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Using the nomogram and charts (FIGURES 1 and 2) on pages 4 and 5,determine the diameter of a suitable air main for the distribution of6 m3 min–1 FAD. System pressure is 6 bar, the length of the pipe run is 175 metres, and a maximum pressure drop of 0.3 bar is allowed.
The distribution main must also include the following :
• 4 bends (r = 2d)
• 2 elbow fittings
• 6 tee connectors
• 2 diaphragm valves.

2. Describe the relevant advantages and disadvantages of using steel, copper or plastic pipe for a distribution main.

3. What problems may occur if a pipe is specified :
• larger
• smaller
than the optimum value?

4. A compressor delivers 300 s–1 of free air into a pipe at a pressure of 6 bar gauge. Using the pressure drop formula:
calculate the minimum diameter of pipe if the pressure drop in a system is to be limited to 0.3 bar when is delivered through a pipe of equivalent length 160 m

Explain the difference in operation between a ‘macro’ and a ‘micro’ oil mist lubricator. State where you would use a ‘micro’ in preference to a ‘macro’ lubricator.

6. List four possible causes of overheating on a multi-stage reciprocating compressor.

7. Produce a system monitoring chart that could be used by a machine operator to maintain a daily record of hydraulic system performance details.

8. State the precautions to be taken when changing the fluid in a hydraulic system from a mineral oil based hydraulic fluid to a fire resistant fluid

Advantages and disadvantages of using steel, copper or plastic pipes for a distribution main

Using the nomogram and charts (FIGURES 1 and 2) on pages 4 and 5, determine the diameter of a suitable air main for the distribution of 6 m3 min–1 FAD.  System pressure is 6 bar, the length of the pipe run is 175 metres, and a maximum pressure drop of 0.3 bar is allowed.

The distribution main must also include the following:

  • 4 bends (r = 2d)
  • 2 elbow fittings
  • 6 tee connectors
  • 2 diaphragm valves.

Pipe flow, Q = 6m3/min

                  Q = 6 × 1000/60

                  Q = 100dm3/s

Let the initial diameter of the pipe be dmm

We, therefore, assume that the initial velocity is 15m/s

Air density at 6 bar = 1.225 × 6

                                = 7.36kg/m3

The initial area, A = Q/60/15

                              = 0.006667m2

Initial diameter, d = (4 × A/π)0.5

                                             = 0.092m

                              = say 100mm

The total equivalent length LT = Lpipe + Lbends + Ltee + Lelbow + Ldiaphragm

LT = 175 + (4 × 0.8) + (2 × 7) + (6 × 10) + (2 × 6)

LT = 264.2m

From the nomogram (refer image) we get 2.2mbar/m pressure drop. Hence, total pressure drop is 2.2 × 10-3×264.2 = 0.58bar > 0.3bar

Hence we need to go for 150mm diameter pipe which is beyond the nomogram.

Describe the relevant advantages and disadvantages of using steel, copper or plastic pipe for a distribution main.

Steel pipes are available in various types depending on the percentage of carbon present.

Advantages.

  • High resistance to abrasion
  • Highly durable and practically indestructible, providing a long lifespan
  • There are minimal noise and vibration between the fittings
  • Cheaper and easier to install in comparison to other materials due to convenient lengths

Disadvantages.

  • They are heavy and require special means of transport
  • They not advisable to use in areas that are inaccessible
  • Due to heavyweight, these pipes are usually shorter in lengths thus leading to many joints (De Volder and Reynaerts, 2010).
  • They are prone to corrosion due to rusting

Copper pipes

Advantages

  • They are easy to install
  • They are more flexible than most of the other materials
  • They are more resistant to natural forces and disasters
  • They have a high resale value
  • They have high fire resistant properties/qualities.

Disadvantages

  • High acquisition and installation costs
  • It is prone to corrosion in acidic conditions
  • They cannot handle extreme temperatures. This is because the pipes crack when heated and bursts when exposed to a lot of colds.
  • The water gets a metallic taste due to mild reaction to the copper
  • Due to high thermal conductivity thorough lagging is needed (De Volder and Reynaerts, 2010).
  • Due to good electrical conductivity, the pipe should be grounded to avoid short-circuiting

Plastic pipes

Plastic pipes are available in various forms such as PVC, PEX, CPVC, and PP amongst others.

Advantages

  • They have rust resistant properties
  • They have a smooth surface which leads to reduced friction between the fluid and the walls of the pipe.
  • They are light in weight
  • They are available in long lengths
  • They have good elastic properties
  • They are cheap to acquire and install
  • They are easy to carry therefore have reduced cost of transportation and handling
  • They are less prone to transmission leakages

Disadvantages

  • Plastic lose its strength at high temperatures
  • Plastic is non-biodegradable therefore harmful to the environment (not environmentally friendly).
  • Not suitable for high-pressure applications
  • They are brittle in nature
  1. What problems may occur if a pipe is specified?
  • Larger
  • Smaller

Than the optimum value?

