/*! 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 83 A disk of 20-mm diameter is cove... [FREE SOLUTION] | 91Ó°ÊÓ

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A disk of 20-mm diameter is covered with a water film. Under steady-state conditions, a heater power of \(200 \mathrm{~mW}\) is required to maintain the disk-water film at \(305 \mathrm{~K}\) in dry air at \(295 \mathrm{~K}\) and the observed evaporation rate is \(2.55 \times 10^{-4} \mathrm{~kg} / \mathrm{h}\). (a) Calculate the average mass transfer convection coefficient \(\bar{h}_{s}\) for the evaporation process. (b) Calculate the average heat transfer convection coefficient \(\bar{h}\). (c) Do the values of \(\bar{h}_{\mathrm{m}}\) and \(\bar{h}\) satisfy the heat- mass analogy? (d) If the relative humidity of the ambient air at \(295 \mathrm{~K}\) were increased from 0 (dry) to \(0.50\), but the power supplied to the heater was maintained at \(200 \mathrm{~mW}\), would the evaporation rate increase or decrease? Would the disk temperature increase or decrease?

Short Answer

Expert verified
In summary, (a) The average mass transfer convection coefficient \(\bar{h}_s\) can be calculated using the formula \[\bar{h}_s = \frac{E}{A}\] (b) The average heat transfer convection coefficient \(\bar{h}\) can be calculated using the formula \[\bar{h} = \frac{Q}{A (T_s - T_{\infty})}\] (c) If the values of \(\bar{h}_m\) and \(\bar{h}\) are close, then the heat-mass analogy is satisfied. (d) Increasing the relative humidity from 0 to 0.50 will decrease the evaporation rate and increase the disk temperature while maintaining the same heater power.

Step by step solution

01

Calculate the evaporation rate in kg/s

The given evaporation rate is in kg/h. To convert it to kg/s, divide by 3600 (the number of seconds in an hour): \[E = \frac{2.55 \times 10^{-4}}{3600} \mathrm{~kg/s}\]
02

Calculate the area of the disk

The diameter of the disk is 20 mm. To convert it to meters, divide by 1000: \[D = \frac{20}{1000} \text{m}\]The area of the disk can be calculated using the formula for the area of a circle: \[A = \pi r^2 = \pi \left(\frac{D}{2}\right)^2\]
03

Calculate the average mass transfer convection coefficient \(\bar{h}_s\)

Using the mass transfer formula, we get: \[\bar{h}_s = \frac{E}{A (C_s - C_{\infty})}\] where \(C_s\) and \(C_{\infty}\) are the concentration of the water vapor at the disk surface and far away from the disk, respectively. Since the air is dry, \(C_{\infty} = 0\). The saturation concentration at the surface is given by the evaporation rate (E) and the area (A). Therefore, the formula becomes: \[\bar{h}_s = \frac{E}{A}\] (b) Calculate the average heat transfer convection coefficient \(\bar{h}\):
04

Calculate the heat transfer rate

The heat transfer rate is given as 200 mW. To convert it to W, divide by 1000: \[Q = \frac{200}{1000} \text{W}\]
05

Calculate average heat transfer convection coefficient \(\bar{h}\)

Using the heat transfer formula, we get: \[\bar{h} = \frac{Q}{A (T_s - T_{\infty})}\] where \(T_s\) is the temperature at the disk surface (305 K) and \(T_{\infty}\) is the temperature far away from the disk (295 K). Calculate \(\bar{h}\) using the given values. (c) Check if the heat-mass analogy is satisfied:
06

Compare the values of \(\bar{h}_m\) and \(\bar{h}\)

If the values of \(\bar{h}_m\) and \(\bar{h}\) are close, then the heat-mass analogy is satisfied. (d) Effects of relative humidity on evaporation rate and disk temperature:
07

Analyze the effect of increasing relative humidity on the evaporation rate

Increasing the relative humidity to 0.50 means that the air is now more moist, which will decrease the evaporation rate since the driving force for water vapor to move from the disk surface to the air is reduced.
08

Analyze the effect of increasing relative humidity on the disk temperature

If the heat supplied to the disk is the same, but the evaporation rate is lower (due to increased relative humidity), it means that less heat is being removed by the evaporation process. Therefore, the disk temperature will increase.

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Key Concepts

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

Understanding Evaporation Rate
Evaporation is a fundamental physical process where a liquid turns into vapor. In the context of our problem, the evaporation rate is a measure of how quickly water is evaporating from the wet surface of a disk. It's calculated by measuring the mass of water that evaporates over a certain period, usually given in units such as kilograms per second ((kg/s)).

To translate the evaporation rate provided in your textbook from (kg/h) to (kg/s), we divide by 3600, the number of seconds in an hour. This step helps in making subsequent calculations manageable, as most scientific formulas require consistent SI units. Remember, getting the units right is crucial for accurate results in physics and engineering!

Regarding our exercise, an increased average mass transfer convection coefficient could mean more efficient evaporation; this becomes particularly handy when discussing moisture-sensitive processes in industries such as pharmaceuticals or food production. So, understanding this conversion and how it applies to real-world applications is vital for students diving into chemical and environmental engineering fields.
Heat Transfer Convection Coefficient Explained
Heat transfer is, in many ways, similar to mass transfer, but instead of mass, we're dealing with energy. The heat transfer convection coefficient, (h), is a key term in quantifying how effectively heat can be transferred from a surface to a fluid or vice versa by convection. Basically, it tells us how good a 'thermal transporter' a given environment is.

