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Exercise collection for the exam preparation

Materiale completo di Heating and Cooling Systems - Produzione di Energia Termica e Frigorifera per il corso di Energy Engineering presso Politecnico di Milano. Il documento tratta: HEATING AND COOLING SYSTEMS 1st and 2nd year M.Sc. Energy Engineering Prof. L. Molinaroli ADDITIONAL EXERCISES Notice: All the missing data have to be assumed according to thermodynamic and/or engineering considerations. The thermodynamic consistency of the solution found should…

Heating and Cooling Systems - Produzione di Energia Termica e FrigoriferaCompleti

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Materiale completo di Heating and Cooling Systems - Produzione di Energia Termica e Frigorifera per il corso di Energy Engineering presso Politecnico di Milano. Il documento tratta: HEATING AND COOLING SYSTEMS 1st and 2nd year M.Sc. Energy Engineering Prof. L. Molinaroli ADDITIONAL EXERCISES Notice: All the missing data have to be assumed according to thermodynamic and/or engineering considerations. The thermodynamic consistency of the solution found should…

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Pagina 1

HEATING AND COOLING SYSTEMS 1st and 2nd year M.Sc. Energy Engineering Prof. L. Molinaroli ADDITIONAL EXERCISES Notice: All the missing data have to be assumed according to thermodynamic and/or engineering considerations. The thermodynamic consistency of the solution found should be checked (e.g. 1st law analysis at system level, no crossing of temperature profiles in the heat exchang- ers etc…). The properties of the refrigerants can be computed using Coolprop software. Vapour compression systems 1. Do the thermodynamic design of a vapour compression air-to-water heat pump that has to heat a water flow rate equal to ṁ = 5 kg/s from tWATER,IN = 30 °C to tWATER,OUT = 35 °C knowing that an air flow rate equal to v̇ = 40000 m3/h whose temperature is equal to tAIR,IN = -5 °C is used as secondary fluid at the evaporator. Consider a simple configuration system working with R32 and use the following equation to compute the compressor isentropic efficiency: ηISO = 0,4482 + 0,0637·β - 0,0071·β2 (β is the pressure ratio). 2. Do the thermodynamic design of a vapour compression water-to-water chiller that has to cool a water flow rate equal to ṁ = 25 kg/s from tWATER,IN = 23 °C to tWATER,OUT = 18 °C knowing that a water flow rate equal to ṁ = 30 kg/s, whose temperature is equal to tWATER,IN = 30 °C, is used as secondary fluid at condenser. Consider a simple configuration system working with R32 and use the following equation to compute the compressor isentropic efficiency: ηISO = 0,4482 + 0,0637·β - 0,0071·β2 (β is the pressure ratio). 3. Solve previous exercises considering advanced configurations such as LL-SLHX (εLL-SLHX = 0,5), two stages and, for the heat pump only, cascade with R32 in the bottom system and R1234yf in the top one. 4. Solve previous exercises considering a…

Pagina 7

R32 8000 prof. Luca Molinaroli 7000 Dipartimento di Energia - Politecnico di Milano 80 °C 90 °C 100 °C 110 °C 120 °C 140 °C 160 °C Reference state: h = 200 kJ/kg e s = 1 kJ/(kg·K) for saturated liquid at T = 0 °C 6000 Data generated with Refprop 10.0 5000 70 °C 4000 60 °C 50 °C 3000 40 °C Pressure [kPa] 2000 30 °C 20 °C 10 °C 1000 900 800 0 °C 700 600 -10 °C 500 400 -20 °C 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 300 140 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 480 500 520 540 560 580 600 620 640 Enthalpy [kJ/kg]

Pagina 13

R1234yf 5000 prof. Luca Molinaroli Dipartimento di Energia - Politecnico di Milano 4000 Reference state: h = 200 kJ/kg e s = 1 kJ/(kg·K) 100 °C 110 °C 120 °C 130 °C for saturated liquid at T = 0 °C Data generated with Refprop 10.0 90 °C 3000 80 °C 70 °C 2000 60 °C 50 °C Pressure [kPa] 1000 40 °C 900 800 30 °C 700 600 20 °C 500 10 °C 400 0 °C 300 -10 °C 200 -20 °C 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 100 160 180 200 220 240 260 280 300 320 340 360 380 400 420 440 460 Enthalpy [kJ/kg]

Pagina 19

R290 7000 prof. Luca Molinaroli 6000 Dipartimento di Energia - Politecnico di Milano 100 °C 110 °C 120 °C 130 °C 140 °C 150 °C Reference state: h = 200 kJ/kg e s = 1 kJ/(kg·K) 5000 for saturated liquid at T = 0 °C Data generated with Refprop 10.0 4000 90 °C 80 °C 3000 70 °C 60 °C 2000 50 °C 40 °C Pressure [kPa] 30 °C 1000 900 20 °C 800 700 10 °C 600 500 0 °C 400 -10 °C 300 -20 °C 200 -30 °C 0.10 0.20 0.30 0.40 -40 °C 0.50 0.60 0.70 0.80 0.90 100 80 120 160 200 240 280 320 360 400 440 480 520 560 600 640 680 720 760 800 Enthalpy [kJ/kg]

Pagina 25

R744 15000 prof. Luca Molinaroli 14000 Dipartimento di Energia - Politecnico di Milano 40 °C 50 °C 60 °C 70 °C 80 °C 90 °C 100 °C 120 °C 140 °C 160 °C 13000 Reference state: h = 200 kJ/kg e s = 1 kJ/(kg·K) 12000 for saturated liquid at T = 0 °C Data generated with Refprop 10.0 11000 10000 9000 8000 30 °C 7000 6000 20 °C Pressure [kPa] 5000 10 °C 4000 0 °C 3000 -10 °C 2000 -20 °C -30 °C 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1000 100 150 200 250 300 350 400 450 500 550 600 Enthalpy [kJ/kg]

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Prima pagina: Exercise collection for the exam preparation