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Consider a \(0.3\)-m-diameter and \(1.8-\mathrm{m}\)-long horizontal cylinder in a room at \(20^{\circ} \mathrm{C}\). If the outer surface temperature of the cylinder is \(40^{\circ} \mathrm{C}\), the natural convection heat transfer coefficient is (a) \(3.0 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(3.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(3.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(4.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(5.7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Short Answer

Expert verified
Solution: 1. Determine the given values. 2. Calculate the temperature difference. 3. Calculate the characteristic length. 4. Calculate the Grashof number. 5. Calculate the Prandtl number. 6. Calculate the Nusselt number. 7. Calculate the heat transfer coefficient. 8. Determine the closest matching value. After following the steps, we found that the natural convection heat transfer coefficient for the given horizontal cylinder is approximately \(1.381\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}\), which is closest to \(1.4\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}\). However, none of the options provided in the problem statement match this result.

Step by step solution

01

Determine the given values

The given values are as follows: - Diameter (d) of the cylinder: \(0.3\,\mathrm{m}\) - Length (L) of the cylinder: \(1.8\,\mathrm{m}\) - Room temperature (Tr): \(20^{\circ} \mathrm{C}\) - Outer surface temperature (Tc) of the cylinder: \(40^{\circ} \mathrm{C}\)
02

Calculate the temperature difference

Calculate the temperature difference between the cylinder surface and the room: $$T_{diff} = T_c - T_r = 40 - 20 = 20\,\text{K}$$
03

Calculate the characteristic length

For a horizontal cylinder, the characteristic length (Lc) is given by the diameter (d): $$L_c = d = 0.3\,\mathrm{m}$$
04

Calculate the Grashof number

The Grashof number (Gr) is a dimensionless number that measures the influence of buoyancy forces relative to viscous forces in natural convection. We assume the given parameters are for air, and use the following properties at the film temperature (\(\frac{T_r + T_c}{2}\)): - Coefficient of thermal expansion (beta): \(3.41 \times 10^{-3}\,\mathrm{K^{-1}}\) - Dynamic viscosity (mu): \(1.97 \times 10^{-5}\,\mathrm{kg \cdot m^{-1} \cdot s^{-1}}\) - Kinematic viscosity (nu): \(1.56 \times 10^{-5}\,\mathrm{m^2 \cdot s^{-1}}\) - Acceleration due to gravity (g): \(9.81\,\mathrm{m \cdot s^{-2}}\) $$Gr = \frac{g \cdot \beta \cdot T_{diff} \cdot L_c^3}{\nu^2} = \frac{9.81 \cdot 3.41 \times 10^{-3} \cdot 20 \cdot (0.3)^3}{(1.56 \times 10^{-5})^2} \approx 5.77 \times 10^8$$
05

Calculate the Prandtl number

The Prandtl number (Pr) is a dimensionless number that measures the ratio of momentum diffusivity to the thermal diffusivity. For air, at the film temperature, the thermal conductivity (k) is \(0.027\,\mathrm{W \cdot m^{-1} \cdot K^{-1}}\). The Prandtl number is given by the following formula: $$Pr = \frac{\mu \cdot c_p}{k} = \frac{1.97 \times 10^{-5} \cdot 1004}{0.027} \approx 0.72$$
06

Calculate the Nusselt number

We can use the Churchill-Chu correlation for the Nusselt number (Nu) for a horizontal cylinder in a natural convection situation: $$Nu = 0.60 + \frac{0.387 \cdot Gr^{1/6}}{[1+ (0.559/Pr)^{9/16}]^{8/27}}$$ $$Nu = 0.60 + \frac{0.387 \cdot (5.77 \times 10^8)^{1/6}}{[1+ (0.559/0.72)^{9/16}]^{8/27}} \approx 12.47$$
07

Calculate the heat transfer coefficient

The heat transfer coefficient (h) can be calculated using the Nusselt number and the thermal conductivity of air: $$h = \frac{Nu \cdot k}{L_c} = \frac{12.47 \cdot 0.027}{0.3} \approx 1.381\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}$$
08

Determine the closest matching value

The calculated heat transfer coefficient value is closest to \(1.4\,\mathrm{W \cdot m^{-2} \cdot K^{-1}}\), but none of the options provided in the problem statement match this result. It seems there is an issue with either the problem statement or the given answer options, as the calculated value does not correspond to any of the options provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Grashof Number
In natural convection, the Grashof number is crucial because it helps us evaluate the influence of buoyancy forces compared to viscous forces. These forces are key when it comes to the movement of fluid around an object like our horizontal cylinder. Essentially, the Grashof number helps predict whether flow will be turbulent or laminar, which affects heat transfer rates.

