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Värme- och strömningsteknik Thermal and flow engineering Refrigeration Kylteknik Ron Zevenhoven Exam 24-2-2017 4 questions, max.

An ideal vapour-compression cycle heat pump with R-134a as the working fluid (refrigerant) is used to achieve an indoor temperature of 21.3ºC while the outdoor temperature is -6ºC.

1 Värme- och strömningsteknik Thermal and flow engineering Refrigeration Kylteknik Ron Zevenhoven Exam questions, max. points = = 30 All support material is allowed except for telecommunication devices. Last exam / sista tent An ideal vapour-compression cycle heat pump with R-134a as the working fluid (refrigerant) is used to achieve an indoor temperature of 21.3ºC while the outdoor temperature is -6ºC. For the condenser and the evaporator the temperature difference ΔT between the saturation temperature of the working fluid in each of the heat exchanger and the outside and inside air flowing through the heat exchangers is ΔT=10ºC. Assume that saturated vapour leaves the evaporator while saturated liquid leaves the condenser. See the Figure below. a. Determine the pressures in the high pressure and low pressure parts of the cycle (MPa). (1p.) b. Determine the mass flow rate ṁ of the working fluid (kg/s) and compressor power (kw). (2p.) c. Determine the COP of the heat pump. (1p.) d. If during summer the unit is used as for cooling the indoor space, determine the cooling capacity (i.e. the heat removed from the indoor space) in kw. (2p.) e. Describe the problems that arise if air enters the refrigerant cycle, filling up, say, 5% of the volume (2p.) Use tabelised data for R-134a, otherwise the log p, h diagram given below (next page). page 1 of 5

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The absolute humidities ω (kg moisture / kg dry air) of the incoming and outgoing air streams. (2 p.) b. The amount of liquid water that is removed, ṁ w, as kg / kg dry air (2 p.). c.

3 274. Consider the air conditioning process given schematically in the figure. A wet air stream is cooled, with a small pressure drop, while the relative humidity (ϕ) increases although some liquid water is removed. Determine / calculate: a. The absolute humidities ω (kg moisture / kg dry air) of the incoming and outgoing air streams. (2 p.) b. The amount of liquid water that is removed, ṁ w, as kg / kg dry air (2 p.). c. The heat Q (kj/kg dry air) that is removed using the cooling coil. (4 p.) Data: For the enthalpy of liquid water at 15 C (and p ~ 100 kpa): h w = 62,99 kj/kg. More data can be found from psychrometric chart given on the previous page. 15 C: saturation pressure water = 1,705 kpa, enthalpy saturated vapour = 2528,9 kj/kg 30 C: saturation pressure water = 4,246 kpa, enthalpy saturated vapour = 2556,3 kj/kg Specific heat dry air: c p = 1,004 kj/(kg K) 275. The chilling room of a meat plant is 18 m 20 m 5.5 m in size and has a capacity of 450 beef carcasses. The power consumed by the fans and the lights of the chilling room are 26 and 3 kw, respectively, and the room gains heat through its walls, floor and roof at a rate of 13 kw. The average mass of beef carcasses is 285 kg. The carcasses enter the chilling room at 36 C after they are washed to facilitate evaporative cooling and are cooled to 15 C in 10 h. The water is expected to evaporate at a rate of kg/s. The air enters the evaporator section of the refrigeration system at 0.7 C and leaves at -2 C. Also, the average temperature difference between the air and the refrigerant in the evaporator is 5.5 C. See the Figure. For a chilling room of a meat plant with a capacity of 450 beef carcasses calculate the cooling load (in kw), assuming that the specific heat of beef carcass is 3.14 kj/kg C. (6 p.) Note: more data on the next page. Data: The heat of fusion and the heat of vaporization of water at 0 C are kj/kg and 2501 kj/kg. The density and specific heat of air at 0 C are kg/m 3 and kj/kg C. page 3 of 5

4 276. A special Claude process is used for air liquefaction according to the process scheme given in the figure below. Ambient air at pin = 100 kpa, Tin =280 K is isothermally compressed to p0 =17 MPa, followed by isentropic expansion to p1 = 4.8 MPa, followed by heat exchange with gas from the gas/liquid separator, throttling to p2 = 100 kpa and separation of the liquefied air product, which is obtained at p3 = 100 kpa. The cold gas from the separator is, after heat exchange to temperature T4 = T1, emitted to the surroundings. 3 2 Heat exchange Throttle 1 (1-γ) ṁ AIR OUT 4 0 Q ṁṁ AIR IN 3 P expansion P compressor 3 γ ṁ LIQUEFIED AIR Gas/liquid separator a. Calculate the amount γ = liquefied air / air input (kg/kg) (3 p.) and b. Calculate the overall mechanical power input as kj/kg liquefied air. (3 p.) c. Draw the cycle in the enclosed pressure -enthalpy for air (last page); hand it in with your answers). (2 p.) Pressure drop effects in tube sections and the heat exchanger may be neglected. Note: temperature T2 before the throttling valve is found from the diagram when γ is known! Please give your course evaluation for this page 4 of 5

5 ÅA Refrigeration / Kylteknik , Question 276 Name: Matriculate number: page 5 of 5

1 /35 2 /35 3 /30 Total /100

1 /35 2 /35 3 /30 Total /100 Test is open book and notes. Answer all questions and sign honor code statement: I have neither given nor received unauthorized assistance during this exam. Signed Remember to show your work partial credit