Answer:
heat loss per 1-m length of this insulation is 4368.145 W
Explanation:
given data
inside radius r1 = 6 cm
outside radius r2 = 8 cm
thermal conductivity k = 0.5 W/m°C
inside temperature t1 = 430°C
outside temperature t2 = 30°C
to find out
Determine the heat loss per 1-m length of this insulation
solution
we know thermal resistance formula for cylinder that is express as
Rth = 
.................1
here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity
so
heat loss is change in temperature divide thermal resistance
Q = 
Q = 
Q = 4368.145 W
so heat loss per 1-m length of this insulation is 4368.145 W
Answer:
230.4W
Explanation:
Heat transfer by conduction consists of the transport of energy in the form of heat through solids, in this case a jacket.
the equation is as follows
Where
Q=heat
k=conductivity=0.04
A=Area=1.8m^2
T2=33C
T1=1C
L=thickness=1cm=0.01m
Q=230.4W
the skier loses heat at the rate of 230.4W
Answer:
|W|=169.28 KJ/kg
ΔS = -0.544 KJ/Kg.K
Explanation:
Given that
T= 100°F
We know that
1 °F = 255.92 K
100°F = 310 .92 K
We know that work for isothermal process
Lets take mass is 1 kg.
So work per unit mass
We know that for air R=0.287KJ/kg.K
W= - 169.28 KJ/kg
Negative sign indicates compression
|W|=169.28 KJ/kg
We know that change in entropy at constant volume
ΔS = -0.544 KJ/Kg.K
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Explanation:
Production planning: Planning is ideal so that there are the right resources, at the right time and in the right quantity that can meet the production needs of a period.
Strategies: The strategic development of production is the area that will assist in organizational competitiveness and in meeting consumer demand and needs.
Product and service design: Development of new products and services and their improvement, innovations and greater benefits
Production systems: Study of physical arrangements so that production takes place effectively according to the ideal layout for each type of product or service.
Production capacity planning: Analysis of the short, medium and long term related to production, and identification if necessary to obtain more resources, increase in staff, machinery, etc., to meet present and future demands.
<em>Each area of knowledge acquired will assist in the development of a professional career, as technical knowledge is essential in decision-making, provision, problem solving, the development of new ideas and innovation.</em>
