2. In the oil sands, before mined bitumen can be transported by pipeline it is d

Question

2. In the oil sands, before mined bitumen can be transported by pipeline it is diluted, in some cases with naphtha. Assume a y-junction as shown in the figure below is used to accomplish this dilution Bitumen (API gravity of 7, Q-0.43 m/s) and naphtha (p 647 kg/m) enter the junction in separate pipes and leave as a mixture. If the mixture leaving the junction must have a density of 925 kg/m and a flowrate of 0.96 m/s, what volumetric flowrate of naphtha is required? Bitumen & naphtha mixture Bitumen Naphtha Text, Figure P5.15 (7th ed.; modified)

Explanation / Answer

API gravity = 141.5/G- 131.5

given API gravity = 7

G= specific gravity

7= 141,5/G- 131.5

G= 141.5/(131.5+7)= 1.021

density of bitumen= specic gravity* density of water= 1.02*1000 =1020 kg/m3

let x= mass flow rate of naptha, mass flow rate of bitumen = volumetric flow rate of bitumen* density = 0.43 m3/s *1020 kg/m3=438.6 kg/s

since mass is conserved, mass flow rate entering = mass flow rate of bitumen+ mass flow rate of naptha = 438.6+x

this is also the mass flow rate leaving

volumetric flow rate of mixture leaving = 0.96 m3/s and density = 925 kg/m3, mass flow rate leaving = 925*0.96 kg/s=888 kg/s

hence 438.6+x= 888

x= 888-438.6 =449.4 kg/s

volumetric flow rate of naptha = mass flow rate/ density = 449.4kg/s / 647 kg/m3 =0.69 m3/s

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