The Thermal and Fluid Systems

The figure shows that a hydraulic system is used to shape a metal sheet to a bowl, as illustrated by the inset. In the practical operation, a wrong metal sheet, a thick one for example, may inadvertently be placed between the two deformation molds. It will overload the system and cause potentially severe damages. In order to prevent that, this system uses a self-protection mechanism. When facing overload, the cylinder will automatically retract. Now the directional valve is actuated and the cylinder is just going to carry out the work. Please describe in the order of reactions how the self-protection mechanism works when the cylinder is overloaded. Please draw the movement directions of the valves, the direction of the flow, and the motion of the cylinder on the figure. Valve 3 800 psi Valve 4 Valve 2 Valve 1 100 psi Inset 850 psi 1,000 psi Metal Sheet 1 2. A double-acting cylinder is controlled by an electrical means. When Pushbutton 2 (START) is pushed and released, analyze and describe in detail the reactions of EACH COMPONENT in the hydraulic and electrical diagrams in the logic sequence. Then, DRAW A FINAL CONCLUSION WHAT MOVEMENT THE CYLINDER IS DOING. You need to mark the reactions on the diagrams. 1-LS L1 2-LS 5 6 6 7 L2 START STOP 2-PB 1-CR 1-PB 1 1 3 2 4 2 1-CRA SOL A 8 2 3 1-CRB 5 1 4 6 2-LS 2-CR 7 2 2-CRA Oil In (a) 1-LS 8 5 SOL A 2 2-CRB (b) 2 3. a) Please describe the three heat transfer mechanisms. b) A heat exchanger is used to transfer heat from one matter to another. Using right material and design is critical for its heat-exchanging efficiency. It often uses Al or Cu, black surface, and multi-layer structure. In addition, a mechanical fan is often used. The following are two typical examples. A heat exchanger involves all the three heat transfer modes. Please use the three equations: Q1→2 = − Ak dT dx Q1→2 = Ah(T1 − T2 ) Q1→2 = A (T2 − T1 ) 4 4 to explain why a heat exchanger uses such a material and a design. Please draw a simple sketch illustrating how the heat flux flows inside one fin or film of a heat exchanger. Multi-layer Al fins Multi-layer Al films 3 IET 260: Hydraulics and Pneumatics Chapter 15: Basic Electrical Controls for Fluid Power Circuits Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 LC Room 105E Electrically-controlled fluid power systems are most powerful and useful in industry. Electrical Control Devices: Push-Button Switch Push-button switch is the switch can be opened or closed by a push. There are four common types: 1) Single-pole, single-throw, normally open SPST-NO 3) Double-pole, single-throw DPST-NO/NC 2) Single-pole, single-throw, normally closed SPST-NC 4) Double-pole, double-throw DPDT-NO/NC Electrical Control Devices: Limit Switch Limit switch is the switch used to open and/or close circuit when actuated, often at the end of extension and/or retraction of a cylinder. Hydraulic Symbol Limit Switch This limit switch is closed to shut off the hydraulic circuit. This limit switch is closed to shut off the hydraulic circuit. Electrical Control Devices: Limit Switch 1) Normally open 2) Normally closed LS-NO When activated LS-NC When activated Electrical Control Devices: Pressure Switch Pressure switch is the switch that open or close, based on the pilot pressure. PS-NO PS-NC Hydraulic Symbol Electrical Symbol Electrical Control Devices: Solenoid Solenoid uses electromagnet to generate a push or pull force to operate hydraulic valve. Hydraulic Symbol Electrical Symbol Electrical Control Devices: Relay Relay is a switch whose contacts open or close when their coils are energized. High Voltage Normally Closed Contacts Normally Open Contacts Low Voltage Low Voltage High Voltage Normally Closed Contacts Relay Coils Normally Open Contacts Electrical Control Devices: Timer A timer is used to control the time duration of a working cycle. 1) Normally open (closed when energized) 2) Normally closed (open when energized) 3) Normally open (open when energized) 4) Normally closed (closed when energized) Ladder Diagram: Five Basic Rules L1 1. Place control (input) devices on the left and load (output) devices on the right. 