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3 years ago

The natural variation of a process relative to the variation allowed by the design specifications is known as

Engineering

1 answer:

Nimfa-mama [501]3 years ago

3 0

Answer:

"Process capability" is the correct answer.

Explanation:

The Process Capability seems to be a method of measuring of how and why the framework performs concerning something like the successful objectives. This same capacity is characterized as that of the client's voice over procedure speech.
Through using functionality indicators it analyses the performance with an in-control process with the permissible range.

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On-site oil storage containers must be marked "Used Oil."<br><br> True. <br><br> False.

Scrat [10]

Answer:

How must used oil storage containers be marked? Containers and aboveground tanks used to store used oil at generator facilities must be labeled or marked clearly with the words “Used Oil" (40 CFR Section 279.22(c)).

Explanation:

i think it will help you

7 0

3 years ago

A sheet of steel 1.5 mm thick has nitrogen (N2) atmospheres on both sides at 1200°C and is permitted to achieve steady-state dif

Gelneren [198K]

Answer:

do the wam wam

Explanation:

3 0

3 years ago

Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent

Aleksandr [31]

Answer:

The maximum load the bar can withstand = 35.43 KN

Explanation:

Ultimate tensile strength of the given aluminium bar $\sigma$ = 540 M pa

Cross section area of the bar = $8.1^{2}$ = 65.61 $mm^{2}$

We know that the ultimate strength of the bar is calculated from

$\sigma = \frac{P_{max} }{A}$

$540 = \frac{P_{max} }{65.61}$

$P_{max}$ = 540 × 65.61

$P_{max}$ = 35.43 KN

Therefore the maximum load the bar can withstand = 35.43 KN

6 0

4 years ago

An industrial load with an operating voltage of 480/0° V is connected to the power system. The load absorbs 120 kW with a laggin

Leni [432]

Answer:

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

Explanation:

From the question we are told that:

Voltage $V=480/0 \textdegree V$

Power $P=120kW$

Initial Power factor $p.f_1=0.77 lagging$

Final Power factor $p.f_2=0.9 lagging$

Generally the equation for Reactive Power is mathematically given by

Q=P(tan \theta_2-tan \theta_1)

Since

$p.f_1=0.77$

$cos \theta_1 =0.77$

$\theta_1=cos^{-1}0.77$

$\theta_1=39.65 \textdegree$

And

$p.f_2=0.9$

$cos \theta_2 =0.9$

$\theta_2=cos^{-1}0.9$

$\theta_2=25.84 \textdegree$

Therefore

$Q=P(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=120*10^3(tan 25.84 \textdegree-tan 39.65 \textdegree)$

$Q=-41.33VAR$

Therefore

The size of the capacitor in vars that is necessary to raise the power factor to 0.9 lagging is

$Q=41.33 KVAR\ \\at\\\ 480 Vrms$

6 0

3 years ago

As part of a heat treatment process, cylindrical, 304 stainless steel rods of 100-mm diameter are cooled from an initial tempera

saveliy_v [14]

Answer:

Explanation:

Given that:

diameter = 100 mm

initial temperature = 500 ° C

Conventional coefficient = 500 W/m^2 K

length = 1 m

We obtain the following data from the tables A-1;

For the stainless steel of the rod $\overline T = 548 \ K$

$\rho = 7900 \ kg/m^3$

$K = 19.0 \ W/mk \\ \\ C_p = 545 \ J/kg.K$

$\alpha = 4.40 \times 10^{-6} \ m^2/s \\ \\ B_i = \dfrac{h(\rho/4)}{K} \\ \\ =0.657$

Here, we can't apply the lumped capacitance method, since Bi > 0.1

$\theta_o = \dfrac{T_o-T_{\infty}}{T_i -T_\infty}} \\ \\ \theta_o = \dfrac{50-30}{500 -30}} \\ \\ \theta_o = 0.0426\\$

$0.0426 = c_1 \ exp (- E^2_1 F_o_)\\ \\ \\ 0.0426 = 1.1382 \ exp (-10.9287)^2 \ f_o \\ \\ = f_o = \dfrac{In(0.0374)}{0.863} \\ \\ f_o = 3.81$

$t_f = \dfrac{f_o r^2}{\alpha} \\ \\ t_f = \dfrac{3.81 \times (0.05)^2}{4.40 \times 10^{-6}} \\ \\ t_f= 2162.5 \\ \\ t_f = 36 mins$

However, on a single rod, the energy extracted is:

$\theta = pcv (T_i - T_{\infty} )(1 - \dfrac{2 \theta}{c} J_1 (\zeta) ) \\ \\ = 7900 \\times 546 \times 0.007854 \times (500 -300) (1 - \dfrac{2 \times 0.0426}{1.3643}) \\ \\ \theta = 1.54 \times 10^7 \ J$

Hence, for centerline temperature at 50 °C;

The surface temperature is:

$T(r_o,t) = T_{\infty} +(T_1 -T_{\infty}) \theta_o \ J_o(\zeta_1) \\ \\ = 30 + (500-30) \times 0.0426 \times 0.5386 \\ \\ \mathbf{T(r_o,t) = 41.69 ^0 \ C}$

5 0

3 years ago