/*! 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 104 A bayonet cooler is used to redu... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 104

A bayonet cooler is used to reduce the temperature of a pharmaceutical fluid. The pharmaceutical fluid flows through the cooler, which is fabricated of \(10-\mathrm{mm}-\) diameter, thin-walled tubing with two 250 -mm-long straight sections and a coil with six and a half turns and a coil diameter of \(75 \mathrm{~mm}\). A coolant flows outside the cooler, with a convection coefficient at the outside surface of \(h_{o}=500 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) and a coolant temperature of \(20^{\circ} \mathrm{C}\). Consider the situation where the pharmaceutical fluid enters at \(90^{\circ} \mathrm{C}\) with a mass flow rate of \(0.005 \mathrm{~kg} / \mathrm{s}\). The pharmaceutical has the following properties: \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, \quad \mu=4 \times 10^{-3} \mathrm{~N} \cdot \mathrm{s} / \mathrm{m}^{2}\), \(c_{p}=2000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Short Answer

Expert verified
The short answer to this question is as follows: After calculating the length and surface area of the tubing in contact with the pharmaceutical fluid, we can determine the heat transfer rate, temperature change experienced by the fluid, and the overall heat transfer coefficient of the bayonet cooler. Using Newton's law of cooling and the convection coefficient at the outside surface, we can solve for the outer temperature of the tubing (T_out). The overall heat transfer coefficient (U) is then calculated using the formula \(U = \cfrac{Q}{A\Delta T}\) and can be used to further analyze the performance of the bayonet cooler.

Step by step solution

01

Calculate the length and surface area of the tubing

First, we need to determine the total length and surface area of the tubing that is in contact with the pharmaceutical fluid. There are two straight sections of length 250 mm each and a coil with six and a half turns. We'll start by finding the length of one full turn of the coil: Length of one full turn of the coil = Coil diameter 脳 蟺 The total length of the coil = (Length of one full turn of the coil 脳 Number of turns) The length of the tubing is the sum of the lengths of the straight sections and the coil.
02

Determine the heat transfer rate

The heat transfer rate (Q) can be calculated using the specific heat transfer rate (q) and the surface area (A) of the tubing: Q = q 脳 A Where q can be calculated as: q = h_o (T_out - T_c) Here, T_out is the temperature of the fluid on the outer surface of the tubing, T_c is the coolant temperature, and h_o is the convection coefficient at the outside surface of the cooler. We do not know T_out yet, but it will be calculated later.
03

Calculate the temperature change of the fluid

To determine the temperature change of the pharmaceutical fluid (螖T), we need to use the mass flow rate of the fluid (m_dot), its specific heat capacity (c_p), and the heat transfer rate (Q): 螖T = Q / (m_dot * c_p)
04

Calculate the overall heat transfer coefficient

Once the temperature change is known, we can determine the overall heat transfer coefficient (U) of the bayonet cooler using Newton's law of cooling: U = (Q / A) / 螖T
05

Solve for T_out

Now, we have all the information needed to calculate the outer temperature of the tubing (T_out) that is in contact with the coolant. We use the following equation: q = h_o (T_out - T_c) Solve for T_out. The solution to this exercise involves various heat transfer equations and physical properties of the pharmaceutical fluid. Through a step-by-step approach, we can determine the temperature change experienced by the fluid, the overall heat transfer coefficient of the bayonet cooler, and other related parameters.

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Key Concepts

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

Convection
Convection is a mode of heat transfer that occurs in fluids (liquids and gases) where fluid particles move from one place to another, transporting heat energy. This process can be either natural or forced. In the context of the bayonet cooler, convection plays a crucial role in transferring heat between the pharmaceutical fluid in the tubing and the coolant outside.
  • Natural Convection: Occurs due to temperature differences in the fluid, leading to a natural flow.
  • Forced Convection: External sources like pumps or fans drive the fluid motion.
In this exercise, the coolant is responsible for convective heat transfer at the outer surface of the tubing, with a given convection coefficient (\( h_{o} = 500 \, \text{W/m}^{2}\text{K} \)). This coefficient quantifies the efficiency of heat transfer in the convection process, influenced by the fluid's properties, flow characteristics, and surface conditions.
Specific Heat Capacity
Specific heat capacity (\( c_{p} \)) defines how much heat energy is required to change the temperature of a substance. It is a crucial property in understanding how substances respond to energy input or loss. Measured in Joules per kilogram per Kelvin (\( \text{J/kgK} \)), it signifies the amount of energy needed to increase the temperature of one kilogram of a substance by one degree Celsius.
For the pharmaceutical fluid in the cooler, the specific heat capacity is 2000 \( \text{J/kg} \cdot \text{K} \). This tells us that a relatively moderate amount of energy is needed to change its temperature. This property is used when calculating the temperature change (\(\Delta T\)) of the fluid as it transfers heat to the coolant. The mass flow rate (\( m_{\text{dot}} \)) of the fluid and its specific heat capacity help us determine how quickly and efficiently the fluid's temperature can be reduced.
Heat Transfer Rate
The heat transfer rate, often denoted as Q, is a measure of how much heat is being transferred per unit time from the pharmaceutical fluid to the coolant. It is a critical parameter in assessing the performance of any heat exchange system. In this exercise, it links to the surface area of the tubing and the specific heat transfer rate (\( q \)), which itself involves the convection coefficient, coolant temperature, and the temperature at the outer surface of the tubing.
The basic formula for calculating the heat transfer rate is:\[ Q = q \times A \]Where:
  • \( q \) represents the specific heat transfer rate, a function of the convection coefficient and the temperature difference between the outer surface (\( T_{\text{out}} \)) and the coolant (\( T_{c} \)).
  • \( A \) is the surface area through which the heat is transferred.
By calculating Q, students can understand how effectively the cooler reduces the fluid's temperature, an important aspect of maintaining the desired state of the pharmaceutical product.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient (\( U \)) is a vital metric that combines the resistances to heat flow through all components in the heat transfer system, including the materials and convection fluids. It provides an overall measure of heat transfer efficiency against all barriers and is expressed in \( \text{W/m}^{2}\text{K} \).
In the bayonet cooler, after determining the temperature change (\(\Delta T\)) of the pharmaceutical fluid, calculating the overall heat transfer coefficient helps assess the system's effectiveness.Newton鈥檚 law of cooling is used to compute \( U \):\[ U = \frac{Q}{A \times \Delta T} \]Where:
  • \( Q \) is the known heat transfer rate.
  • \( A \) is the heat transfer area.
  • \( \Delta T \) is the temperature difference across the system.
Knowing \( U \) allows us to evaluate and optimize the design and operation of heat exchangers, ensuring that pharmaceutical products are effectively cooled without compromising their quality or consistency.