2. Place one, and only one, load per rung. START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 3. Number all the connectors in order from left to right and top to bottom. 2 4. Number all the rungs in order from top to bottom. 3 5. Label all the components. L2 POWER LINE 1-LS 1-CRA 1 6 1-CRB 2 SOL A Ladder Diagram: Voltage Drop L1 0V START STOP 2. Place one, and only one, load per rung. 1 1-PB L2 24 V 2-PB 1-CR 4 3 5 2 1 1-LS 2 1-CRA 3 1 6 1-CRB Control Devices 2 SOL A Loads Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB Hydraulic Diagram Electrical Diagram Power Diagram Ladder Diagram 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 Relay coil engaged 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 Relay coil engaged 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Solenoid engaged Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 Relay coil disengaged 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Solenoid disengaged Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 Relay coil disengaged 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Solenoid disengaged Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A The cylinder remains retracted. When the “START” button is pushed: 1) 2) 3) 4) 5) 1-CR is engaged. 1-CRA and 1-CRB are closed. SOL A is engaged. DCV is actuated/its position is shifted. The cylinder starts to extend. 6) When 1-LS is actuated, the current in Rung 1 is off/1-CR is disengaged. 7) 1-CRA and 1-CRB are opened. 8) SOL A is disengaged. 9) DCV is deactivated and return to its initial position. The cylinder retracts. Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 5 2 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 6 1-CRB 2 SOL A Electrical Control of A Cylinder Using A Single Limit Switch 4 5 L1 L2 POWER LINE 1-LS START STOP 1 1-PB 2-PB 1-CR 4 3 2 5 1 SOL A 1-LS 2 6 2 1-CRA Oil In 3 1 2 6 1-CRB SOL A When the cylinder is moving. if the “STOP” button is push: 1) 2) 3) 4) The current in Rung 1 is off and1-CR is disengaged. 1-CRA and 1-CRB are opened. The current in Rung 2 is shut off. SOL A is disengaged. DCV is deactuated and return to its initial position. The cylinder retracts. Reciprocation of A Cylinder Using Pressure or Limited Switches POWER LINE 1-PS 2-PS 1-PS SOL A SOL B SOL B Oil In 2-PS SOL A Reciprocation of A Cylinder Using Pressure or Limited Switches When pressure builds up to 800 psi POWER LINE 800 psi 800 psi 1-PS 2-PS 2-PS is closed SOL A 1-PS SOL B SOL B Oil In 2-PS SOL A Reciprocation of A Cylinder Using Pressure or Limited Switches POWER LINE 800 psi 800 psi 1-PS 2-PS 1-PS SOL A SOL B SOL B Oil In 2-PS SOL A SOL A engaged Reciprocation of A Cylinder Using Pressure or Limited Switches When pressure releases POWER LINE 800 psi 800 psi 1-PS 2-PS SOL A Oil In 2-PS opens SOL B 1-PS 2-PS SOL B SOL A Reciprocation of A Cylinder Using Pressure or Limited Switches When pressure builds up to 800 psi POWER LINE SOL B engaged 800 psi 800 psi 1-PS 2-PS 1-PS 1-PS is closed SOL A Oil In SOL B SOL B 2-PS SOL A Reciprocation of A Cylinder Using Pressure or Limited Switches POWER LINE 1-PS 2-PS 1-PS SOL A SOL B SOL B Oil In 2-PS SOL A Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A At the beginning, both cylinders are fully retracted. 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START Sol A engaged SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A SOL B SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START Sol A disengaged SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C 1-LS switched on Sol C engaged Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START SOL A 2-LS SOL B 1-LS SOL C Double-Cylinder Sequencing Circuit 1-LS 2-LS Cylinder 1 SOL A Cylinder 2 SOL B SOL C Air In START One push on “START” button results in: SOL A 1) 2) 3) 4) Cylinder Cylinder Cylinder Cylinder 1 fully extends; 2 fully extends; 1 fully retracts; 2 fully retracts. 