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stepan [7]
3 years ago
11

Is my paper's main idea, or thesis, clearly stated early on (within the first paragraph, ideally)?

Engineering
1 answer:
Burka [1]3 years ago
5 0
I dont know is your papers main idea stated clearly?
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Briefly discuss if it would be better to operate with pumps in parallel or series and how your answer would change as the steepn
Aleksandr [31]

Answer:

1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.

Explanation:

1) Pump in series:

When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.

In the first diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3.

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.

- Point 3 is the point where the system is operating when both pumps are running.

Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.

2) Pump in parallel:

When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.

In the second diagram, we have 3 curves namely:

- system curve

- single pump curve

- 2 pump in series curve

Also, we have points labeled 1, 2 and 3

- Point 1 represents the point that the system operates with one pump running.

- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.

- Point 3 is the point where the system is operating when both pumps are running.

In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.

5 0
3 years ago
The Acme tool is aligned to the work with: A. A square B. The eye C. An Acme tool gage D. A center gage
kotegsom [21]

Answer:

(c) an acme tool gage

Explanation:

with the help of an acme tool we measure the pitch of a screw thread it is used as reference for finding the pitch of the thread it is also called an inspection tool is aligned to work with an acme tool gage it is also used for gage the internal and external threaded product. it is also used for grinding and setting tools it is mainly work 29° thread angle

8 0
4 years ago
The customer told you that being competitive in their market is a moving target. They asked you to add several items to the scop
Naddik [55]

Answer:

jjjdhcjdu eheifbsjosv.

Explanation:

7 0
3 years ago
Select the correct answer. Felix aspires to be an engineer working for the government. What credentials will Felix require to ap
Aneli [31]

Answer:

OB

Explanation:

5 0
3 years ago
Consider liquid n-hexane in a 50-mm diameter graduated cylinder. Air blows across the top of the cylinder. The distance from the
ra1l [238]

The evaporation rate of the n-Hexane is 7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

<u>Explanation</u>:

This is a situation regarding diffusing A through non-diffusing B.

A = n-Hexane B=Air

Where the molar flux is provided by,

N_{A}=D_{A B} P_{T}\left(P_{A 1}-P_{A 2}\right) / R T z P_{b m}

\mathrm{D}_{\mathrm{AB}}=8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}

P_{t}=1 a t m=101325 P a\\

\text { so, } P_{A 1}= the vapor pressure at hexane 25 \mathrm{C} =20158.2 \mathrm{Pa}

For wind, assume negligible hexane is present, hence P_{A 2}=0

Now,

\mathrm{P}_{\mathrm{B} 1}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 1}=101325-20158.2 \mathrm{P}_{\mathrm{a}}

\mathrm{P}_{\mathrm{B} 2}=\mathrm{P}_{\mathrm{T}}-\mathrm{P}_{\mathrm{A} 2}=\mathrm{P}_{\mathrm{T}}=101325 \mathrm{Pa}

P_{B M}=\frac{\left(P_{B 2}-P_{B 1}\right)}{\log _{e}\left(P_{B 2} / P_{B 1}\right)}\\

=\frac{101325-81166.8}{\ln \left(\frac{101325}{81166.8}\right) \mathrm{Pa}}

=90873.57 \mathrm{Pa}

R=8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K}

z=\text { distance }=20 \mathrm{cm}=0.2 \mathrm{m}\\

where T = 298 K

substituting all in the equation, we get

\begin{aligned}&\mathrm{N}_{\mathrm{A}}=\\&\left(8.8 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\right) \times 101325 \mathrm{Pa} \times(20158.2 \mathrm{Pa}) /(8.314 \mathrm{J} / \mathrm{mol}-\mathrm{K} \times 0.2 \mathrm{m} \times 298 \mathrm{K}\\&\times 90873.57 \mathrm{Pa})\end{aligned}

=0.004 \mathrm{mol} / \mathrm{m}^{2} \mathrm{s}\\

Now,Flux \times area  = Molar rate of evaporation

Evaporation rate = 0.004 \mathrm{mol} / \mathrm{m}^{2}-5 \mathrm{x}\left(\pi \mathrm{d}^{2} / 4 \mathrm{m}^{2}\right)=0.004 \times(3.14 \times 0.05 \times 0.05 / 4)

Evaporation rate =7.85 \times 10^{-6} \mathrm{mol} / \mathrm{s}

6 0
3 years ago
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