Air Cooled Pressurized Bed Essay

MINOR PROJECT REPORT

ON

PRESSURIZED SORES BED

Submitted to: Submitted by:

Rahul Sindhwani Saksham Gandhi A2305412097

Assistant professor Shivam Agarwal A2305412097

ASET Tushar Garg A23054120113

AMITY UNIVERSITY

UTTAR PRADESH

Abstract of Project :

Air-cooled pressurized bed is a type of bed consists of pressurized mattress which consists of certain number of compartments supported by metal rods. The metal rods, which in turn are supported by small bladders. These bladders are connected to pump control, which consists of alternating pressure mode to control pressure

require continues movement 5- Hip region - moves alternatively i.e. require continues movement 6-8 – leg region – moves alternatively i.e. require continues movement 9- Heel region - moves alternatively i.e. require continues movement

The alternative motion is achieved by the help of rods (steel rods of diameter 3 cm) which are further moved by the inflation and deflation of bladders respectively.

The steels rod are cushioned with the help of a foam and by using silicon sheets to give patient an extra comfort (the breadth of a foam used will be 2 cm)

Cross-section of a single component –

Here, we have a single compartment, where the height of the compartment is 50cm.

So, we place a position of 2 cm at a height of 33 cm from the base level at a height of bladder of 5 inches (12.7cm) and we placed a rod in the rod holder of 3.1 cm wide and 10 cm long supported by bladder. So as shown in fig. 4 the holder is round at bottom to support the rod and open at top, or full extension when bladder is fully inflated.

Bladeless fan

Volume occupied by the cushioned layer =

A=π(D^2-d^2)×L

=3.14*(2.5^2 – 1.5^2)*60 =753.6cm^3

So, total volume occupied = 1178cm^3

First we have 6 grooves of rod holders which air loss takes place,

Now considering a single rod holder

Volume of holder without rod = (πr^2)/2 *t + l*b*t =3.14/2*(3.1/2)^2*2+ (8.4*3.1*2) = 64.47cm^3

Now if we have 6 rods placed in it

Volume left = 68 – 14.3 =54cm^3

This is the free space from each iron holder

Now from 6 iron rod holder = 6 * 54 = 324 cm^3

Total free space in upper box = 490000 – 17071 -4712- 5000 = 463217 cm^3 Similarly in lower box = 49000 – 17071 – 5890 – 5000 = 462039 cm^3

Mattress ( Specialized TPU material ) It contains air pores of 2cm diameter each It will be palced above the rods Area of each air pore = π *