Industrial Machinery Repair Part Episode 2 Part 9 pptx

434 Steam Traps Figure 22.2 Float-and-thermostatic trap it critical to select a trap that can handle the specific pressure, capacity, and size requirements of the system. The key advantage of float-and-thermostatic traps is their ability for quick steam-system startup because they continuously purge the system of air and other noncondensable gases. One disadvantage is the sensitivity of the float ball to damage by hydraulic hammer. Float-and-thermostatic traps are an economical solution for lighter conden- sate loads and lower pressures. However, when the pressure and capacity requirements increase, the physical size of the unit increases and its cost rises. It also becomes more difficult to handle. Thermodynamic or Disk-Type Thermodynamic, or disk-type, steam traps use a flat disk that moves between a cap and seat (see Figure 22.3). Upon startup, condensate flow raises the disk and opens the discharge port. Steam, or very hot condensate entering the trap, seats the disk. It remains seated, closing the discharge port, as long as pressure is maintained above it. Heat radiates out through the cap, thus diminishing the pressure over the disk, opening the trap to discharge condensate. Steam Traps 435 Cap Disk Figure 22.3 Thermodynamic steam trap Wear and dirt are particular problems with a disk-type trap. Because of the large, flat seating surfaces, any particulate contamination, such as dirt or sand, will lodge between the disk and the valve seat. This prevents the valve from sealing and permits live steam to flow through the discharge port. If pressure is not maintained above the disk, the trap will cycle frequently. This wastes steam and can cause the device to fail prematurely. The key advantage of these traps is that one trap can handle a complete range of pressures. In addition, they are relatively compact for the amount of condensate they discharge. The chief disadvantage is difficulty in handling air and other noncondensable gases. Bimetallic A bimetallic steam trap, which is shown in Figure 22.4, operates on the same principle as a residential-heating thermostat. A bimetallic strip, or wafer, connected to a valve disk bends or distorts when subjected to a change in temperature. When properly calibrated, the disk closes tightly against a seat when steam is present and opens when condensate, air, and other gases are present. Two key advantages of bimetallic traps are: (1) compact size relative to their condensate load-handling capabilities, and (2) immunity to hydraulic- hammer damage. 436 Steam Traps Bimetal strip Figure 22.4 Bimetal trap Their biggest disadvantage is the need for constant adjustment or calibra- tion, which is usually done at the factory for the intended steam operating pressure. If the trap is used at a lower pressure, it may discharge live steam. If used at a higher pressure, condensate may back up into the steam system. Thermostatic or Thermal Element Thermostatic, or thermal-element, traps are thermally actuated using an assembly constructed of high-strength, corrosion-resistant stainless steel plates that are seam-welded together. Figure 22.5 shows this type of trap. Upon startup, the thermal element is positioned to open the valve and purge condensate, air, and other gases. As the system warms up, heat generates pressure in the thermal element, causing it to expand and throttle the flow of hot condensate through the discharge valve. The steam that follows the hot condensate into the trap expands the thermal element with great force, which causes the trap to close. Condensate that enters the trap during sys- tem operation cools the element. As the thermal element cools, it lifts the valve off the seat and allows condensate to discharge quickly. Thermal elements can be designed to operate at any steam temperature. In steam-tracing applications, it may be desirable to allow controlled amounts of condensate to back up in the lines in order to extract more heat from Steam Traps 437 Figure 22.5 Thermostatic trap the condensate. In other applications, any hint of condensate in the system is undesirable. The thermostatic trap can handle either of these conditions, but the thermal element must be properly selected to accommodate the specific temperature range of the application. Thermostatic traps are compact, and a given trap operates over a wide range of pressures and capacities. However, they are not recommended for condensate loads over 15,000 pounds per hour. Performance When properly selected, installed, and maintained, steam traps are relatively trouble-free and highly efficient. The critical factors that affect efficiency include capacity and pressure ratings, steam quality, mechanical damage, and calibration. Capacity Rating Each type and size of steam trap has a specified capacity for the amount of condensate and noncompressible gas that it can handle. Care must be taken to ensure that the proper steam trap is selected to meet the application’s capacity needs. 