/*! 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 17 A flat-plate solar collector is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 17

A flat-plate solar collector is used to heat atmospheric air flowing through a rectangular channel. The bottom surface of the channel is well insulated, while the top surface is subjected to a uniform heat flux \(q_{o}^{\prime \prime}\), which is due to the net effect of solar radiation absorption and heat exchange between the absorber and cover plates. (a) Beginning with an appropriate differential control volume, obtain an equation that could be used to determine the mean air temperature \(T_{m}(x)\) as a function of distance along the channel. Solve this equation to obtain an expression for the mean temperature of the air leaving the collector. (b) With air inlet conditions of \(\dot{m}=0.1 \mathrm{~kg} / \mathrm{s}\) and \(T_{m, i}=40^{\circ} \mathrm{C}\), what is the air outlet temperature if \(L=3 \mathrm{~m}, w=1 \mathrm{~m}\), and \(q_{o}^{\prime \prime}=700 \mathrm{~W} / \mathrm{m}^{2}\) ? The specific heat of air is \(c_{p}=1008 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
The mean air temperature Tm(x) as a function of the distance along the channel is given by the expression \(T_m(x) - T_{m, i} = \frac{wq_{o}^{\prime \prime}}{\dot{m}c_p}x\). Using the given values in the problem, the air outlet temperature is calculated to be approximately 60.83掳C.

Step by step solution

01

Define the control volume and perform an energy balance

Let's choose a differential control volume of length dx within the channel. The energy that enters the control volume comes from the heat flux through the top surface of the channel, and the energy that leaves the control volume comes from the heated air exiting the control volume. Using the given information, we can now perform an energy balance on the control volume: Energy in = Energy out + Energy stored Since steady-state conditions are assumed in the problem, Energy stored is zero.
02

Express energy in and energy out terms

Let's define the energy in and energy out terms: Energy in = Heat flux entering through the top surface = \(wq_{o}^{\prime \prime}dx\) Energy out = Enthalpy increase of the air moving through the control volume = \(\dot{m}c_p(T_m(x+dx)-T_m(x))\) Again, since steady-state conditions are assumed, we can equate energy in and energy out terms.
03

Equate energy in and energy out, and simplify the equation

Equating energy in and out terms and simplifying the equation: \(wq_{o}^{\prime \prime}dx = \dot{m}c_p(T_m(x+dx)-T_m(x))\) Dividing by dx: \(wq_{o}^{\prime \prime} = \dot{m}c_p\frac{dT_m}{dx}\) Now, we have an equation relating the mean air temperature Tm and the distance x along the channel.
04

Integrate the equation to obtain an expression for the mean temperature

Integrating the equation with respect to x: \(\int \frac{dT_m}{dx}dx = \int\frac{wq_{o}^{\prime \prime}}{\dot{m}c_p}dx\) Integrating both sides, we get: \(T_m(x) - T_{m, i} = \frac{wq_{o}^{\prime \prime}}{\dot{m}c_p}x\) Now, we have an expression for the mean air temperature Tm(x) as a function of the distance along the channel.
05

Calculate the air outlet temperature

Now, let's use this expression for the mean air temperature and the given values of m_dot, T_{m, i}, L, w, and \(q_{o}^{\prime \prime}\) to find the air outlet temperature: \(T_m(L) = T_{m, i} + \frac{wq_{o}^{\prime \prime}}{\dot{m}c_p}L\) Substitute the given values: \(T_m(3) = 40^\circ C + \frac{(1)(700 W/m^2)}{(0.1 kg/s)(1008 J/kg\cdot K)}(3m)\) Calculating the result, we get: \(T_m(3) = 40^\circ C + 20.83^\circ C\) \(T_m(3) = 60.83^\circ C\) Hence, the air outlet temperature is approximately 60.83掳C.

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 91影视!

Key Concepts

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

Energy Balance Equation
The energy balance equation is fundamental to understanding how solar collector systems, such as the flat-plate solar collector described in the exercise, function to heat air or other fluids. The equation is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another.

When applying this principle to a control volume, such as the channel within the solar collector, the energy balance can be simplified to:\[\text{Energy in} = \text{Energy out} + \text{Energy stored}\]However, if we consider the steady-state conditions as mentioned in the exercise, the energy stored in the control volume becomes negligible (in other words, zero). Therefore, the equation becomes:\[\text{Energy in} = \text{Energy out}\]Understanding this balance is crucial when determining the mean air temperature along the length of the channel within the collector. Any energy input from the heat flux must equal the energy taken away by the air as it increases in temperature.
Mean Air Temperature
The mean air temperature is a measure of the average temperature of the air flowing through the solar collector's channel. In the context of the exercise, finding the mean air temperature as a function of distance along the channel is essential for predicting the performance of the solar collector system.

