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In a manufacturing plant that produces cosmetic products, glycerin is being heated by flowing through a \(25-\mathrm{mm}\)-diameter and \(10-\mathrm{m}\)-long tube. With a mass flow rate of \(0.5 \mathrm{~kg} / \mathrm{s}\), the flow of glycerin enters the tube at \(25^{\circ} \mathrm{C}\). The tube surface is maintained at a constant surface temperature of \(140^{\circ} \mathrm{C}\). Determine the outlet mean temperature and the total rate of heat transfer for the tube. Evaluate the properties for glycerin at \(30^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Question: Determine the outlet mean temperature and total rate of heat transfer for a glycerin flow inside a heated tube with given parameters. Answer: To determine the outlet mean temperature and total rate of heat transfer for the glycerin flow inside the heated tube, follow these steps: 1. Calculate the Reynolds number using the mass flow rate, density, diameter, and dynamic viscosity of glycerin. 2. Calculate the Nusselt number using the Reynolds number and Prandtl number. 3. Calculate the heat transfer coefficient using the Nusselt number, thermal conductivity of glycerin, and diameter of the tube. 4. Calculate the outlet mean temperature and total rate of heat transfer using the heat transfer coefficient, area of the tube, and logarithmic mean temperature difference.

Step by step solution

01

Calculate Reynolds number

Assuming the properties of glycerin at \(30^{\circ}\text{C}\) and using the mass flow rate, density, diameter, and dynamic viscosity of glycerin, we can calculate the Reynolds number (\(Re\)): \(Re = \frac{\rho v D}{\mu}\) Where: - \(\rho\) is the density of glycerin - \(v\) is the velocity of glycerin flow (calculated using mass flow rate) - \(D\) is the diameter of the tube - \(\mu\) is the dynamic viscosity of glycerin First, we need to calculate the velocity of glycerin flow using the mass flow rate, the density, and the area of the tube: \(v = \frac{\dot{m}}{\rho \pi (\frac{D}{2})^2}\) Now, we can substitute the values of density, velocity, diameter, and dynamic viscosity to calculate the Reynolds number.
02

Calculate Nusselt number

Using the Reynolds number calculated above, we can determine the Nusselt number (\(Nu\)). The Nusselt number is a dimensionless quantity that represents the ratio of convective heat transfer to conductive heat transfer. For laminar flow inside a heated tube, we can use the following equation (Gnielinski correlation): \(Nu = \frac{(f/8)(Re - 1000) Pr}{1 + 12.7\sqrt{(f/8)}(Pr^{2/3} - 1)}\) Here, \(f\) is the Darcy friction factor, \(Re\) is the Reynolds number, and \(Pr\) is the Prandtl number. The Darcy friction factor can be calculated using the following equation: \(f = 0.0791 Re^{-0.25}\)
03

Calculate heat transfer coefficient

The next step is to calculate the heat transfer coefficient (\(h\)) using the calculated Nusselt number: \(h = \frac{Nu k}{D}\) Where \(k\) is the thermal conductivity of glycerin.
04

Calculate outlet mean temperature and total rate of heat transfer

Finally, we can calculate the outlet mean temperature(\(T_{out}\)) using the logarithmic mean temperature difference(LMTD) method: \(\Delta T_{\text{ln}} = \frac{(T_s - T_{in}) - (T_s - T_{out})}{\ln\frac{(T_s - T_{in})}{(T_s - T_{out})}}\) Now, we can calculate the total rate of heat transfer(\(\dot{Q}\)) using the heat transfer coefficient, the area of the tube, and the logarithmic mean temperature difference: \(\dot{Q} = h A \Delta T_{\text{ln}}\) By solving for \(T_{out}\) in the LMTD equation above and substituting it in the heat transfer rate equation, we can find the outlet mean temperature and total rate of heat transfer for the glycerin flow inside the heated tube.

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Key Concepts

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

Reynolds Number
The Reynolds number (Re) is a dimensionless quantity used in fluid mechanics to predict flow patterns in different fluid flow situations. It compares the inertial forces to the viscous forces within the fluid and is an important parameter for determining whether a flow will be laminar or turbulent.

Calculated by the equation,
\(Re = \frac{\rho v D}{\mu}\),
where \(\rho\) is the fluid density, \(v\) the velocity of the fluid, \(D\) the characteristic diameter (in this case, the diameter of the tube), and \(\mu\) the dynamic viscosity of the fluid.

