A liquid is contained in a reactor vessel at 115 bar absolute pressure. It is tr

Question

A liquid is contained in a reactor vessel at 115 bar absolute pressure. It is transferred to a storage vessel through a 50 mm internal diameter commercial steel pipe. The storage vessel is nitrogen blanketed, and pressure above the liquid surface is kept constant at 1500 N/m2 gauge. The total run of pipe between the two vessels is 200 m. The miscellaneous losses due to entry and exit losses, fittings, valves, etc., amount to 800 equivalent pipe diameters. The liquid level in the storage vessel is at an elevation 20 m below the level in the reactor. A turbine is fitted in the pipeline to recover the excess energy that is available, over that required to transfer the liquid from one vessel to the other. Estimate the power that can be taken from the turbine, when the liquid transfer rate is 5000 kg/h. Take the efficiency of the turbine as 70%. The properties of the fluid are density 895 kg=m3 , viscosity 0:76 mNm2s.

Explanation / Answer

Answer:

Given

Pipeline dia D = 50 mm = 0.05 m

Cross sectional area of pipe (A)= pi*D2/4 =3.14 * 0.052 /4 = 0.00196 m2

liquid mass transfer rate (M) = 5000 kg/h

Density of liquid (rho) = 895 kg/m3

liquid velocity in pipe (v) = M/rho*A*3600 = 0.792 m/s

Liquid viscosity (mue) = 0.76 mNm-2s

Reynolds number Re = rho*D*v/mue = 895 * 0.05 * 0.792/0.76*10-3 = 46634.2

Absolute roughness of commercial steel pipe r = 0.046 mm

Relative roughness rr = r/D = 0.046/50 = 0.00092

Friction factor f for the above Re and rr is 0.003. This value is obtained from graph relating these values.

Total length of pipeline including miscellaneous losses L = 200 + (800*0.05) = 240 m

Friction loss in pipeline delPf = 8*f*L*rho.v2/2D = 8*0.003*240*895*0.7922/2*0.05 = 32336.7 N/m2

The liquid level in the storage vessel is at an elevation 20 m below the level in the reactor.

Difference in elevation delZ= -20 m

Pressure P1 = 115 bar = 115*105 = N/m2

Pressure P2 = 1500 = N/m2

Pressure difference delP= P1 - P2 = 115*105 - 1500 = 11498500 N/m2

Energy balance

g*delZ + delP/rho - delPf/rho -W = 0

g - acceleration due to gravity 9.81 m/s2

W - Work done J/kg

9.81(-20) + (11498500/895) - (32336.7/895) - W = 0

W = 12615.15 J/kg

Power = W*M/eff*3600

eff is efficiency = 70% = 0.7

Power = 12615.15 * 5000/0.7*3600 = 25030 W = 25.03 kW

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