The change in the mean air temperature along the channel can be described with the differential equation derived from the energy balance. By integrating this equation, as shown in the solution, we can express the mean air temperature at any point along the length of the channel.The mathematical expression derived from integrating is:\[T_m(x) = T_{m, i} + \frac{wq_{o}^{\prime \prime}}{\dot{m}c_p}x\]This formula allows the calculation of the air temperature at the channel's outlet, which is crucial for determining the efficiency of the solar collector and the feasibility of the system for heating purposes.
Specific Heat Capacity
Specific heat capacity, denoted as \(c_p\) in the textbook exercise, is the amount of energy required to raise the temperature of a unit mass of a substance by one degree in temperature. It's a property that varies from one material to another and, for a gas like air, can depend on the pressure and temperature conditions.

In the given exercise, the specific heat of air is assumed constant, and its value is pivotal when calculating the increase in the air's temperature as it receives heat from the heated surface of the solar collector. The specific heat of air at constant pressure, \(c_p\), is used in the energy balance equation to relate the amount of energy transferred to the air to the subsequent increase in temperature. This relationship is expressed by the equation:\[\text{Energy out} = \dot{m}c_p(T_m(x+dx)-T_m(x))\]The value of specific heat capacity must be factored into the design and analysis of a solar collector system to ensure accurate predictions and efficient operation.

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 air heater for an industrial application consists of an insulated, concentric tube annulus, for which air flows through a thin-walled inner tube. Saturated steam flows through the outer annulus, and condensation of the steam maintains a uniform temperature \(T_{s}\) on the tube surface. Consider conditions for which air enters a 50 -mmdiameter tube at a pressure of \(5 \mathrm{~atm}\), a temperature of \(T_{m, i}=17^{\circ} \mathrm{C}\), and a flow rate of \(\dot{m}=0.03 \mathrm{~kg} / \mathrm{s}\), while saturated steam at \(2.455\) bars condenses on the outer surface of the tube. If the length of the annulus is \(L=5 \mathrm{~m}\), what are the outlet temperature \(T_{m, o}\) and pressure \(p_{o}\) of the air? What is the mass rate at which condensate leaves the annulus?

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 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 \(P r=10\).

Heated air required for a food-drying process is generated by passing ambient air at \(20^{\circ} \mathrm{C}\) through long, circular tubes \((D=50 \mathrm{~mm}, L=5 \mathrm{~m})\) housed in a steam condenser. Saturated steam at atmospheric pressure condenses on the outer surface of the tubes, maintaining a uniform surface temperature of \(100^{\circ} \mathrm{C}\). (a) If an airflow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\) is maintained in each tube, determine the air outlet temperature \(T_{m, o}\) and the total heat rate \(q\) for the tube. (b) The air outlet temperature may be controlled by adjusting the tube mass flow rate. Compute and plot \(T_{m \rho}\) as a function of \(\dot{m}\) for \(0.005 \leq \dot{m} \leq\) \(0.050 \mathrm{~kg} / \mathrm{s}\). If a particular drying process requires approximately \(1 \mathrm{~kg} / \mathrm{s}\) of air at \(75^{\circ} \mathrm{C}\), what design and operating conditions should be prescribed for the air heater, subject to the constraint that the tube diameter and length be fixed at \(50 \mathrm{~mm}\) and \(5 \mathrm{~m}\), respectively?

8.106 Consider the pharmaceutical product of Problem 8.27. Prior to finalizing the manufacturing process, test trials are performed to experimentally determine the dependence of the shelf life of the drug as a function of the sterilization temperature. Hence, the sterilization temperature must be carefully controlled in the trials. To promote good mixing of the pharmaceutical and, in turn, relatively uniform outlet temperatures across the exit tube area, experiments are performed using a device that is constructed of two interwoven coiled tubes, each of 10 -mm diameter. The thin-walled tubing is welded to a solid high thermal conductivity rod of diameter \(D_{r}=40 \mathrm{~mm}\). One tube carries the pharmaceutical product at a mean velocity of \(u_{p}=0.1 \mathrm{~m} / \mathrm{s}\) and inlet temperature of \(25^{\circ} \mathrm{C}\), while the second tube carries pressurized liquid water at \(u_{w}=0.12 \mathrm{~m} / \mathrm{s}\) with an inlet temperature of \(127^{\circ} \mathrm{C}\). The tubes do not contact each other but are each welded to the solid metal rod, with each tube making 20 turns around the rod. The exterior of the apparatus is well insulated. (a) Determine the outlet temperature of the pharmaceutical product. Evaluate the liquid water properties at \(380 \mathrm{~K}\). (b) Investigate the sensitivity of the pharmaceutical's outlet temperature to the velocity of the pressurized water over the range \(0.10
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Force

Read Explanation

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Electromagnetism

Read Explanation

Torque and Rotational Motion

Read Explanation
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.