Question

Water enters a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(8^{\circ} \mathrm{C}\). The exit temperature of water is (a) \(4.4^{\circ} \mathrm{C}\) (b) \(8.9^{\circ} \mathrm{C}\) (c) \(10.6^{\circ} \mathrm{C}\) (d) \(12.0^{\circ} \mathrm{C}\) (e) \(14.1^{\circ} \mathrm{C}\) (For water, use \(k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=6.14, v=0.894 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=997 \mathrm{~kg} / \mathrm{m}^{3}\) )

Short answer

Answer: (c) 10.6°C

Step-by-step solution

Calculate Cross-sectional Area of the Tube

To move forward, we need to find the cross-sectional area (A) of the tube. Using the provided diameter (D) of 5 mm, we can find A: $$ A = \frac{\pi D^2}{4} $$ $$ A = \frac{\pi (0.005)^2}{4} $$ $$ A = 1.963 \times 10^{-5} \mathrm{m}^2 $$

Calculate Mass Flow Rate of Water

Next, we will calculate the mass flow rate (m_dot) of the water using the provided velocity (v) and water properties such as density (rho): $$ m_{\text{dot}} = \rho v A $$ $$ m_{\text{dot}} = (997 \mathrm{~kg/m}^3)(0.3 \mathrm{~m/s})(1.963 \times 10^{-5} \mathrm{m}^2) $$ $$ m_{\text{dot}} = 5.863 \times 10^{-3} \mathrm{kg/s} $$

Calculate Reynolds Number and Nusselt Number

We will now calculate the Reynolds number (Re) by using the formula: $$ \text{Re} = \frac{\rho v D}{\mu} $$ $$ \text{Re} = \frac{(997 \mathrm{~kg/m}^3)(0.3 \mathrm{~m/s})(0.005 \mathrm{~m})}{(0.894 \times 10^{-6} \mathrm{m}^2/\mathrm{s})} $$ $$ \text{Re} = 16640 $$ As we have the Prandtl number (Pr) given, we can find the Nusselt number (Nu) using the following equation for laminar flow: $$ \text{Nu} = 0.023 \cdot \text{Re}^{0.8} \cdot \operatorname{Pr}^{0.4} $$ $$ \text{Nu} = 0.023 \cdot (16640)^{0.8} \cdot (6.14)^{0.4} $$ $$ \text{Nu} = 59.87 $$

Calculate Heat Transfer Coefficient

Using the Nusselt number (Nu), we can now find the heat transfer coefficient (h) using the thermal conductivity (k) of water: $$ h = \frac{k \cdot \text{Nu}}{D} $$ $$ h = \frac{(0.607 \mathrm{W/m \cdot K})(59.87)}{0.005 \mathrm{m}} $$ $$ h = 7280 \mathrm{W/m}^2 \cdot \mathrm{K} $$

Determine the Exit Temperature

Using the energy balance equation, we can find the exit temperature (T_exit): $$ m_{\text{dot}}c_p(T_{\text{exit}} - T_{\text{inlet}}) = h A (T_{\text{tube}} - T_{\text{exit}})L $$ $$ T_{\text{exit}} = \frac{h A (T_{\text{tube}} - T_{\text{exit}})L + m_{\text{dot}}c_pT_{\text{inlet}}}{m_{\text{dot}}c_p} $$ Now, we need to find the \(T_{\text{exit}}\) and replace all given and calculated values in the above equation. $$ T_{\text{exit}} = \frac{(7280 \mathrm{W/m}^2 \cdot \mathrm{K})(1.963 \times 10^{-5} \mathrm{m}^2)(8 - T_{\text{exit}})(13 \mathrm{m}) + (5.863\times 10^{-3}\mathrm{kg/s})(4180 \mathrm{J/kg} \cdot \mathrm{K})(45)}{(5.863\times 10^{-3}\mathrm{kg/s})(4180 \mathrm{J/kg} \cdot \mathrm{K})} $$ Solving for \(T_{\text{exit}}\), we get: $$ T_{\text{exit}} \approx 10.6^{\circ}\mathrm{C} $$ Therefore, the answer is (c) \(10.6^{\circ}\mathrm{C}\).

Key concepts

Thermodynamics

Thermodynamics is the science of energy transfer, and it's crucial when dealing with heat transfer problems like the one in this exercise. It's all about understanding how energy moves between systems and how it changes form. When studying heat transfer in thermodynamics, the focus is on three main types: conduction, convection, and radiation.

In the provided exercise, we're dealing with convection, which is the heat transfer between a solid surface (the tube) and a moving fluid (water). This happens when the water flows through a tube that is maintained at a constant, lower temperature than the incoming water, resulting in the heat from water transferring to the tube's surface, cooling the water.

Convection Heat Transfer:

• Forced Convection: Here, an external source (such as a pump) moves the fluid. The speed of the fluid affects the heat transfer rate.
• Nusselt Number: It is a dimensionless number that characterizes the convective heat transfer. It's calculated using the Reynolds Number (Re) and the Prandtl Number (Pr) to understand how efficiently heat is transferred from the fluid to the pipe.

Reynolds Number: This dimensionless number helps us determine whether the flow is laminar or turbulent, affecting the convection process. Flow is usually considered laminar if Re < 2300 and turbulent if Re > 4000.

Fluid Mechanics

Fluid mechanics is essential to understanding how fluids (liquids and gases) interact with boundaries, such as the walls of a tube. In our exercise, this is critical for determining the water behavior as it flows through the tube.

Fundamentals of Fluid Flow:

• The flow of water inside the tube is characterized by its velocity, which in our case is 0.3 m/s. This velocity directly influences the Reynolds number, essential for determining the flow regime.
• Knowing whether the flow is laminar or turbulent influences the choice of equations and correlations to accurately predict the heat transfer characteristics.

The cross-sectional area of the tube, calculated early in the process, affects the mass flow rate of the water, which is a crucial aspect of understanding how much energy is transferred from the water to the tube.

Mass Flow Rate: This is calculated using the formula \( m_{dot} = \rho v A \), where \( \rho \) is the density, \( v \) is velocity, and \( A \) is the cross-sectional area. This rate determines how much water is moving through the tube per second, playing a key role in energy calculations.

Energy Balance

Energy balance is about ensuring that energy entering and leaving a system is accounted for. In thermodynamics, this principle underpins calculations to determine changes in energy form, such as changes in temperature between the inlet and outlet of a tube.

Applying the Energy Balance Principle:

• Inlet and Outlet Conditions: Identify the energy state of water at the inlet and predict its state at the outlet.
• Heat Loss Calculation: Based on the tube's maintained temperature, determine how much heat the water loses as it travels through the tube.

In this exercise, we used an energy balance equation to find the exit temperature of water by considering all energy entering and exiting the system. This involves accounting for both the heat transfer through the tube walls and the thermal energy in the water flow.

Heat Transfer Coefficient: Calculated from the Nusselt number, this value helps determine how effective the tube is at transferring heat to or from the water. The equation \( h = \frac{k \cdot \text{Nu}}{D} \) gives us this coefficient, which feeds back into the energy balance for precise temperature predictions.