How do information systems support the activities in a supply chain?

A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______

lbvjy [14]

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.

5 0

3 years ago

Name the main classes of polymer and define their characteristic properties

Svetlanka [38]

Answer:

Polymers are the naturally occurring or synthetic macromolecules that are composed of repeating subunits, called monomers.

The three main classes of polymers are: thermoplastic, thermosetting, and the elastomers.

Thermoplastic polymers have linear bonding. These polymers can be melted again and thus can recycled.

Thermosetting polymers have cross-linked bonding. These polymers decompose when heated and thus can not be remelted and recycled.

Elastomers have linear bonding with some cross-linking. These polymers extreme elastic extensibility and thus can revert back to its original shape after deformation, without causing any permanent damage.

8 0

4 years ago

Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is

ryzh [129]

Answer:

a) $\dot W = 0.978\,kW$ , b) $I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}$

$COP_{HP} = 14.198$

The reversible work input is:

$\dot W = \frac{\dot Q_{H}}{COP_{HP}}$

$\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s} \right)$

$\dot W = 0.978\,kW$

b) The irreversibility is given by the difference between real work and ideal work inputs:

$I = \dot W_{real} - \dot W_{ideal}$

$I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW$

7 0

4 years ago

What is the function of a fixed resistor?

agasfer [191]

Answer:

add resistance to a circuit

Explanation:

It depends on the design in which it is incorporated. A fixed resistor has many uses, including, but not limited to ...

dropping voltage
limiting current
contributing to a time delay
adjusting frequency response
eliminating (or creating) signal reflections
acting as a fuse
calibrating or trimming a response
providing protection against electrical shock or ESD
acting as a reference when measuring variable resistors

8 0

4 years ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K ; R = 0.05

convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

Replacing the given data; we have:

$\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)}$ $= \pi (0.1)*2000(276-273)$

$\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96$

$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

$\frac {0.05}{r} =e^{0.0970}$

$\frac {0.05}{r} =1.102$

$r=\frac{0.05}{1.102}$

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0

4 years ago