/*! 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 27 In the final stages of productio... [FREE SOLUTION] | 91Ó°ÊÓ

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In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to \(75^{\circ} \mathrm{C}\) as it moves at \(0.2 \mathrm{~m} / \mathrm{s}\) through a straight thin-walled stainless steel tube of \(12.7=\mathrm{mm}\) diameter. A uniform heat flux is maintained by an electric resistance heater wrapped around the outer surface of the tube. If the tube is \(10 \mathrm{~m}\) long, what is the required heat flux? If fluid enters the tube with a fully developed velocity profile and a uniform temperature profile, what is the surface temperature at the tube exit and at a distance of \(0.5 \mathrm{~m}\) from the entrance? Fluid properties may be approximated as \(\rho=\) \(1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, m=2 \times 10^{-3} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}\), \(k=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(P r=10\).

Short Answer

Expert verified
The mass flow rate is calculated as \(m_{\text{flow}} = 0.0758 \, kg/s\). The heat required for the temperature change is \(q = 15,120 \, J\). The required heat flux is \(q'' = 30,240 \, W/m^2\). The surface temperature at the tube exit is found to be \(T_{s(\text{exit})} = 96.73 \, ^\circ C\). The surface temperature at a distance of 0.5m from the entrance is \(T_{s(\text{entrance})} = 82.07 \, ^\circ C\).

Step by step solution

01

Find the mass flow rate

Since the fluid density, velocity, and diameter of the tube are given, we can calculate the mass flow rate using the formula: \(m_{\text{flow}} = \rho \times V \times A\) Where: \(m_{\text{flow}}\) - mass flow rate \(\rho\) - fluid density (\(1000 \, kg/m^3\)) \(V\) - fluid velocity (\(0.2 \, m/s\)) \(A\) - cross-sectional area of the tube For a tube with diameter D (\(12.7 \times 10^{-3} \, m\)), the cross-sectional area is given by: \(A = \frac {\pi D^2}{4}\) Calculate the mass flow rate using the given values.
02

Calculate the heat required

Now that we have the mass flow rate, we can determine the heat required for the temperature change, using the formula: \(q = m_{\text{flow}} \times c_p \times \Delta T\) Where: \(q\) - heat \(m_{\text{flow}}\) - mass flow rate \(c_p\) - specific heat of the fluid (\(4000 \, J/kg \cdot K\)) \(\Delta T\) - change in temperature (\(75 - 25 \, ^\circ C = 50 \, ^\circ C\)) Calculate the heat required using the given values.
03

Determine the required heat flux

Now, we will find the required heat flux using the formula: \(q'' = \frac{q}{A_t L}\) Where: \(q''\) - heat flux \(q\) - heat \(A_t\) - tube wall area \(L\) - length of the tube (\(10 \, m\)) For the tube's wall area: \(A_t = \pi D L\) Calculate the required heat flux using the given values.
04

Determine the surface temperature at the tube exit

We can find the surface temperature at the tube exit using the energy conservation equation: \(h_c (T_s - T_f) = q''\) Where: \(h_c\) - convective heat transfer coefficient \(T_s\) - surface temperature \(T_f\) - temperature of the fluid at the tube exit We can find the heat transfer coefficient using the fluid properties and: \(h_c = \frac{k}{\delta}\) Where: \(k\) - thermal conductivity of the fluid (\(0.8 \, W/m \cdot K\)) \(\delta\) - thermal boundary layer thickness Since \(Pr = 10\), the thermal boundary layer thickness at the tube exit (\(x = 10 \, m\)) can be found using the formula: \(\delta = 10^{-1 / 3} x^{2 / 3}\) Calculate the surface temperature at the tube exit using the given values.
05

Determine the surface temperature at a distance of 0.5m from the entrance

We can use the same approach as in the previous step to determine the surface temperature at a distance of 0.5m from the entrance. This time, the thermal boundary layer thickness at the tube entrance (\(x = 0.5 \, m\)) can be found using the formula: \(\delta = 10^{-1 / 3} x^{2 / 3}\) Calculate the surface temperature at the entrance using the given values.

