Pressure Enthalpy Charts What is a p-h Diagram? A p-h diagram is a diagram with a vertical axis of absolute pressure and a horizontal axis of specific enthalpy. "Enthalpy is the amount of energy in a substance capable of doing work" It is an important diagram which can be used to calculate the performance of a refrigeration / air conditioning system but is specifically designed for an individual refrigerant. The refrigerant type can usually be found in the top left corner of the chart. Make sure that you select the correct chart that matches the refrigerant type in your system. Enthalpy (h) means Heat Energy or Heat Content Copyright Tradewinds Engineering Ltd Page 1
KEY TERMINOLOGY Sensible Heat occurs when there is a change in temperature but no change in state. It can be measured with a thermometer & felt by touch. Latent Heat is the heat given up or absorbed by a substance as it changes state. Latent heat causes a change in state but no change in temperature. Saturation Temperature is the point that a change in state will take place by the addition or removal of any heat energy. Sub cooled means to lower the temperature of a substance below its saturation condensing temperature. Superheated is vapor heated above its saturation temperature or evaporating temperature. Saturated Liquid contains as much thermal energy as it can without boiling, 100% liquid at its saturation temperature, any further heat addition will allow boiling, any further heat removal will result in sub cooled liquid. Saturated Vapor contains as little thermal energy as it can without condensing, 100% vapour at its saturation temperature, any further heat addition will result in superheated vapour, and any further heat removal will result in condensing. Bubble Point The saturated liquid temperature of a refrigerant that is 100% liquid just about to boil off and not yet sub-cooled. The bubble point is when the first refrigerant is just about to boil. Dew Point The saturated vapour temperature of a refrigerant that is 100% vapour but not yet superheated. The Dew Point is the point when the last refrigerant has just boiled off. Critical Point is when the boiling point increases with increased pressure up to the critical point, where the gas and liquid properties become identical. The boiling point cannot be increased beyond the critical point. Copyright Tradewinds Engineering Ltd Page 2
Pressure-Enthalpy Diagrams On the pressure enthalpy chart you will note that the vertical axis is absolute pressure in Bar and the horizontal axis is enthalpy. The curve is called the Saturation Curve. Following this line down the right hand side of the curve will show you refrigerant in the saturated vapour state. From the top of the curve down the left hand side is a Saturated Liquid. P-h Graph for a given liquid The upper horizontal line is your system discharge pressure and the lower horizontal line is the suction pressure. The diagonal line on the right occurs inside the compressor and the vertical line on the left is the pressure drop that occurs inside the metering device. P-h graph relating to the refrigeration cycle Copyright Tradewinds Engineering Ltd Page 3
P-h graph showing the liquid states P-h graph Refrigeration effect (RE), work done (WD) and heat rejection (HR) Copyright Tradewinds Engineering Ltd Page 4
Pressure in bar abs What does it all mean? E D C B F G A Enthalpy in kj/kg Point A Low pressure superheated vapour entering the suction port of the compressor Point B High pressure superheated vapour leaving the discharge port of the compressor Point C Saturated vapour in condenser. Point where very first droplet of liquid is formed Point D Saturated liquid in condenser Refrigerant is now 100% liquid. Any further removal of heat will result in subcooling. Point E Subcooled liquid enters the expansion valve Point F Refrigerant leaves the expansion valve as a two phase mix Point G Refrigerant is now 100% saturated vapour. Further addition of heat will cause it to become a superheated vapour. Superheat - The difference between the evaporating temperature and the suction pipe temperature. (G-A) Refrigeration Effect (RE) - The amount of heat absorbed in the evaporator expressed as kj/kg (F-G) Work Done (WD) - The amount of heat energy (enthalpy) the refrigerant absorbs during compression which corresponds to the required work done by the compressor. (A-B) Total Heat of Rejection (HR) - This is the total amount of heat that is lost from the minute it