2-LS SOL B 1-LS SOL C Box-Sorting Circuit Limited Switch Conveyor Track Box-Sorting Circuit conveyor motor 2 2-LS 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 2-LS High Box 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA Both motors are turned on. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 2-LS High Box 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA Conveyor motor 2 is stopped. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 2-LS High Box 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA Conveyor motor 2 is stopped. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 High Box 2-LS 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA Conveyor motor 2 is stopped. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 High Box 2-LS 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA Conveyor motor 2 is turned on again. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit conveyor motor 2 High Box 2-LS 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA 1-CRB 1-CRA SOL A 2-CRB conveyor motor Box-Sorting Circuit conveyor motor 2 High Box 2-LS 1-LS conveyor motor 1 High Box STOP SOL A START conveyor 2-CR motor 1 2-CRA Air In Low Box 2-LS 1-LS 1-CR 1-CRA This system is waiting for next sorting. 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 Box-Sorting Circuit This ladder diagram is from the textbook but it has one critical fault and one minor fault. Where are the faults and how do you fix them? STOP START conveyor 2-CR motor 1 2-CRA 2-LS 1-LS 1-CR 1-CRA 1-CRB 1-CRA SOL A 2-CRB conveyor motor 2 The End IET 260: Hydraulics and Pneumatics Chapter 10: Hydraulic Conductors and Fittings Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 LC Room 105E There are not too many to talk about. Please read by yourselves. IET 260: Hydraulics and Pneumatics Chapter 11: Ancillary Hydraulic Devices Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 LC Room 105E Oil Reservoir or Tank A tank serves not only as a storage space but also the location where the oil is conditioned. A tank serves the following functions: 1. Permit foreign substances to settle to the bottom 2. Allows entrained air to escape from oil 3. Prevent localized turbulence in reservoir 4. Promote heat dissipation through reservoir Oil Reservoir or Tank Oil to Pump Returning Oil Hot Oil Filter This tank does not function well because the hot oil will flow back to the hydraulic system. Short circuit of the returning oil Oil Reservoir or Tank Baffle Plate Oil to Pump Returning Oil Hot Oil A baffle plate is placed in the center and its height is about 70% of the height of the oil level. The baffle plate is used to prevent the short circuit of the same returning oil, promoting efficient heat dissipation. Oil Reservoir or Tank: Sizing The size of a tank has to be able to contain all the oil of a fluid power system. An empirical relation is a tank having a capacity of three times the flow rate: Tank Volume = 3  Pump Flow Rate (gpm) For example, if the pump’s flow rate is 10 gpm, the tank’s size needs to be ~30 gallon. Accumulator An accumulator is a device that stores potential energy and provides a temporary second source of fluid power for a hydraulic system. It is important for reducing energy loss, decreasing pressure fluctuation, avoiding hydraulic shock and responding to emergency power need. There are three types: 1. Weight-loaded or gravity type 2. Spring-loaded type 3. Gas-loaded type Accumulator: Weight-Loaded Dead Weight Wdead P= Apiston A weight-loaded accumulator is basically a vertical cylinder. It produces a constant pressure regardless of its volume filled with the oil. Its main disadvantage is its large size and heavy weight. Accumulator: Spring-Loaded kx P= Apiston A spring-loaded accumulator is a cylinder, which is operational in any direction. Its output pressure depends on the displacement of the mechanical spring. That is, its output pressure varies with its volume filled with the oil. Accumulator: Gas-Loaded Non-Separator-Type Inert Gas nRT P= V Oil A gas-loaded non-separator-type accumulator is a seamless container. It has to be working vertically. Its output pressure is a function