All categories

2 years ago

If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?

Engineering

1 answer:

sleet_krkn [62]2 years ago

7 0

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

You might be interested in

An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere wherein t

weeeeeb [17]

Answer:

the temperature T at which the treatment is carried is out is 1274.24 K

Explanation:

Fick's second Law posits that the rate of change of concentration of diffusing species is directly proportional to the second derivative of the concentration.

Using the expression of the Fick's second Law:

$\mathbf{\frac{C_x-C_o}{C_s-C_o} = 1- erf(\frac{x}{2\sqrt{Dt} })}$

where;

$C_o$ = initial concentration

$C_x$ = the depth of the concentration

$C_s$ = surface concentration

$erf(\frac{x}{2\sqrt{Dt} })}$ = Gaussian error function.

Let variable z be used for the expression of the Gaussian error function. $erf(\frac{x}{2\sqrt{Dt} })}$

Then, from the above equation: replacing $C_x$ with 0.35 ; $C_o$ with 0.2 and $C_s$ with 1.0; we have:

$\mathbf{\frac{0.35-0.2}{1.0-0.2} = 1- erf( z)}$

erf (z) = 0.8125

we obtain the error function value close to 0.8125 from the error function table and we did the interpolation to obtain the exact value of variable corresponding to 0.8125.

The table below shows the tabular form of the error function value close to 0.8125 .

Value for z Value for erf (z)

0.9 0.797

z 0.8125

0.950 0.8209

From above; we can find the value of variable corresponding to the error function 0.8125 .

i.e

$\frac{z-0.9}{0.95-0.9} =\frac{0.8125-0.797}{0.8209-0.797}$

z = 0.932

However, the temperature dependence relation for the diffusion coefficient D can be expressed as:

$z = \frac{x}{\sqrt{Dt} }$

where;

z = 0.932

x = 3.5 mm = 0.0035 m

t = 50 h = 180000 sec

$0.932 = \frac{0.35}{2\sqrt{D*180000} }$

D = $1.958*10^{-11} m^2/s$

Finally, the temperature T at which the treatment is carried is out is calculated as:

$\mathbf{D=D_o \ exp \ (-\frac{Q_d}{RT}) }$

From the table ‘Diffusion data’, we obtain the values of temperature-independent pre exponential and activation energy for diffusion of carbon in FCC Fe.

$D_o = 2.3*10^{-5} \ m^2/s$

$Q_d = 148, 000 \ J/mol$

Replacing all values needed for the above equation; we have:

$1.958*10^{-11}= (2.3*10^{-5})exp(\frac{-148,000}{(8.31)T})$

$8.51*10^{-7}=exp(\frac{-17,810}{T})$

$In(8.51*10^{-7})=(\frac{-17,810}{T})$

-13.977 = -17,810/T

T = -17,810/ - 13.977

T = 1274.24 K

Hence, the temperature T at which the treatment is carried is out is 1274.24 K

5 0

3 years ago

Which of the following describes fibers? a)- Single crystals with extremely large length-to-diameter ratios. b)- Polycrystalline

ElenaW [278]

Answer:

b)Poly crystalline and amorphous materials with small diameter

Explanation:

Fibers have high length to diameter ratio and also have high strength.Generally length of fibers is very high and diameter is very low as compare to length.

Mostly fibers is used to transfer data from one place to another place with help of fiber optical cables.Fiber optic cables is used in telecommunication.In these cables data covert in to the electric single and reach at define location and after data is decode and covert from electric single in to original data.

Fibers poly crystalline and amorphous materials with small diameter.

6 0

3 years ago

A piston/cylinder contains 1.5 kg of water at 200 kPa, 150°C. It is now heated by a process in which pressure is linearly relate

Fofino [41]

Answer:

final volume V2 = 0.71136 m³

work done in process W = -291.24 kJ

heat transfer Q = 164 kJ

Explanation:

given data

mass = 1.5 kg

pressure p1 = 200 kPa

temperature t1 = 150°C

final pressure p2 = 600 kPa

final temperature t2 = 350°C

solution

we will use here superheated water table that is

for pressure 200 kPa and 150°C temperature

v1 = 0.95964 m³/kg

u1 = 2576.87 kJ/kg

and

for pressure 600 kPa and 350°C temperature

v2 = 0.47424 m³/kg

u2 = 2881.12 kJ/kg

so v1 is express as

V1 = v1 × m ............................1

V1 = 0.95964 × 1.5

V1 = 1.43946 m³

and

V2 = v2 × m ............................2

V2 = 0.47424 × 1.5

final volume V2 = 0.71136 m³

and

W = P(avg) × dV .............................3

P(avg) = $\frac{p1+p2}{2}$ = $\frac{200+600}{2}$ = 400 × 10³

put here value

W = 400 × 10³ × (0.71136 - 1.43946 )

work done in process W = -291.24 kJ

and

heat transfer is

Q = m × (u2 - u1) + W .............................4

Q = 1.5 × (2881.12 - 2576.87) + 292.24

heat transfer Q = 164 kJ

7 0

3 years ago

Air is compressed isothermally from 13 psia and 55°F to 80 psia in a reversible steady-flow device. Calculate the work required,

solong [7]

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

P₂= 80 psia

Temperature =°F Temperature is convert to °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67 x -1.817

=- 64.10Btu/lbm

The required work therefore for this isothermal compression is 64.10 Btu/lbm

8 0

3 years ago

A 2-cm-diameter vertical water jet is injected upward by a nozzle at a speed of 15 m/s. Determine the maximum weight of a flat p

Ede4ka [16]

Answer:58.28 N

Explanation:

Given data

$dia. of nozzle \left ( d\right )=2 cm$

$initial velocity\left ( u\right )=15 m/s$

$height\left ( h\right )=2m$

Now velocity of jet at height of 2m

$v^2-u^2=2gh$

$v^2=15^2-2\left ( 9.81\right )\left ( 2\right )$

$v=\sqrt{185.76}=13.62 m/s$

$Now\ forces\ on\ plate\ are\ weight\left ( Downward\right ) and jet\ force\left ( upward\right )$

equating them

W= $\left ( \rho Av\right )v$

W= $10^{3}\times \frac{\pi}{4}\left ( 0.02\right )^2\times 13.62^2$

W=58.28 N

7 0

3 years ago