Question

A 10 -cm-diameter, 30-cm-high cylindrical bottle contains cold water at \(3^{\circ} \mathrm{C}\). The bottle is placed in windy air at \(27^{\circ} \mathrm{C}\). The water temperature is measured to be \(11^{\circ} \mathrm{C}\) after \(45 \mathrm{~min}\) of cooling. Disregarding radiation effects and heat transfer from the top and bottom surfaces, estimate the average wind velocity.

Short answer

The main factors affecting the convective heat transfer between a cylindrical bottle filled with water and air are the temperature difference between the solid surface and the fluid, the surface area of contact, and a heat transfer coefficient that characterizes the specific system. By following a step by step solution, we can estimate the average wind velocity based on the given data, such as initial and final water temperatures, air temperature, cooling duration, and dimensions of the bottle, and by using empirical correlations for forced convection in air.

Step-by-step solution

Define unknowns and given data

We are given: - Initial water temperature: \(T_{w1} = 3^{\circ}\mathrm{C}\) - Final water temperature: \(T_{w2} = 11^{\circ}\mathrm{C}\) - Air temperature: \(T_\infty = 27^{\circ}\mathrm{C}\) - Cooling duration: \(t = 45\) min - Diameter of the bottle: \(D = 10\) cm - Height of the bottle: \(H = 30\) cm - Neglected radiation effects We are asked to determine the average wind velocity \(V\).

Calculate the heat transferred

The mass of the water in the bottle can be calculated as follows: \(V_{water} = \pi(\frac{D}{2})^2 \times H\) To find the mass of the water, we use the density of water (\(\rho_{water}\)) and the volume of the water (\(V_{water}\)): \(m = \rho_{water} \times V_{water}\) Next, we can calculate the amount of heat transferred (\(Q\)) using the mass of the water, specific heat capacity of water (\(c_p\)), and temperature difference: \(Q = m \times c_p \times \Delta T\) Where \(\Delta T = T_{w2} - T_{w1}\).

Calculate the average heat transfer rate

The average heat transfer rate can be calculated as follows: \(\bar{q} = \dfrac{Q}{t}\)

Calculate the convective heat transfer coefficient

As we consider only convective heat transfer from the lateral surface area of the bottle, we can write the expression for the heat transfer rate as: \(\bar{q} = h A_s (T_\infty - T_s)\) Where \(h\) is the convective heat transfer coefficient, \(A_s\) is the lateral surface area of the bottle, \(T_\infty\) is the temperature of the air and \(T_s\) is the average surface temperature of the bottle. We can take the average surface temperature of the bottle \(T_s = \dfrac{T_{w1}+T_{w2}}{2}\). The lateral surface area of the bottle can be calculated as follows: \(A_s = \pi D \times H\) Solve the equation for the convective heat transfer coefficient \(h\): \(h = \dfrac{\bar{q}}{A_s (T_\infty - T_s)}\)

Estimate the average wind velocity

To solve for the wind velocity, we can use the relationship between the convective heat transfer coefficient (\(h\)) and the wind velocity, which can be given by an empirical correlation for forced convection in air: \(h = C \times V^n\) C and n are constants that depend on the flow regime (laminar or turbulent). For typical wind speeds and cylinder sizes, we can assume a turbulent flow regime and use \(C \approx 0.029\) and \(n = 0.8\). We can now solve for the wind velocity \(V\): \(V = (\dfrac{h}{C})^\frac{1}{n}\) By calculating the required values in each step, we will arrive at an estimate of the average wind velocity during the cooling process.

Key concepts

Convective Heat Transfer Coefficient

The convective heat transfer coefficient, represented as \( h \), plays a crucial role in determining the rate at which heat is transferred between a surface and a fluid flowing over it. In the context of the cylindrical bottle cooling process, \( h \) helps us measure how effectively the heat from the water inside the bottle is being lost to the surrounding windy air.

To calculate \( h \), we use the formula \( h = \frac{\bar{q}}{A_s (T_\infty - T_s)} \). In this equation:

• \( \bar{q} \) is the average heat transfer rate.
• \( A_s \) is the lateral surface area of the cylinder bottle. For a cylinder, \( A_s = \pi D \times H \).
• \( T_\infty \) is the air temperature, and \( T_s \) is the average surface temperature of the bottle.

Knowing \( h \) assists in calculating the wind velocity, as it bridges the information between thermal dynamics and fluid flow.

Forced Convection

Forced convection is a mechanism by which fluid motion enhances heat transfer. Unlike natural convection, where fluid movement is due to buoyancy, forced convection involves an external source such as a fan or wind. In our scenario, the windy air provides the forced convection, increasing the cooling rate of the bottle.

Forced convection can be described using empirical correlations that relate the convective heat transfer coefficient \( h \) to the conditions of the surrounding system. These correlations often take into account factors such as fluid velocity, cylinder dimensions, and the nature of the fluid flow, whether laminar or turbulent.

The increased effectiveness of forced convection over natural convection makes it particularly suitable for situations where rapid heat transfer is desired, like cooling processes.

Empirical Correlation

Empirical correlations are essential tools for estimating the convective heat transfer coefficient \( h \) when theoretical analysis is challenging. These correlations are derived from experiments and observations, providing a practical approach to solving complex heat transfer problems.

In the context of our exercise, the empirical correlation used is \( h = C \times V^n \). This formula allows us to express \( h \) in terms of average wind velocity \( V \). Here:

• \( C \approx 0.029 \) for turbulent air flow around cylinders.
• \( n = 0.8 \), which is typical for cylinders in airflow.

Such correlations simplify the real-world complexity of forced convection heat transfer, making it accessible for engineering calculations.

Cylinder Cooling Process

The cylinder cooling process in this exercise involves the transfer of thermal energy from the cylindrical bottle containing water, to the surrounding windy air, driven by forced convection. The shape and orientation of the cylinder, along with fluid properties, significantly influence the cooling efficiency.

The critical aspects of this process include:

• Determining the initial and final temperatures of the water, \( T_{w1} \) and \( T_{w2} \), to calculate the temperature change.
• Using the bottle's dimensions to find the lateral surface area, crucial for heat transfer calculations.
• Understanding that the neglect of radiation simplifies the model but assumes all energy loss is through convection.

Overall, analyzing this cooling process not only hones problem-solving skills but also illustrates the principles of heat transfer in practical applications.