of the oil-occupying volume and temperature. Its main advantage is convenient to use; its main disadvantage is gas absorption. Accumulator: Gas-Loaded Separator-Type: Piston Gas nRT P= V Oil A gas-loaded piston-type accumulator is a seamless container separated by a piston. It has to be working in any direction. Its output pressure is a function of the oil-occupying volume and temperature. Its main disadvantage is unable to respond to pressure change because of the inertia of the piston and the friction of the seals. Accumulator: Gas-Loaded Separator-Type: Diaphragm Gas nRT P= V Oil A gas-loaded diaphragm-type accumulator is a seamless container separated by a an elastic barrier. It can work in any direction. Its output pressure is a function of the oil-occupying volume and temperature. Its main advantage is light-weight and has no mechanical friction, enabling it for airborne applications. Accumulator: Gas-Loaded Separator-Type: Bladder Gas nRT P= V Oil A gas-loaded bladder-type accumulator uses a elastic bladder in a seamless container. It can work in any direction. Its output pressure is a function of the oil-occupying volume and temperature. Its main advantage is that the light-weight bladder enables quick response to pressure change and hydraulic shock. Accumulator’s Applications: Auxiliary Power Source 850 psi 1. The accumulator is charged after the cylinder fully retracts. 2. The accumulator can then be used as the secondary power to drive the cylinder. Accumulator’s Applications: Auxiliary Power Source 850 psi Because the charged accumulator is always ready when needed, the use of the accumulator allows the intermittent operation of the pump. Accumulator’s Applications: Leakage Compensator 800 ~ 900 psi Because of oil leakage, the pressure will decrease below 800psi some time later. Pump is stopped Accumulator’s Applications: Leakage Compensator 800 ~ 900 psi Because of oil leakage, the pressure will decrease below 800psi some time later. An accumulator can be used to help the pressure in 800~900psi for a long time. Pump is stopped Accumulator’s Applications: Leakage Compensator Pilot-control contact switch: When P < 800 psi, pump on When P > 900 psi, pump off 950 psi 800 ~ 900 psi Accumulator’s Applications: Leakage Compensator 800 ~ 900 psi 950 psi When P < 800 psi, pump on Accumulator’s Applications: Leakage Compensator When P > 900 psi, pump off 950 psi The accumulator will compensate the oil leakage until the system pressure drops below 900 psi. In this way, the pump does not need to keep on. Accumulator’s Applications: Emergency Power Source Because of safety requirement, the cylinder has to be in the retraction position even if the normal supply of oil pressure is lost. 850 psi Such an application needs to use an accumulator as an emergency power source. Accumulator’s Applications: Emergency Power Source Accumulator’s Applications: Emergency Power Source Accumulator’s Applications: Emergency Power Source 850 psi Accumulator’s Applications: Hydraulic Shock Absorber To Power Application 850 psi Accumulator’s Applications: Hydraulic Shock Absorber To Power Application 850 psi Sealing Devices 1. O-ring 2. V-ring 3. Cup-packing Sealing Devices: Single-Acting Cylinder V-type or cup-type ring faces high pressure High Pressure Atmosphere Sealing Devices: Double-Acting Cylinder V-type or cup-type rings face both the high pressures High Pressure High Pressure Sealing Devices: Single-Acting Cylinder Muptiple V-type or cup-type rings can be used to ensure seamless sealing. High Pressure Atmosphere Heat Exchanger Because various forms of energy loss are inevitable, the oil in a hydraulic system can become very hot. Heat exchanger is necessary. Air-Cooled Water-Cooled IET 260: Hydraulics and Pneumatics Chapter 12: Maintenance of Hydraulic Systems Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 LC Room 105E We also skip this chapter. Please read by yourselves. IET 260: Hydraulics and Pneumatics Chapter 13: Pneumatics Air Preparation and Components Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 LC Room 