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andrezito [222]
3 years ago
10

What information should a professional Engineer keep private

Engineering
1 answer:
enyata [817]3 years ago
3 0

Answer:

Work to develop software and related documents that respect the privacy of those who will be affected by that software.

Explanation:

In the Software Engineering Code of Ethics and Professional Practice, under Principle, which is section three (3) of the document, it is stated in clause 12 on what information should a professional Engineer keep private.

In it, it is stated that professionals engineers should keep private information relating to "Work to develop software and related documents that respect the privacy of those who will be affected by that software."

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5 0
4 years ago
Cool water at 15°C is throttled from 5(atm) to 1(atm), as in a kitchen faucet. What is the temperature change of the water? What
Tresset [83]

Answer:

the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

Explanation:

Given that:

Initial Temperature T_1 = 15°C

Initial Pressure P_1 = 5 atm

Final Pressure P_2 = 1 atm

Data obtain from steam tables of saturated water at  15°C are as follows:

Specific volume  v = 1.001 cm³/gm

The change in temperature = 2°C

Specific heat of water = 4.19 J/gm.K

volume expansivity β = 1.5 × 10⁻⁴ K⁻¹

The expression to determine the change in temperature can be given as :

\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \  K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})

Δ T = 0.093 K

Now; we can calculate the lost work bt the formula:

W_{lost} = T_{surr} *S

where ;

T_{surr} is the temperature of the surrounding. = 20°C = (20+273.15)K =  293.15 K

From above the change in entropy is:

\delta \  S = C_p \  In (\frac{T+ \delta \ T }{T}) *  \beta V \delta P

\delta \  S = 4.19*  \  In (\frac{288.15+0.093 }{288.15}) -  1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})

\delta \  S =1.408*10^{-3} \ J/gm.K

W_{lost} = T_{surr} *S

W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K

W_{lost} = 0.413 \  kJ/kg

Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg

6 0
4 years ago
The repair and rehabilitation of damaged or spalled concrete is done by removing the:____________A. Loose concreteB. CollapsingC
Mrac [35]

Answer:

A. Loose concrete

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4 years ago
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3 years ago
Animation can occur before an action
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3 years ago
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