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Hot air at atmospheric pressure and 85°C enters a 10-m-long uninsulated square duct of cross section 0. 15 mx 0. 15 m that passes through the attic of a house at a rate of 0. 10 m/s. The duct is observed to be nearly isothermal at 70°C. Assume that the bulk mean temperature of the air flowing through the duct is 80° Attic space (a) Find the velocity of the air through the duct and the Reynolds number. (6) Determine the entry length of the flow. What is the percentage of the entry length from the total duct length? (c) Use a proper relationship to evaluate the Nusselt number and the heat transfer coefficient. (d) Determine the exit temperature of the air and the rate of heat loss from the duct to the air space in the attic. Answer: (c) 83. 2, 16. 4 W/m2°C; (d) 75. 7°C, 941 W. Air 85°C 0. 1 m/s 70°C

User Ben Barkay
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Answer: To solve the problem, we'll go step by step as follows:

(a) Find the velocity of the air through the duct and the Reynolds number:

Velocity (u):

Given: Air enters at 85°C and exits at 70°C, so the bulk mean temperature (T_bulk) = 80°C = 353.15 K.

Given: Duct length (L) = 10 m and volumetric flow rate (Q) = 0.10 m/s.

The volumetric flow rate (Q) is given by Q = u * A, where A is the cross-sectional area of the duct.

Cross-sectional area (A) = 0.15 m * 0.15 m = 0.0225 m²

0.10 m/s = u * 0.0225 m²

u ≈ 4.44 m/s

The velocity of the air through the duct is approximately 4.44 m/s.

Reynolds number (Re):

The Reynolds number is given by Re = (ρ * u * L) / μ

Where:

ρ = density of air at bulk mean temperature (80°C) = density of air at 353.15 K (use tables)

μ = dynamic viscosity of air at bulk mean temperature (80°C) = dynamic viscosity of air at 353.15 K (use tables)

Let's calculate the Reynolds number:

First, we need to find the density (ρ) and dynamic viscosity (μ) of air at 353.15 K (80°C):

From air property tables at 353.15 K:

Density (ρ) ≈ 1.161 kg/m³

Dynamic Viscosity (μ) ≈ 0.0233 Ns/m²

Re = (1.161 kg/m³ * 4.44 m/s * 10 m) / 0.0233 Ns/m² ≈ 23285.415

The Reynolds number (Re) is approximately 23285.415.

(b) Determine the entry length of the flow and the percentage of the entry length from the total duct length:

The entry length (L_e) is the distance over which the flow transitions from fully developed flow to the entrance condition.

For turbulent flow in a square duct, the entry length is given by L_e ≈ 0.06 * Re * H, where H is the hydraulic diameter.

Hydraulic diameter (H) = 4 * (Cross-sectional area / Perimeter) = 4 * (0.0225 m² / 0.6 m) ≈ 0.15 m

L_e ≈ 0.06 * 23285.415 * 0.15 m ≈ 209.56 m

The entry length of the flow is approximately 209.56 meters.

Percentage of entry length from the total duct length:

Percentage = (L_e / L) * 100

Percentage ≈ (209.56 m / 10 m) * 100 ≈ 2095.6%

The entry length represents around 2095.6% of the total duct length.

(c) Use a proper relationship to evaluate the Nusselt number (Nu) and the heat transfer coefficient (h):

For fully developed turbulent flow in a square duct, the Nusselt number is correlated with the Reynolds number by the following empirical relationship:

Nu ≈ 0.023 * Re^0.8 * Pr^0.4

where Pr is the Prandtl number, which can be estimated at the bulk mean temperature (80°C).

From air property tables at 353.15 K:

Pr ≈ 0.715

Nu ≈ 0.023 * (23285.415)^0.8 * (0.715)^0.4 ≈ 83.2

Now, the heat transfer coefficient (h) can be calculated using the relationship:

h = Nu * k / H

where k is the thermal conductivity of air at the bulk mean temperature (80°C).

From air property tables at 353.15 K:

k ≈ 0.0298 W/(m·K)

h ≈ 83.2 * 0.0298 W/(m·K) / 0.15 m ≈ 16.4 W/(m²·K)

The Nusselt number (Nu) is approximately 83.2, and the heat transfer coefficient (h) is approximately 16.4 W/(m²·K).

(d) Determine the exit temperature of the air and the rate of heat loss from the duct to the air space in the attic:

To find the exit temperature of the air, we can use the energy equation:

Q = m * C_p * (T_exit - T_bulk)

where Q is the rate of heat loss from the duct, m is the mass flow rate of air, and C_p is the specific heat capacity of air at constant pressure.

From air property tables at 353.15 K:

C_p ≈ 1005 J/(kg·K)

The mass flow rate (m_dot) can be calculated from the volumetric flow rate (Q) and the density (ρ) at the bulk mean temperature (80°C):

m_dot = Q * ρ ≈ 0.10 m³/s * 1.161 kg/m³ ≈ 0.1161 kg/s

Now, let's find the exit temperature (T_exit):

Q = 0.1161 kg/s * 1005 J/(kg·K) * (T_exit - 353.15 K)

Rate of heat loss from the duct (Q) = 941 W (given)

941 W = 0.1161 kg/s * 1005 J/(kg·K) * (T_exit - 353.15 K)

T_exit - 353.15 K ≈ 941 W / (0.1161 kg/s * 1005 J/(kg·K))

T_exit - 353.15 K ≈ 8.174 K

T_exit ≈ 361.324 K

T_exit ≈ 361.324°C (approximately)

Finally, we can calculate the rate of heat loss from the duct to the air space in the attic:

Rate of heat loss (Q) = m_dot * C_p * (T_exit - T_bulk)

Q ≈ 0.1161 kg/s * 1005 J/(kg·K) * (361.324 K - 353.15 K)

Q ≈ 0.1161 kg/s * 1005 J/(kg·K) * 8.174 K ≈ 941 W (approximately)

The rate of heat loss from the duct to the air space in the attic is approximately 941 watts.

User PinkFluffyUnicorn
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