438 Steam Traps Pressure Rating As discussed previously, each type of steam trap has a range of steam pres- sures that it can effectively handle. Therefore, each application must be carefully evaluated to determine the normal and maximum pressures that will be generated by the steam system. Traps must be selected for the worst-case scenario. Steam Quality Steam quality determines the amount of condensate to be handled by the steam trap. In addition to an increased volume of condensate, poor steam quality may increase the amount of particulate matter present in the con- densate. High concentrations of solids directly affect the performance of steam traps. If particulate matter is trapped between the purge valve and its seat, the steam trap may not properly shut off the discharge port. This will result in live steam being continuously exhausted through the trap. Mechanical Damage Inverted-bucket and float-type steam traps are highly susceptible to mechan- ical damage. If the level arms or mechanical linkages are damaged or distorted, the trap cannot operate properly. Regular inspection and main- tenance of these types of traps are essential. Calibration Steam traps, such as the bimetallic type, must be periodically recalibrated to ensure proper operation. All steam traps should be adjusted on a regular schedule. Installation Installation of steam traps is relatively straightforward. As long as they are properly sized, the only installation imperative is that they are plumb. If the trap is tilted or cocked, the bucket, float, or thermal valve will not operate properly. In addition, a nonplumb installation may prevent the condensate chamber from fully discharging accumulated liquids. Steam Traps 439 Table 22.1 Common failure modes of steam traps THE PROBLEM THE CAUSES Trap will not discharge Will not shut-off Continuously blows steam Capacity suddenly falls off Condensate will not drain Not enough steam heat Traps freeze in winter Back flow in return line Back-pressure too high • Boiler foaming or priming • • Boiler gauge reads low • Bypass open or leaking • • Condensate load greater than design • Condensate short-circuits • Defective thermostatic elements • Dirt or scale in trap • • Discharge line has long horizontal runs • Flashing in return main • • High-pressure traps discharge into low-pressure return • Incorrect fittings or connectors • • Internal parts of trap broken or damaged • • • • Internal parts of trap plugged • • Kettles or other units increasing condensate load • Leaky steam coils • No cooling leg ahead of thermostatic trap • • Open by-pass or vent in return line • Pressure regulator out of order • Process load greater than design • Plugged return lines • Plugged strainer, valve, or fitting ahead of trap • Scored or out-of-round valve seat in trap • Steam pressure too high • System is air-bound • Trap and piping not insulated • Trap below return main • • Trap blowing steam into return • Trap inlet pressure too low • • Trap too small for load • Source: Integrated Systems Inc. 440 Steam Traps Operating Methods Steam traps are designed for a relatively constant volume, pressure, and condensate load. Operating practices should attempt to maintain these parameters as much as possible. Actual operating practices are determined by the process system, rather than the trap selected for a specific system. The operator should periodically inspect them to ensure proper operation. Special attention should be given to the drain line to ensure that the trap is properly seated when not in the bleed or vent position. Troubleshooting A common failure mode of steam traps is failure of the sealing device (i.e., plunger, disk, or valve) to return to a leak-tight seat when in its normal operating mode. Leakage during normal operation may lead to abnormal operating costs or degradation of the process system. A single 3 4  steam trap that fails to seat properly can increase operating costs by $40,000 to $50,000 per year. Traps that fail to seat properly or are constantly in an unload position should be repaired or replaced as quickly as possible. Regular inspection and adjustment programs should be included in the standard operating procedures (SOPs). Most of the failure modes that affect steam traps can be attributed to vari- ations in operating