Question

A fluid \(\left(\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}, \mu=1.4 \times 10^{-3} \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}\right.\), \(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and \(\left.k=0.58 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) flows with an average velocity of \(0.3 \mathrm{~m} / \mathrm{s}\) through a \(14-\mathrm{m}\) long tube with inside diameter of \(0.01 \mathrm{~m}\). Heat is uniformly added to the entire tube at the rate of \(1500 \mathrm{~W} / \mathrm{m}^{2}\). Determine \((a)\) the value of convection heat transfer coefficient at the exit, \((b)\) the value of \(T_{s}-T_{m}\), and (c) the value of \(T_{e}-T_{i}\).

Short answer

Question: Calculate the convection heat transfer coefficient at the exit, the temperature difference between the tube surface and the fluid's mean temperature, and the temperature difference between the exit and the inlet of the tube. given the fluid's density is 1000 kg/m³, the average velocity is 0.3 m/s, the tube's inside diameter is 0.01 m, the dynamic viscosity is 1.4 x 10⁻³ kg/(m·s), the specific heat capacity is 4.2 x 10³ J/(kg·K), the thermal conductivity is 0.58 W/(m·K), and the heat flux is 1500 W/m². Answer: To calculate the convection heat transfer coefficient (h), the temperature difference between the tube surface and the fluid's mean temperature (Ts - Tm), and the temperature difference between the exit and the inlet (Te - Ti), follow the step-by-step solution provided. Use the given fluid properties, average velocity, tube inside diameter, and heat flux to find the final answer.

Step-by-step solution

Calculate the Reynolds number

To determine the flow regime and the appropriate correlations for the heat transfer coefficient, we'll first need to calculate the Reynolds number (\(Re\)) for the flow. The Reynolds number is given by: \(Re = \frac{ρVD}{μ}\) where \(ρ\) is the fluid density, \(V\) is the average velocity, \(D\) is the inside diameter of the tube, and \(μ\) is the dynamic viscosity of the fluid. Using the given values, we can calculate the Reynolds number: \(Re = \frac{1000\ kg/m^3 × 0.3\ m/s × 0.01\ m}{1.4 × 10^{-3}\ kg/(m·s)}\ \)

Calculate the Prandtl number

The Prandtl number (\(Pr\)) is a dimensionless number that relates fluid properties to heat transfer and is given by: \(Pr = \frac{c_pμ}{k}\) where \(c_p\) is the specific heat capacity, \(μ\) is the dynamic viscosity, and \(k\) is the thermal conductivity of the fluid. Using the given values: \(Pr = \frac{4.2 × 10^3\ J/(kg·K) × 1.4 × 10^{-3}\ kg/(m·s)}{0.58\ W/(m·K)}\)

Determine the Nusselt number using the Dittus-Boelter equation

Since the flow is turbulent (\(Re > 2000\)), we can use the Dittus-Boelter equation to find the Nusselt number (\(Nu\)): \(Nu = 0.023 Re^{0.8} Pr^{n}\) For heating, the exponent \(n\) is equal to 0.3. We can plug in the calculated \(Re\) and \(Pr\) values: \(Nu = 0.023 × Re^{0.8} × Pr^{0.3}\)

Calculate the convection heat transfer coefficient (h)

The Nusselt number can be used to find the convection heat transfer coefficient, h, using the following equation: \(h = \frac{Nu × k}{D}\) Using the calculated Nusselt number, thermal conductivity, and tube diameter, we can find the value for h.

Determine the value of Ts - Tm (temperature difference)

To find the temperature difference between the tube surface and the fluid's mean temperature, we'll use the relationship between heat flux and heat transfer coefficient: \(q = h(T_s - T_m)\) The value of heat flux, q, is given as \(1500\ W/m^2\). Using the calculated h value, we can find the value of \((T_s - T_m)\).

