1. Teknologi dan Rekayasa Basic Cooling Techniques (013.TPTU.KK.03)

2. Principles and Settings Refrigerasi air condition  Refrigerasi is kalor switching from low temperatures to the temperature in the higher  Refrigerasi room temperature is lower or maintain the property and so temperaturnya lower than the temperature is lower than the environmental temperature  The air condition is the condition of the air which include: temperature, humidity, Quality, and Circulation

3. How to reduce the temperature 1.Raising the temperature cooling media Rate movements kalor absorbed by the cooling media m.Cp.DT Q = (Kw) 2. Phase change cooling media a. Solid to liquid, the rate movement kalor fusion, a. Solid to liquid, the rate movement kalor fusion, Q=m.h sf. Air pada 0 0 C, h sf =335 kJ/kg Q=m.h sf. Air pada 0 0 C, h sf =335 kJ/kg b. Liquid to gas, the rate movement kalor evaporation, Q=m.h fg b. Liquid to gas, the rate movement kalor evaporation, Q=m.h fg c. Solid to gas, the rate movement kalor sublime, Q=m.h sg. CO 2 dense in -78.5 o C, h sg =573 kJ/kg c. Solid to gas, the rate movement kalor sublime, Q=m.h sg. CO 2 dense in -78.5 o C, h sg =573 kJ/kg 3.Mix the salt into a water - CaCl 2 solution can decrease the water temperature to- 50 0 C - NaCl solution can lower the temperature to aoi-20 0 C - CaCl 2 solution can decrease the water temperature to- 50 0 C - NaCl solution can lower the temperature to aoi-20 0 C

4. Expansion of the liquid accompanied by flashing (process 1-2) produced a decrease in temperature (DT1) larger than the decrease in temperature (DT2) resulting from the expansion in the area of liquid (process 3-4) Expansion of the liquid accompanied by flashing (process 1-2) produced a decrease in temperature (DT1) larger than the decrease in temperature (DT2) resulting from the expansion in the area of liquid (process 3-4) Expansion of the liquid accompanied by flashing 4. Expansion of the liquid accompanied by flashing

5. Basic principles on Refrigerasi  Fusion changes in the solid phase into liquid  Boiling point the temperature changes in the early phases of liquid into vapor phase.  Related to the boiling point pressure  Kalor delitescence kalor namely the vaporization needed to evaporate 1 kg of liquid from the liquid into vapor saturated saturated  Heat up the situation after the steam exceeds the saturated vapor  Cold state that is more liquid after passing through the liquid saturated  Condensation of the phase change from the saturated vapor into saturated liquid.  In the process of condensation occurs release kalor  Ireversible process throtlling held constant but entalpi

6. Machine Kalor No engine kalor that have effisiensi 100% Hukum I : Q1-Q2=W Efisiensi mesin kalor

7. Machine Refrigerasi Work machine refrigerasi expressed in koefesien performance or COP

8. Steam Cycle Compression

9. In the four Refrigerasi Compression Cycle Steam  Isentropic compression (1 to 2)  Constant pressure condensation (2 to 3)  Isenhalpic expansion (3 to 4)  Constant pressure evaporation (4 to 1)

10. Reverse Carnot Cycle

11. Refrigeration cycles are often displayed on pressure-enthalpy diagrams

12. Performance of refrigeration cycles is measured by the Coefficient of Performance For Refrigerators and Air Conditioners:

13. Teknologi Dan Rekayasa Section 1: Theory of Heat Unit 1: Theory

14. Teknologi dan Rekayasa UNIT OBJECTIVES  Define temperature and convert between temperature scales  Define the British Thermal Unit, btu  Explain heat transfer by conduction, convection and radiation  Understand sensible heat, latent heat and specific heat  Explain the concept of pressure  Explain the difference between psia and psig After studying this unit, the reader should be able to

