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KonstantinChe [14]
4 years ago
15

Draw an ERD for each of the following situations. (If you believe that you need to make additional assumptions, clearly state th

em for each situation.) Draw the same situation using the tool you have been told to use in the course. a. A company has a number of employees. The attributes of EMPLOYEE include Employee ID (identifier), Name, Address, and Birthdate. The company also has several projects. Attributes of PROJECT include Project ID (identifier), Project Name, and Start Date. Each employee may be assigned to one or more projects or may not be assigned to a project. A project must have at least one employee assigned and may have any number of employees assigned. An employee's billing rate may vary by project, and the company wishes to record the applicable billing rate (Billing Rate) for each employee when assigned to a particular project. Do the attribute names in this description follow the guidelines for naming attributes? If not, suggest better names. Do you have any associative entities on your ERD? If so, what are the identifiers for those associative entities? Does your ERD allow a project to be created before it has any employees assigned to it? Explain. How would you change your ERD if the Billing Rate could change in the middle of a project? b. A laboratory has several chemists who work on one or more projects. Chemists also may use certain kinds of equipment on each project. Attributes of CHEMIST include Employee ID (identifier), Name, and Phone No. Attributes of PROJECT include Project ID (identifier) and Start Date. Attributes of EQUIPMENT include Serial No and Cost The organization wishes to record Assign Date—that is, the date when a given equipment item was assigned to a particular chemist working on a specified project A chemist must be assigned to at least one project and one equipment item. A given equipment item need not be assigned, and a given project need not be assigned either a chemist or an equipment item. Provide good definitions for all of the relationships in this situation. c. A college course may have one or more scheduled sections or may not have a scheduled section. Attributes of COURSE include Course ID, Course Name, and Units. Attributes of SECTION include Section Number and Semester ID. Semester ID is composed of two parts: Semester and Year. Section Number is an integer (such as 1 or 2) that distinguishes one section from another for the same course but does not uniquely identify a section. How did you model SECTION? Why did you choose this way versus alternative ways to model SECTION? d. A hospital has a large number of registered physicians. Attributes of PHYSICIAN include Physician ID (the identifier) and Specialty. Patients are admitted to the hospital by physicians. Attributes of PATIENT include Patient ID (the identifier) and Patient Name. Any patient who is admitted must have exactly one admitting physician. A physician may optionally admit any number of patients. Once admitted, a given patient must be treated by at least one physician. A particular physician may treat any number of patients, or may not treat any patient& Whenever a patient is treated by a physician, the hospital wishes to record the details of the treatment (Treatment Detail). Components of Treatment Detail include Date, Time, and Results. Did you draw more than one relationship between physician and patient? Why or why not? Did you include hncnithi ac an antitv type? Why or why not?
Engineering
1 answer:
Julli [10]4 years ago
3 0

Answer:

it wqas red

Explanation:

red

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1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati
alexandr1967 [171]

Answer:

Gs = 2.647

e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}

where,  \gamma_m = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

\gamma_w = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation,\gamma_m = 16.62 kN/m3

puttng all value to get Gs value;

16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}

Gs - 1.055*e = 1.805 .........(2)

solving both  equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

6 0
3 years ago
Heating of Oil by Air. A flow of 2200 lbm/h of hydrocarbon oil at 100°F enters a heat exchanger, where it is heated to 150°F by
irakobra [83]

Answer:

2062 lbm/h

Explanation:

The air will lose heat and the oil will gain heat.

These heats will be equal in magnitude.

qo = -qa

They will be of different signs because one is entering iits system and the other is exiting.

The heat exchanged by oil is:

qo = Gp * Cpo * (tof - toi)

The heat exchanged by air is:

qa = Ga * Cpa * (taf - tai)

The specific heat capacity of air at constant pressure is:

Cpa = 0.24 BTU/(lbm*F)

Therefore:

Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)

Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))

Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h

5 0
3 years ago
Question 8(Multiple Choice Worth 2 points)
8090 [49]

Answer:

4 number answer is correct.

8 0
3 years ago
Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced
bekas [8.4K]

Answer:

a) 0.684

b) 0.90

Explanation:

Catalyst

EO + W → EG

<u>a) calculate the conversion exiting the first reactor </u>

CAo = 16.1 / 2   mol/dm^3

Given that there are two stream one  contains 16.1 mol/dm^3 while the other contains   0.9 wt% catalyst

Vo = 7.24 dm^3/s

Vm = 800 gal = 3028 dm^3

hence Im = Vin/ Vo = (3028 dm^3) / (7.24dm^3/s) = 418.232 secs = 6.97 mins

next determine the value of conversion exiting the reactor ( Xai ) using the relation below

KIm = \frac{Xai}{1-Xai}  ------ ( 1 )

make Xai subject of the relation

Xai = KIm / 1 + KIm  ---  ( 2 )

<em>where : K = 0.311 ,  Im = 6.97   ( input values into equation 2 )</em>

Xai = 0.684

<u>B) calculate the conversion exiting the second reactor</u>

CA1 = CA0 ( 1 - Xai )

therefore CA1 = 2.5438 mol/dm^3

Vo = 7.24 dm^3/s

To determine the value of the conversion exiting the second reactor  ( Xa2 ) we will use the relation below

XA2 = ( Xai + Im K ) / ( Im K + 1 ) ----- ( 3 )

<em> where : Xai = 0.684 , Im = 6.97,  and K = 0.311  ( input values into equation 3 )</em>

XA2 = 0.90

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4 0
3 years ago
Hot water at an average temperature of 88°C and an average velocity of 0.4 m/s is flowing through a 5-m section of a thin-walled
SIZIF [17.4K]

Answer:

a) The rate of heat transfer will be 19.58 Watts.

b) The temperature drop of the hot water will be 0.024 Degree Celcius.

Explanation:

4 0
4 years ago
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