Engineering, 11.02.2020 18:19 mnaiachickadee

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperature of 70°C and flow in subsea surrounding of 5°C. As a result of the temperature difference between the reservoir and the subsea surrounding, the knowledge of heat transfer is critical to prevent gas hydrate and wax deposition blockages. Consider a subsea pipeline with inner diameter of 0.5 m and wall thickness of 8 mm is used for transporting liquid hydrocarbon at an average temperature of 70°C, and the average convection heat transfer coefficient on the inner pipeline surface is estimated to be 250 W/m^2 K. The subsea surrounding has a temperature of 5°C and the average convection heat transfer coefficient on the outer pipeline surface is estimated to be 150 W/m^2 K. If the pipeline is made of material with thermal conductivity of 60 W/m K, by using the heat conduction equation.

a. Obtain the temperature Variation in the pipeline wall.
b. Determine the inner surface temperature of the pipeline.
c. Obtain the mathematical expression for the rate of heat loss from the liquid hydrocarbon in the pipeline.
d. Determine the heat flux through the outer pipeline surface.

Answers: 3

Show answers

Another question on Engineering

Engineering, 04.07.2019 18:10

Afluid flows with a velocity field given by v=(x/t)i.. determine the local and convective accelerations when x=3 and t=1.

Answers: 2

Answer

Engineering, 04.07.2019 18:10

For the closed feedwater heater below, feedwater enters state 3 at a pressure of 2000 psia and temperature of 420 °f at a rate of ix10 ibhr. the feedwat extracted steam enters state 1 at a pressure of 1000 psia and enthalpy of 1500 btu/lbm. the extracted er leaves at an enthalpy of 528.7 btu/lbm steam leaves as a saturated liquid. (16) a) determine the mass flow rate of the extraction steam used to heat the feedwater (10) b) determine the terminal temperature difference of the closed feedwater heater

Answers: 3

Answer

Engineering, 04.07.2019 18:10

Slip occurs via two partial dislocations because of (a) the shorter path of the partial dislocation lines; (b) the lower energy state through partial dislocations; (c) the charge balance.

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Aturning operation is performed with following conditions: rake angle of 12°, a feed of 0.35 mm/rev, and a depth of cut of 1.1 mm. the work piece is aluminum alloy 6061 with t6 heat treatment (a16061-t6). the resultant chip thickness was measured to be 1.0 mm. estimate the cutting force, fc. use shear stress of 207 mpa and coefficient of friction on the tool face of 0.6.

Answers: 1

Answer

You know the right answer?

In subsea oil and natural gas production, hydrocarbon fluids may leave the reservoir with a temperat...

Questions