All categories

2 years ago

What are the indicators of ineffective systems engineering?

Engineering

1 answer:

liberstina [14]2 years ago

7 0

Answer:

Indicators for ineffective system engineering are as follows

1.Requirement trends

2.System definition change backlog trends

3.interface trends

4.Requirement validation trends

5.Requirement verification trends

6.Work product approval trends

7.Review action closure trends

8.Risk exposure trends

9.Risk handling trends

10.Technology maturity trends

11.Technical measurement trends

12.System engineering skills trends

13.Process compliance trends

You might be interested in

A 1000 W iron utilizes a resistance wire which is 20 inches long and has a diameter of 0.08 inches. Determine the rate of heat g

SSSSS [86.1K]

Answer:

The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3

Heat flux is 9.67×10^7 Btu/hrft^2

Explanation:

Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr

Area (A) = πD^2/4

Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft

A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2

Volume (V) = A × Length

L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft

V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3

Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3

Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2

3 0

2 years ago

Which of following are coding languages used in controlling a robot? *

Bess [88]

Answer:

C/C++

Explanation:

C/C++

7 0

2 years ago

A pin fin of uniform cross-sectional area is fabricated of an aluminum alloy (k = 160W m-1 K-1 ). The fin diameter is D = 4 mm,

disa [49]

Answer: (a) 36.18mm

(b) 23.52

Explanation: see attachment

4 0

3 years ago

Four of the minterms of the completely specified function f(a, b, c, d) are m0, m1, m4, and m5.

Sveta_85 [38]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms for f so that f has eight primary implicants with two literals and no other prime implicant are $m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14}$ and $m_{15}$

b) The essential prime implicant are $c' d',a'b',ab$ and $cd$

c) The minimum sum-of-product expression for f are

$a'b' +ab +c'd'+cd+a'c',\\ a'b'+ab+c'd'+cd+a'd,\\a'b'+ab+c'd'+cd+bc' and \\ a'b'+ab+c'd' +cd+bd$

Explanation:

The explanation is shown on the second third and fourth image

8 0

2 years ago

At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r

ludmilkaskok [199]

Answer:

a) $COP_{R} = 25.014$ , b) $T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

$COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}$

$COP_{R} = 25.014$

b) The respective coefficient of performance is determined:

$COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}$

$COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}$

$COP_{R} = 5$

But:

$COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}$

The temperature at hot reservoir is found with some algebraic help:

$COP_{R} \cdot (T_{H}-T_{L})=T_{L}$

$T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}$

$T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}} \right)$

$T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5} \right)$

$T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)$

8 0

2 years ago

Read 2 more answers