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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

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An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal effici
Oksana_A [137]

Answer Explanation:

the efficiency of the the engine is given by=1-\frac{T_2}{T_1}

where T₂= lower temperature

           T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1-\frac{T_2}{T_1}

                =1-\frac{300}{900}

                 =1-0.3333

                 =0.6666

                 =66%

66% is less than 70% so so inventor claim is wrong

3 0
3 years ago
To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws
ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\

E=52000Hp.h

E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\

E=38724920Wh

E=52000Hph(\frac{1977378.4  ft lb}{1Hph}

E=1.028x10^11 ftlb

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