Insulation Thickness of Hot Cup of Tea
Heat and Mass Transfer Lab Project Report
Bachelor of Science Mechanical Engineering Department
Muhammad Nabeel Khaliq Syed Umer Ali Shah Raja Arslan Shamshad Saad Bin Sarfraz
BME143038 BME143029 BME143039 BME143004
Submitted to: Miss.Shumaila Rasheed Department of Mechanical Engineering CUST, Islamabad
9thJune, 2017
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Abstract
The project that has been assigned to our group was to find the critical thickness of insulation of a cup using different insulating materials. We use two different materials i.e. aluminum foil and polyurethane. polyurethane. We first first find the insulation insulation thickness thickness and heat transfer transfer rate of both both materials materials and then experimentally check that which material is a better insulator. Thus by experimental results we conclude that polyurethane is a better insulator than aluminum foil.
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Contents 1.0 Problem Statement ............................................... ........................................................ ..................................................................... ............. 5 2.0 Introduction Introduction ................................................... ........................................................ ............................................................................ .................... 5 3.0 Literature Review……………………………………………………………………………...5 4.0 Calculations............................................................................................................................... 6 4.1 Aluminum foil…………………………………………………………………………6 4.2 Polyurethane…………………………………………………………………………..8 5.0 Methodology ............................................... ....................................................... .............................................................................. ....................... 9 6.0 Results and Discussion Discussion ............................................................ ................................................. 9 7.0 References .................................................... ............................................................................................................ ........................................................................... ................... 10 8.0 Appendices .................................................... ........................................................................................................... .......................................................................... ................... 11
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List of Figures Figure 1 Aluminum foil foil insulation insulation ............................................................................................. . 11 11 Figure 2 Polyurethane Polyurethane insulation insulation ................................................... .............................................. 12
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1.0 Problem Statement Find the critical insulation thickness of cup using different insulating materials like aluminum foil, polystyrene (Styrofoam), Polyurethane etc. Also, find the heat transfer rate and then compare that which material is a better insulator.
2.0 Introduction In this project, we have to find the critical insulation thickness of cup by insulating the cup with a suitable insulating material. The critical insulation thickness of can be easily find by knowing the thermal conductivity of the insulating material and convective coefficient.
3.0 Literature Review The critical radius gives rise to a maximum radial heat flow for a given overall temperature difference or to a minimum overall temperature difference for a given radial heat flow.[1] The standard theory of the critical radius, as set forth in all heat transfer textbooks, contains numerous simplifying assumptions. Among these, the most significant is the assumption that the heat transfer coefficient at the outer surface of the body is independent of the body radius and is circumferentially uniform.[2] Circumferential variations of the heat transfer coefficient are a reality of nature. The effect of such variations will be assessed here with the aid of numerical solutions of the differential equation of energy conservation. The physical situation to be considered is the cylinder in crossflow. The theory of critical thickness of insulation for cylindrical and spherical geometries is discussed in a comprehensive and methodical manner. [3]The article begins with the introduction of the classical analysis, which assumes surface convection with a constant heat transfer coefficient, h. Next, various various modifications modifications of the basic basic theory are are discussed discussed in detail. These These modifications modifications allow for: the variation of h with outside radius, (constant-property forced convection); the variation of h with temperature-dependent fluid properties (variable-property forced convection); circumferential variation of h with forced convection; the variation of h with outside radius and the temperature difference between the outside surface and the surroundings (natural convection); surface radiation; combined radiation and natural convection; combined radiation, natural convection, and surface heat release due to change of phase of the surrounding fluid; and effects of radiation, inclination, and insulation opacity.[4] The last three sections of the article consider the optimum configurations for a circular pipe covered with insulations of equilateral polygonal, polygonal, rectangular, and eccentric circular shapes. For polygonal polygonal and rectangular rectangular geometries, the two-dimensional conduction analysis reveals that the conduction for the critical configuration is the same whether the outside surface boundary condition is that of convection or constant
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temperature. This permits the use of conduction shape factors to determine the optimum configuration for polygonal and rectangular insulations.[5]‘ Heat flow is an inevitable consequence of contact between objects of differing temperature. Thermal insulation provides a region for insulation in which thermal conduction is reduced or thermal radiation is reflected rather than absorbed by the lower temperature body. To change the temperature of an object, energy is required in the form of heat generation to increase the temperature, or heat extraction to reduce the temperature. Once the heat generation or heat extraction is terminated a reverse flow of heat occurs to reverse the temperature back to ambient. To maintain a given temperature considerable continuous energy is required. Insulation will reduce this energy loss.
