Question

A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8

Answer 1

The rate of heat loss per unit length for a calm day is 0.000916 W/m. The rate of heat loss on a breezy day with a wind speed of 8 m/s is 96 W/m.

Step-by-step explanation:

(a) To calculate the rate of heat loss per unit length for a calm day, we can use the formula for heat transfer by radiation:

Q = emissivity × Stefan-Boltzmann constant × surface area × (T14 - T24)

Where Q is the rate of heat transfer, emissivity is the surface emissivity, and T1 and T2 are the temperatures of the steam and the surroundings, respectively.

Plugging in the values:

Q = 0.8 × (5.67 × 10-8 W/(m2 ×10 K4)) × π × (0.1 m)2 × (150 + 273)4 × (150 - 20)

Q = 0.000916 W/m

(b) To calculate the rate of heat loss on a breezy day with a wind speed of 8 m/s, we can use the formula:

Q = hA(T1 - T2)

Where Q is the rate of heat transfer, h is the convective heat transfer coefficient, A is the surface area, and T1 and T2 are the temperatures of the steam and the surroundings, respectively.

Plugging in the values:

Q = 8 × (0.1 m)2 × (150 - 20)

Q = 96 W/m

Answer 2

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Step-by-step explanation:

From the information given:

Diameter D
`= 100 mm = 0.1 m`

Surface emissivity ε = 0.8

Temperature of steam
`T_s` = 150° C = 423K

Atmospheric air temperature
`T_(\infty) = 20^0 \ C = 293 \ K`

Velocity of wind V = 8 m/s

To calculate average film temperature:

`T_f = (T_s+T_(\infty))/(2)`

`T_f = (423+293)/(2)`

`T_f = (716)/(2)`

`T_f = 358 \ K`

To calculate volume expansion coefficient

`\beta= (1)/(T_f) \\ \\ \beta= (1)/(358) \\ \\ \beta= 2.79 * 10^(-3) \ K^(-1)`

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

`Ra_(D) = (g * \beta (T_s-T_(\infty)) * D_b^3)/(\alpha* v)`

`Ra_(D) = (9.8 * (2.79 *10^(-3))(150-20) * (0.1)^3)/((31.244* 10^(-6)) * (21.7984* 10^(-6)))`

`Ra_(D) = 5.224 * 10^6`

The average Nusselt number is:

`Nu_D = \Big \{ 0.60 + (0.387(Ra_D)^(1/6))/([ 1+ (0.559/Pr)^(9/16)]^(8/27)) \Big \}^2`

`Nu_D = \Big \{ 0.60 + (0.387(5.224* 10^6)^(1/6))/([ 1+ (0.559/0.698)^(9/16)]^(8/27)) \Big \}^2`

`Nu_D = \Big \{ 0.60 + (5.0977)/([ 1.8826]^(8/27))\Big \}^2`

`Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2`

`Nu_D = 23.29`

However, for the heat transfer coefficient; we have:

`h_D = (Nu_D* k)/(D_b) \\ \\ h_D = ((23.29) * (30.608 * 10^(-3) ))/(0.1)`

`h_D = 7.129 \ Wm^2 .K`

Hence, Stefan-Boltzmann constant
`\sigma = 5.67 * 10^(-8) \ W/m^2.K^4`

Now;

To determine the heat loss using the formula:

`q'_b = q'_(ev) + q'_(rad) \\ \\ q'_b = h_D (\pi D_o) (T_t-T_(\infty))+\varepsilon(\pi D_b)\sigma (T_t^4-T_(\infty )^4)`

`q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^(-8)) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}`

Now; here we need to determine the Reynold no and the average Nusselt number:

`Re_D = (VD_b)/(v ) \\ \\ Re_D = (8 *0.1)/(21.7984 * 10^(-6)) \\ \\ Re_D = 3.6699 * 10^4`

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

`Nu_D = 0.3 + (0.62 * Re_D^(1/2)* Pr^(1/3))/([1+(0.4/Pr)^(2/3)]^(1/4)) [1+ ((Re_D)/(282000))^(5/8)]^(4/5)`

`Nu_D = 0.3 + (0.62 * (3.6699*10^4)^(1/2)* (0.698)^(1/3))/([1+(0.4/0.698)^(2/3)]^(1/4)) [1+ ((3.669*10^4)/(282000))^(5/8)]^(4/5)`

`Nu_D = (0.3 +(105.359)/(1.140)* 1.218) \\ \\ Nu_D = 112.86`

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

`h_D = (Nu_(D)* k)/(D_b) \\ \ h_D = (112.86*30.608 *10^(-3))/(0.1) \\ \\ h_D = 34.5 \ W/m^2 .K`

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

`q'b = h_D (\pi D_b) (T_s-T_(\infty)) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^(-8)) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}`