105E Equation of State for Ideal Gas Idea gas is defined as a body of randomly-moving, non-interacting (perfect elasticity), point (negligible size) particles. PV = nRT m = RT mmolar Where P is the absolute pressure, T is the absolute temperature, and R is the universal gas constant. By re-arranging: PV m = R T mmolar For a confined gas volume: PV = const . T That is: P1V1 P2V2 = T1 T2 Air Compressors Like hydraulic pressure pumps, there are three types of compressors: gear, vane and piston. A compressor increases the pressure of a gas by reducing its volume. Because gas is compressible, a compressor always has a gas tank with it. Thus, the prime mover of a compressor can work in an intermittent way. Electrical Motor Compressor Gas Tank Air Compressors: Sizing of Gas Tank The tank size of a compressor depends on the pressures and flow rate requirements of a power system and the compressor’s flow rate. It can be decided by the following empirical formula: VTank = 14.7t (QTank − QCompressor ) Pmax − Pmin ( ft3 ) Where t is the time during which the tank needs to supply the required amount of air (minute), Pmax is the maximum pressure level and Pmin is the minimum pressure level of the tank during the time of period. Air Compressors: Capacity Rating The capacity of a compressor is generally rated in the cfm of free air, which is defined as the air at one atmosphere (14.7 psi) and 68°F. The unit is also called a standard cubic feet per minute (scfm). P1V1 P2V2 = T1 T2  P1 V2 =   P2  T2   V1  T1  V2  P1  T2  V1 =    t  P2  T1  t  P1  T2 Q2 =    P2  T1  Q1  Air Compressors: Capacity Rating The capacity of a compressor is rated 300 scfm. What is its flow rate at 250 psi and 90°F?  P1  T2 Q2 =    P2  T1 At the standard condition: P1 = 14.7 psi T1 = 68 F + 460 = 528 F Q1 = 300cfm  Q1  At the pressured condition: P2 = 250 psi + 14.7 psi = 264.7 psi T2 = 90 F + 460 = 550 F Q2 = ?  14.7 psi  550 F    300cfm = 17.35cfm  Q2 =    264.7 psi  528 F  Air Conditioners: Air Filter A filter is used to remove contaminants in the air before it reaches pneumatic components, such as valves and actuators. Air Conditioners: Pressure Regulator A pressure regulator is used to achieve a constant pressure for a pneumatic system. Air Conditioners: Pressure Indicator A pressure indicator is used to indicate the pressure. Pressure Indicator FRL Combined Unit of Filter, Regulator and Lubricator (FRL) Air Control Valves B A B A P P Exhaust Exhaust A B A B P Ex P Ex A B Four-way, two-position directional control valve P Ex Air Control Valves A B A B 2 4 P Ex P 1 3 In pneumatics, the pressure port is numbered (1). The exhaust port is numbered (3) and the second exhaust port is numbered (5) if existing. The other two are numbered (2) or (4). Air Control Valves A P Exhaust 1 A B Ex1 P Ex2 A B Exhaust 2 A Exhaust 1 B P Exhaust 2 B 2 4 P 3 1 5 Air Control Valves A B A Ex1 P Ex2 Ex1 P Ex2 A B B Ex1 P Ex2 A B Ex1 P Ex2 Air Motors One-Directional Bi-Directional IET 260: Hydraulics and Pneumatics Chapter 14: Pneumatics Circuits and Applications Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 SSC Room 212F Air Pressure Loss in Pipeline As in the case of liquid, an air flow also suffers energy loss due to friction. The energy loss is calculated by the following empirical formula: cLQ 2 Pf = 3600  CR  d 5 where c is an experimentally determined coefficient, CR is the compression ratio of the pressure in pipe over the atmospheric pressure. Pneumatic Circuits: Operation of Single-Acting Cylinder No Pump Shown. The triangle represents pressured air source Supply Line No Return Line Release to Air FRL Pneumatic Circuits: Operation of Single-Acting Cylinder Pneumatic Circuits: Operation of Single-Acting Cylinder Pneumatic Circuits: Operation of Double-Acting Cylinder Pneumatic Circuits: Operation of Double-Acting Cylinder Pneumatic Circuits: Operation of Double-Acting Cylinder 100 psi 10 psi Pneumatic Circuits: Operation of Double-Acting Cylinder 100 psi 10 psi Pneumatic Circuits: Operation of Double-Acting Cylinder 100 psi 10 psi The End ETM 260: Thermal and