Question

. A long 10-cm-diameter steam pipe whose external surface temperature is 110oC passes through some open area that is not protected against the winds. Determine the rate of heat loss from the pipe per unit of its length when the air is at 1 atm pressure and 10oC and the wind is blowing across the pipe at a velocity of 8 m/s. Use the following relation to calculate the Nusselt number.

Answer 1

The Nusselt number = 124

Step-by-step explanation:

Our assumption is that air is an ideal gas and that the radiation effect is negligible

Surface temperature,
`T_(s) = 110^(0) C`

`T_(\infty) = 10^(0) C`

Velocity, v = 8 m/s

The film temperature can be calculated as,
`T_(f) = (T_(s) + T_(\infty) )/(2)`

`T_(f) = (110 +10 )/(2) \\T_(f) = 60^(0) C`

At the film temperature,
`T_(f) = 60^(0) C` and 1 atm pressure, air has the following properties:

`K = 0.02808 W/m-k`

`P_(r) = 0.7202`

Reynold number,
`Re = (vD)/(V)`

D = 10 cm = 0.1 m

V = 1.896 * 10⁻⁵ m²/s

`Re = (0.1 * 8)/(1.896 * 10^(-5) )`

Re = 4.2194 * 10⁴

The Nusselt number will be calculated using the relation:

`Nu = 0.3 + (0.62 Re^(1/2) Pr^(1/3) )/([1 +( 0.41 Pr)^(2/3)] ^(1/4) ) + [1 + ((Re)/(282000) )^(5/8) ]^(4/5)`

Substituting Re = 4.2194 * 10⁴ and
`P_(r) = 0.7202` into the equation above

the Nusselt number, Nu = 124

Answer 2

`Nu = 30.311`

Step-by-step explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

`Nu = \left\{0.6+\frac{0.387\cdot Ra_(D)^{(1)/(6) }}{[1+\left((0.559)/(Pr) \right)^{(9)/(16)} ]^{(8)/(27) }} \right\}^(2)`, for
`Ra_(D) \le 10^(12)`.

Where
`Ra_(D)` is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

`Ra_(D) = (g\cdot \beta\cdot (T_(pipe)-T_(air))\cdot D^(3))/(\\u^(2))\cdot Pr`

By assuming that air behaves ideally, the coefficient of volume expansion is:

`\beta = (1)/(T)`

`\beta = (1)/(283.15\,K)`

`\beta = 3.532* 10^(-3)\,(1)/(K)`

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

`\\u = 1.426* 10^(-5)\,(m^(2))/(s)`

`\mu = 1.778* 10^(-5)\,(kg)/(m\cdot s)`

`k = 0.02439\,(W)/(m\cdot ^(\textdegree)C)`

`c_(p) = 1006\,(J)/(kg\cdot ^(\textdegree)C)`

The Prandtl number is:

`Pr = (\mu\cdot c_(p))/(k)`

`Pr = ((1.778* 10^(-5)\,(kg)/(m\cdot s) )\cdot (1006\,(J)/(kg\cdot ^(\textdegree)C) ))/(0.02439\,(W)/(m\cdot ^(\textdegree)C) )`

`Pr = 0.733`

Likewise, the Rayleigh number is:

`Ra_(D) = ((9.807\,(m)/(s^(2)) )\cdot (3.532* 10^(-3)\,(1)/(K) )\cdot (110^(\textdegree)C-10^(\textdegree)C)\cdot (0.1\,m)^(3))/((1.426* 10^(-5)\,(m^(2))/(s))^(2) )\cdot (0.733)`

`Ra_(D) = 12.486* 10^(6)`

Finally, the Nusselt number is:

`Nu = \left\{0.6+\frac{0.387\cdot (12.486* 10^(6))^{(1)/(6) }}{\left[1 + \left((0.559)/(0.733)\right)^{(9)/(16) }\right]^{(8)/(27) }} \right\}^(2)`

`Nu = 30.311`