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Elis [28]
3 years ago
13

A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput

e
A) its dry unit weight. B) its porosity. C) Its void ratio. D) its degree of saturation.
Engineering
1 answer:
olchik [2.2K]3 years ago
5 0

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity  = 0.37

C). Void ratio  = 0.593 

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

 

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

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The phase sequence of a 3-phase system for which VAN = 120 /90o V and VBN = 120 /210o V is:_______
Vadim26 [7]

Answer:

b. abc

Explanation:

The phase rotation and its sequence is order in which voltage is reached to their respective sources. The wave forms of a polyphase flow through AC source to the panel. The three phase system has two possible sequence which can be a-b-c or c-b-a. The correct answer is therefore b.

4 0
3 years ago
If the efficiency of the boiler is 91.2 % , the overall efficiency of the turbine, which includes the Carnot efficiency and its
Tju [1.3M]

Answer:

Net efficiency of generating unit = 42.08 - 5 = 37.08 %

Explanation:

We have given that efficiency of the boiler = 91.2 % = 0.912

Carnot efficiency = 46.9 % = 0.469

Efficiency of generator = 98.4% =0.984

We have to find the efficiency of total generating unit

For finding the efficiency of total generating unit we have to multiply all the efficiencies

So efficiency of generating unit = 0.912×0.469×0.984 = 0.4208 = 42.08 %

For plant losses we have to subtract 5%

So net efficiency of generating unit = 42.08 - 5 = 37.08 %

4 0
3 years ago
Describe the algorithm you use for looking up a person’s telephone number in the phone book. The input is person’s name; the out
Stella [2.4K]

Answer:

The Algorithm for finding a number from a phone book with the person's name as the input and the phone number as output is as follows:

1. Try to remember the name, i.e last name first and first name last, Also make sure you get the spelling right.

2. Using the first letter of the last name, locate the appropriate alphabetical section in which the name should appear.

3. Using the second letter of the last name, find the subsection of first and second letters combined, in the appropriate order, in which the name should appear. (If the last name consists of only two letters, find the appropriate first name.)

4. Using the third letter, find the possible names in a subsection of the first three letters in the correct order. Continue this step with x+1 letters of the name until you have a subsection of names exactly matching the last name of the person whose number you are trying to locate. (x is the number of letters used in the previous step, consistently.) If there is only one of the last name, (check for duplicates) identify the number, and return phone number information.

5. Begin the second step using the first letter of the first name, but limit the section to only those exactly matching the last name. Continue to step 4, again focusing on the first name only within the set of exactly matching last names.

6. When both first and last name match the name you are locating, check for duplicates. IF there are no duplicates, return phone number information.

Explanation:

People's names are generally arranged in phone books in alphabetical order by the last name of the person. The first name of the person is listed after the last name so that people of the same last name can be differentiated.

7 0
3 years ago
Read 2 more answers
What does polarity give you information about?
Brrunno [24]

Well, I do know that polarity affects the voltage.

6 0
3 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
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