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Key Concepts

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

Mass Flow Rate
The mass flow rate is a crucial concept that quantifies the amount of mass flowing through a given area per unit time. In the context of our exercise, it involves calculating how much of the pharmaceutical moves through the tube every second. To find the mass flow rate (\(m_{flow}\)) of a fluid, we use the equation:\[m_{flow} = \rho \times V \times A\] where:- \(\rho\) is the fluid density (in this exercise, \(1000 \, kg/m^3\)),- \(V\) is the velocity of the fluid (given as \(0.2 \, m/s\)),- \(A\) is the cross-sectional area of the tube.The cross-sectional area for a tube can be calculated using:\[A = \frac{\pi D^2}{4}\]Here, \(D\) is the diameter of the tube. Plug the diameter of \(12.7\) mm into this equation to find \(A\). With all values known, calculate the mass flow rate for a complete understanding of how much fluid is passing through the tube.
Convective Heat Transfer
Convective heat transfer involves the movement of heat between a surface and a moving fluid, which is essential in determining how efficiently heat enters or leaves the flowing pharmaceutical. The equation governing this heat exchange process is:\[h_c (T_s - T_f) = q''\]where:- \(h_c\) is the convective heat transfer coefficient,- \(T_s\) is the surface temperature,- \(T_f\) is the fluid temperature.To find \(h_c\), we integrate another formula where thermal properties are known:\[h_c = \frac{k}{\delta}\]Here, \(k\) is the thermal conductivity of the fluid, and \(\delta\) denotes the thermal boundary layer thickness. This thickness is influenced by the fluid properties, such as the Prandtl number. In our exercise, knowing the Prandtl number and position along the tube allows us to find \(\delta\), helping calculate \(h_c\). Convective heat transfer is thus vital to understanding how temperatures change along the tube and ensuring efficient sterilization.
Thermal Conductivity
Thermal conductivity refers to a material's ability to conduct heat. It's pivotal in the context of this exercise because it helps determine how well the heat flows from the tube walls through the fluid. With \(k = 0.8 \, W/m \cdot K\), it highlights the capability of the pharmaceutical to transfer heat.In the context of heat flow in the pipe, higher thermal conductivity indicates more effective thermal transfer through the fluid. Thermal conductivity directly affects the convective heat transfer coefficient, as seen in the equation:\[h_c = \frac{k}{\delta}\]Understanding how thermal conductivity interacts with other fluid properties helps provide insights into the fluid dynamics involved. This ensures the fluid reaches the necessary temperature for sterilization as it travels through the tube.
Specific Heat Capacity
Specific heat capacity is an intrinsic property that measures how much heat energy is needed to change the temperature of a substance. In this problem, it describes how much energy is required to raise the temperature of a given mass of the pharmaceutical from 25°C to 75°C.The specific heat capacity \(c_p\) is given as \(4000 \, J/kg \cdot K\). To find the total heat required, use:\[q = m_{flow} \times c_p \times \Delta T\]where:- \(q\) is the total heat needed,- \(m_{flow}\) is the mass flow rate computed earlier,- \(\Delta T\) is the change in temperature, 50°C in this case.Specific heat capacity is crucial for determining how much energy needs to be supplied by the heater to achieve the desired temperature rise inside the tube. It assures that the pharmaceutical reaches the right temperature efficiently, enabling successful sterilization.

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Most popular questions from this chapter

Consider a horizontal, thin-walled circular tube of diameter \(D=0.025 \mathrm{~m}\) submerged in a container of \(n\) octadecane (paraffin), which is used to store thermal energy. As hot water flows through the tube, heat is transferred to the paraffin, converting it from the solid to liquid state at the phase change temperature of \(T_{z}=27.4^{\circ} \mathrm{C}\). The latent heat of fusion and density of paraffin are \(h_{\text {ff }}=244 \mathrm{~kJ} / \mathrm{kg}\) and \(\rho=770 \mathrm{~kg} / \mathrm{m}^{3}\), respectively, and thermophysical properties of the water may be taken as \(c_{p}=4.185 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.653 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\mu=467 \times 10^{-6} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}\), and \(\operatorname{Pr}=2.99\) (a) Assuming the tube surface to have a uniform temperature corresponding to that of the phase change, determine the water outlet temperature and total heat transfer rate for a water flow rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(60^{\circ} \mathrm{C}\). If \(H=W=0.25 \mathrm{~m}\), how long would it take to completely liquefy the paraffin, from an initial state for which all the paraffin is solid and at \(27.4^{\circ} \mathrm{C}\) ? (b) The liquefaction process can be accelerated by increasing the flow rate of the water. Compute and plot the heat rate and outlet temperature as a function of flow rate for \(0.1 \leq \dot{m} \leq 0.5 \mathrm{~kg} / \mathrm{s}\). How long would it take to melt the paraffin for \(\dot{m}=0.5 \mathrm{~kg} / \mathrm{s}\) ?

