Throttling Thermodynamics Professor Lee Carkner Lecture 22
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 =
PAL #21 Refrigeration Mass flow rate if W’ in = 0.45 kW W’ in = m’(h 2 -h 1 ) m’ = (0.45)/( ) = Find COP from W’ in and Q’ L Q’ L = m’(h 1 -h 4 ) = ( )( ) = kW COP = Q’ L /W’ in = (1.091)/(0.45) =
Cascade Systems For larger commercial systems, efficiency becomes more important e.g., deep freeze Called cascade cycles
Two-Stage Cascade
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 )]
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
Gas Refrigeration We can also us a reverse Brayton cycle Isentropic compression Isentropic expansion in turbine
Reversed Brayton Cycle
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 )]
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
Reversible Heat Pump
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?
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
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
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
Heat Exchanger How is gas liquefied? Throttled and cooled Cold gas runs back through the heat exchanger cooling the incoming gas Cycle starts over
Next Time Read: Homework: Ch 11, P: 35, 56, Ch 12, P: 8, 27