Fluid Systems Heat Transfer Chapter 1: Introduction Qingzhou Xu q.xu@moreheadstate.edu (606) 783 9598 This Chapter introduces the three basic modes of heat transfer: 1. Heat Conduction 2. Heat Convection 3. Heat Radiation Heat Transfer Heat is the collective mean effect of the kinetic energy of individual microscopic particles (electrons, atoms and molecules) in random movement. There are three heat transfer modes: conduction, convection and radiation. Chapter 1.3 Modes of Heat Transfer: Heat Conduction The heat flux resulting from thermal conduction is proportional to the magnitude of the temperature gradient and opposite to it in sign. q = −k dT dx The constant of proportionality, k, is called the thermal conductivity with a unit of W/mK or Btu/hftF. The heat flux represents the thermal power passing through a unit area (or the thermal energy per second per a unit area). Thus, it has a unit of W/m2. The heat flux is a vector quantity. In a three-dimensional condition:  q = −kT Chapter 1.3 Modes of Heat Transfer: Heat Conduction Heat conduction is a form of energy exchange through electron, atomic or molecular direct collision in a substance that has temperature differences. For gases and liquids, thermal conduction occurs primarily due to molecular collisions. For ionic and covalent materials, thermal conduction happens mainly because of atomic core collisions. For metals, the collisions of free-moving electrons, other than those of vibrating atomic cores, are the dominant contributor to thermal conduction. So, for metallic materials, a good electrical conductor is also a good heat conductor. Chapter 1.3 Modes of Heat Transfer: Heat Conduction Gas, due to low density and few molecular collisions, has very low conductivity. For the same solid material, it is less thermally conductive when it is in a porous form than when it is in a solid form. This explains why good insulation materials are fiber, foam or porous materials. Pure metals are good thermal conductors. When containing alloying elements, the thermal conductivity of a pure metal will decrease due to impurity atom scattering. In general, k solid  k liquid  k gas Chapter 1.3 Modes of Heat Transfer: Heat Conduction Example 1.2: A copper slab (k = 372 W/mK) is 3 mm thick. A 2-mm-thick stainless steel (k = 17 W/mK) is placed on the both sides of the copper slab. Find the temperature distribution in the copper slab and the heat flux through the wall in the steady state. Solution: Assume the steady-state heat flux is q. Then  400 C T1 T2 100  C 400  C − T1 q = −17W / m  K  0.002 m T1 − T2 q = −372W / m  K  0.003 m T2 − 100  C q = −17W / m  K  0.002 m T1 = 255 C T2 = 245 C q = 1.233106 W / m2 Chapter 1.3 Modes of Heat Transfer: 1-D Heat Conduction Equation Over the time period of t, the net thermal energy received by the finite element are: A0 tqx − A0 tqx + x If the temperature in the finite element is increased by T over the time period of t, because of energy conservation: A0 tqx − A0 tqx + x = c( xA0 )T − x x + x A0 qx q x + x q T = c x t q T = c lim x → 0 x t → 0 t − lim − q T = c x t   T  T k  = c x  x  t Continuum equation for heat conduction Chapter 1.3 Modes of Heat Transfer: 1-D Heat Conduction Equation   T  T k  = c x  x  t If the k is a location-independent constant, the equation can be written: 1 T  2T c T  =  t k t x 2 The  is called the thermal diffusivity. It has a unit of m2/s. Chapter 1.3 Modes of Heat Transfer: Heat Convection Convection is a heat transfer mechanism in which heat is conveyed away from a hot surface or to a cold surface through a moving gas or liquid fluid. Convection does not take place in solid materials. There are two convection methods: natural convection and forced convection. Natural convection: Natural convection is also called free convection. It occurs due to the spontaneous fluidic circulation resulting from the different buoyancies along a hot or cold surface. Because natural convection needs a cold, heavier fluid to replace a hot, lighter fluid, it can only occur in a gravitational field. Chapter 1.3 Modes of Heat