Here's a simple breakdown:
  • When the Grashof number is high, buoyancy forces dominate, often leading to turbulent flow. This can enhance heat transfer.
  • When it's low, viscous forces are stronger, generally resulting in smooth, laminar flow, which might reduce heat transfer efficiency.
To calculate it:\[ Gr = \frac{g \cdot \beta \cdot T_{\text{diff}} \cdot L_c^3}{u^2} \]Where:
  • \(g\) is the gravitational acceleration, theoretical driving force of convection.
  • \(\beta\) is the thermal expansion coefficient, indicating how much the fluid expands or contracts with temperature changes.
  • \(T_{\text{diff}}\) is the temperature difference between the surface and surrounding fluid.
  • \(L_c\) is the characteristic length, in this exercise being the diameter of the cylinder.
  • \(u\) is the kinematic viscosity, showcasing the fluid's resistance to flow.
Prandtl Number
The Prandtl number answers a key question: How does the momentum transfer compare to heat transfer within a fluid? It's a ratio that combines two types of diffusivities: how momentum spreads (viscosity), and how heat spreads (thermal conductivity). This number is essential for understanding fluid flow patterns and heat exchange efficiency in natural convection.

To compute the Prandtl number:
  • \(Pr = \frac{\mu \cdot c_p}{k}\)
  • \(\mu\) is the dynamic viscosity, representing the fluid's internal resistance to flow.
  • \(c_p\) is the specific heat at constant pressure, denoting the heat required to change the fluid's temperature.
  • \(k\) is the thermal conductivity of the fluid, showing its capacity to conduct heat.
A low Prandtl number, like in gases, means the thermal diffusivity is dominant, allowing heat to spread quickly. A higher number, common in liquids, implies the opposite, where heat conduction isn't as swift as momentum diffusion.
Nusselt Number
The Nusselt number is one of those terms you'll see popping up often in heat transfer studies. It bridges the gap between the theoretical and practical numbers by representing the enhancement of heat transfer through convection over mere conduction.

This number is vital as it tells us how well heat is being transferred across a surface—that surface being a horizontal cylinder in our scenario:
To determine it, use the Churchill-Chu correlation for natural convection around a cylinder:\[Nu = 0.60 + \frac{0.387 \cdot Gr^{1/6}}{[1+ (0.559/Pr)^{9/16}]^{8/27}}\]Where you're mixing in the Grashof number and the Prandtl number:
  • A high Nusselt number indicates an efficient transfer of heat.
  • Conversely, a lower number suggests that heat is more reliant on conduction, which is slower compared to convection.
Once you have the Nusselt number, it is then used to calculate the heat transfer coefficient, providing an understanding of the actual heat exchange capabilities in the setup. It's super practical for engineers trying to design efficient heating or cooling systems!

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Most popular questions from this chapter

The two concentric spheres of diameters \(D_{i}=20 \mathrm{~cm}\) and \(D_{o}=30 \mathrm{~cm}\) are separated by air at \(1 \mathrm{~atm}\) pressure. The surface temperatures of the two spheres enclosing the air are \(T_{i}=320 \mathrm{~K}\) and \(T_{o}=280 \mathrm{~K}\), respectively. Determine the rate of heat transfer from the inner sphere to the outer sphere by natural convection.

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

A hot fluid \(\left(k_{\text {fluid }}=0.72 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) is flowing as a laminar fully-developed flow inside a pipe with an inner diameter of \(35 \mathrm{~mm}\) and a wall thickness of \(5 \mathrm{~mm}\). The pipe is \(10 \mathrm{~m}\) long and the outer surface is exposed to air at \(10^{\circ} \mathrm{C}\). The average temperature difference between the hot fluid and the pipe inner surface is \(\Delta T_{\text {avg }}=10^{\circ} \mathrm{C}\), and the inner and outer surface temperatures are constant. Determine the outer surface temperature of the pipe. Evaluate the air properties at \(50^{\circ} \mathrm{C}\). Is this a good assumption?

Exhaust gases from a manufacturing plant are being discharged through a \(10-\mathrm{m}-\) tall exhaust stack with outer diameter of \(1 \mathrm{~m}\). The exhaust gases are discharged at a rate of \(0.125 \mathrm{~kg} / \mathrm{s}\), while temperature drop between inlet and exit of the exhaust stack is \(30^{\circ} \mathrm{C}\), and the constant pressure-specific heat of the exhaust gases is \(1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). On a particular calm day, the surrounding quiescent air temperature is \(33^{\circ} \mathrm{C}\). Solar radiation is incident on the exhaust stack outer surface at a rate of \(500 \mathrm{~W} / \mathrm{m}^{2}\), and both the emissivity and solar absorptivity of the outer surface are \(0.9\). Determine the exhaust stack outer surface temperature. Assume the film temperature is \(60^{\circ} \mathrm{C}\).

A solar collector consists of a horizontal copper tube of outer diameter \(5 \mathrm{~cm}\) enclosed in a concentric thin glass tube of \(9 \mathrm{~cm}\) diameter. Water is heated as it flows through the tube, and the annular space between the copper and glass tube is filled with air at 1 atm pressure. During a clear day, the temperatures of the tube surface and the glass cover are measured to be \(60^{\circ} \mathrm{C}\) and \(32^{\circ} \mathrm{C}\), respectively. Determine the rate of heat loss from the collector by natural convection per meter length of the tube.

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