1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
faust18 [17]
3 years ago
7

In a 1D compression test with double drainage, the pore pressure readings are practically zero after 8 minutes for a clay sample

of 2 cm thickness. What will be the duration (in minutes) of the consolidation process for the same clay, if the test is done under single drainage condition?
Engineering
1 answer:
frosja888 [35]3 years ago
3 0

Answer:

The duration of the consolidation process for the same clay is 32 min

Explanation:

for clay 1:

t1=0

H1=thickness=2 cm

for the clay 2:

t2=?

H2=2 cm

The time factor is equal to:

T=(\frac{Cv}{d^{2} })t

where Cv is the coefficient of consolidation

(\frac{Cvt}{d^{2} })_{1}=  (\frac{Cvt}{d^{2} })_{2}

if Cv is constant, we have:

(\frac{t1}{(\frac{H1}{2}) ^{2} })_{1}=(\frac{t2}{H2^{2} })_{2}\\\frac{0}{(\frac{2}{2})^{2})  }=\frac{t2}{2^{2} }

Clearing t2:

t2=32 min

You might be interested in
What type of cartilage provides support and shock absorption?.
Katena32 [7]

Answer:

fibrocartilage

7 0
2 years ago
When were dresses made
klio [65]

Answer:

The world's oldest dress called the Tarkhan Dress is at 5,100 to 5,500 years of age.

Does that help? Or do you need something else? I can change my answer if this is not what you need! :D

Explanation:

6 0
3 years ago
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
3 years ago
Exhaust gas from a furnace is used to preheat the combustion air supplied to the furnace burners. The gas, which has a flow rate
Monica [59]

Answer:

The total tube surface area in m² required to achieve an air outlet temperature of 850 K is 192.3 m²

Explanation:

Here we have the heat Q given as follows;

Q = 15 × 1075 × (1100 - t_{A2}) = 10 × 1075 × (850 - 300) = 5912500 J

∴ 1100 - t_{A2} = 1100/3

t_{A2}  = 733.33 K

\Delta \bar{t}_{a} =\frac{t_{A_{1}}+t_{A_{2}}}{2} - \frac{t_{B_{1}}+t_{B_{2}}}{2}

Where

\Delta \bar{t}_{a} = Arithmetic mean temperature difference

t_{A_{1} = Inlet temperature of the gas = 1100 K

t_{A_{2} = Outlet temperature of the gas = 733.33 K

t_{B_{1} =  Inlet temperature of the air = 300 K

t_{B_{2} = Outlet temperature of the air = 850 K

Hence, plugging in the values, we have;

\Delta \bar{t}_{a} =\frac{1100+733.33}{2} - \frac{300+850}{2} = 341\tfrac{2}{3} \, K = 341.67 \, K

Hence, from;

\dot{Q} = UA\Delta \bar{t}_{a}, we have

5912500  = 90 × A × 341.67

A = \frac{5912500  }{90 \times 341.67} = 192.3 \, m^2

Hence, the total tube surface area in m² required to achieve an air outlet temperature of 850 K = 192.3 m².

4 0
3 years ago
You must signal [blank] before any turn or lane change.
Soloha48 [4]
A. 5 seconds :) Good luck!
5 0
3 years ago
Other questions:
  • The air contained in a room loses heat to the surroundings at a rate of 60 kJ/min while work is supplied to the room by computer
    7·2 answers
  • Which investigative process is most helpful for learning about past societies?
    10·1 answer
  • Let CFG G be the following grammar.
    7·2 answers
  • An intelligence signal is amplified by a 65% efficient amplifier before being combined with a 250W carrier to generate an AM sig
    5·1 answer
  • 2. What is the original length of the rectangular bar if the deformation is 0.005 in with a force of 1000 lbs and an area of 0.7
    9·1 answer
  • I need solution fast plesss​
    9·1 answer
  • How many millimeters are there in a centimeter?
    10·1 answer
  • When was solar power envold ​
    8·2 answers
  • A force is a push or pull in A.a circle B.an arc C.a straight line
    6·1 answer
  • A(n) _____ is an apparatus that changes alternating current (AC) to direct current (DC)
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!