Transfer: Heat Convection Forced convection: It results from the artificially forced fluid movement due to an external force, such as a fan or a pump. Forced convection is typically used to increase the rate of heat exchange. Chapter 1.3 Modes of Heat Transfer: Heat Convection The heat flux resulting from convection can be generalized to be: q = h(Tbody − T ) The h is called the film coefficient or heat transfer coefficient. It has a unit of W/m2K. Chapter 1.3 Modes of Heat Transfer: Heat Convection Heat convection is complicated. As a result, the heat transfer coefficient is not really a constant but depends on the boundary condition and the temperature change across the surface layer (T). 1) Under forced convection and if T is not too large, the h is relatively constant. 2) Under natural convection (that is, when fluid buoys up from a hot body or down from a cold one): h  T 1/ 4 or T 1/ 3 3. If the body is hot enough to boil a liquid surrounding it:. h  T 2 Chapter 1.3 Modes of Heat Transfer: Lumped-Capacity Solution An object is cooled down by heat convection. The object is so small that the temperature inside it is virtually uniform. Find the temperature change with time. Solution: The heat flux releasing from the object by heat convection is: q = h (T − T ) The energy of the object lost per t through convection is: Ti Elost = qAt = Ath (T − T ) The decrease of the object’s internal energy during t is: Einternal = −( V )cT According to the energy conservation: −( V )cT = Ath (T − T ) Let’s assume: = cV hA Ah lim T = − (T − T ) lim t T →0 t → 0 cV 1 dT = − (T − T )dt  All the physical parameters are lumped to be a time constant T Chapter 1.3 Modes of Heat Transfer: Lumped-Capacity Solution An object is cooled down by heat convection. The object is so small that the temperature inside it is virtually uniform. Find the temperature change with time. dT 1 = − dt T − T  1 dT = − (T − T )dt  d (T − T ) 1 = − dt T − T  1 d ln(T − T ) = − dt  T − T = Ce−t / Because T = Ti at t =0 Ti − T = Ce −0 / The final solution is determined to be: C = Ti − T T − T = e −t /  Ti − T Chapter 1.3 Modes of Heat Transfer: Biot Number Biot number is defined as: hL Bi = k L : The length of heat conduction 2L 0.1 1. If Bi1, the heat conduction inside the object is the limiting step for the overall heat transfer. The conduction inside the object is the one to be considered. The problem needs to solve the continuum equation:  2T 1 T = 2  t x Chapter 1.3 Modes of Heat Transfer: Heat Convection Example 1.4: A thermocouple is a round bead 1mm in diameter. It is initially at room temperature and is suddenly placed in a 200C gas flow. The heat transfer coefficient is 250 W/m2K. The values of k,  and c are 45 W/mK, 9,300 kg/m3 and 0.18 kJ/kgK, respectively. Evaluate the temperature response of the thermocouple. Solution: h L (250W / m 2 K )(0.001m) / 2 The Biot number is: Bi = = 0.00278  0.1 = 45W / mK k Thus, the lumped-capacity method can be used to solve this problem. The time constant is: 4 = cV hA = c  R 3 3 h 4R 2 cD 9,300 kg / m  0.18 kJ / kgK  0.001 m = = 6  250W / m 2 K 6h = 1.116 s Using the lumped-capacity solution: T − T = e −t /  Ti − T T − 200  C −t / 1.116 s = e 20  C − 200  C T = 200 − 180 e −t / 1.116 s  C Chapter 1.3 Modes of Heat Transfer: Radiation Chapter 1.3 Modes of Heat Transfer: Thermal Radiation Thermal radiation is an emission of electromagnetic waves from a substance’s surface whose temperature is greater than the absolute zero. Electrons and atoms in a substance are in constant movement. The charged particles, such as electrons and protons, make random movements, generating a spectrum of electromagnetic waves, that is, lights. This process represents a conversion of thermal energy into electromagnetic or light energy. Some of these lights are visible; most are invisible. When these lights leave the substance, they carry away energy. In contrast to conduction and convection, thermal radiation does not require a medium. The electromagnetic waves travel easily through vacuum at the speed of light. Gases are transparent