Laboratory Report Experiment No. 5

Technological Institute of the Philippines – Manila CHE409-Unit Operations Laboratory I, 1st Semester 2015-2016

Heat and Radiation Experiments Charles Bonn Kirby F. Mayo, Zeny Naranjo, Gliezel Panopio, Klinneth Joy Samillano, Eazyl D. Salazar Abstract - This experiment focuses on the effects of various variables on the transmission of radiation from a source to a body and the emission of radiation absorbed by a body to the environment. This experiment also focuses on the different laws that simplifies how the transmission of energy works like Stefan-Boltzman law and Kirchoff’s law. This experiment demonstrates how the energy absorbed by a body is affected by other variables like distance, light intensity, type of material, etc. Index Terms – Radiation is the emission of energy as electromagnetic waves or as moving subatomic particles, Radiometer which is also called roentgenometer is a device used for measuring the radiant flax of electromagnetic radiation, Black body is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, Temperature is the degree or intensity of heat present in a substance or object.

I.

R2 sin   R 2  L2 2

The radiation intensity in a surface is inversely proportional to the square of the distance to the radiation source surface. The total energy dQ of an element dA flows through a hemisphere with radius r. An element of the surface, dA, creates a solid angle deliminated by dA.

d o  dAa / r 2 The energy flow through dA, is,

dQ  i d  dA

OBJECTIVE

To determine the transmission principle for radiation and the different factors affecting on it.

II. INTENDED LEARNING OUTCOMES The students shall be able to: 2.1 demonstrate the inverse of the radiation distance square law.

Where

i

is the radiation intensity on the

2.4 inderstand the Kirchoff's law 2.5 determine the effect of radiation area to the measurement of radiatiometer

III. DISCUSSION

,

Stefan-Boltzman's Law The radiation intensity changes with the fourth power of source's temperature.

Qemitted   (TS4  TA4 )

2.2 understand the Stefan-Boltzman Law 2.3 observe the influence of different materials in the irradiation



where,

Qemitted = the energy transmitted of a surface of a black body.



= Stefan-Boltzman constant =5.67x104 W/m2-K4

Incident Radiation and Emitted Radiation

T= temperature of the source

The radiometer shows the intensity of the radiation but not emitted radiation by the hot surface.

TA = temperature of the radiometer and surroundings

The relation between the received radiation by the sensor and the radiation emitted by the hot source is the following.

Qincident = Qemitted xsinθ2

In the case the rectangular badge,

Emission Power The emission is the proportion of the total energy emitted by a surface in relation to the total energy emitted by a black surface with the same temperature. For a real body, the emission is function of the wavelength of the radiation, the incident angle, the surface temperature and geometry of the surface.

Heat and Radiation Experiments

Qemitted   E (TS4  TA4 ) 4. where E is the emittance of the surface. Kirchoff's Law The emittance of the gray surface is equal to the absorption received by another surface in thermal balance. For a gray body surface with an area A1, Temperature T1, when the body reaches the thermal equilibrium, it must absorb the same radiation than it emits. Thereupon,

A1 T14  A1E1 T14

  E1

5.

polished surface of the sphere has an emissivity of 0.1. Calculate the heat loss per m2 by radiation? A plane surface having an area of 1.0 m 2 is insulated on the bottom side and is placed on the ground exposed to air at 290 K, and the convective heat transfer coefficient from the air is 12 W/m2-K. The plane radiates to clear sky. The effective radiation temperature of the sky can be assumed as 80 K. If the plane is a black body, calculate the temperature of the plane at equilibrium. A horizontal oxidized steel pipe carrying steam and having an OD of 0.1638 m has a surface temperature of 374.9 K and is exposed to air at 297.1 K. in a large enclosure. Calculate the heat loss for 0.305 m of pipe from the natural convection plus radiation. For the steel pipe, use an ε of 0.79. X. ANSWERS

Area Factors The radiant energy exchange from a surface to another surface depends on the interconnection geometry and this is function of the confirmed surface. In this case we introduced the area factor, F, which is the fraction of emitted energy for the unit of a surface confronted with another surface. The time rate of the radiant heat transference (Q12) between two black surfaces with areas A1 and A2, temperatures T1 and T2 is

1.

