Answered: QUESTION-) 236 °C steam flows in a…

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN:9781305387102

Author:Kreith, Frank; Manglik, Raj M.

Publisher:Kreith, Frank; Manglik, Raj M.

Chapter2: Steady Heat Conduction

Section: Chapter Questions

Problem 2.21P

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QUESTION-)

236 °C steam flows in a pipe which features are given below. Inner and outer diameters of pipe are 300 mm and 320 mm. Coefficient of heat transmission is 40 W/mK. It's external ambient is air and it's temperature is 20 °C. Take the temperature of the pipe's external surface 180 °C. It is known that the internal temp convection coefficient is 600 W/m^2K and external heat transfer coefficient is 5,9109 W/m^2K The pipe's length is 500 meter. You need to add the radiation in the calculation. The rate of the radiation emissivity is 0.85. You can take the environmental surface temp directly.

a-) Please draw this question on the resistance network.

b-) Please calculate the heat loss of pipe.

c-) By applying insulation to this pipe, it is desired to reduce the heat loss to 20% of the uninsulated loss (option a). Calculate the insulation thickness accordingly. Take glass wool as insulation material.(Conductivity of glass wool = 0.040 W/mK.)

Given,

The temperature of steam inside the pipe, Ti=236 °C=236+273.15K =509.15 K

The inner diameter of the pipe, di=300 mm=0.3 m

The outer diameter of the pipe, do=320 mm=0.32 m

The conductivity of the pipe, k=40 W/m.K

The convective coefficient at the inner surface, hi=600 W/m2.K

The convective coefficient at the outer surface, ho=5,9109 W/m2.K

The temperature at the pipe's outer surface, T°=180 °C=180+273.15=453.15 K

The ambient temperature, Ta=20 °C=20+273.15=293.15 K

The conductivity of the insulation material, kins=0.040 W/m.K

The radiation emissivity is, ε=0.85

The length of the pipe, L=500 m

The inner radius of the pipe, ri=di2=0.15 m

The outer radius of the pipe, ro=do2=0.16 m

Expert Solution

Step 1

Outer surface area (A₀)= $π \times d_{o} \times L$ = 502.65m²

Inner surface area (A_i)= $π \times d_{i} \times L$ =471.2388m²

Internal convective resistance R_ir= $\frac{1}{h_{i} A_{i}} = \frac{1}{600 \times 471.2388} = 3.5367 + 10^{- 6}$

Conduction resistance R_cond= $\frac{\ln (\frac{r_{o}}{r_{i}})}{2 πLk} = \frac{\ln (\frac{0.32}{0.30})}{2 \times π \times 500 \times 40} = 5.13581 \times 10^{- 7}$

Outer convective resistance R_or= $\frac{1}{h_{o} \times A_{o}} = \frac{1}{59109 \times 502.65} = 3.371 \times 10^{- 9}$

To radiative heat transfer coefficient (h_r), we equate,

$\begin{array}{rcl} h_{r} (T_{s} - T_{\infty}) & = & ε \times σ \times ({T_{S}}^{4} - {T_{\infty}}^{4}) \\ h_{r} & = & \frac{ε \times σ \times ({T_{s}}^{4} - {T_{\infty}}^{4})}{(T_{s} - T_{\infty})} \\ = & \frac{0.85 \times 5.67 \times 10^{- 8} \times (453 . 15^{4} - 293 . 15^{4})}{(453.15 - 293.15)} \\ = & 10.476 W / m^{2} k \end{array}$

Therefore Radiative resistance (R_r)= $\frac{1}{h_{r} \times A_{o}} = \frac{1}{10.476 \times 502.65} = 1.899 \times 10^{- 4}$

We can draw the resistance diagram as shown below

To find R_eq,

$\frac{1}{R_{e q}} = \frac{1}{R_{o r}} + \frac{1}{R_{r}}$

Therefore R_eq=3.370 $\times 10^{- 9}$

Total Resistance $\begin{array}{rcl} R & = & R_{i r} + R_{c o n d} + R_{e q} \\ = & 3.5367 \times 10^{- 6} + 5.13581 \times 10^{- 7} + 3.370 \times 10^{- 9} \\ = & 4.0536 \times 10^{- 6} \end{array}$

Step 2

To find heat loss,

Use formula,

$\begin{array}{rcl} Q & = & \frac{T_{i} - T_{\infty}}{R} \\ = & \frac{509.15 - 293.15}{4.0536 \times 10^{- 6}} \\ = & 53.285 \times 10^{6} \\ = & 53.285 M W \end{array}$

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