15. Teknologi dan Rekayasa TEMPERATURE  The level of heat or heat intensity  Measured with thermometers  English system – Fahrenheit (°F)  Metric system – Celsius (°C)  Fahrenheit Absolute scale – Rankine (°R)  Celsius Absolute scale - Kelvin (°K)  Absolute zero – Temperature at which all molecular movement stops (-460°F)

16. Teknologi dan Rekayasa 212°F 32°F 0°F - 40°F - 460°F 100 °C 0 °C - 17.8 °C - 40 °C - 273 °C 672°R 492°R 460°R 420°R 0°R 373°K 273°K 255.2°K 233°K 0°K FAHRENHEIT CELSIUS RANKINE KELVIN

17. Teknologi dan Rekayasa FAHRENHEIT TO CELSIUS CONVERSIONS °C = (5/9) (°F – 32 ) EXAMPLE: CONVERT 212°F TO CELSIUS °C = (5/9) (212 – 32 ) °C = (5/9) (180 ) °C = 5 x 20 °C = 100

18. Teknologi dan Rekayasa CELSIUS TO FAHRENHEIT CONVERSION °F = (9/5)°C + 32 EXAMPLE: CONVERT 10°C TO FAHRENHEIT °F = (9/5)(10) + 32 °F = (9 x 2) + 32 °F = 18 + 32 °F = 50

19. Teknologi dan Rekayasa INTRODUCTION TO HEAT  Heat is the motion of molecules  Heat cannot be created or destroyed  Heat can be measured and accounted for  Heat can be transferred from one substance to another  Heat travels from a warmer substance to a cooler substance  Quantity of heat in a substance is measured in British Thermal Units, BTUs

20. Teknologi dan Rekayasa 68°F 69°F ONE POUND OF WATER IDENTICAL POUND OF WATER ONE BTU OF HEAT ENERGY HAS BEEN ADDED TO ONE POUND OF WATER

21. Teknologi dan Rekayasa HEAT TRANSFER BY CONDUCTION  Heat energy travels from one molecule to molecule within a substance  Heat energy travels from one substance to another  Heat does not conduct at the same rate in all materials  Example of conduction: Heat will travel through a copper rod when placed near fire

22. Teknologi dan Rekayasa HEAT TRANSFER BY CONVECTION  Heat transfers through a fluid from one substance to another  Natural convection utilizes natural fluid flow, such as the rising of warm air and the falling of cooler air  Forced convection uses fans or pumps to move fluids from one point to another  Example of convection: Baseboard Heating

23. Teknologi dan Rekayasa SECTION OF BASEBOARD HEAT

24. Teknologi dan Rekayasa HEAT TRANSFER BY RADIATION  Radiant heat passes through air, heating the first solid object the heat comes in contact with  These heated objects, in turn, heat the surrounding area  Radiant heat can travel through a vacuum  Radiant heat can travel through space without heating it  Example of radiation: An electric heater that glows red

25. Teknologi dan Rekayasa HEAT INTENSITY = 400°F 10’ 20’ HEAT INTENSITY = 100°F

26. Teknologi dan Rekayasa  Heat transfer that results in a change in temperature of a substance  Sensible heat transfers can be measured with a thermometer  Example of a sensible heat transfer: Changing the temperature of a sample of water from 68°F to 69°F SENSIBLE HEAT

27. Teknologi dan Rekayasa  Also referred to as hidden heat  Latent heat transfers result in a change of state of a substance with no change in temperature  Latent heat transfers cannot be measured with a thermometer  Example of a latent heat transfer: Changing 1 pound of ice at 32°F to 1 pound of water at 32°F LATENT HEAT

28. Teknologi dan Rekayasa SPECIFIC HEAT  Defined as the number of btus required to raise the temperature of 1 pound of a substance 1 degree Fahrenheit  Specific heat of water is 1.00  Specific heat of ice is approximately 0.50  Specific heat of steam is approximately 0.50  Specific heat of air is approximately 0.24