a. Absorptivity – the fraction of irradiation absorbed by a body. It varies with wavelength and is defined as the absorbance of the body per unit path length and concentration. For a black body, its value is 1.0. In opaque bodies, absorptivity plus transmissivity equals 1.0. Absorptivity is represented by the Greek letter alpha, α. b. Reflectivity – the fraction of irradiation transmitted by the surface of a body. Reflectivity deviates from the other properties in that it is bidirectional in nature. In other words, this property depends on the direction of the incident of radiation as well as the direction of the reflection. For a black body, the value of reflectivity is zero. Reflectivity is represented by the Greek letter rho, ρ. c. Transmissivty – the fraction of irradiation transmitted by the surface of the body. It is also considered as a material’s effectiveness in transmitting radiant energy. In opaque bodies, its value is 0. Transmissivity is represented by the letter T. d. Emissivity - is the effectiveness of a material’s surface in emitting thermal radiation. Materials that have an emissivity value less than 1.0 are called gray bodies. Emissivity is represented by the greek letter epsilon, ε.

2.

Given: DS = D Required: F21 LC = D Solution: A1F12 = A2F21 Since the sphere only sees the cubical box: view factor F12 = 1.0

Q12  A1 F12  T14  T24  The area factor is found by analysis, numeric approximation and analogy. IV. RESOURCES AND MATERIALS V. PROCEDURE VI. DATA AND RESULTS VII. CALCULATIONS VIII.

CONCLUSION

IX. QUESTION/PROBLEMS 1.

2. 3.

Define the following: a. absorptivity b. reflectivity c. transmissivity d. emissivity What is the view factor F21 of a sphere (1) of diameter D inside a cubical box (2) of length L = D? A space satellite in the shape of a sphere is travelling in outer space, where its surface temperature is held at 283.2 K. The sphere only sees outer space, which can be considered as a black body with a temperature of 0 K. The

A1 = AS =

πD 2

A2 = AC = 6 LC2 = 6D2

π 2 D (1.0) = 6D2 (F21) 4

Technological Institute of the Philippines – Manila CHE409-Unit Operations Laboratory I, 1st Semester 2015-2016

F21 =

= (6.12 W/m2-K)( π (0.1683m)(0.305m))(374.9 –

π 6

297.1)K QC = 76.79 W

3.

Given: TS = 283.2 K TA = 0 K F12 = 1.0 ε = 0.1 Solution: Q = αε(TS4 – TA4)

Required: Q

START QR = εAα(TS4 – TA4) = 0.79(

QR = 86.42 W Q= (76.79 + 86.42)W = 163.21 W

W 2 4 m −K )(0.1)(283.2 – 0)K

Q = (5.67*10-8

π (0.1683m)(0.305m))((374.9)4 – (297.1)4)K4

XI. FLOWCHARTS Q = 36.47 W/m 4.

2 2

Given: A = 1.0 m TS = 290 K TA = 80 K ε = 1.0 hc = 12 W/m2-K Solution: qr = qc qc = hcA(TS – Tf) qr = εAα(Tf4 – TA4) εAα(Tf4 – TA4) = hcA(TS – Tf) εα(Tf4 – TA4) = hc(TS – Tf)

Tf = 266.40 K Given: OD = 0.1683 m TS = 374.9 K TA = 297.1 K L = 0.305 m ε = 0.79 Solution:

Required: Q

L3 gβ ρ2 ∆ T =313,449,412 ≈ 108 μ2

NGrNPr = 313,449,412(0.709) = 222,235,633.1 ≈ 108

∆ T / D )1/4

hc = 1.32((374.9 – 297.1)K/(0.1683 m))1/4 hc = 6.12 W/m2-K QC = hcA(TS – TA)

Stefan-Boltzman’s Law Emission Power I Emission Power II Kirchoff’s Law Area Factors

XIII.

Properties of air at 297.1 K: PR = 0.709 gβρ2/µ2 = 1.42*108/K-m3

At NGrNPr ≈ 108: hc = 1.32(

II. III. IV. V. VI.

XII. HAZARDS AND COUNTERMEASURES

Q = QC + QR

N Gr=

Inverse of the Distance Square Law

Required: Tf

1.0(5.67*10-8 W/m2-K4)(Tf4 – (80 K)4) = (12 W/m2-K)((290 K) – Tf)

5.

I.

WASTE DISPOSAL

Heat and Radiation Experiments