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Fed [463]
3 years ago
14

The time factor for a doubly drained clay layer

Engineering
1 answer:
Margarita [4]3 years ago
5 0

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

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Answer:

See the explanation below.

Explanation:

First find the enthalpies h₁, h₂, h₃, h₄, h₅, and h₆.

Find h₁:

Using Saturated Water Table and given pressure p₁ = 100 kPa

h₁ = 417.5 kJ/kg

Find h₂:

In order to find h₂, add the w_{p} to h₁, where  w_{p}  is the work done by pump and h₁ is the enthalpy computed above h₁ = 417.5 kJ/kg.

But first we need to compute  w_{p} To computer  

Pressures:

p₁ = 100 kPa

p₂ = 15,000 kPa

and

Using saturated water pressure table, the volume of water v_{f} = 1.0432

Dividing 1.0432/1000 gives us:

Volume of water = v₁ =  0.001043 m³/kg

Compute the value of h₂:

h₂ = h₁ + v₁ (p₂ - p₁)

    = 417.5 kJ/kg + 0.001043 m³/kg ( 15,000 kPa - 100 kPa)

    =  417.5 + 0.001043 (14900)

    = 417.5 + 15.5407

    = 433.04 kJ/kg

Find h₃  

Using steam table:

At pressure p₃ = 15000 kPa

and Temperature = T₃ = 450°C

Then h₃ = 3159 kJ/kg

The entropy s₃ = 6.14 kJ/ kg K

Find h₄

Since entropy s₃ is equal to s₄ So

s₄ = 6.14 kJ/kgK

To compute h₄

s₄ = s_{f} + x_{4} s_{fg}

x_{4} = s_{4} -s_{f} /s_{fg}

x_{4} = 6.14 -  2.45 / 3.89

x_{4}   = 0.9497

The enthalpy h₄:

h₄ = h_{f} +x_{4} h_{fg}

    = 908.4 + 0.9497(1889.8)

    =  908.4 + 1794.7430

    = 2703 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₃ = 6.14 kJ/ kg K and pressure p₄ = 2000 kPa

Find h₅

Using steam table:

At pressure p₅ = 2000 kPa

and Temperature = T₅ = 450°C

Then h₅  = 3358 kJ/kg

Find h₆:

Since the entropy s₅ = 7.286 kJ/kgK is equal s₆ to  So

s₆ = 7.286 kJ/kgK = 7.29 kJ/kgK

To compute h₆

s₆ = s_{f} + x_{6} s_{fg}

x_{6} = s_{6} -s_{f} /s_{fg}

x_{6} = 7.29 - 1.3028 / 6.0562

x_{6}   = 0.988

The enthalpy h₆:

h₆ = h_{f} +x_{6} h_{fg}

    = 417.51 + 0.988 (2257.5)

    = 417.51 + 2230.41

  h₆ =  2648 kJ/kg

This can simply be computed using the software for steam tables online. Just use the entropy s₅ = 7.286 kJ/kgK and pressure p₅ = 2000 kPa

Compute power used by pump:

P_{p} is found by using:

mass flow rate = m =  1.74 kg/s

Volume of water = v₁ =  0.001043 m³/kg

p₁ = 100 kPa

p₂ = 15,000 kPa

P_{p}  = ( m ) ( v₁ ) ( p₂ - p₁ )

     = (1.74 kg/s) (0.001043 m³/kg) (15,000 kPa - 100 kPa)

     = (1.74 kg/s) (0.001043 m³/kg) (14900)

     = 27.04

P_{p} = 27 kW

Compute heat added q_{a} and heat rejected q_{r}  from boiler using computed enthalpies:

q_{a} = ( h₃ - h₂ ) + ( h₅ - h₄ )

      = ( 3159 kJ/kg - 433.04 kJ/kg ) + ( 3358 kJ/kg - 2703 kJ/kg )

      = 2726 + 655

      = 3381  kJ/kg

q_{r} =  h₆ - h₁

  = 2648 kJ/kg - 417.5 kJ/kg

  = 2232 kJ/kg

Compute net work

W_{net} = q_{a} - q_{r}

       = 3381  kJ/kg - 2232 kJ/kg

       = 1150 kJ/kg

Compute power produced by the cycle

mass flow rate = m =  1.74 kg/s

W_{net} = 1150 kJ/kg

P = m * W_{net}

  = 1.74 kg/s * 1150 kJ/kg

  = 2001 kW

Compute rate of heat transfer in the reheater

Q = m * ( h₅ - h₄ )

   =  1.74 kg/s * 655

   =  1140 kW

Compute Thermal efficiency of this system

μ_{t} = 1 - q_{r} /  q_{a}

   = 1 - 2232 kJ/kg / 3381  kJ/kg

   = 1 - 0.6601

   = 0.34

   = 34%

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