and final states, in and out for the entry and exit points. State at the Inlet: State at the Outlet (isentropic): State at the Outlet (actual): Isentropic efficiency = Win kJ/kg %3D As = kJ/kg-K
Expert Solution
steps

Step by step

Solved in 5 steps

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions