/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 A hot fluid passes through a thi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hot fluid passes through a thin-walled tube of \(10-\mathrm{mm}\) diameter and 1-m length, and a coolant at \(T_{\infty}=25^{\circ} \mathrm{C}\) is in cross flow over the tube. When the flow rate is \(\dot{m}=18 \mathrm{~kg} / \mathrm{h}\) and the inlet temperature is \(T_{m, i}=85^{\circ} \mathrm{C}\), the outlet temperature is \(T_{m \rho}=78^{\circ} \mathrm{C}\). Assuming fully developed flow and thermal conditions in the tube, determine the outlet temperature, \(T_{m, a}\) if the flow rate is increased by a factor of 2 . That is, \(\dot{m}=36 \mathrm{~kg} / \mathrm{h}\), with all other conditions the same. The thermophysical properties of the hot fluid are \(\rho=\) \(1079 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=2637 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \mu=0.0034 \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), and \(k=0.261 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
When the flow rate of the hot fluid is doubled from 18 kg/h to 36 kg/h, the new outlet temperature, \(T_{m,a}\), can be determined by following these steps: 1. Determine the heat transfer rate in the original scenario using Newton's law of cooling. 2. Calculate the log mean temperature difference (LMTD) in the original scenario. 3. Determine the overall heat transfer coefficient (\(U \times A\)) using the heat transfer rate and LMTD. 4. Calculate the new heat transfer rate with the doubled flow rate. 5. Determine the new outlet temperature based on the new heat transfer rate, LMTD, and overall heat transfer coefficient. After following these steps and solving the equations, we can find the new outlet temperature, \(T_{m,a}\).

Step by step solution

01

Determine the heat transfer rate

For a fully developed flow and thermal conditions, we can use Newton's law of cooling to find the heat transfer rate, \(Q\): \[Q = \dot{m}c_p(T_{m,i} - T_{m \rho})\] Using the given values, we can calculate the heat transfer rate: \[Q = 18\frac{\mathrm{kg}}{\mathrm{h}}\times2637\frac{\mathrm{J}}{\mathrm{kg}\cdot\mathrm{K}}(85^{\circ}\mathrm{C} - 78^{\circ}\mathrm{C})\]
02

Calculate the log mean temperature difference

Next, we need to calculate the log mean temperature difference (LMTD), with the coolant temperature remaining constant at \(T_{\infty}=25^{\circ}\mathrm{C}\): \[\text{LMTD} = \frac{(T_{m,i} - T_{\infty}) - (T_{m \rho} - T_{\infty})}{\ln{\frac{T_{m,i} - T_{\infty}}{T_{m \rho} - T_{\infty}}}}\]
03

Determine the overall heat transfer coefficient

Now, we can use the heat transfer rate and the LMTD to find the overall heat transfer coefficient (\(U \times A\)), using the formula: \[Q = (U \times A) \times \text{LMTD}\] Solve this equation for (\(U \times A\)).
04

Calculate the new heat transfer rate

Next, we need to find the new heat transfer rate, \(Q_{new}\), given the doubled flow rate: \[Q_{new} = 2\dot{m}c_p(T_{m,i} - T_{m,a})\]
05

Determine the new outlet temperature

Finally, we need to calculate the new outlet temperature, given the new heat transfer rate, the LMTD, and the (\(U \times A\)) value obtained in step 3. We can rearrange the heat transfer rate formula, considering all values known except for the new LMTD: \[Q_{new} = (U \times A) \times \text{LMTD}_{new}\] Then we can use a similar LMTD formula to solve \(T_{m,a}\): \[\text{LMTD}_{new} = \frac{(T_{m,i} - T_{\infty}) - (T_{m,a} - T_{\infty})}{\ln{\frac{T_{m,i} - T_{\infty}}{T_{m,a} - T_{\infty}}}}\] With all other values given or previously calculated, we can now solve for the new outlet temperature \(T_{m,a}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Law of Cooling
Newton's Law of Cooling helps us calculate the rate at which heat is transferred in a cooling process. This law states that the rate of heat transfer, or how quickly temperature changes, is proportional to the difference in temperature between the object and its surroundings. In simpler terms, the greater the temperature difference, the faster the cooling. This concept is essential when dealing with heat exchangers or cooling systems.
\( Q = \dot{m} c_p (T_{m,i} - T_{m,\rho}) \) is the formula we use to calculate the heat transfer rate, where \( Q \) represents the heat transfer rate, \( \dot{m} \) is the mass flow rate, \( c_p \) is the specific heat capacity, and \( T_{m,i} - T_{m,\rho} \) is the temperature change.
To understand the underlying principle, imagine holding a hot cup of coffee in a cold room. The heat from the coffee will transfer to the cooler air around it until both the coffee and the room reach the same temperature. This is precisely what Newton's Law of Cooling describes.
Log Mean Temperature Difference (LMTD)
The Log Mean Temperature Difference, or LMTD, is an important concept in heat transfer calculations. It is a mathematical average that allows us to consider the varying temperature difference between two fluids along a heat exchanger.
When calculating LMTD, you must consider the inlet and outlet temperatures of both the fluid being cooled and the coolant. The formula is \[ \text{LMTD} = \frac{(T_{m,i} - T_{\infty}) - (T_{m,\rho} - T_{\infty})}{\ln{\frac{T_{m,i} - T_{\infty}}{T_{m,\rho} - T_{\infty}}}} \].
This equation helps us understand how temperature differences change across a heat exchanger, enabling accurate design and performance predictions. Think of LMTD as a way to capture the average temperature difference in systems where it isn't constant.
Overall Heat Transfer Coefficient
The Overall Heat Transfer Coefficient, denoted as \( U \), provides insights into how efficiently heat is transferred across a surface. It's like giving a grade to a heat exchanger's ability to transfer heat. This coefficient considers all modes of heat transfer: conduction, convection, and radiation.
The formula \( Q = (U \times A) \times \text{LMTD} \) is where the overall heat transfer coefficient comes into play. Here, \( A \) denotes the area for heat transfer, and \( \text{LMTD} \) is the log mean temperature difference.
By rearranging this formula, we can determine \( U \), effectively measuring the efficiency of the heat exchanger. A higher \( U \) value means the system is better at transferring heat. This coefficient allows engineers to optimize designs, ensuring systems operate effectively and economically.
Fully Developed Flow
Fully Developed Flow in fluid dynamics is when the velocity profile of the fluid remains constant along the flow direction. This concept is crucial in simplifying calculations because it implies that the fluid motion has reached a steady state.
In this exercise, mentioning fully developed flow means that the fluid characteristics, like velocity and temperature profiles, do not change as it moves through the tube. This stability provides a predictable environment to apply Newton's Law of Cooling and other heat transfer equations.
  • It assumes no further mixing or change in flow characteristics.
  • Simplifies mathematical models, aiding in designing efficient heat exchangers.
  • Helps accurately predict performance under given conditions.
When dealing with engineering problems, assuming fully developed flow allows for more straightforward equations and calculations, as transient effects and complex flow behaviors are minimized.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electronic circuit board dissipating \(50 \mathrm{~W}\) is sandwiched between two ducted, forced-air-cooled heat sinks. The sinks are \(150 \mathrm{~mm}\) in length and have 20 rectangular passages \(6 \mathrm{~mm} \times 25 \mathrm{~mm}\). Atmospheric air at a volumetric flow rate of \(0.060 \mathrm{~m}^{3} / \mathrm{s}\) and \(27^{\circ} \mathrm{C}\) is drawn through the sinks by a blower. Estimate the operating temperature of the board and the pressure drop across the sinks.