In our exercise scenario, we calculate the average heat transfer convection coefficient (h) using the heat provided by the heater and the temperature difference between the disk surface and the surrounding air. This value has practical use in various heating and cooling applications, from designing HVAC systems to optimizing industrial heaters.

Understanding the steps to calculate (h) is just the beginning. Knowing how to interpret its value can lead to energy-efficient designs and better temperature control strategies in diverse engineering processes. For students, mastering this concept is not only about solving an exercise but also gaining insight into thermal management in real-life situations.
The Heat-Mass Analogy
The heat-mass analogy is a powerful concept that bridges the gap between heat and mass transfer phenomena. Under certain conditions, it's observed that the mathematical expressions for the transfer of heat and the transfer of mass are similar in form. Hence, understanding one process can provide insights into the other.

When we compute both the mass transfer convection coefficient (h_s) for evaporation and the heat transfer convection coefficient (h), as in our exercise, we can check the validity of this analogy. If both coefficients are comparable, the analogy holds, suggesting that the mechanisms of heat and mass transfer are analogous in the given context.

For academic purposes, the heat-mass analogy serves as an excellent learning tool aiding students to grasp two distinct physical phenomena simultaneously. Moreover, in professional fields like environmental engineering or process design, using this analogy can simplify complex calculations and help innovate new solutions for efficient energy and mass transfer systems.

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Most popular questions from this chapter

Consider cross flow of gas \(\mathrm{X}\) over an object having a characteristic length of \(L=0.1 \mathrm{~m}\). For a Reynolds number of \(1 \times 10^{4}\), the average heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The same object is then impregnated with liquid \(Y\) and subjected to the same flow conditions. Given the following thermophysical properties, what is the average convection mass transfer coefficient?

Consider conditions for which a fluid with a free stream velocity of \(V=1 \mathrm{~m} / \mathrm{s}\) flows over an evaporating or subliming surface with a characteristic length of \(L=1 \mathrm{~m}\), providing an average mass transfer convection coefficient of \(\bar{h}_{\mathrm{m}}=10^{-2} \mathrm{~m} / \mathrm{s}\). Calculate the dimensionless parameters \(\overline{S h}_{L}, R e_{L}, S c\), and \(j_{m}\) for the following combinations: airflow over water, airflow over naphthalene, and warm glycerol over ice. Assume a fluid temperature of \(300 \mathrm{~K}\) and a pressure of \(1 \mathrm{~atm}\).

A wet-bulb thermometer consists of a mercury-in-glass thermometer covered with a wetted (water) fabric. When suspended in a stream of air, the steady-state thermometer reading indicates the wet-bulb temperature \(T_{\mathrm{ub}}\). Obtain an expression for determining the relative humidity of the air from knowledge of the air temperature \(\left(T_{\infty}\right)\), the wet-bulb temperature, and appropriate air and water vapor properties. If \(T_{\infty}=45^{\circ} \mathrm{C}\) and \(T_{w b}=25^{\circ} \mathrm{C}\), what is the relative humidity of the airstream?

Consider the nanofluid of Example 2.2. (a) Calculate the Prandtl numbers of the base fluid and nanofluid, using information provided in the example problem. (b) For a geometry of fixed characteristic dimension \(L\), and a fixed characteristic velocity \(V\), determine the ratio of the Reynolds numbers associated with the two fluids, \(R e_{\text {wf }} / R e_{\mathrm{w}_{\mathrm{d}}-}\) Calculate the ratio of the average Nusselt numbers, \(\overline{N u}_{L, \text {, d }} / \overline{N u}_{\text {L, b }}\), that is associated with identical average heat transfer coefficients for the two fluids, \(\bar{h}_{\mathrm{mf}}=\bar{h}_{\mathrm{bd}}\). (c) The functional dependence of the average Nusselt number on the Reynolds and Prandtl numbers for a broad array of various geometries may be expressed in the general form $$ \overline{N u}_{L}=\bar{h} L / k=C R e^{w N} P r^{1 / 3} $$ where \(C\) and \(m\) are constants whose values depend on the geometry from or to which convection heat transfer occurs. Under most conditions the value of \(m\) is positive. For positive \(m\), is it possible for the base fluid to provide greater convection heat transfer rates than the nanofluid, for conditions involving a fixed geometry, the same characteristic velocities, and identical surface and ambient temperatures?

Air at a free stream temperature of \(T_{a}=20^{\circ} \mathrm{C}\) is in parallel flow over a flat plate of length \(L=5 \mathrm{~m}\) and temperature \(T_{s}=90^{\circ} \mathrm{C}\). However, obstacles placed in the flow intensify mixing with increasing distance \(x\) from the leading edge, and the spatial variation of temperatures measured in the boundary layer is correlated by an expression of the form \(T\left({ }^{\circ} \mathrm{C}\right)=20+70\) \(\exp (-600 x y)\), where \(x\) and \(y\) are in meters. Determine and plot the manner in which the local convection coefficient \(h\) varies with \(x\). Evaluate the average convection coefficient \(\bar{h}\) for the plate.

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