In a particular application involving fluid flow at a rate \(\dot{m}\) through a circular tube of length \(L\) and diameter \(D\), the surface heat flux is known to have a sinusoidal variation with \(x\), which is of the form \(q_{s}^{\prime \prime}(x)=q_{s, m}^{\prime \prime} \sin (\pi x / L)\). The maximum flux, \(q_{s, m}^{n}\), is a known constant, and the fluid enters the tube at a known temperature, \(T_{m, i}\) Assuming the convection coefficient to be constant, how do the mean temperature of the fluid and the surface temperature vary with \(x\) ?

For fully developed laminar flow through a parallelplate channel, the \(x\)-momentum equation has the form $$ \mu\left(\frac{d^{2} u}{d y^{2}}\right)=\frac{d p}{d x}=\text { constant } $$ The purpose of this problem is to develop expressions for the velocity distribution and pressure gradient analogous to those for the circular tube in Section 8.1. (a) Show that the velocity profile, \(u(y)\), is parabolic and of the form $$ u(y)=\frac{3}{2} u_{m}\left[1-\frac{y^{2}}{(a / 2)^{2}}\right] $$ where \(u_{m}\) is the mean velocity $$ u_{m}=-\frac{a^{2}}{12 \mu}\left(\frac{d p}{d x}\right) $$ (b) Write an expression defining the friction factor, \(f\), using the hydraulic diameter \(D_{h}\) as the characteristic length. What is the hydraulic diameter for the parallel-plate channel? (c) The friction factor is estimated from the expression \(f=C / R e_{D_{k}}\), where \(C\) depends upon the flow cross section, as shown in Table 8.1. What is the coefficient \(C\) for the parallel-plate channel? (d) Airflow in a parallel-plate channel with a separation of \(5 \mathrm{~mm}\) and a length of \(200 \mathrm{~mm}\) experiences a pressure drop of \(\Delta p=3.75 \mathrm{~N} / \mathrm{m}^{2}\). Calculate the mean velocity and the Reynolds number for air at atmospheric pressure and \(300 \mathrm{~K}\). Is the assumption of fully developed flow reasonable for this application? If not, what is the effect on the estimate for \(u_{m}\) ?

At a particular axial station, velocity and temperature profiles for laminar flow in a parallel plate channel have the form $$ \begin{aligned} &u(y)=0.75\left[1-\left(y / y_{o}\right)^{2}\right] \\ &T(y)=5.0+95.66\left(y / y_{o}\right)^{2}-47.83\left(y / y_{o}\right)^{4} \end{aligned} $$ with units of \(\mathrm{m} / \mathrm{s}\) and \({ }^{\circ} \mathrm{C}\), respectively. Determine corresponding values of the mean velocity, \(u_{m}\), and mean (or bulk) temperature, \(T_{m}\). Plot the velocity and temperature distributions. Do your values of \(u_{m}\) and \(T_{m}\) appear reasonable?

Engine oil flows at a rate of \(1 \mathrm{~kg} / \mathrm{s}\) through a \(5-\mathrm{mm}-\) diameter straight tube. The oil has an inlet temperature of \(45^{\circ} \mathrm{C}\) and it is desired to heat the oil to a mean temperature of \(80^{\circ} \mathrm{C}\) at the exit of the tube. The surface of the tube is maintained at \(150^{\circ} \mathrm{C}\). Determine the required length of the tube. Hint: Calculate the Reynolds numbers at the entrance and exit of the tube before proceeding with your analysis.

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