29. Teknologi dan Rekayasa SPECIFIC HEAT FORMULA Q = Weight x Specific Heat x Temperature Difference Where Q = Quantity of heat needed for the temperature change Example: 1000 pounds of steel must be heated from 0°F to 70°F. How much heat is required to accomplish this? The specific heat of steel is 0.116 btu/lb Substituting in the above formula gives us Q = 1000 pounds x 0.116 btu/lb x (70°F - 0°F) Q = 1,000 x 0.116 x 70 = 8,120 btu

30. Teknologi dan Rekayasa 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 HEAT CONTENT (BTUs) - 50°F 0°F 50°F 100°F 150°F 200°F 250°F 300°F 350°F ICE AT 0°F ICE AT 32°F 16 btu EXAMPLE USING 1 POUND OF ICE (32-0) x (0.5) = 16 btu

31. Teknologi dan Rekayasa 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 HEAT CONTENT (BTUs) - 50°F 0°F 50°F 100°F 150°F 200°F 250°F 300°F 350°F ICE AT 32°F EXAMPLE USING 1 POUND OF ICE 16 + 144(1.0) = 160 btu WATER AT 32°F 160 btu

32. Teknologi dan Rekayasa HEAT CONTENT (BTUs) - 50°F 0°F 50°F 100°F 150°F 200°F 250°F 300°F 350°F EXAMPLE USING 1 POUND OF ICE 160 + 212-32(1.0) = 340 btu WATER AT 32°F WATER AT 212°F 340 btu

33. Teknologi dan Rekayasa HEAT CONTENT (BTUs) 0°F 50°F 100°F 200°F 250°F 300°F 350°F EXAMPLE USING 1 POUND OF ICE 340 + 970(1.0) = 1310 btu WATER AT 212°F STEAM AT 212°F 1310 btu 0 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400

34. Teknologi dan Rekayasa SUMMARY OF ICE EXAMPLE Ice at 0°F to Ice at 32°F (32 – 0) (0.5) = 16 btu Ice at 32°F to Water at 32°F = 144 btu Water at 32°F to Water at 212°F (212 – 32) (1.0) = 180 btu Water at 212°F to Steam at 212°F = 970 btu Steam at 212°F to Steam at 350°F (350-212)(0.5) = 69 btu TOTAL HEAT TRANSFER = 1,379 btu

35. Teknologi dan Rekayasa PRESSURE  Defined as the force per unit area  Often expressed in pounds per square inch  Example: If a 100-pound weight rests on a surface of 1 square inch, the pressure is 100 psi  Example: If a 100-pound weight rests on a surface of 100 square inches, the pressure is only 1 psi

36. Teknologi dan Rekayasa 1 cubic inch block with a weight of 1 pound 1 square inch Pressure = 1 psi 100 pound block Pressure = 100 psi 100 square inches Pressure = 1 psi

37. Teknologi dan Rekayasa ATMOSPHERIC PRESSURE  The atmosphere we live in has weight  The atmosphere exerts a pressure of 14.696 psi at sea level (often rounded off to 15 psi)  14.696 psi at sea level is known as the standard condition  Measured with a barometer

38. Teknologi dan Rekayasa THE BAROMETER  Used to measure atmospheric pressure  Constructed as a 36” glass tube  Tube is sealed at one end and filled with mercury  The tube is inverted and placed mercury  As atmospheric pressure drops, so does the level of mercury in the tube  At atmospheric pressure, the height of the mercury will be 29.92”

39. Teknologi dan Rekayasa INCHES OF MERCURY AND PSI  The column of mercury is 29.2” at atmospheric condition of 14.696 psi  One psi is equal to approximately 2” Hg  Example: If the barometer reads 20”Hg, then the atmospheric pressure is approximately equal to 10 psi  Absolute pressures are measured in pounds per square inch absolute, psia

40. Teknologi dan Rekayasa Mercury puddle Glass tube Column of mercury Height of mercury column is 29.92” at standard condition Atmospheric pressure pushes down on the mercury As atmospheric pressure drops, so does the level of mercury in the tube