When dealing with internal flow in a pipe, if the Reynolds number is less than 2000, the flow is typically laminar; if it's greater than 4000, the flow is turbulent, and the range in-between is known as the transition zone. In our exercise, calculating the Reynolds number helps in the determination of the Nusselt number and consequently the convective heat transfer rate.
Nusselt Number
The Nusselt number (Nu) is another dimensionless parameter, representing the ratio of convective to conduction heat transfer at a boundary in a fluid. It is a vital number for engineers to predict the heat transfer from a fluid to a solid surface or vice versa.

\(Nu = \frac{h L}{k}\),
where \(h\) is the convective heat transfer coefficient, \(L\) a characteristic length, usually the diameter of the pipe in the case of internal flow, and \(k\) the thermal conductivity of the fluid. High Nusselt numbers indicate dominant convective heat transfer, whereas low Nusselt numbers suggest that conduction is prevailing.
  • In our exercise, we used the Gnielinski correlation to determine Nu, which is suitable for calculating the convective heat transfer for a flow within the transition or turbulent region.
  • The calculation involved using the Reynolds number and Darcy's friction factor, highlighting how the different dimensionless numbers interplay to solve heat transfer problems.
Convective Heat Transfer
Convective heat transfer is the transport of thermal energy due to fluid motion. It occurs when a fluid at a different temperature interacts with a surface or when the fluid mixes internally. Convective heat transfer is crucial in engineering applications such as cooling or heating systems.

The heat transfer coefficient (\(h\)), a key variable in convective heat transfer, quantifies how well heat is transported from a solid surface to the fluid or vice versa and is typically measured in watts per square meter-kelvin (\(W/m^2K\)).

Our exercise specifically looked at the process of heating glycerin by calculating the heat transfer coefficient using the Nusselt number, which involves understanding the specific properties of glycerin, such as its thermal conductivity, to ensure accurate calculations.
Logarithmic Mean Temperature Difference (LMTD)
The Logarithmic Mean Temperature Difference (LMTD) is an average temperature difference between the hot and cold streams in heat exchangers. It's important for designing and analyzing the performance of heat exchangers and is particularly effective when there is a significant temperature change throughout the exchanger.

An LMTD calculation is expressed as:
\(\Delta T_{\text{ln}} = \frac{(T_s - T_{in}) - (T_s - T_{out})}{\ln\frac{(T_s - T_{in})}{(T_s - T_{out})}}\),
where \(T_s\) is the surface temperature, \(T_{in}\) is the inlet fluid temperature and \(T_{out}\) is the outlet fluid temperature. LMTD provides a measure of the driving force behind heat transfer in flow systems.

In the exercise example, by using LMTD in conjunction with the heat transfer rate equation, one can solve for the outlet temperature of the glycerin and determine the heat exchanger's performance.

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Most popular questions from this chapter

Liquid water enters a 10 - \(\mathrm{m}\)-long smooth rectangular tube with \(a=50 \mathrm{~mm}\) and \(b=25 \mathrm{~mm}\). The surface temperature is maintained constant, and water enters the tube at \(20^{\circ} \mathrm{C}\) with a mass flow rate of \(0.25 \mathrm{~kg} / \mathrm{s}\). Determine the tube surface temperature necessary to heat the water to the desired outlet temperature of \(80^{\circ} \mathrm{C}\).

WWhich fluid at room temperature requires a larger pump to move at a specified velocity in a given tube: water or engine oil? Why?

Air at \(10^{\circ} \mathrm{C}\) enters an \(18-\mathrm{m}\)-long rectangular duct of cross section \(0.15 \mathrm{~m} \times 0.20 \mathrm{~m}\) at a velocity of \(4.5 \mathrm{~m} / \mathrm{s}\). The duct is subjected to uniform radiation heating throughout the surface at a rate of \(400 \mathrm{~W} / \mathrm{m}^{3}\). The wall temperature at the exit of the duct is (a) \(58.8^{\circ} \mathrm{C}\) (b) \(61.9^{\circ} \mathrm{C}\) (c) \(64.6^{\circ} \mathrm{C}\) (d) \(69.1^{\circ} \mathrm{C}\) (e) \(75.5^{\circ} \mathrm{C}\) (For air, use \(k=0.02551 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=0.7296, v=1.562 \times\) \(10^{-5} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=1.184 \mathrm{~kg} / \mathrm{m}^{3}\).)

In the fully developed region of flow in a circular tube, will the velocity profile change in the flow direction? How about the temperature profile?

Consider a fluid with a Prandtl number of 7 flowing through a smooth circular tube. Using the Colburn, Petukhov, and Gnielinski equations, determine the Nusselt numbers for Reynolds numbers at \(3500,10^{4}\), and \(5 \times 10^{5}\). Compare and discuss the results.

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