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2 years ago

14

An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current trave

ling through the wire sets up a magnetic field around the wire. TRUE or FALSE

1 answer:

german2 years ago

3 0

Answer:

true

Explanation:

true An electromagnet is formed when a coil of wire wrapped around an iron core is hooked up to a dry cell battery. The current traveling through the wire sets up a magnetic field around the wire. TRUE or FALSE

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Natural ventilation uses primarily

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Natural ventilation unlike fan forced ventilation uses the natural forces of wind and buoyancy to deliver fresh air into buildings

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2 years ago

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2 years ago

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Using the degree day method calculate the annual kwh use in springfiled il with a heat loss of 12kwh an inddor twmperature of 7

vladimir2022 [97]

Answer:

The correct solution is "21024 KWh/degree day".

Explanation:

The given query is incomplete. Below is the attachment of complete query is provided.

The given values are:

Indoor design temperature:

= 70°F

Now,

According to the question,

The heat loss annually will be:

= $12\times 24\times 365$

= $105120 \ KW$

Degree days will be:

= $75-65$

= $5$

Hence,

Annual KWh use will be:

= $\frac{Heat \ loss \ annually}{degree \ days}$

On substituting the values, we get

= $\frac{105120}{5}$

= $21024 \ KWh/degree \ day$

8 0

3 years ago

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

Vikki [24]

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

3 0

2 years ago

The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str

Dimas [21]

Answer:

Fractional conversion=0.749

Percentage by which the other reactant is in excess=16.56%

extent = 56.23 mol/s

Explanation:

reaction

$C_{2}H_{4} + HBr - - -> C_{2}H_{5}Br$

molar composition of the product stream:

$C_{2}H_{5}Br$ = 51.7%

HBr = 17.3 %

when you add these two compositions you don't get 100% (69%) so there is also ethylene in the product stream.

$C_{2}H_{4}$ = 100% - 69% = 31%

We are going to suppose a flow rate in the product stream of 100 mol/s. Although you have a flow rate for the feed stream since you don't know the molar composition in the entrance is easier suppose a molar flow in the outlet.

with this information we can estimate the number of moles in the product stream as:

$mol_{i}=MF_{T}*molar-percentaje$

MF= total molar flow.

$C_{2}H_{5}Br = 100 \frac{m}{s}*0.517=51.7 mol/s$

HBr =17.3 mol/s

$C_{2}H_{4}$ = 31 mol/s

The stoichiometry for this reaction is 1: 1. With this information we can estimate the moles of each reagent in the feed stream. To produce 51.7 mol of $C_{2}H_{5}Br$ we needed 51.7 moles of HBr and $C_{2}H_{4}$ respectively.

$mol_{in}=mol_{used}+mol_{out}$

$HBr = 51.7 mol + 17.3 mol =69 mol$

$C_{2}H_{4}$ = 82.7 mol

with this we can see that the limit reagent in this reaction is HBr.

82.7 mol of $C_{2}H_{4}$ need the same number of mol of HBr but there is just 69 moles of this.

The molar composition in the inlet is:

HBr = $\frac{mol-HBr}{total-mol}*100=\frac{69}{69+82.7}*100$ = 45.4%

$C_{2}H_{4}$ = 100% - 45.4% = 54.51%

Fractional conversion of HBr (FC) = $\frac{mol_{used}}{mol_{in}}$

$=\frac{69 mol-17.3mol}{69 mol}= 0.749$

percentage of $C_{2}H_{4}$ in excess = $\frac{mol_{in}-mol_{needed}}{mol_{in}}*100=\frac{69 mol}{82.7}*100$ =16.56%

b. The extent of reaction (e) is defined as:

$e=\frac{n_{out}-n_{in}}{(+-)v}$

where n indicate the moles of a compound in the inlet (in) and in outlet (out) respectively and v is the stoichiometric coefficient for that compound in the reaction. It can be negative if it's a reagent or positive if it's a product.

In this case we have 165 mol/s a flow rate in the inlet. Assuming that the composition in the inlet and the fractional conversion are the same we have.

mol HBr (inlet) = 165 mol/s * 45.4% = 75.04 mol

$FC_{HBr}=\frac{mol_{in}-mol_{out}}{mol_{in}}$

mol HBr_{out}= $mol_{in}-mol_{in}*FC=75.04-75.04*0.749=18.81 mol$

v=-1

$e=\frac{18.81mol-75.04mol}{-1}=56.23 mol$

5 0

3 years ago