Question Details

Consider a hydrodynamically and thermally fully-developed, steady fluid flow of 1 kg/s in a uniformly heated pipe with diameter of 0.1 m and length of 40 m. A constant heat flux of magnitude 15000 W/m2 is imposed on the outer surface of the pipe. The bulk-mean temperature of the fluid at the entrance to the pipe is 200 °C. The Reynolds number (Re) of the flow is 85000, and the Prandtl number (Pr) of the fluid is 5. The thermal conductivity and the specific heat of the fluid are 0.08 w.m1K-1 and 2600 J-kg1K-1, respectively. The correlation Nu = 0.023 Re0.8 Pr0.4 is applicable, where the Nusselt Number (Nu) is defined on the basis of the pipe diameter. The pipe surface temperature at the exit is _______ °C (round off to the nearest integer).

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Correct Answer :

Correct answer is : 321.267

Solution :

Correct Answer: The pipe surface temperature at the exit of the pipe is 321.267 °C (which rounds to 321 °C as the nearest integer).

Analysis of the Given Data and Image:
Based on the schematic diagram and parameters shown in the provided image:

  • Fluid mass flow rate: m˙=1kg/s
  • Pipe diameter: D=0.1m
  • Pipe length: L=40m
  • Uniform surface heat flux: q=15000W/m2
  • Fluid entrance bulk-mean temperature: Tb,inlet=200C
  • Reynolds number: Re=85000
  • Prandtl number: Pr=5
  • Thermal conductivity of the fluid: k=0.08W/(mK)
  • Specific heat of the fluid: Cp=2600J/(kgK)
  • Step 1: Calculate the bulk-mean temperature of the fluid at the exit (Tb,exit)
    We apply the energy conservation equation over the entire length of the pipe:

    Q = q ( π D L ) = m˙ Cp ( Tb,exit - Tb,inlet )

    Substituting the given values into the equation:

    15000 π 0.1 40 = 1 2600 ( Tb,exit - 200 )

    60000 π = 2600 ( Tb,exit - 200 )

    188495.559 = 2600 ( Tb,exit - 200 )

    Tb,exit - 200 = 188495.5592600 72.498

    Tb,exit 272.498 °C

    Step 2: Calculate the heat transfer coefficient (h)
    We use the Nusselt number correlation defined on the basis of pipe diameter:

    Nu = hDk = 0.023 Re0.8 Pr0.4

    Substituting the values of Re and Pr:

    h0.10.08 = 0.023 (85000)0.8 50.4

    h 1.25 = 0.023 8783.568 1.90365

    h 1.25 384.46

    h 307.5678W/(m2K)

    Step 3: Calculate the pipe surface wall temperature at the exit (Tw,exit)
    Using Newton's law of cooling for convection at the exit location:

    q = h ( Tw,exit - Tb,exit )

    Substituting the values:

    15000 = 307.5678 ( Tw,exit - 272.498 )

    Tw,exit - 272.498 = 15000307.5678 48.769

    Tw,exit 272.498 + 48.769 = 321.267 °C

    Rounding to the nearest integer, we find the pipe surface temperature at the exit is 321 °C.

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