Solution.

Let us consider that for a given application the flow rate, Q is constant.

Flow rate, Q = Area × Velocity

Area, A = πr2 where r is the radius of the pipe.

Suppose, the pipe being used is small in size:

To maintain constant flow rate, Q, if the radius is less than the optimum radius the velocity of flow in the pipe will be more.

As the velocity increases the head losses due to friction in the pipe will increase. (Head loss V2)

Hence the difference in total energy at the entry and exit of the pipe will be more, that is, energy losses are higher in case of undersized pipes and the external power required to maintain the flow will be more as compared to oversized pipes (Brusselman et al, 2010).

If a pipe has a diameter less than the optimum diameter then the pressure inside the pipe will increase which may be too much for the pipe to handle and the pipe may crack or burst. So optimum value is very important while constructing a pipe.

Calculation of minimum diameter of the pipe for a pressure drop of 0.3 bar over 160 m

Suppose, the value of the pipe being used is bigger than the optimum value:

Larger pipe means that diameter is more than the optimum diameter value but still carrying fluid under pressure only and always running full (Gandossi, 2013).

If flow rate, Q is the same in both the larger pipe and the optimum one and radius, R increases then velocity, V decreases. Hence, head losses caused by friction decreases.

If a pipe has a diameter exceeding the optimum diameter then the pressure inside the pipe will drop and more work has to be done to recover that lost pressure. This will lead to increase in power consumption of the pump which in turn will increase the total expenditure.

In Conclusion, the transmission losses with the use of undersized pipes are higher as compared to oversized pipes for the same length of pipes.

  1. A compressor delivers 300  s–1 of free air into a pipe at a pressure of 6 bar gauge.   Using the pressure drop formula:

Calculate the minimum diameter of the pipe if the pressure drop in a system is to be limited to 0.3 bar when is delivered through a pipe of equivalent length 160 m.

Pressure drop =800 2 5 31 lQ Rd.

Given the following data

                              Flow rate, Q = 300lt/s

                              Pipe length, L = 160m

                              Pressure drop in the system = 0.3 bar

Solution.

Consider the pressure drop equation

                              Pdrop = 800lQ2/Rd531

Write the Reynold’s number equation.

                             Re = ρvd/µ

Here,

               ρ = density of air

               v = velocity

               µ = viscosity

               v (velocity) = Q/a

              Friction factor = 64/ Re

Calculation of velocity.

                v = 0.3/0.25πd2

Substitute the above values in the Reynolds equation.

Re = {1.225 × (0.3 ÷ π/4d2)}/{16.96 × 10-6}

Re = (0.0276 × 106)/d

R = 64/ Re

R = (64 × d)/ (0.0276 × 106)

R = 2318.8 × 10-6 d

Substitute the values in the pressure drop equation

Pdrop = (800 × 160 × 0.32)/ (2318.8 × 10-6)d5.31

d5.31 = (4.968 × 106)/ (0.3 × 105)

d = 2.25m

This is the minimum diameter of the pipe.

  1. Explain the difference in operation between a ‘macro’ and a ‘micro’ oil mist lubricator.  State where you would use a ‘micro’ in preference to a ‘macro’ lubricator.

Solution.

Macro oil mist lubricator

  • It contains more oil of up to 80%
  • They contain very large and irregular shaped particles that are weakly dispersed in the fluid.
  • Emulsions form when they are combined with water and the oil droplets spread over a wide area of the water surface forming a diameter of over 1 micron.
  • They have a milky appearance.

Micro oil mist lubricator

  • It contains less oil content of only 40%
  • They have very small and regular shaped particles that strongly are dispersed in the fluid.
  • Similarly to macro oil mist lubricator, emulsions form when added to water and the oil droplets spread over a wide area of the water but with a diameter of less than 1 micron.
  • They have a translucent appearance.

Micro mist oil lubrication is preferable in the intake manifold of an engine and a situation where the pneumatic piping system is used. In addition, micro-mist lubrication is preferred for long pipes and cylinders with complicated shapes (Izquierdo et al, 2010).

  1. List four possible causes of overheating on a multi-stage reciprocating compressor.

Solution.

There are several reasons for a compressor to run hot or overheat. Such reasons include high compression ratio, high return gas temperatures, and lack of external cooling.

High compression ratio

High compression ratios are the result of either lower than normal suction pressure or higher than normal discharge pressures. Change in suction pressure will affect the compression ratio more rapidly than changes in the discharge pressure. For this reason, it is important to keep the suction pressure at its high possible value and also keeping the discharge pressure within normal operating conditions is still important (Brusselman et al, 2010). 

Reasons for compressor overheating

When the returning gas temperatures are high the, an overheating will occur in a multi-stage reciprocating compressor.

Low suction pressure.

