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Throttling Thermodynamics Professor Lee Carkner Lecture 22

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PAL #21 Refrigeration Refrigerator cycle where P = 120 kPa and x = 0.3 before the evaporator and 60 C after the compressor Start at point 4, P 4 = 120 kPa, x = 0.3, look up h 4 = h 3 = h 4 = 86.83, for a saturated liquid this means P 3 = Since P 2 = P 3 and T 2 = 60, look up h for superheated vapor, h 2 = At point 1, P 1 = P 4 = 120 kPa, saturated vapor, h 4 =

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PAL #21 Refrigeration Mass flow rate if W’ in = 0.45 kW W’ in = m’(h 2 -h 1 ) m’ = (0.45)/(298.87-236.97) = Find COP from W’ in and Q’ L Q’ L = m’(h 1 -h 4 ) = (0.000727)(236.97-86.83) = 1.091 kW COP = Q’ L /W’ in = (1.091)/(0.45) =

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Cascade Systems For larger commercial systems, efficiency becomes more important e.g., deep freeze Called cascade cycles

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Two-Stage Cascade

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Cascade Efficiency The condenser of cycle B (points 1-4) is connected to the evaporator of cycle A (points 5-8) m’ A (h 5 -h 8 ) = m’ B (h 2 -h 3 ) COP Cascde = m’ B (h 1 -h 4 )/[m’ A (h 6 -h 5 )+m’ B (h 2 -h 1 )]

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Multistage Compression Some fluid is vaporized and is sent back to the high pressure compressor Can also use just one compressor and multiple throttle valves and evaporators for multiple temperatures

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Gas Refrigeration We can also us a reverse Brayton cycle Isentropic compression Isentropic expansion in turbine

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Reversed Brayton Cycle

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Gas Refrigeration Efficiency w net,in = w comp – w turb = (h 2 -h 1 )-(h 3 -h 4 ) COP = q L /w net,in = (h 1 -h 4 ) / [(h 2 -h 1 )-(h 3 -h 4 )]

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Heat Pumps COP HP = Q H /W net,in = Q H / (Q H – Q L ) COP HP,Carnot = 1 / (1 – T L /T H ) Often designed as dual heat pump/air conditioners Low COP if the outside temperature is very cold Can also push the heat extraction underground

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Reversible Heat Pump

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Joule-Thompson Expansion Can be achieved by a pump circulating fluid through a pipe with an expansion valve in the middle We know that in this case, h i = h f What will be the final properties of the fluid?

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Isenthalpic Curve If the apparatus is changed a little, a new P f and T f are produced The curve represents possible beginning and ending points for a throttling process A series of isenthalpic curves can be produced for a substance

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Inversion Curve Each curve has two regions T f >T i T f < T i In between, the slope, or Joule-Thompson coefficient ( ), is zero: For a series of isenthalpic curves, a curve connecting =0 points is the inversion curve

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Liquefying Gasses In order to cool a gas, its temperature must start below the maximum inversion temperature T M.I. is near room temperature for many gasses Some gasses have to be pre-cooled

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Heat Exchanger How is gas liquefied? Throttled and cooled Cold gas runs back through the heat exchanger cooling the incoming gas Cycle starts over

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Next Time Read: 12.1-12.3 Homework: Ch 11, P: 35, 56, Ch 12, P: 8, 27

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