Engine oil flows through a \(25-\mathrm{mm}\)-diameter tube at a rate of \(0.5 \mathrm{~kg} / \mathrm{s}\). The oil enters the tube at a temperature of \(25^{\circ} \mathrm{C}\), while the tube surface temperature is maintained at \(100^{\circ} \mathrm{C}\). (a) Determine the oil outlet temperature for a \(5-\mathrm{m}\) and for a 100 -m long tube. For each case, compare the log mean temperature difference to the arithmetic mean temperature difference. (b) For \(5 \leq L \leq 100 \mathrm{~m}\), compute and plot the average Nusselt number \(\overline{N u}_{D}\) and the oil outlet temperature as a function of \(L\).

A liquid food product is processed in a continuousflow sterilizer. The liquid enters the sterilizer at a temperature and flow rate of \(T_{m, i, h}=20^{\circ} \mathrm{C}, \dot{m}=1 \mathrm{~kg} / \mathrm{s}\), respectively. A time-at-temperature constraint requires that the product be held at a mean temperature of \(T_{m}=90^{\circ} \mathrm{C}\) for \(10 \mathrm{~s}\) to kill bacteria, while a second constraint is that the local product temperature cannot exceed \(T_{\max }=230^{\circ} \mathrm{C}\) in order to preserve a pleasing taste. The sterilizer consists of an upstream, \(L_{k}=5 \mathrm{~m}\) heating section characterized by a uniform heat flux, an intermediate insulated sterilizing section, and a downstream cooling section of length \(L_{c}=10 \mathrm{~m}\). The cooling section is composed of an uninsulated tube exposed to a quiescent environment at \(T_{\infty}=20^{\circ} \mathrm{C}\). The thin-walled tubing is of diameter \(D=40 \mathrm{~mm}\). Food properties are similar to those of liquid water at \(T=330 \mathrm{~K}\). (a) What heat flux is required in the heating section to ensure a maximum mean product temperature of \(T_{m}=90^{\circ} \mathrm{C}\) ? (b) Determine the location and value of the maximum local product temperature. Is the second constraint satisfied? (c) Determine the minimum length of the sterilizing section needed to satisfy the time-at-temperature constraint. (d) Sketch the axial distribution of the mean, surface, and centerline temperatures from the inlet of the heating section to the outlet of the cooling section.

Fully developed conditions are known to exist for water flowing through a \(25-\mathrm{mm}\)-diameter tube at \(0.01 \mathrm{~kg} / \mathrm{s}\) and \(27^{\circ} \mathrm{C}\). What is the maximum velocity of the water in the tube? What is the pressure gradient associated with the flow?

Exhaust gases from a wire processing oven are discharged into a tall stack, and the gas and stack surface temperatures at the outlet of the stack must be estimated. Knowledge of the outlet gas temperature \(T_{m, o}\) is useful for predicting the dispersion of effluents in the thermal plume, while knowledge of the outlet stack surface temperature \(T_{s, a}\) indicates whether condensation of the gas products will occur. The thin-walled, cylindrical stack is \(0.5 \mathrm{~m}\) in diameter and \(6.0 \mathrm{~m}\) high. The exhaust gas flow rate is \(0.5 \mathrm{~kg} / \mathrm{s}\), and the inlet temperature is \(600^{\circ} \mathrm{C}\). (a) Consider conditions for which the ambient air temperature and wind velocity are \(4^{\circ} \mathrm{C}\) and \(5 \mathrm{~m} / \mathrm{s}\), respectively. Approximating the thermophysical properties of the gas as those of atmospheric air, estimate the outlet gas and stack surface temperatures for the given conditions. (b) The gas outlet temperature is sensitive to variations in the ambient air temperature and wind velocity. For \(T_{\infty}=-25^{\circ} \mathrm{C}, 5^{\circ} \mathrm{C}\), and \(35^{\circ} \mathrm{C}\), compute and plot the gas outlet temperature as a function of wind velocity for \(2 \leq V \leq 10 \mathrm{~m} / \mathrm{s}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.