parameters or improper maintenance. Table 22.1 lists the more common causes of steam trap failures. Operation outside the trap’s design envelope results in loss of efficiency and may result in premature failure. In many cases, changes in the conden- sate load, steam pressure or temperature, and other related parameters are the root causes of poor performance or reliability problems. Careful atten- tion should be given to the actual versus design system parameters. Such deviations are often the root causes of problems under investigation. Poor maintenance practices or the lack of a regular inspection program may be the primary source of steam trap problems. It is important for steam traps to be routinely inspected and repaired to assure proper operation. 23 V-Belt Drives “Only Permanent Repairs Made Here” V-belt drives are widely used in industry and commercial applications. V-belts are utilized to transfer energy from a driver to the driven and usu- ally transfer one speed ratio to another through the use of different sheave sizes. V-belts are constructed for three basic components, which vary from maker to maker: 1 Load carrying section to transfer power. 2 Rubber compression section to expand sideways in the groove. 3 Cover of cotton or synthetic fiber to resist abrasion. Understanding the construction of V-belts assists in the understanding of belt maintenance. The standard V-belt must ride in the sheave properly. If the belt is worn or the sheave is worn, then you will have slippage of the belt and transfer of power, and speed will change resulting in a speed change to a piece of equipment. If a V-belt drive is located near oil, grease, or chemicals the V-belts could lose their capability through the deterioration of the belt material, again resulting in the reduction of energy transfer and quickly resulting in belt breakage or massive belt slippage. Introduction Belt drives are an important part of a conveyor system. They are used to transmit needed power from the drive unit to a portion of the conveyor system. This chapter will cover: 1 Various types of belts that are used to transmit power; 2 The advantages and disadvantage of using belt drives; 3 The correct installation procedure for belt drives; 442 V-Belt Drives Driven roll Drive motor Figure 23.1 Belt drive 4 How to maintain belt drives; 5 How to calculate speeds and ratios that will enable you to make corrections or adjustments to belt drive speeds; 6 How to determine belt length and sheave sizes when making speed adjustments. Belt Drives Belt drives are used to transmit power between a drive unit and a driven unit. For example, if we have an electric motor and a contact roll on a conveyor, we need a way to transmit the power from the electric motor to the roll. This can be done easily and efficiently with a belt drive unit. See Figure 23.1. Belt drives can consist of one or multiple belts, depending on the load that the unit must transmit. The belts need to be the matched with the sheave type, and they must be tight enough to prevent slippage. Examples of the different belt and sheave sizes are as follows: 1 Fractional horsepower V-belts: 2L, 3L, 4L, and 5L; 2 Conventional V-belts: A, B, A-B, C, D, and E; Conventional cogged V-belts: AX, BX, and CX; 3 Narrow V-belts: 3V, 5V, and 8V; Narrow cogged V-belts: 3VX and 5VX; 4 Power band belts: these use the same top width designations as the above belts, but the number of bands is designated by the number preceding V-Belt Drives 443 A B B A Figure 23.2 Examples of V-belts the top width designation. For example, a 3-ribbed 5V belt would be labeled 3/5V; 5 Positive-drive belts: XL, L, H, XH, and XXH. The size of the belt must match the sheave size. If they do not match, then the belt will not make proper contact with the sheave and will decrease the amount of load it can transmit. They may look something like the illustration in Figure 23.2. Usually a set of numbers will follow the belt designation. These numbers represent the actual length of the belt in inches. On conventional belts, the length is given for the inside length of the belt, and on narrow belts it is given for the outside length. An example of this would be a 5V750 belt; the size of the belt gives it the 5V and the outside length of 75.0" gives it the 750. More information about the specific belt dimensions can be found in the Goodyear Power Transmission Belt Drives manual. Belt Selection V-Belts V-belts are best suited for transmitting light loads between short range sheaves. They are excellent at absorbing shock. When an overload occurs, they will act as an overload device and slip, thereby protecting [...]