We have 22 years experience

Joda Logo
Joda Logo
aerogel news
joda icon Home > Product News

NEWS

Performance Test Data Of Silica Aerogel Insulation Blanket

Pubdate: 2018-03-19 17:38:29
From: Joda Insulation Material Aerogel

Performance test data of silica aerogel insulating blanket

The core of building energy efficiency in China is the renovation of building envelope and heating system. The wall insulation construction technology plays an increasingly important role in building energy efficiency.

In today's society, the developed countries attach great importance to energy conservation and heat preservation. The energy consumption per unit building area of our country is equivalent to 3 to 4 times that of the developed countries with similar climate conditions. How to make rational use of energy and improve the utilization of energy is a fundamental plan for China's development. And the improvement of energy saving technology and water in building exterior wall enclosure has great significance for the reduction of social and building energy consumption.

What is aerogel?

The aerogel is the product of drying the gel in the gel when the gel skeleton is kept intact. Silica aerogel is referred to as the "blue smoke", "solid smoke", is now known as the lightest solid material, is by far the best performance of thermal insulation material.

Silica aerogel insulation blanket is a thermal insulation materials made of silica aerogel as the main material. It has many advantages, such as low thermal conductivity, small density, high strength, high utilization rate of space, sound insulation, green environmental protection, waterproof and non flammable.

So what is the thermal conductivity of the silica aerogel insulating blanket?

The coefficient of thermal conductivity refers to the thickness of the material on the two sides of the 1m thick material under the condition of stable heat transfer. The temperature difference between the two sides of the surface is 1 degrees (K, c), and the heat transferred through the area of 1 square meters in one second (1s), the unit is watt / meter (W/ (M. K), here K is available instead of C).

We all know that the thermal conductivity of this kind of heat insulating blanket is low, that is, the coefficient of thermal conductivity is low, and it has good thermal insulation effect. Then how can the thermal conductivity be calculated?

We need to set up a mathematical model for experiment. Suppose a column cylinder is surrounded by a silica aerogel insulating blanket. The temperature field inside the cylinder is steady heat transfer, that is, every temperature in the body has nothing to do with time. The outer radius of the cylinder is R1, the outer radius R2 is wrapped, the temperature of the cylinder wall is TW1, and the exterior temperature of the thermal insulation materials is TW2. The L=0.87M of the cylinder plant can be considered as a constant in this experiment. By controlling the temperature of the heating tube, the temperature field in the cylinder can reach the steady state heat transfer.

According to Fu Liye's law, we have to:

=2 PI lambda l* (tw1-tw2) /ln (R2 /R1)

The power to control the temperature of a cylinder heating tube to achieve steady heat transfer (heater power) = the natural logarithm of the ratio of thermal conductivity *2 PI l* (tw1-tw2) to the ratio of internal and external temperature / outer radius to inner radius.

Through the reverse solution:

Lambda phi = (R2 /R1) / 2 pi l* (tw1-tw2)]

During the process of temperature measurement, 8 points of the cylinder outer wall 101~108 were arranged, and 8 points of the silica aerogel insulating blanket 111~118 wrapped around the cylinder were tested for different temperature changes.

Test data can be obtained: the power of the heater is 30W, R2=10.02, R1=8.01, and the internal and external temperature difference is (117.5 degrees centigrade -38.4 degrees centigrade). The above data can be obtained according to Fu Liye's law.

Lambda phi = (R2 /R1) / 2 pi l* (tw1-tw2)]

=30*ln (10.02/8.01) /[2*3.14*0.87* (117.5-38.4)]

=0.016W/ (M. K)

The coefficient of thermal conductivity will be higher according to the temperature difference, the lower the thermal conductivity. The thermal conductivity of silica aerogel and geuda production technology of insulation mat test result is 0.016 W/ (M - K), this is the leader in thermal insulation materials industry.

Number 101 102 103 104 105 106 107 108
Temperature/℃ 116.5 118.7 118.7 118.7 118.2 117.2 110.7 113.5
Number 111 112 113 114 115 116 117 118
Temperature/℃ 50.1 32.5 42.7 39.1 35 39.8 41.5 44.3

Comparison with the thermal conductivity of traditional thermal insulation materials

The thermal conductivity of the silica aerogel insulating blanket is approximately equal to the thermal conductivity of the polyurethane insulation board. Considering the fire resistance, the polyurethane is flammable, and the combustion produces toxic gases, causing safety hazards to the building. The test materials with muskets firing and keep not burning, low thermal conductivity, is ideal for high temperature insulation wall insulation materials.

Silica gel glass fiber insulation mat is one of the main products of geuda. The hydrophobicity of the product is very high, the water repellent rate at normal temperature can reach more than 99%, the flame retardancy is excellent and the fire protection grade reaches the national A1 standard. And the product has a wide range of use temperature, which can be used in the range of -200~650 centigrade according to different requirements. It is easy to install and convenient for construction. It can be widely used in pipes, tanks, furnace bodies, buildings and so on, which are adiabatic under 650 C.

Silica Aerogel Insulation Blanket Performance Test Data, Thermal Insulation Materials   Thermal Insulation Materials, Insulating Blanket

Joda Top