41. Teknologi dan Rekayasa INCHES OF MERCURY AND PSI  The column of mercury is 29.2” at atmospheric condition of 14.696 psi  One psi is equal to approximately 2” Hg  Example: If the barometer reads 20”Hg, then the atmospheric pressure is approximately equal to 10 psi  Absolute pressures are measured in pounds per square inch absolute, psia

42. Teknologi dan Rekayasa PRESSURE GAGES  Bourden tube – measures pressure in a closed system  Used to measure the pressures in an air conditioning or refrigeration system  Gages read 0 psi when opened to the atmosphere  Gage pressures are measured in pounds per square inch gage, psig

43. Teknologi dan Rekayasa PRESSURE CONVERSIONS  To convert gage pressure to absolute pressure, we add 15 (14.696) psi to the gage reading  To convert absolute pressure to gage pressure, we subtract 15 (14.696) from the absolute pressure  Example: 0 psig = 15 psia  Example: 70 psig = 85 psia

44. Teknologi Dan Rekayasa The Psychrometric Chart

45. Teknologi dan Rekayasa

46. Properties of Air The psychrometric chart contains five physical properties to describe the characteristics of air:  Dry-bulb temperature  Wet-bulb temperature  Dew-point temperature  Relative humidity  Humidity ratio

47. Teknologi dan Rekayasa Dry-bulb temperatures are read from an ordinary thermometer that has a dry bulb Wet-bulb temperatures are read from a thermometer whose bulb is covered by a wet wick. The difference between the wet-bulb temperature and the drybulb temperature is caused by the cooling effect produced by the evaporation of moisture from the wick. This evaporation effect reduces the temperature of the bulb and, therefore, the thermometer reading Wet-bulb temperatures are read from a thermometer whose bulb is covered by a wet wick. The difference between the wet-bulb temperature and the drybulb temperature is caused by the cooling effect produced by the evaporation of moisture from the wick. This evaporation effect reduces the temperature of the bulb and, therefore, the thermometer reading. Consequently, the difference between dry-bulb and wet-bulb temperature readings is a measure of the dryness of air. The drier the air, the greater the difference between the dry-bulb and wet-bulb readings

48. Teknologi dan Rekayasa dew-point temperature, is the temperature at which moisture leaves the air and condenses on objects, just as dew forms on grass and plant leaves saturated When the dry-bulb, wet-bulb, and dew-point temperatures are the same, the air is saturated

49. Teknologi dan Rekayasa Properties of Saturated Air 25°F 30°F 35°F 40°F 45°F 50°F 55°F 19.14 24.19 29.94 36.51 44.34 53.63 64.63 60°F 65°F 70°F 75°F 80°F 85°F 90°F 77.61 92.89 110.82 131.83 156.38 185.03 218.42 dry-bulb temp. humidity ratio dry-bulb temp. humidity ratio

50. Teknologi dan Rekayasa Plotting Saturated Points 0 30 Dry-Bulb Temperature (°F) Humidity Ratio (grains/lb of dry air) 2535404550556065707580859095100105110 20 40 60 80 100 120 140 160 180 200 220 218.4 2 185.0 3 156.3 8 131.8 3 110.8 2 92.8 9 77.6 1 64.6 3 53.6 3 44.3 4 36.5 1 29.9 4 24.1 9 19.1 4

51. Teknologi dan Rekayasa Saturation Curve 100% relative humidity curve 30 Dry-Bulb Temperature (°F) Humidity Ratio (grains/lb of dry air) 2535404550556065707580859095100105110 20 40 60 80 100 120 140 160 180 200 220

52. Teknologi dan Rekayasa Relative Humidity Curves relative humidity 0% 10% 20% 30% 40% 50% 60% 70% 80% 90% 100% 30 Dry-Bulb Temperature (°F) Humidity Ratio (grains/lb of dry air) 2535404550556065707580859095100105110 20 40 60 80 100 120 140 160 180 200 220

