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Rom4ik [11]
3 years ago
6

Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe

Engineering
1 answer:
lys-0071 [83]3 years ago
4 0

Answer:

Dalton

The part of the close-out report that would describe how he would plan and manage projects in the future is:

summary of project management effectiveness

Explanation:

The Project Close-out Report is a project management document, which identifies the variances from the baseline plans.  These variances are specified in terms of project performance, project cost, and schedule.  The project close-out report records the completion of the project and the subsequent handover of project deliverables to others.  The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.

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A boiler is used to heat steam at a brewery to be used in various applications such as heating water to brew the beer and saniti
Natalija [7]

Answer:

net boiler heat = 301.94 kW

Explanation:

given data

saturated steam = 6.0 bars

temperature = 18°C

flow rate = 115 m³/h = 0.03194 m³/s

heat use by boiler = 90 %

to find out

rate of heat does the boiler output

solution

we can say saturated steam is produce at 6 bar from liquid water 18°C

we know at 6 bar from steam table

hg = 2756 kJ/kg

and

enthalpy of water at 18°C

hf = 75.64 kJ/kg

so heat required for 1 kg is

=hg - hf

= 2680.36 kJ/kg

and

from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg

so here mass flow rate is

mass flow rate = \frac{0.03194}{0.315}

mass flow rate m = 0.10139 kg/s

so heat required is

H = h × m  

here h is heat required and m is mass flow rate

H = 2680.36  × 0.10139

H =  271.75 kJ/s = 271.75 kW

now 90 % of boiler heat is used for generate saturated stream

so net boiler heat = \frac{H}{0.90}

net boiler heat = \frac{271.75}{0.90}

net boiler heat = 301.94 kW

5 0
3 years ago
The flow rate in the pipe system below is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. Point 1 is 0.60 m higher
DedPeter [7]

Answer:

Explanation:

The rate of flow in the pipe system in Figure P4.5.2 is 0.05 m3/s. The pressure at point 1 is measured to be 260 kPa. All the pipes are galvanized iron with roughness value of 0.15 mm. Determine the pressure at point 2. Take the loss coefficient for the sudden contraction as 0.05 and v = 1.141 × 10−6 m2/s.

The answer to the above question is

The pressure at point 2 = 75.959 kPa

Explanation:

Bernoulli's equation with losses gives

hL = z₁ - z₃ +(P₁-P₃)/(ρ×g) + (v₁²-v₃²)/(2×g)

Between points 1 and 2, z₁ = z₃ + 0.6 m therefore

hL = 0.6 m +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

hL = (f₁×L₁×v₁²)/(D₁×2×g) + (f₂×L₂×v₂²)/(D₂×2×g) + (f₃×L₃×v₃²)/(D₃×2×g) + k×V₃₂/(2×g) = 0.6 +(P₁-P₂)/(ρ×g) + (v₁²-v₃²)/(2×g)

But v = Q/A

or  since A = π×D²/4 we have

A₁ = 1.77×10-2 m² , A₂ = 5.73×10-2 m², A₃ = 3.8×10-2 m²  

Therefore from v = Q/A we have v₁ = 2.83 m/s v₂ = 0.87 m/s and v₃  = 1.315 m/s  from there we find the friction coefficient from Moody Diagram as follows

ε = \frac{Roughness _. value}{ Diameter} Which gives

the friction coefficients as f₁ = 0.02, f₂ = 0.017 and f₃ =0.0175

Substituting he above values into the h_{l} equation we get h_{l} = 19.761 m

Combined head loss = 19.761 m

Hence 19.743 m  = 0.6 m +(260 kPa-P₃)/(ρ×9.81) + (6.276)/(2×9.81)

or 260 kPa-18.82 m × 9.81 m/s²×ρ=  P₃

Where ρ = density of water, we have

260000 Pa - 18.82 m×9.81 m/s²×997 kg/m³ = 75958.598 kg/m·s² = 75.959 kPa

6 0
3 years ago
use engineering judgement to estimate the size of cooling and heating equipment that is needed for a three story, 31,000 square
Blizzard [7]

Answer: operating cost = 22820.736 $

Energy = 32.291 KBtu/sf.

Explanation:

Total heating load is given as = 31000 * 31.25  

 = 968.75* 103 Btu

From the cooling capacity application;

If 1000ftsquare = 2.8 TR

Therefore 30000ftsquare = x

Where x is the total cooling load,

Therefore x = (30000 * 2.8) / 1000

x = 84TR.

Therefore, the total cooling load, x = 84TR

Using conversion factor;

i.e. converting Btu/hr to MBH

1 MBH = 1000 Btu per hour

we have 968.75* 103 Btu/hr = 968.75 MBH

let us proceed.

Estimating the “Annual operating cost” we first calculate the maximum operating cost (Avg/Annum)

For cooling load:

Max. operatn. Cost = 14 * 84 = 1176$

For heating load:

The average heating loading hours = 1000 hrs.

Conversion to Btu/yr gives = 968.75* 103 * 1000  

 = 968.75* 106 Btu/yr.

Conversion of 968.75* 103Btu to Kwh gives  

1Btu = 0.000293

968.75* 103 Btu = 283.84 Kwh  

Therefore, average cost gives;

(8.04/100) $ = 1Kwh

X = 283.84*103

X = 22820.736 $

Operating cost = 22820.736 $

Energy use in KBtu/sf is given as  

E = 968.75* 103 / 30000  

E = 32.291 KBtu/sf.

8 0
4 years ago
Which of the following statements about cylinder placement are true?
Alenkinab [10]
What is the following?
3 0
3 years ago
Teaching how to characterize and implement high speed power devices for tomorrow's engineers
Simora [160]

Answer: Teaching how to characterize and implement high speed power devices for tomorrow's engineers

Explanation:

6 0
3 years ago
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