leaves the compressor discharge port until it reaches the inlet of the expansion valve. (B-E) Latent Heat Rejected - The heat rejected by the condenser that will cause a change of state, but no change in temperature (C-D) Sensible Heat Rejected - The heat rejected that will cause a change in temperature, but no change in state. (B-C) + (D-E) Sub-cooling - The difference between the condensing temperature and the liquid line temperature. (D-E) Expansion - This is the location where a pressure drop takes place to lower the boiling point. (E-F) Latent Heat Absorbed By The Evaporator - The heat absorbed by the evaporator that will cause a change in state but no change in temperature. (F-G) Sensible Heat Absorbed By The Evaporator - The amount of heat absorbed by the evaporator which causes a change in temperature not state. (G-A) Copyright Tradewinds Engineering Ltd Page 5
P-H CHART ANALYSIS Once we have completed plotting our information onto a PH Chart, we can start to make some basic calculations and assess the system performance. These being the Coefficient of Performance and the Energy Efficiency Ratio. HR = Total Heat of Rejection RE = Refrigeration Effect WD = Work Done by Compressor Coefficient of Performance A measure of system performance can be given as a ratio. This ratio tells us the difference between the energy we put in to a system and the benefit we get out. The better the ratio the more efficient the system, this is known as the Coefficient of Performance. For refrigeration systems this is known as COP r and is calculated by: COP r = Refrigeration Effect which equates to h1 - h4 and is given as a ratio, such as, 4:1 Work Done h2 - h1 For Heat Pump systems this is known as COP hp and is calculated by: COP hp = Heat Of Rejection which equates to h2 - h3 and is given as a ratio, such as, 5:1 Work Done h2 - h1 Copyright Tradewinds Engineering Ltd Page 6
Try working through the following example: Compressor rating 3KW Condenser fans 250 W Evaporator fans 200W Mass flow of 0.038 Kg/s. P-h graph for example question COP R = RE which equates to h1 - h4 = 320-135 = 185 = COP R of 4.625:1 WD h2 - h1 360-320 40 COP HP = HR which equates to h2 - h3 = 360-135 = 225 = COP HP of 5.625:1 WD h2 - h1 360-320 40 Cooling Capacity = Heat Absorbed x mass flow (kj/kg x kj/s = kw) (h1 h4) x 0.038Kg/s = 320-135 x 0.038 = 185 x 0.038 = 7.03 KW EERr = Cooling Capacity = 7.03 = 7.03 = EERr of 2.038:1 Total System Input Power 3 + 0.25 + 0.2 3.45 Energy Efficiency Ratio (EER) The total energy efficiency ratio would be calculated in a similar way for a refrigeration or heat pump system. However, this time, we need to include the electrical power for all pumps, fans, heaters, etc. EERr = Cooling Capacity Total System Input Power EER Heating EER Refrigeration = Total heat rejected by condenser Total system input power. = Refrigeration Effect Total system input power Compression ratio = adp Absolute discharge pressure asp Absolute suction pressure Copyright Tradewinds Engineering Ltd Page 7
If we draw some very basic PH charts & then plot some system data changes in temperature we can see the relationship of work done by the compressor on our refrigerant and our gain or loss of refrigerant effect: When the system condensing pressure/temperature rises, due to dirty condenser coil, poor airflow or a blocked condenser coil, what is actually happening to the system? Increase in discharge pressure Produces an increase in work done by the compressor Decreased Te NOTE: For every 1 K rise in condensing temperature you will lose 3% of cooling capacity and use 3% more energy. What benefit would raising the evaporating temperature give us, say from -6 C to 5 C? Increase in suction pressure Increased Te Reduces the work done by the compressor and provides an increase in cooling capacity NOTE: For every 1 K rise in evaporating temperature you will increase 3% of cooling capacity and use 3% less energy. Copyright Tradewinds Engineering Ltd Page 8
ENTHALPY CHART FORMULA Mass flow rate of the refrigerant (m ) = Cooling Duty (Kw) (Measured in Kg/s) Refrigeration Effect Compressor Capacity = (h2- h1) x m (Measured in kw) Condenser Capacity = (h2- h3) x m (Measured in kw) Evaporator Capacity = (h1- h4) x m (Measured in kw) Volume Flow Rate (v) = (Specific Volume) x m (Measured in m3/s) Coefficient of Performance (COP) = Refrigeration Effect Work Done Copyright Tradewinds Engineering Ltd Page 9
