Let's break down the problem into clear steps and solve both parts (a) and (b): --- ## **Given Data** - **Diameter of pipe, \( D \):** 89 mm = .089 m - **Emissivity, \( \epsilon \):** .8 - **Steam (surface) temperature, \( T_s \):** 200°C = 473 K - **Surrounding temperature, \( T_\infty \):** 20°C = 293 K - **Stefan-Boltzmann constant, \( \sigma \):** \( 5.67 \times 10^{-8} \) W/m²·K⁴ - **Wind speed (for part b):** 8 m/s --- ## **Step 1: Surface Area per Unit Length** For a pipe of diameter \( D \), the surface area per meter length is: \[ A = \pi D \times 1 \] \[ A = \pi \times .089 \approx .28 \text{ m}^2/\text{m} \] --- ## **Step 2: Radiative Heat Loss** The heat loss by radiation per unit length: \[ q_{rad} = \epsilon \sigma A (T_s^4 - T_\infty^4) \] Plug in the values: \[ T_s = 473 \text{ K} \] \[ T_\infty = 293 \text{ K} \] \[ q_{rad} = .8 \times 5.67 \times 10^{-8} \times .28 \times (473^4 - 293^4) \] Calculate \( 473^4 \) and \( 293^4 \): \[ 473^4 = 5.0062 \times 10^{10} \] \[ 293^4 = 7.3866 \times 10^9 \] \[ 473^4 - 293^4 = 4.2676 \times 10^{10} \] Now, substitute: \[ q_{rad} = .8 \times 5.67 \times 10^{-8} \times .28 \times 4.2676 \times 10^{10} \] \[ = .8 \times 5.67 \times .28 \times 4.2676 \times 10^{2} \] \[ = .8 \times 5.67 \times .28 \times 426.76 \] Calculate step by step: - \( 5.67 \times .28 = 1.5876 \) - \( 1.5876 \times 426.76 = 677.39 \) - \( .8 \times 677.39 = 541.91 \) So, \[ \boxed{q_{rad} \approx 542\, \text{W/m}} \] --- ## **Step 3: Convective Heat Loss** The convective heat loss is given by: \[ q_{conv} = h_c \cdot A \cdot (T_s - T_\infty) \] Where \( h_c \) is the convective heat transfer coefficient. ### **(a) Calm Day (Natural Convection)** For a horizontal cylinder, the empirical correlation for natural convection is: \[ Nu = .36 + \frac{.518 Ra_D^{1/4}}{[1 + (.559/Pr)^{9/16}]^{4/9}} \] But a simpler approximation for air: \[ Nu = C \cdot Ra_D^n \] For \( 10^4 < Ra_D < 10^9 \), \( C = .53 \), \( n = .25 \). #### **Calculate Properties at Film Temperature** Film temperature: \[ T_f = \frac{T_s + T_\infty}{2} = \frac{200 + 20}{2} = 110^\circ C = 383\,\text{K} \] **Air properties at \( 110^\circ C \) (approx):** - \( k = .030\,\text{W/m·K} \) - \( Pr = .7 \) - \( \nu = 27 \times 10^{-6}\,\text{m}^2/\text{s} \) - \( \beta = 1/T_f = 1/383 = .00261\,\text{K}^{-1} \) #### **Calculate Grashof Number (\( Gr_D \)):** \[ Gr_D = \frac{g \beta (T_s - T_\infty) D^3}{\nu^2} \] Where \( g = 9.81\,\text{m/s}^2 \) \[ Gr_D = \frac{9.81 \times .00261 \times (200 - 20) \times (.089)^3}{(27 \times 10^{-6})^2} \] \[ = \frac{9.81 \times .00261 \times 180 \times .000704}{7.29 \times 10^{-10}} \] \[ = \frac{9.81 \times .00261 \times .12672}{7.29 \times 10^{-10}} \] \[ = \frac{9.81 \times .0003307}{7.29 \times 10^{-10}} \] \[ = \frac{.003243}{7.29 \times 10^{-10}} \] \[ = 4.45 \times 10^6 \] #### **Rayleigh Number (\( Ra_D \)):** \[ Ra_D = Gr_D \cdot Pr = 4.45 \times 10^6 \times .7 = 3.12 \times 10^6 \] #### **Nusselt Number (\( Nu \)):** \[ Nu = .53 Ra_D^{.25} = .53 \times (3.12 \times 10^6)^{.25} \] \[ = .53 \times (3.12 \times 10^6)^{.25} \] \[ = .53 \times 42.2 = 22.4 \] #### **Heat Transfer Coefficient (\( h_c \)):** \[ h_c = \frac{Nu \cdot k}{D} = \frac{22.4 \times .030}{.089} = 7.55\,\text{W/m}^2\text{K} \] #### **Convective Heat Loss:** \[ q_{conv} = h_c \cdot A \cdot (T_s - T_\infty) = 7.55 \times .28 \times 180 = 380.88\,\text{W/m} \] --- ### **Total Heat Loss (Calm Day):** \[ q_{total} = q_{rad} + q_{conv} = 542 + 381 = \boxed{923\,\text{W/m}} \] --- ## **Step 4: Forced Convection (Windy Day, \( v = 8\,\text{m/s} \))** For a cylinder in cross-flow: \[ Nu = .193 \cdot Re_D^{.618} \cdot Pr^{1/3} \] Where: \[ Re_D = \frac{v D}{\nu} = \frac{8 \times .089}{27 \times 10^{-6}} = 26363 \] \[ Nu = .193 \times (26363)^{.618} \times (.7)^{1/3} \] First, \( (26363)^{.618} \approx 502 \) \[ Nu = .193 \times 502 \times .887 = .193 \times 445 = 85.9 \] \[ h_c = \frac{Nu \cdot k}{D} = \frac{85.9 \times .030}{.089} = 28.94\,\text{W/m}^2\text{K} \] \[ q_{conv} = h_c \cdot A \cdot (T_s - T_\infty) = 28.94 \times .28 \times 180 = 1458.5\,\text{W/m} \] --- ### **Total Heat Loss (Windy Day):** \[ q_{total} = q_{rad} + q_{conv} = 542 + 1459 = \boxed{2001\,\text{W/m}} \] --- ## **Summary Table** | Condition | Conv. Loss (W/m) | Rad. Loss (W/m) | Total Loss (W/m) | |---------------------|------------------|-----------------|------------------| | Calm (natural) | 381 | 542 | 923 | | Windy (forced, 8 m/s)| 1459 | 542 | 2001 | --- ## **Final Answers** **(a) Heat loss per unit length on a calm day:** \[ \boxed{923\,\text{W/m}} \] **(b) Heat loss per unit length on a breezy day (8 m/s wind):** \[ \boxed{2001\,\text{W/m}} \] --- If you need clarification on any step or want more details, just ask!