Usually, the compression process roughly follows the constant entropy line on the pressure-enthalpy diagram. With a constant discharge pressure, when the process of compression begins on a constant entropy line of a higher value, the resultant temperature of discharge is higher. Starting the process of compression on a higher constant entropy line occurs in two ways, that is, lower suction pressure and higher suction temperature. It is therefore important to operate the system with the highest possible suction pressure (Comba et al, 2011).

An increase in suction temperature results in a corresponding increase in discharge temperature. This means that the higher the suction temperature the higher the discharge temperature thus leading to overheating.

  1. Produce a system monitoring chart that could be used by a machine operator to maintain a daily record of hydraulic system performance details.

Hydraulic system performance checklist

No.

Checklist.

Good condition.

Poor condition.

Remarks.

1

Check the lubricant tank oil level

2

Check the hydraulic tank level

3

Check the pressure of the hydraulic system.

4

Check the clamping stock

5

Check the clamping devices

6

Check the clamping pressure of the hydraulic system for chuck/fixture

7

Check the chuck function

8

Lubricate the clamping device if any manual.

9

Check the coolant supply.

10

Check the selectable zero offset entries

11

Clean the oil filters

12

Observe the oil pipe connections.

  1. State the precautions to be taken when changing the fluid in a hydraulic system from a mineral oil based hydraulic fluid to a fire-resistant fluid.

Solution.

The unique characteristics of fire-resistant fluids make it necessary to carefully consider the modifications required before converting to a fire-resistant fluid or when converting from one type of fire-resistant fluid to another.

When changing the fluid in a hydraulic system from a mineral oil based hydraulic fluid to a fire resistant fluid, the following precautions should be taken as discussed.

All of these fluids have a higher specific gravity than does mineral oil based fluid. As a result, the pump inlet may need to be enlarged to avoid cavitation. Long inlet lines must be avoided, and inlet fluid strainers should be no finer than 60 mesh.

It may also be necessary to design and install additional filtration circuits on the discharge and return lines because fire-resistant fluids tend to suspend contamination much more readily than mineral base oils (Izquierdo et al, 2010).

Before installing the fire-resistant fluid, the entire system must be flushed thoroughly to remove any residual contamination. Adding Varsol to the system and operating all of the circuits before draining the original hydraulic oil can help to dislodge sediment and sludge, which should be removed prior to the installation of the new fire-resistant fluid.

Make certain that compatibility tests are performed to ensure that seals, packing material and hoses will not deteriorate prematurely after the conversion is completed.

Synthetic fire-resistant fluids, such as phosphate ester, are not compatible with most standard sealing materials. Also, keep in mind that water glycol fire-resistant fluid will attack components with aluminium, zinc or magnesium materials (Comba et al, 2011).

The most important thing to remember when considering a conversion to a fire-resistant fluid is to become aware of all of the potential problems that can occur. Then carefully investigate all of the available options before rushing into a decision that could prove to be very costly.

Changing to a fire resistant fluid generally requires draining of existing fluid and thorough cleaning and flushing of the system to assume minimal contamination.

If the system contains painted surfaces, compatibility of the fluid with the paint should be verified as well as compatibility with seals, holes and other components of the system (Izquierdo et al, 2010).

In some cases, it is necessary to change the suction pipe size and filters to accommodate the fluid being used.

For any additional precautions, the specific manufacturers of both the system components and the fluid should be consulted.

Maintenance procedure objectives should include the following;

  • Keeping fluid clean and controlling contamination.
  • Maintaining proper temperatures.
  • Maintaining proper oil levels.
  • Periodic oil analysis.
  • Routine inspections:-noise levels, vibration, pressures, shock loads, leakage, temperatures, and foaming. 

References.

Brusselman, E., Moens, M., Steurbaut, W., and Nuyttens, D., 2010. Evaluation of hydraulic, pneumatic and mechanical agitation for the spray application of Steinernema carpocapsae (Rhabditida: Steinernematidae). Biocontrol Science and Technology, 20(4), pp.339-351. 

Comba, S., Di Molfetta, A. and Sethi, R., 2011. A comparison between field applications of nano-, micro-, and millimetric zero-valent iron for the remediation of contaminated aquifers. Water, Air, & Soil Pollution, 215(1-4), pp.595-607. 

De Volder, M. and Reynaerts, D., 2010. Pneumatic and hydraulic microactuators: a review. Journal of Micromechanics and microengineering, 20(4), p.043001.

Gandossi, L., 2013. An overview of hydraulic fracturing and other formation stimulation technologies for shale gas production. Eur. Commisison Jt. Res. Cent. Tech. Reports. 

Izquierdo, D., Azcona, R., Del Cerro, F.L., Fernandez, C. and Delicado, B., 2010, February. Electrical power distribution system (HV270DC), for application in more electric aircraft. In Applied Power Electronics Conference and Exposition (APEC), 2010 Twenty-Fifth Annual IEEE (pp. 1300-1305). IEEE.

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[Accessed 25 May 2024].

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