Although fires occur very infrequently in chemical operations, when they do occu

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Although fires occur very infrequently in chemical operations, when they do occur, they can be very destructive. The damage caused by a fire can be kept to a minimum if a properly-designed fire protection system is installed. There are many facets to a well-designed fire protection system, including the detection systems, the first protection hardware, and the fire fighters. The fire protection systems may rely on several kinds of fire extinguishing and control agents, including dry chemicals, foams, carbon dioxide or and water. Water is included as the major fire protection agent in almost all situations (there are a few cases where water may be incompatible with the equipment or materials being protected). Water is generally cheaper than other agents, and it is generally available in large quantities. The high heat capacity and high heat of vaporization of water also contribute to its suitability as a fire control agent. Water is also inert (with respect to most other materials and chemicals) and can be stored and delivered relatively easily. Thus, water is supplied to most plant locations for fire control. There are some problems to be considered when using water as a fire control agent. One of the first things to consider is that the quantity of water used to fight a fire is usually many times the theoretical minimum needed for extinguishment of control. Thus, there is a large amount of runoff water that must be disposed of. At first thought, it would seem that the runoff water could just be allowed to flow through drains and be disposed of as though it were rain water. In some cases, that can be done, but in many cases the nm off water will be contaminated with chemicals that are either process chemical s spilled during the fire or products of or combustion during the fire. In those cases, the water may be so contaminated that it cannot simply be discharged to a storm drain or sewage treatment system. It must be collected and treated before discharge. In almost all cases, the rate at which water is applied for fire protection is much greater than the rate at which water will fall during a rainstorm. Thus, special drainage systems must be provided to assure that the runoff water can be disposed of properly. The design of fire water distribution systems is performed in the same way as any set of hydraulic calculations, although variations in methodology are sometimes used to reduce the amount of work. In addition, the fire water system will usually have several branches and loops so that the flow calculations arc more complicated than they would be for a single pipeline leading from one point to another. The design calculations may also have to account for the fact that the flow may differ from time to time because a single fire water supply system will be used for an entire plant, and it is not likely that the entire system will be in operation at once. PROBLEM: In our case, water is supplied to a pump for distribution to a fire water spraying system. The water is taken from an outdoor reservoir 100 m from the pump inlet, and the water surface is 5 m above the inlet to the pump. The pump is to deliver water to a water spray system on top of several liquefied gas storage tanks 500 m from the pump, at a discharge elevation of 35 m. The pressure at the nozzle manifold must be at least 700 kPa and water is to be delivered to the nozzle manifold for the spraying system at a rate of 900 mVhr. The net positive suction head of the pump is 10 m. The pipeline from the pump to the nozzle manifold is 10-inch Schedule 40 steel pipe. Pressure losses caused by fining and valves upstream of the pump arc equivalent to 15 m of piping, and pressure losses caused by fittings and valves downstream of the pump arc equivalent to 50 m of piping. A) What is the minimum diameter for the supply piping? B) What is the power required to drive the pump? Assume the pump has 80 percent efficiency and give your answer in kW. C) For what outlet pressure should the pump be designed?

Explanation / Answer

A) We know that

Flow = Area * velocity

Hence 900 m3/hr = Pi/4*d^2 * 3 ( Assumed velocity in discharge is 3 m/s)

900/3600 m3/s = pi/4*d^2*3

Therefore (0.25/3)*4= pi*d^2

0.33 = 3.14* d^2

0.33/3.14 = d^2

d^2 = 0.1051

d = sqrt (0.1051) = 0.324 * 1000 = 324

Hence 324 is the minimum diameter for the supply piping.

b) We know that

Ph(kW) = q g h / (3.6 106)

Here q = flow rate = 900 m3/hr

= density = 1000 kg/m3

g = 9.8 m/s^2

h in m = 700 KPa = 71 m

Eff of pump = 0.80

By substituting all the value in above equation, we get Power required to drive the pump = 217.66 KW

c) Pump Outlet design pressure = 700 * 1.5 = 1050 kPa. (because design pressure = 1.5* operating pressure)

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