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Most popular questions from this chapter

Air at \(1 \mathrm{~atm}\) and \(285 \mathrm{~K}\) enters a 2 -m-long rectangular duct with cross section \(75 \mathrm{~mm} \times 150 \mathrm{~mm}\). The duct is maintained at a constant surface temperature of \(400 \mathrm{~K}\), and the air mass flow rate is \(0.10 \mathrm{~kg} / \mathrm{s}\). Determine the heat transfer rate from the duct to the air and the air outlet temperature.

A thin-walled tube with a diameter of \(6 \mathrm{~mm}\) and length of \(20 \mathrm{~m}\) is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at \(200^{\circ} \mathrm{C}\) and with a mass flow rate of \(0.003 \mathrm{~kg} / \mathrm{s}\). Autumn winds at a temperature of \(15^{\circ} \mathrm{C}\) blow directly across the tube at a velocity of \(5 \mathrm{~m} / \mathrm{s}\). Assume the thermophysical properties of the exhaust gas are those of air. (a) Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube. (b) Estimate the heat transfer coefficient for the air flowing across the outside of the tube. (c) Estimate the overall heat transfer coefficient \(U\) and the temperature of the exhaust gas when it reaches the laboratory.

Air flowing through a tube of \(75-\mathrm{mm}\) diameter passes over a 150 -mm- long roughened section that is constructed from naphthalene having the properties \(M=128.16 \mathrm{~kg} / \mathrm{kmol}\) and \(p_{\mathrm{sat}}(300 \mathrm{~K})=1.31 \times 10^{-4} \mathrm{bar}\). The air is at \(1 \mathrm{~atm}\) and \(300 \mathrm{~K}\), and the Reynolds number is \(R e_{D}=35,000\). In an experiment for which flow was maintained for \(3 \mathrm{~h}\), mass loss due to sublimation from the roughened surface was determined to be \(0.01 \mathrm{~kg}\). What is the associated convection mass transfer coefficient? What would be the corresponding convection heat transfer coefficient? Contrast these results with those predicted by conventional smooth tube correlations.

When viscous dissipation is included, Equation \(8.48\) (multiplied by \(\rho c_{p}\) ) becomes $$ \rho c_{p} u \frac{\partial T}{\partial x}=\frac{k}{r} \frac{\partial}{\partial r}\left(r \frac{\partial T}{\partial r}\right)+\mu\left(\frac{d u}{d r}\right)^{2} $$ This problem explores the importance of viscous dissipation. The conditions under consideration are laminar, fully developed flow in a circular pipe, with \(u\) given by Equation 8.15. (a) By integrating the left-hand side over a section of a pipe of length \(L\) and radius \(r_{o}\), show that this term yields the right-hand side of Equation 8.34. (b) Integrate the viscous dissipation term over the same volume. (c) Find the temperature rise caused by viscous dissipation by equating the two terms calculated above. Use the same conditions as in Problem 8.9.

One way to cool chips mounted on the circuit boards of a computer is to encapsulate the boards in metal frames that provide efficient pathways for conduction to supporting cold plates. Heat generated by the chips is then dissipated by transfer to water flowing through passages drilled in the plates. Because the plates are made from a metal of large thermal conductivity (typically aluminium or copper), they may be assumed to be at a temperature, \(T_{s, c p^{-}}\) (a) Consider circuit boards attached to cold plates of height \(H=750 \mathrm{~mm}\) and width \(L=600 \mathrm{~mm}\), each with \(N=10\) holes of diameter \(D=10 \mathrm{~mm}\). If operating conditions maintain plate temperatures of \(T_{\text {s.tp }}=32^{\circ} \mathrm{C}\) with water flow at \(\dot{m}_{1}=0.2 \mathrm{~kg} / \mathrm{s}\) per passage and \(T_{m, i}=7^{\circ} \mathrm{C}\), how much heat may be dissipated by the circuit boards? (b) To enhance cooling, thereby allowing increased power generation without an attendant increase in system temperatures, a hybrid cooling scheme may be used. The scheme involves forced airflow over the encapsulated circuit boards, as well as water flow through the cold plates. Consider conditions for which \(N_{\mathrm{cb}}=10\) circuit boards of width \(W=350 \mathrm{~mm}\) are attached to the cold plates and their average surface temperature is \(T_{s, \text { do }}=47^{\circ} \mathrm{C}\) when \(T_{s, \text { ep }}=32^{\circ} \mathrm{C}\). If air is in parallel flow over the plates with \(u_{\infty}=10 \mathrm{~m} / \mathrm{s}\) and \(T_{\infty}=7^{\circ} \mathrm{C}\), how much of the heat generated by the circuit boards is transferred to the air?
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