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Margarita [4]
3 years ago
10

An operator decided to develop a subsea drill center of 4 oil wells. The production from the drill center is to be transported t

o a receiving facility by a single pipeline. What do you recommend the operator to install for production commingling from the 4 wells,
Engineering
1 answer:
Vikki [24]3 years ago
3 0

Answer:

The possible options the operator can choose from for the manifold system are;

1) Cluster manifold system

2) Template manifold system

Explanation:

Given that a subsea drill center is to be used for the development of the field that have four wells, we have;

1) Clustered system

In the clustered system design, the wells are situated and drilled around the designated area where the manifold will eventually be installed, such that there is increased flexibility in investment such that the economy of the development can be factored in as the field is being developed favoring the suitability of the cluster field system for a field with a few number of wells such as the one in question.

However for shallow wells and in a situation of high financial uncertainties such that the wells can be drilled at the same time the the manifolds are being constructed saving costs of waiting for the template the cluster manifold system will be more appropriate

Also as there are few wells the cluster manifold system can be more cost effective in terms of scheduling and resource allocation.

2) Template System

In the template manifold system design, the wells are drilled in a prefabricated well template housing which will hold the completion tools of the well, As such the well completion are well arranged and interconnected within the template design

Whereby the aim is for a fast an economic as well as a well built system, then the right choice is the template manifold design where there is direct flow from the wells to the template manifold improving flow assurance, and reducing installation costs as the system does not require jumper installation which is costly

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Solution :

Given

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Now,

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Given,   $t_1=2+x_1$

                 = 2 + 0.368

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At user equilibrium, $t_2=t_1$

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$t_2=1+x_2$

$2.368=1+x_2$

$x_2 = 1.368$

$x_2 = 1.368 \times 1000$

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7 0
3 years ago
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Answer:

Tech A

Explanation:

The amount of energy required to apply the same force with a 1:1 ratio is divided into 4, so you can apply 4 times as much force than a 1:1 ratio. efficiency and speed come into play here, but assuming the machine powering the gear can run at a unlimited RPM, 4:1 will have more force and a slower output speed than a 2:1 ratio.

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3 years ago
How does the two-stroke Otto cycle differ from the four-stroke Otto cycle?
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Answer:

Two stroke cycle                                               Four stroke cycle

1.Have on power stroke in one revolution.   1.have one power  

                                                                   stroke in two  revolution                                                                            

2.Complete the cycle in 2 stroke                 2.Complete the cycle in 4 stroke    

3.It have ports                                                3.It have vales

                                                                         

4.Greater requirement of cooling              4.Lesser requirement of cooling  

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3 0
3 years ago
A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

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c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

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Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

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