... Figure 23 . 12) : 1 Diameter of the drive sheave 2 Diameter of the driven sheave 3 Center-to-center distance between the shafts Now use the following formula to solve the equation: Belt length = drive diameter × 3.14 driven diameter × 3.14 + 2 2 + center to center × 2 Use the formula above to find the belt length 6" × 3.14 12" × 3.14 + + 35" × 2 2 2 6" × 3.14 12" × 3.14 + + 35" × 2 98 .26 " or 98 " = 2 2 Belt... Over 16.0 O.D 42 deg 5V Under 10.0 O.D 38 deg 456 V-Belt Drives 10.01–16.0 O.D 40 deg Figure 23 .20 Sheave gauge Figure 23 .21 Final alignment 50 100 150 V-Belt Drives 457 35" Figure 23 .22 Belt tensioning tighten the belts to the manufacturer’s recommendation Be sure to measure deflection and tension This information can be found in a belt tension gauge’s information sheet See Figure 23 .22 Check for parallel... that the belt is riding on 2 The shaft rpm of the sheave With this information, we can use the following formula: FPM = diameter × 3.14 × rpm 12 Use this formula to find the speed of the following belt (see Figure 23 .13): diameter × 3.14 × rpm 12 6" × 3 14 × 1800 28 26 = 1" FPM = V-Belt Drives 451 Driven Drive 12" 6" 90 0 rpm 1800 rpm Figure 23 .13 Belt speed calculation Figure 23 .14 Belt maintenance Belt... the shaft is level See Figure 23 .18 454 V-Belt Drives Figure 23 .17 Sheave inspection gauges Figure 23 .18 Shaft alignment V-Belt Drives 455 25 " 25 " Figure 23 . 19 Sheave alignment 2 Next, make sure the shafts are parallel This is done by measuring at different points on the shaft and adjusting the shafts until they are an equal distance apart Make sure that the shafts are pulled in as close as possible... Figure 23 .10 Let’s change the speed of the driven shaft to 90 0 rpm (see Figure 23 .11): Driven shaft rpm = 6 × 1800 90 0 12 = 6 × 1800 90 0 Belt Length Many times when a mechanic has to change out belts, the numbers on the belts cannot be read So what should be done? Take a tape measure and wrap it around the sheaves to get the belt length? This is not a very accurate V-Belt Drives 4 49 Driven Drive _ 6" 90 0... Belt drive speed ratio of the sheaves, and now we can calculate the speed of the driven shaft by using the following formula (see Figure 23 .9) : Drive sheave diameter in inches × drive shaft rpm Driven sheave diameter 6 × 1800 Driven shaft rpm = 12 6 × 1800 90 0 = 12 Driven shaft rpm = Now we understand how changing the size of a sheave will also change the shaft speed Knowing this, we could also assume... sheave as the driven, then the speed will be equal See Figure 23 .8 If we change the size of the driven sheave, then the speed of the shaft will also change We know what the speed is of the electric motor and the size V-Belt Drives 447 Driven Drive 6" 6" 1800 rpm 1800 rpm Figure 23 .8 Shaft speed Driven Drive 12" 6" rpm 1800 rpm Figure 23 .9 Belt drive speed ratio of the sheaves, and now we can calculate... are also useful when the size of the sheave cannot be found stamped on it See Figure 23 .20 3 Install the sheaves on the shafts following the manufacturer’s recommendations Locate and install the first sheave, then use a straightedge or a string to line the other one up with the one previously installed See Figure 23 .21 Belt Installation Install the belt on the sheaves Never force a belt on with a screwdriver... replace one belt of a multibelt drive, it will be tighter than the others See Figure 23 .15 A belt that is tighter than the others in a set will pull all the load Store the old belts as a set You may be able to use part of the set on a drive requiring fewer belts Figure 23 .15 Belt tensioning V-Belt Drives 453 Figure 23 .16 Belt tension gauges Sheave and Belt Installation Proper tools must be selected... See Figures 23 .16 and 23 .17 When the proper procedures are followed for installing V-belts, they will yield years of trouble-free service Shaft and Sheave Alignment 1 The shafts must be parallel or the life of the belt will be shortened The first step is to level the shafts; this is done by placing a level on each of the shafts Then shim the low side until the shaft is level See Figure 23 .18 454 V-Belt . × 3.14 2 + driven diameter × 3.14 2 + center to center × 2 Use the formula above to find the belt length. Belt length = 6" × 3.14 2 + 12& quot; × 3.14 2 + 35" × 2 98 .26 " or 98 ". 2 98 .26 " or 98 " = 6" × 3.14 2 + 12& quot; × 3.14 2 + 35" × 2 450 V-Belt Drives 12& quot; 6" Driven Drive 35" Figure 23 . 12 Belt length example Multiple Sheaves When. = diameter × 3.14 × rpm 12 2 826 = 6" × 3. 14 × 1800 1" V-Belt Drives 451 12& quot; 6" Driven Drive 1800 rpm 90 0 rpm Figure 23 .13 Belt speed calculation Figure 23 .14 Belt maintenance Belt