A flat-plate solar collector is used to heat water by having water flow through tubes attached at the back of the thin solar absorber plate. The absorber plate has a surface area of \(2 \mathrm{~m}^{2}\) with emissivity and absorptivity of \(0.9\). The surface temperature of the absorber is \(35^{\circ} \mathrm{C}\), and solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) with a surrounding temperature of \(0^{\circ} \mathrm{C}\). Convection heat transfer coefficient at the absorber surface is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the ambient temperature is \(25^{\circ} \mathrm{C}\). Net heat rate absorbed by the solar collector heats the water from an inlet temperature \(\left(T_{\text {in }}\right)\) to an outlet temperature \(\left(T_{\text {out }}\right)\). If the water flow rate is \(5 \mathrm{~g} / \mathrm{s}\) with a specific heat of \(4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the temperature rise of the water.

Short Answer

Expert verified
Answer: The temperature rise of the water in the flat-plate solar collector is approximately 0.022 K.

Step by step solution

01

Calculate the Incident Solar Radiation on the Surface of the Absorber

As given, solar radiation is incident on the absorber at \(500 \mathrm{~W} / \mathrm{m}^{2}\) and the surface area of the absorber plate is \(2 \mathrm{~m}^{2}\). So the total solar radiation energy falling on the surface of the absorber can be calculated as: Solar Radiation Energy = (Radiation Intensity) x (Surface Area) Solar Radiation Energy = \( 500\, \frac{\text{W}}{\text{m}^2} \times 2\, \text{m}^2 =1000\, \text{W}\)
02

Calculate the Absorbed Solar Radiation

The absorber plate has an absorptivity of \(0.9\). Use this to calculate the absorbed solar radiation energy: Absorbed Energy = (Absorptivity) x (Solar Radiation Energy) Absorbed_Energy = \(0.9 \times 1000\, \text{W} = 900\, \text{W}\)
03

Calculate the Heat Loss due to Radiation and Convection

To determine the heat loss, we will use the given emissivity, absorber surface temperature, surrounding temperature, and convection heat transfer coefficient. Calculate the heat loss due to radiation energy using the Stefan-Boltzmann Law (\(E = \epsilon \sigma A (T^4 - T_\text{amb}^4)\)): Heat Loss (Radiation) = Emissivity × (Stefan-Boltzmann Constant) × Surface Area × (Absorber Temperature^4 - Surrounding Temperature^4) The Stefan-Boltzmann Constant is approximately \(5.67 \times 10^{-8} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^4\). Convert the temperatures to Kelvin: \(T_\text{absorber} = 35 + 273.15 = 308.15\,\text{K}\) \(T_\text{surrounding} = 0 + 273.15 = 273.15\,\text{K}\) Heat Loss (Radiation) = \(0.9 \times 5.67 \times 10^{-8}\, \frac{\text{W}}{\text{m}^{2} \cdot \mathrm{K}^4} \times 2\, \text{m}^{2} \times (308.15^{4} - 273.15^{4}) = 331.24\, \text{W}\) Now, calculate the heat loss due to convection using: Heat Loss (Convection) = (Convection Heat Transfer Coefficient) × Surface Area × (Absorber Temperature - Ambient Temperature) Convert the temperatures to Celsius: Heat Loss (Convection) = \(5\, \frac{\text{W}}{\text{m}^2\cdot \mathrm{K}} \times 2\, \text{m}^{2} \times (35 - 25) = 100\, \text{W}\)
04

Calculate the Net Heat Rate Absorbed by the Solar Collector

Combine the absorbed solar radiation energy and the heat loss due to radiation and convection to find the net heat rate absorbed by the solar collector: Net Heat Rate = Absorbed Energy - Heat Loss (Radiation) - Heat Loss (Convection) Net Heat Rate = \(900\, \text{W} - 331.24\, \text{W} - 100\, \text{W} = 468.76\,\text{W}\)
05

Calculate the Temperature Rise of the Water

Now, use the given water flow rate, specific heat, and net heat rate absorbed to determine the temperature rise of the water: Water Flow Rate = \(5\, \frac{\text{g}}{\text{s}} = 0.005\,\frac{\text{kg}}{\text{s}}\) Specific Heat of Water = \(4.2\, \frac{\text{kJ}}{\text{kg} \cdot \mathrm{K}} = 4200\, \frac{\text{J}}{\text{kg} \cdot \mathrm{K}}\) We will use the formula: Heat Rate = (Mass Flow Rate) × Specific Heat × Temperature Rise Temperature Rise = \(\frac{\text{Heat Rate}}{(\text{Mass Flow Rate}) × (\text{Specific Heat})}\) Temperature Rise = \(\frac{468.76\, \text{W}}{0.005\, \frac{\text{kg}}{\text{s}} \times 4200\, \frac{\text{J}}{\text{kg} \cdot \mathrm{K}}} = 0.022\,\text{K}\) The temperature rise of the water in the flat-plate solar collector is approximately \(0.022\,\text{K}\).

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 Vaia!

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

A 300-ft-long section of a steam pipe whose outer diameter is 4 in passes through an open space at \(50^{\circ} \mathrm{F}\). The average temperature of the outer surface of the pipe is measured to be \(280^{\circ} \mathrm{F}\), and the average heat transfer coefficient on that surface is determined to be \(6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine \((a)\) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $$\$ 1.10 /$$ therm ( 1 therm \(=100,000\) Btu).

Solar radiation is incident on a \(5 \mathrm{~m}^{2}\) solar absorber plate surface at a rate of \(800 \mathrm{~W} / \mathrm{m}^{2}\). Ninety-three percent of the solar radiation is absorbed by the absorber plate, while the remaining 7 percent is reflected away. The solar absorber plate has a surface temperature of \(40^{\circ} \mathrm{C}\) with an emissivity of \(0.9\) that experiences radiation exchange with the surrounding temperature of \(-5^{\circ} \mathrm{C}\). In addition, convective heat transfer occurs between the absorber plate surface and the ambient air of \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the efficiency of the solar absorber, which is defined as the ratio of the usable heat collected by the absorber to the incident solar radiation on the absorber.

Air at \(20^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) blows over a pond. The surface temperature of the pond is at \(40^{\circ} \mathrm{C}\). Determine the heat flux between the surface of the pond and the air.

We often turn the fan on in summer to help us cool. Explain how a fan makes us feel cooler in the summer. Also explain why some people use ceiling fans also in winter.

Consider a person standing in a room maintained at \(20^{\circ} \mathrm{C}\) at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of \(12^{\circ} \mathrm{C}\) in winter and \(23^{\circ} \mathrm{C}\) in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person are \(1.6 \mathrm{~m}^{2}, 0.95\), and \(32^{\circ} \mathrm{C}\), respectively.

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.

Sign-up for free