The flow rate of a hot coal/oil slurry in a pipeline is measured by injecting a

Question

The flow rate of a hot coal/oil slurry in a pipeline is measured by injecting a small side stream of cool oil and measuring the resulting temperature change downstream in the pipeline. The slurry is initially at 300F and has a density of 1.2 g/cm3 and a specific heat of 0.7 Btu/(lbm F). With no side stream injected, the temperature downstream of the mixing point is 298F. With a side stream at 60F and a flow rate of 1 lbm/s, the temperature at this point is 295F. The side stream has a density of 0.8 g/cm3 and a cp of 0.6 Btu/(lbm F). What is the mass flow rate of the slurry?

Explanation / Answer

Inlet
mass flow rate slurry = 'Ws' lbm/s
Tsi = 300F
cps = 0.7 Btu/(lbm F)
density, rhos = 1.2 gm/cm3

Sidestream:
mass flowrate of side stream = Wss = 1 lbm/s
Tssi = 60F
density, rhoss = 0.8 gm/cm3
cpss = 0.6 Btu/(lbm F)
Tssf = 295F

Downstream:
Temperature without side stream, Tf0= 298F
Temperature with side stream, Tssf= 295F

Enthalpy change of slurry without sidestream = DHs0 = Ws*cps*(Tf0-Ti)

Enthalpy change of slurry with sidestream = DHs = Ws*cps*(Tssf-Ti)

Enthalpy transferred to the side stream = DHs0 - DHs0 = Ws*cps*(Tf0 - Tssf)

Enthalpy change of the side stream, DHss = Wss*cpss*(Tssf-Tssi)

Energy balance:

Enthalpy transferred to the side stream = Enthalpy change of the side stream
DHs0 - DHs0 = DHss
Ws*cps*(Tf0 - Tssf) = Wss*cpss*(Tssf-Tssi)
Mass flow rate of slurry, Ws = Wss*cpss*(Tssf-Tssi)/{cps*(Tf0 - Tssf)}
Ws = 1*0.6(295-60)/{0.7*(298-295)} = 67.1 lbm/s

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