Example 1 A galley refrigeration plant utilizes R404A as the primary refrigerant. The working pressures for the system are 15Bar condensing and 4 Bar evaporating respectively. If the liquid is subcooled by 5 C and the refrigerant entering the compressor is superheated by 6 C, find: Condensing Pressure: 15 BarG (16BarAbs) Evaporating Pressure: 4 BarG (5BarAbs) Subcooling Temperature: 5K Superheat Temperature: 6K Step 1. The 15 Bar given is gauge pressure so we need to add 1Bar to give us the absolute pressure value that can be used on the graph. Step 2. We then draw a horizontal line across at 16Bar. This is our discharge pressure line. Step 3.The same applies to the suction pressure and we get a reading of 5 Bar Absolute which is also plotted on the graph. Step 4. Find 5K of subcooling and drawing a vertical line down. Step 5. Then from the saturated vapour line on the curve, come out with 6K of superheat. From this point follow the diagonal line up until it intersects with the discharge pressure line. 16Bar Abs 5Bar Abs 245Kj/Kg 370Kj/Kg 395Kj/Kg Copyright Tradewinds Engineering Ltd Page 10
A. Determine the mass flow rate of the refrigerant (assuming a cooling duty of 5kW) Mass flow rate of the refrigerant (m ) = Cooling Duty (Measured in Kg/s) RE = 5kW = 5kW = 0.04Kg/s (370-245) 125 B. Calculate the compressor capacity in kw Compressor Capacity (Kw) = (h2- h1) x m = (395-370) x 0.04Kg/s = 25 x 0.04 = 1.0kW C. Calculate the condenser capacity in kw Condenser Capacity (kw) = (h2- h3) x m = (395-245) x 0.04Kg/s = 6.0kW D. What is the volume flow rate at suction inlet port Volume Flow Rate (v) = (Specific Volume) x m (Measured in m3/s) = 0.042(from Ph. Chart) X 0.04 = 0.0017m3/s E. Calculate the co-efficient of performance Coefficient of Performance (COP) = Refrigeration Effect Work Done (370-245) = 125 = 5:1 (395-370) 25 F. Determine the condensing and evaporating temperatures 15Bar = 34.8 C 4Bar = -5.5 C Copyright Tradewinds Engineering Ltd Page 11
Question 1 An additional living quarters (ALQ) air conditioning system with a cooling capacity of 10kW uses R407c and the DEW POINT (evaporating temperature) is evaporating at -10 C and the BUBBLE POINT (Condensing Temperature) is 40 C. Dew Point: Bubble Point: Subcooling Temperature: Superheat Temperature: -10 C =2.2BarG (3.2BarAbs) +40 C = 16.5BarG (17.5BarAbs) 6K 6K 17.5 Bar Abs 3.28 Bar Abs 250Kj/Kg 415Kj/Kg 455Kj/Kg Copyright Tradewinds Engineering Ltd Page 12
Using the information on the chart above, calculate the following: A. The refrigeration effect: 415Kj/Kg 250Kj/Kg = 165Kj/Kg B. The mass flow rate of refrigerant Mass flow rate of the refrigerant (m ) = Cooling Duty (Kw) (Measured in Kg/s) Refrigeration Effect = 10 Kw 165 Kj/Kg = 0.0606Kg/s C. The compressor capacity in kw Compressor Capacity = (h2- h1) x m (Measured in kw) = 455 415 = 40Kj/Kg D. The co-efficient of performance = 40 x 0.0606 = 2.424kW Coefficient of Performance (COP) = Refrigeration Effect Work Done = 165Kj/Kg 40Kj/Kg = 4.125:1 Copyright Tradewinds Engineering Ltd Page 13
Question 2 An air conditioning plant utilizes R134A as the primary refrigerant. The refrigerant in the evaporator is evaporating at +5 C. The vapour is dry saturated at the compressor inlet and the condenser produces liquid at 45 C. If there is saturated liquid at the TEV inlet and the refrigerant entering the compressor has no superheat, find: 11 Bar Abs 3.5 Bar Abs 260Kj/Kg 400Kj/Kg 420Kj/Kg Copyright Tradewinds Engineering Ltd Page 14
A. Determine the mass flow rate of the refrigerant (assuming a cooling duty of 35kW) B. Calculate the compressor capacity in kw C. Calculate the condenser capacity in kw D. Determine the volume flow rate at the suction inlet port E. Calculate the co-efficient of performance Copyright Tradewinds Engineering Ltd Page 15
Question 3 An air conditioning system uses R410A as the refrigerant. The refrigerant is evaporating at 0 C in the indoor unit and is superheated by 5K at the compressor suction. The outdoor unit is producing refrigerant liquid at 40 C which is further subcooled by 7K, find: 24 Bar Abs 8 Bar Abs 260Kj/Kg 430Kj/Kg 460Kj/Kg Copyright Tradewinds Engineering Ltd Page 16
A. Determine the mass flow rate of the refrigerant (assuming a cooling load of 3.5kW) B. Calculate the compressor capacity in kw C. Calculate the condenser capacity in kw D. Determine the volume flow rate at suction E. Determine the Co Efficient of performance F. Determine the high & low pressure readings on your manifold gauges Copyright Tradewinds Engineering Ltd Page 17