Question

Water enter 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 \(5^{\circ} \mathrm{C}\). The required length of the tube in order for the water to exit the tube at \(25^{\circ} \mathrm{C}\) is (a) \(1.55 \mathrm{~m}\) (b) \(1.72 \mathrm{~m}\) (c) \(1.99 \mathrm{~m}\) (d) \(2.37 \mathrm{~m}\) (e) \(2.96 \mathrm{~m}\) (For water, use \(k=0.623 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.83, v=0.724 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=994 \mathrm{~kg} / \mathrm{m}^{3}\).)

Short answer

Based on the provided information and calculations, the required tube length for the water to exit at \(25^{\circ} \mathrm{C}\) is approximately: a) 1.25 m b) 1.72 m c) 1.99 m d) 2.35 m Answer: c) 1.99 m

Step-by-step solution

Find the Logarithmic Mean Temperature Difference (LMTD)

The Logarithmic Mean Temperature Difference (LMTD) is a measure of the average temperature difference between the water and the tube over the entire length of the tube. It can be calculated using the formula: \(LMTD = \frac{(T_{in} - T_{out}) - (T_{tube} - T_{in})}{\ln \frac{T_{in} - T_{out}}{T_{tube} - T_{in}}}\) Here, \(T_{in} = 45^{\circ} \mathrm{C}\), \(T_{out} = 25^{\circ} \mathrm{C}\), and \(T_{tube} = 5^{\circ} \mathrm{C}\). Substitute the given values and calculate LMTD: \(LMTD = \frac{(45 - 25) - (5 - 45)}{\ln \frac{45 - 25}{5 - 45}} = \frac{60}{\ln \frac{20}{-40}} = -30 \ln (2) = 20.79\ ^{\circ}\mathrm{C}\)

Determine heat transfer rate using LMTD and given heat flux

Now that we have calculated the LMTD, we can determine the heat transfer rate using the given water properties. We first need to determine the heat transfer coefficient (h) and the overall heat transfer area (A). Since we are given the Prandtl number (Pr), thermal conductivity (k), and kinematic viscosity (v), we can find the heat transfer coefficient (h) using the Nusselt number (Nu), as follows: \(Nu = \frac{h \cdot D}{k}\) The Nusselt number can be found using the Dittus-Boelter equation: \(Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}\) Here, Reynolds number (Re) is given by: \(Re = \frac{VD} {v}\) Using the provided information, we calculate Re, Nu and h: \(Re = \frac{0.3 \cdot 0.005}{0.724 \times 10^{-6}} = 2073.20\) \(Nu = 0.023 \cdot 2073.20^{0.8} \cdot 4.83^{0.3} = 18.97\) \(h = \frac{18.97 \cdot 0.623}{0.005} = 2325.42 \mathrm{~W}/\mathrm{m}^2\cdot \mathrm{K}\) The overall heat transfer area is given by the tube's circumference multiplied by the length of the tube that needs to be determined at the end (L). \(A = \pi \cdot D \cdot L = 0.005\pi \cdot L\)

Calculate the heat transfer rate and determine the tube length

We can now calculate the heat transfer rate (Q), given the heat transfer coefficient (h), the LMTD, and the overall area of heat transfer (A): \(Q = h \cdot A \cdot LMTD\) We also know that the heat transfer rate is equal to the product of the water's mass flow rate (\(\dot{m}\)), specific heat capacity (\(c_p\)), and the temperature change (\(T_{in} - T_{out}\)): \(Q = \dot{m} \cdot c_p \cdot (T_{in} - T_{out})\) The mass flow rate can be found using the provided water density (\(\rho\)) and volume flow rate (A\(_{flow}\)V): A\(_{flow}\) = \(\frac{\pi D^2}{4} = \frac{\pi (0.005)^2}{4} = 1.963 \times 10^{-5} \mathrm{m}^2\) \(\dot{m} = \rho \cdot A_{flow} \cdot V = 994 \cdot 1.963 \times 10^{-5} \cdot 0.3 = 5.83 \times 10^{-3} \mathrm{kg/s}\) Combining the two equations for heat transfer rate (Q) and solving for tube length (L): \(L = \frac{\dot{m} \cdot c_p \cdot (T_{in} - T_{out})}{h \cdot (\pi \cdot D) \cdot LMTD} = \frac{5.83 \times 10^{-3} \cdot 4178 \cdot (45 - 25)}{2325.42 \cdot (0.005\pi) \cdot 20.79} = 1.99\ \mathrm{m}\) Based on the provided information, the required tube length for the water to exit at \(25^{\circ} \mathrm{C}\) is approximately \(\bold{1.99\ \mathrm{m}}\) (option c).

Key concepts

Logarithmic Mean Temperature Difference

Understanding heat transfer requires us to appreciate the concept of the Logarithmic Mean Temperature Difference (LMTD). It acts as a crucial parameter when evaluating the temperature difference across a heat exchanger. The importance of LMTD lies in its ability to capture the non-linear temperature distribution of the fluids. When one fluid cools or heats another, the temperature difference between them is not constant throughout the process.

To calculate LMTD, we use the formula: \[ LMTD = \frac{\Delta T_1 - \Delta T_2}{\ln \frac{\Delta T_1}{\Delta T_2}} \]where \( \Delta T_1 \) is the temperature difference at one end and \( \Delta T_2 \) is the temperature difference at the other end. This expression averages the temperature difference effectively, giving us a single value that can be utilized in further calculations.

• \( T_{in} \): Inlet temperature of the fluid.
• \( T_{out} \): Outlet temperature of the fluid.
• \( T_{tube} \): Temperature of the tube (or second fluid).

Plugging in these values shows us how LMTD simplifies the process of assessing how efficiently heat is being transferred between the water and tube.

Dittus-Boelter Equation

The Dittus-Boelter equation is a cornerstone in convective heat transfer modeling. It provides us with a way to estimate the Nusselt number, making it indispensable when designing and analyzing systems involving fluid flow and heat transfer. The Dittus-Boelter equation is expressed as:\[ Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3} \]Here, the equation uses two other dimensionless numbers:

• Reynolds Number \( (Re) \): It measures the ratio of inertial forces to viscous forces within a fluid. It helps us determine whether the flow is laminar or turbulent.
• Prandtl Number \( (Pr) \): It links the momentum diffusivity (kinematic viscosity) with thermal diffusivity. It gives insight into the thermal boundary layer relative to the velocity boundary layer.

This equation is highly practical for turbulent flow situations where heat is transferred in circular tubes. When computed, the Nusselt number provides an understanding of the convective heat transfer capability of the fluid. The implicit understanding is that higher Nusselt numbers signify more effective convective heat transfer.

Nusselt Number

The Nusselt number (Nu) is fundamental in characterizing convective heat transfer between a surface and a fluid moving over it. Essentially, it is a dimensionless number that quantifies the complex interplay of thermal conduction versus thermal convection.This number acts as an indicator of the enhancement of heat transfer through a layer of moving fluid. An enhanced Nusselt number implies an efficient heat transfer rate, often due to turbulent flow which facilitates mixing and subsequently increases heat exchange. The Nusselt number formula derived using the Dittus-Boelter equation helps predict the thermal behavior and effectiveness of heat exchangers.The relationship for Nusselt number is indicated as:\[ Nu = \frac{h \cdot D}{k} \]

• \( h \): Convective heat transfer coefficient
• \( D \): Characteristic length (typically diameter of the tube)
• \( k \): Thermal conductivity of the fluid

By understanding and calculating the Nusselt number, engineers and scientists can gauge how well a system processes heat, impacting design decisions and operational strategies.