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lisov135 [29]
2 years ago
11

A factory hires a consultant to recommend ways to improve its productivity. The consultant notices that the production floor is

strewn with tools and machines that are not continually in use. Which of the 5S philosophies should the consultant recommend to the management to establish better productivity?
A.
set in place
B.
sort
C.
standardize
D.
shine
E.
sustain
Engineering
1 answer:
Romashka [77]2 years ago
8 0

Answer:

B

Explanation:

Keep only what is used. Remove all unnecessary items. Red tag these items for review.

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nevsk [136]

Answer:

Explanation:

Usage: flip [-t|-u|-d|-m] filename[s]

  Converts ASCII files between Unix, MS-DOS/Windows, or Macintosh newline formats

  Options:

     -u  =  convert file(s) to Unix newline format (newline)

     -d  =  convert file(s) to MS-DOS/Windows newline format (linefeed + newline)

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8 0
3 years ago
Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb
guajiro [1.7K]

Answer:

\eta_{turbine} = 0.603 = 60.3\%

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in  vapor state:

h₂ = h_{g\ at\ 125KPa} = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than s_g and greater than s_f at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88

Now, we will find h_{2s}(enthalpy at the outlet for the isentropic process):

h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg

Now, the isentropic efficiency of the turbine can be given as follows:

\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%

3 0
3 years ago
1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit​
elena55 [62]
Yeah it’s true. Good luck!!
3 0
3 years ago
Design a half-wave recti er which provides a peak voltage of 15 V, and anaverage voltage of 3.8 V when driven by a 120 V (rms) a
nirvana33 [79]

Answer:

You need a 120V to 24V commercial transformer  (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)

Step by step design:

  1. Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer.  120 Vrms = 85 V and 24 Vrms = 17V = Vin
  2. Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
  3. Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA

Our circuit meet the average voltage (Va) specification:

Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it

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3 years ago
Best use for a suspension bridge
babunello [35]

Answer:

over a rive or fast moving water or canyon

Explanation: you would use a suspension bridge in an area where you can't put supports down.

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3 years ago
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