to most of electromagnetic waves except certain-frequency waves. Liquids, even clean water, quickly attenuate radiation. Most solids, except glasses and clear plastics, completely block radiation. Chapter 1.3 Modes of Heat Transfer: Thermal Radiation Black body: This is an object that absorbs all the energies of electromagnetic waves that it receives and reflects nothing. A black body is a perfect absorber and radiator. Chapter 1.3 Modes of Heat Transfer: Thermal Radiation The radiation spectrum of a black body depends on temperature. The locus of its maximum radiation is shifting to a higher frequency with increasing temperature. The total heat flux radiated by a standing-alone blackbody is described by the StefanBoltzmann law: qb (T ) = T 4 The Stefan-Boltzmann constant,  , is 5.670400 x 10-8 W/m2K4 and the T is the absolute temperature. Chapter 1.3 Modes of Heat Transfer: Thermal Radiation     Radiant heat exchange: Two objects are both black bodies. Object 1 radiates only to Object 2 and Object 2 radiates only to Object 1. The net heat transfer from Object 1 and Object 2 is: T2 T1 Q1→2 (T ) = A (T1 − T2 ) 4     4 Chapter 1.3 Modes of Heat Transfer: Thermal Radiation T2 Radiant heat exchange: Two objects are both black bodies. Object 1 radiates not only to Object 2 but also to other objects. In this case, the geometrical factor plays a role at the net heat transfer from Object 1 and Object 2: T1 Q1→2 (T ) = A1 F1−2 (T1 − T2 ) 4 4 Q2→1 (T ) = A2 F2−1 (T2 − T1 ) 4 T3 4 F1− 2  F2−1 A1 F1− 2 = A2 F2−1 Reciprocity Relation Chapter 1.3 Modes of Heat Transfer: Thermal Radiation Example 1.5: A black thermocouple measures the temperature in a chamber with black walls. If the air around the thermocouple is at 20C, the walls are at 100C, and the heat transfer coefficient between the thermocouple and the air is 75 W/m2K, what temperature will the thermocouple read? Solution: Assume the temperature is Ttc. The thermocouple loses heat by convection at a rate: q = h (T − T ) = 75W / m2 K  (Ttc − 20) K The thermocouple gains heat by radiation at a rate: q =  (T1 − T2 ) = 5.670410−8W / m2 K 4 [(100 + 273) 4 − (Ttc + 273) 4 ]K 4 4 4 At equilibrium, heat loses and gains at the same rate. Then, we have: 75W / m2 K  (Ttc − 20) K = 5.670410−8W / m2 K 4 [(100 + 273) 4 − (Ttc + 273) 4 ]K 4 Ttc = 28.4 Chapter 1.3 Modes of Heat Transfer: Thermal Radiation There exist no black bodies in reality. Non-black bodies absorb and emit less radiation than black bodies. For non-black bodies, the heat flux resulting from radiation is related to the emittance, . For a standing-alone non-black body, the Stefan-Boltzmann law becomes: q(T ) = T 4 The Stefan-Boltzmann constant,  , is 5.670400 x 10-8 W/m2K4 and the T is the absolute temperature. Chapter 1.3 Modes of Heat Transfer: Thermal Radiation     T2 T1     Two objects are non-black bodies. Object 1 radiates only to Object 2 and Object 2 radiates only to Object 1. The net heat from Object 1 and Object 2 is: Q1→2 (T ) = A1 (T1 − T2 ) 4 4 Chapter 1.3 Modes of Heat Transfer: Thermal Radiation T2 Two objects are non-black bodies. Object 1 radiates not only to Object 2 but also to other objects. In this case, the geometrical factor plays a role at the net heat from Object 1 and Object 2: T1 Q1→2 (T ) = A1F1−2 (T1 − T2 ) 4 T3 4 Chapter 1.3 Modes of Heat Transfer: Thermal Radiation A theoretical blackbody has an emittance value of 1. A material with an emittnce constant < 1 is called a gray body. For real materials, their emittance varies between 0 and 1 and is a function of wave frequency, surface temperature and surface features (roughness, texture, color, oxidation, etc.). Metals Al Cu Fe Ni Surface Condition Emittance Polished 0.04~0.06 Black anodized 0.8 Polished 0.04~0.05 Commercial sheet 0.15 Polished 0.05~0.07 Oxidized 0.64~0.78 Polished 0.07~0.17 Oxidized 0.37~0.57 Chapter 1.3 Modes of Heat Transfer: Radiation Chapter 1.3 Modes of Heat Transfer: Thermal Radiation How to prevent radiation? Heat Shielding Black or light color one, rough or smooth one? Does the thickness of the sheet matter? The End