All categories

3 years ago

List two reasons why machined parts often require a high degree of precision.

Engineering

1 answer:

Gekata [30.6K]3 years ago

3 0

Explanation:

Precision machining is a subtractive process used in cases where material needs to be removed from a raw product to create the finished product. Precision machining can be used to create a wide variety of products, items, and parts for any number of different objects and materials. These parts usually require tight tolerances variation from nominal dimensions and from part to part, which means that there is not much room for error in the production of the piece. Repeatability and well-controlled tolerances are hallmarks of precision machining. Components, parts and finished durable products that are designed to maintain extremely tight tolerance margins and a high degree of durability are essential and common drivers for utilization of precision machining. For example, parts that need to work together as part of a machine may need to always align within a certain margin of 0.01mm to 0.05mm. Precision engineering and machining help to ensure these parts can not only be made precisely but can be produced with this level of accuracy over and over again.

You might be interested in

Which of the following is NOT an ASE certification? Select one:

stiks02 [169]

The option that is not an ASE certification is . A/C and Refrigerants handling certification (609).

<h3>What is ASE certification?</h3>

The term ASE is known to be a body that tends to promotes excellence in regards to vehicle repair, service as well as parts distribution.

Note that in the world today more than a quarter of million of people are known to possess ASE certifications.

Since ASE Certified professionals work in in all areas of the transportation industry. one can say that The option that is not an ASE certification is. A/C and Refrigerants handling certification (609).

Learn more about ASE certification from

brainly.com/question/5533417

#SPJ1

8 0

2 years ago

A long corridor has a single light bulb and two doors with light switch at each door.

Tpy6a [65]

Answer:

Light = A xor B

Explanation:

If switches A and B produce True or False, then Light will be True for ...

Light = A xor B

8 0

3 years ago

Which term represents an object that has a round or oval base and is connected at every point by lines at a corresponding point

raketka [301]

Answer:

it is a polyhedron

Explanation:

if I am wrong I am sorry

8 0

3 years ago

Read 2 more answers

A plane wall of thickness 0.1 m and thermal conductivity 25 W/m·K having uniform volumetric heat generation of 0.3 MW/m3 is insu

Contact [7]

Answer:

T = 167 ° C

Explanation:

To solve the question we have the following known variables

Type of surface = plane wall ,

Thermal conductivity k = 25.0 W/m·K,

Thickness L = 0.1 m,

Heat generation rate q' = 0.300 MW/m³,

Heat transfer coefficient hc = 400 W/m² ·K,

Ambient temperature T∞ = 32.0 °C

We are to determine the maximum temperature in the wall

Assumptions for the calculation are as follows

Negligible heat loss through the insulation
Steady state system
One dimensional conduction across the wall

Therefore by the one dimensional conduction equation we have

$k\frac{d^{2}T }{dx^{2} } +q'_{G} = \rho c\frac{dT}{dt}$

During steady state

$\frac{dT}{dt}$ = 0 which gives $k\frac{d^{2}T }{dx^{2} } +q'_{G} = 0$

From which we have $\frac{d^{2}T }{dx^{2} } = -\frac{q'_{G}}{k}$

Considering the boundary condition at x =0 where there is no heat loss

$\frac{dT}{dt}$ = 0 also at the other end of the plane wall we have

$-k\frac{dT }{dx } =$ hc (T - T∞) at point x = L

Integrating the equation we have

$\frac{dT }{dx } = \frac{q'_{G}}{k} x+ C_{1}$ from which C₁ is evaluated from the first boundary condition thus

0 = $\frac{q'_{G}}{k} (0)+ C_{1}$ from which C₁ = 0

From the second integration we have

$T = -\frac{q'_{G}}{2k} x^{2} + C_{2}$

From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows

$-k\frac{q'_{G}L}{k} = h_{c}( -\frac{q'_{G}L^{2} }{k} + C_{2}-T∞)$ → C₂ = $q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$

T(x) = $\frac{q'_{G}}{2k} x^{2} + q'_{G}L(\frac{1}{h_{c} }+ \frac{L}{2k} } )+T∞$ and T(x) = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} )-x^{2} )$

∴ Tmax → when x = 0 = T∞ + $\frac{q'_{G}}{2k} (L^{2}+(\frac{2kL}{h_{c} }} ))$

Substituting the values we get

T = 167 ° C

4 0

3 years ago

Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres

natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

$h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}$