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3 years ago

Explain with examples: What are the reasons of a successful and unsuccessful software project?

Engineering

1 answer:

malfutka [58]3 years ago

3 0

Answer: Not Enough Time

Often, the deadline date is decided before the project starts and is non-negotiable. This deadline results in a headlong rush to get started on the assumption, the sooner you begin coding, the sooner you'll finish.

A rush to start coding is almost always the wrong approach. It is important to spend the time to create a good design. Not having a good design leads to continuing changes throughout the development phase. When this happens, time and budget are consumed at a rapid rate.

Solution: Make time to create a good design. Don't be tempted to jump straight in and begin coding. Assign time to this task and the rest of the project will run much better. It will improve your reputation when you deliver something that fulfils the customers' expectations and works the first time correctly.

Explanation:

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You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

Katen [24]

Answer:

The required wall thickness is $1.506 \times 10^{-3}$ m

Explanation:

Given:

Fluid density $\rho = 1200$ $\frac{kg}{m^{3} }$

Diameter of tank $d = 8$ m

Length of tank $l = 4$ m

F.S = 4

For A-36 steel yield stress $\sigma = 250$ MPa,

Allowable stress $\sigma _{allow} = \frac{\sigma}{F.S}$

$\sigma _{allow} = \frac{250}{4} = 62.5$ MPa

Pressure force is given by,

$P = \rho gh$

$P = 1200 \times 9.8 \times 4$

$P = 47088$ Pa

Now for a vertical pipe,

$\sigma _{allow} = \frac{Pd}{4t}$

Where $t =$ required thickness

$t = \frac{Pd}{4 \sigma _{allow} }$

$t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }$

$t = 1.506 \times 10^{-3}$ m

Therefore, the required wall thickness is $1.506 \times 10^{-3}$ m

8 0

3 years ago

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 3

Katarina [22]

Answer: Option D) 298 g/mol is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol ≈ 298 g/mol

Therefore Option D) 298 g/mol is the correct answer

4 0

3 years ago

simple Brayton cycle using air as the working fluid has a pressure ratio of 10. The minimum and maximum temperatures in the cycl

Irina18 [472]

Answer:

a) 764.45K

b) 210.48 kJ/kg

c) 30.14%

Explanation:

pressure ratio = 10

minimum temperature = 295 k

maximum temperature = 1240 k

isentropic efficiency for compressor = 83%

Isentropic efficiency for turbine = 87%

a) Air temperature at turbine exit

we can achieve this by interpolating for enthalpy

h4 = 783.05 kJ/kg ( calculated in the background ) at state 4 using Table A-17 for Ideal gas properties of air

T4 ( temperature at Turbine exit ) = 760 + ( 780 - 760 ) $(\frac{783.05-778.18}{800.13-778.18} )$ = 764.45K

b) The net work output

first we determine the actual work input to compressor

Wc = h2 - h1 ( calculated values )

= 626.57 - 295.17 = 331.4 kJ/kg

next determine the actual work done by Turbine

Wt = h3 - h4 ( calculated values )

= 1324.93 - 783.05 = 541.88 kJ/kg

finally determine the network output of the cycle

Wnet = Wt - Wc

= 541.88 - 331.4 = 210.48 kJ/kg

c) determine thermal efficiency

лth = Wnet / qin ------ ( 1 )

where ; qin = h3 - h2

equation 1 becomes

лth = Wnet / ( h3 - h2 )

= 210.48 / ( 1324.93 - 626.57 )

= 0.3014 = 30.14%

6 0

3 years ago

A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0

Gnoma [55]

Answer: power required to maintain level flight=82.20hp

Explanation:

Given

Area = 200 ft^2

Weight = 2000 lb

Cl( Lift coefficient)= 0.39

Cd( Drag coefficient) = 0.06

The density ρ of air at standard atmospheric pressure = 2.38 X 10^-3 slugs/ft^3

For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that

Lift force = Weight of aircraft

(1/2)ρAU²Cl= W

1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000

U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39

U= $\sqrt{21,547.08}$

Velocity, U= 146.7892ft/s

Drag force of the velocity can be deduced from the formulae

Cd= Drag force(D) /1/2 ρU²A

Drag force=1/2 ρU²ACd

D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06

D=307.69

Drag force= 308lb

power required to maintain level flight is given as

P = Drag force x Velocity = D x U

=308lb X 146.7892ft/s

=45,211.0736lb.ft/s

Changing to hp we have that

1 Horsepower, hp = 550 ft lbf/s

??=45,211.0736lb.ft/s

45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp

6 0

3 years ago

You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall

choli [55]

Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are $\rightarrow 120ft5in+2(10ft)$

The two long are $\rightarrow 122ft1in=122.08ft$

The two shors are $\rightarrow 86ft4.5in = 86.375ft$

The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

$Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3$

That in cubic yards is equal to $180.15 (1cy=27ft^3)$

Hence, we need order 5% plus that represent with the quantity

$Q_{ordered}=1.05*180.15=189.15cy$

8 0

3 years ago