Determine the value of Te - Ti (temperature difference between exit and inlet)

To find the temperature difference between the exit and the inlet, we'll use the energy balance equation: \(\dot{Q} = \dot{m}c_p(T_e - T_i)\) We can rearrange the equation to solve for \((T_e - T_i)\): \(T_e - T_i = \frac{\dot{Q}}{\dot{m}c_p}\) We need to find the mass flow rate (\(\dot{m}\)), which can be calculated using the flow properties and the cross-sectional area of the tube: \(\dot{m} = ρAV\) The cross-sectional area A can be calculated as: \(A = \frac{πD^2}{4}\) Now, we can substitute the given values and the calculated mass flow rate into the energy balance equation to find the value of \((T_e - T_i)\). The final results are: (a)\(h\), (b)\(T_s - T_m\), and (c)\(T_e - T_i\) which will be obtained by following the steps above and using the given data.

Key concepts

Convection

Convection is a crucial mechanism of heat transfer that occurs in fluids, such as liquids and gases. In this process, heat energy is transferred through the motion of fluid molecules. This can happen naturally, like the warming of air by a heater, or it can be forced, such as when a fan blows air through a heating coil.

In the given exercise, convection plays a key role as heat is uniformly added to the fluid flowing through a tube. To calculate the convection heat transfer coefficient (\(h\)) at the exit, it is essential to understand how this coefficient relates to the heat transfer mechanisms within the fluid.

The convection heat transfer coefficient essentially describes the efficiency of heat transfer between the solid surface and the fluid. It depends on various factors such as fluid velocity, fluid properties, and temperature difference between the surface and the fluid.

Understanding convection is vital for determining how effectively heat can be transferred in pipes, radiators, or any system where fluid movement is involved.

Reynolds Number

The Reynolds number (\(Re\)) is a dimensionless value used in fluid mechanics to predict flow patterns in different fluid flow situations. It indicates whether the flow is laminar or turbulent.

In this exercise, calculating the Reynolds number helps in assessing the flow regime of the fluid in the tube. A Reynolds number greater than 2000 suggests turbulent flow, while a number less than 2000 indicates laminar flow.

The formula for determining the Reynolds number is:\[ Re = \frac{ρVD}{μ} \]where:

• \(ρ\) is the fluid density,
• \(V\) is the fluid velocity,
• \(D\) is the characteristic length (diameter of the tube in this case),
• \(μ\) is the dynamic viscosity.

For the given exercise, calculating \(Re\) helps identify that the flow is turbulent, allowing us to use proper relations to calculate the Nusselt number and subsequently the heat transfer coefficient.

Nusselt Number

The Nusselt number (\(Nu\)) is another dimensionless number crucial in heat transfer analysis. It describes the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction.

For a better visualization, imagine a fluid flowing through a tube: the Nusselt number tells you how much better this fluid flow is at transferring heat compared to the same fluid amount that isn't moving (conduction only).

In turbulent flow, the Nusselt number can be estimated using empirical correlations like the Dittus-Boelter equation:\[ Nu = 0.023 Re^{0.8} Pr^{n} \]where:

• \(Re\) is the Reynolds number,
• \(Pr\) is the Prandtl number,
• \(n\) is 0.3 for heating.

Calculating the Nusselt number allows us to find the convection heat transfer coefficient.\[ h = \frac{Nu \times k}{D} \]This coefficient is critical for determining the temperature differences needed in the exercise.

Prandtl Number

The Prandtl number (\(Pr\)) is dimensionless and relates the viscosity of a fluid to its thermal diffusivity, essentially comparing the momentum diffusivity and thermal diffusivity of a fluid.

In simple terms, the Prandtl number indicates whether heat conduction or convection is more significant in energy transfer processes involving the fluid. A high Prandtl number indicates that convection is more dominant than conduction, which is often the case for many fluids.

It is defined as:\[ Pr = \frac{c_p μ}{k} \]where:

• \(c_p\) is the specific heat,
• \(μ\) is the dynamic viscosity,
• \(k\) is the thermal conductivity of the fluid.

In the context of the exercise, the Prandtl number helps in calculating the Nusselt number, contributing to understanding how the fluid's thermal properties interact with its velocity. This interaction is critical in determining heat transfer rates in systems.