4.0 Calculations 4.1 Aluminum Foil T1= 300C= 303K, T∞=240C= 297K
Tm =
303 303 + 297 297 2
Tm= 300K
At 300K, k=0.032W/m. K, h= 50W/m 2. K
Rc = =
ℎ
0.032 50
R c= 0.64mm
At 300K, h= 100W/m 2. K
Rc = =
ℎ
0.032 100
R c= 0.32mm
For h=50W//m 2. K, L=77mm, D1=71mm, D2=72.28mm 2 Rins = ln /2 1 = ln
. .
/2(0.032)(0.077)
R ins ins=1.154K/W 6|Page
Rconv = =
1 2Lh
1 2(0.036)(0.077)(50)
R conv conv= 1.15K/W
Q=
1−∞ + =
303 303 − 297 297 1.154 1.154 + 1.15 1.15
Q = 2.604W
For h=100W//m 2. K, L=77mm, D1=71mm, D2=71.64mm 2 Rins = ln /2 1 = ln
. .
/2(0.032)(0.077)
R ins ins=0.582K/W
Rconv =
1 2Lh 1
=
2(0.0358)(0.077)(100) R conv conv = 0.5774K/W Q=
1−∞ + =
Q = 5.175W
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303 303 − 297 297 0.582 0.582 + 0.5774 0.5774
4.2 Polyurethane T1= 300C= 303K, T∞=240C= 297K
Tm =
303 303 + 297 297 2
Tm = 300K
At 300K, k=0.026W/m. K, h= 50W/m 2. K
Rc = =
ℎ
0.026 50
R c= 0.52mm
At 300K, h= 100W/m 2. K
Rc = =
ℎ
.
R c = 0.26mm
For h=50W//m 2. K, L=87mm, D1=84mm, D2=85.04mm 2 Rins = ln /2 1 = ln
.
/2(0.026)(0.087)
R ins ins=0.866K/W
Rconv =
1 2Lh 1
=
2(0.0425)(0.087)(50) R conv conv = 0.861K/W Q=
1−∞ + =
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303 303 − 297 297 0.866 0.866 + 0.861 0.861
Q = 3.47W For h=100W//m 2. K, L=87mm, D1=84mm, D2=84.52mm 2 Rins = ln /2 1
= ln
.
/2(0.026)(0.087)
R ins ins=0.434K/W
Rconv =
1 2Lh 1
=
2(0.042)(0.087)(100) R conv conv = 2.30K/W Q= =
1−∞ +
Q = 2.1096W
..
5.0 Methodology
First of all, we choose two different materials i.e. Aluminum foil and Polyurethane. After that with the help of k (thermal conductivity of material)and h (convective coefficient) we find out the critical insulation thickness. Then by changing h we can get different critical insulation thickness. Then we find out the heat transfer rate. After that with the appropriate value of thickness, we insulate the cup and then experimentally check that which material is a better insulator.
6.0 Results and Discussion The critical insulation thickness of aluminum foil at h=50W/m2. K is 0.64mm and at h=100W/m2. K is 0.32mm. Similarly, for polyurethane it is 0.52mm and 0.26mm. Although the value of thickness for polyurethane material is low as compared to aluminum foil but it can be increased by increasing h. By experimental results we conclude that polyurethane is a better insulator than aluminum foil.
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7.0 References References [1] E. M. Sparrow, "Reexamination and Correction of the Critical Radius for Radial Heat Conduction," AIChE Conduction," AIChE Journal, Journal, vol. 16, p. 149, 1970. [2] E. R. G. a. S. E. Eckert, "Distribution of Heat Transfer Coefficients Coefficients Around Circular Cylinders in Crossflow at Reynolds Numbers From 20 to 500," Trans. ASME, vol. 74, pp. 343-347, 1952. [3] L. D. Simmons, "Critical Thickness of Insulation Accounting for Variable Convection Coefficient and Radiation Loss," J Loss," J . Heat Transfer, Transfer, vol. 98, p. 150, 1976. [4] E. M. Sparrow, "Re-examinationand Correction of the Critical Radius for Radial Heat Conduction," AfChE Conduction," AfChE J., vol. 16, p. 149, 1970. [5] T. Yuge, "Experiments on Heat Transfer from Spheres Including Combined Natural and Forced Convection," J. Convection," J. Heat Transfer, Transfer, vol. 82, p. 214, 1960.
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8.0 Appendices
Figure 1 Aluminum foil insulation
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Figure 2 Polyurethane insulation
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