53. Teknologi dan Rekayasa Properties wet bulb dew point 30 Dry-Bulb Temperature (°F) Humidity Ratio (grains/lb of dry air) 2535404550556065707580859095100105110 20 40 60 80 100 120 140 160 180 200 220 humidity ratio dry bulb relative humidity

54. Teknologi dan Rekayasa Specific Volume 14.0 13.0 13.5 14.5 specific volume lines (cubic feet / pound of dry air) specific volume lines (cubic feet / pound of dry air) 30 Dry-Bulb Temperature (°F) Humidity Ratio (grains/lb of dry air) 2535404550556065707580859095100105110 20 40 60 80 100 120 140 160 180 200 220

55. Teknologi dan Rekayasa Sensible Heating

56. Teknologi dan Rekayasa Sensible Cooling

57. Teknologi dan Rekayasa Adding Moisture

58. Teknologi dan Rekayasa Removing Moisture

59. Teknologi dan Rekayasa Sensible & Latent Energy

60. Teknologi dan Rekayasa State Point 95°F 78°F 50% 40% 72°F

61. Teknologi dan Rekayasa Mixed Air supplyfan mixture outdoor air (OA) Re-circulated air (RA) A A C C B B coolingcoil

62. Teknologi dan Rekayasa Outside & Return Air 80°F 95°F B B A A outdoor air Re-circulated air

63. Teknologi dan Rekayasa Determining Mix 1,000 cfm 4,000 cfm = 0.25 OARAmixture === 25%75%100% A A C C B B 4,000 cfm mixed air 3,000 cfm RA

64. Teknologi dan Rekayasa Mixed Air Temperature 95°F × 0.25 = 23.75°F 80°F × 0.75 = 60.00°F mixture = 83.75°F

65. Teknologi dan Rekayasa Mixed Air State Point 83.75°F 70°F B B 80°F 95°F C C A A

66. Teknologi dan Rekayasa Space Sensible & Latent Heat supply air sensible heat sensible heat latent heat latent heat return air

67. Teknologi dan Rekayasa Sensible Heat Ratio SHR Sensible Heat Gain Sensible Heat Gain + Latent Heat Gain =

68. Teknologi dan Rekayasa SHR Scale 78°F DB 65°F WB 78°F DB 65°F WB index point

69. Teknologi dan Rekayasa Drawing SHR Line 0.80 SHR line index point A A

70. Teknologi dan Rekayasa Drawing SHR Line 0.60 SHR line C C D D B B 80°F index point 0.60 SHR line

71. Teknologi dan Rekayasa Coil Curves

72. Teknologi dan Rekayasa Determining Airflow STEP 1: Calculate the sensible heat ratio (SHR) = 0.80 80,000 Btu/hr 100,000 Btu/hr SHR = 80,000 Btu/hr sensible heat gain 20,000 Btu/hr latent heat gain

73. Teknologi dan Rekayasa Design Parameters room — 78°F DB, 50% RH outdoor air (OA) — 95°F DB, 78°F WB ventilation — 25% OA

74. Teknologi dan Rekayasa Plotting Design Points STEP 2: Plot room, outdoor, and entering conditions 95°F × 0.25 = 23.75°F 78°F × 0.75 = 58.50°F mixture = 82.25°F

75. Teknologi dan Rekayasa Determining Supply Airflow STEP 3: Identify supply air conditions

76. Teknologi dan Rekayasa Determining Airflow Sensible Heat Gain 1.085 × (Room DB – Supply DB) Supply Airflow = STEP 4: Solve the supply airflow equation 80,000 Btu/hr 1.085 × (78°F – 56.5°F) 3,430 cfm =

77. Teknologi dan Rekayasa Completing the Loop 56.5° F D D A A C C B B D A C B 3,430 cfm 56.5°F DB 55.2°F WB RA OA 82.2°F DB 68.6°F WB mixture SA

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