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3 years ago

The dry unit weight of a soil sample is 14.8 kN/m3.

Engineering

1 answer:

Jlenok [28]3 years ago

4 0

Answer:

See attachment for completed question

Explanation:

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College Engineering 5+3 pts

The dry unit weight of a soil sample is 14.8 kN/m3.

Given that G_s = 2.72 and w = 17%, determine:

(a) Void ratio

(b) Moist unit weight

(d) Unit weight when the sample is fully saturated

See complete solving at attachment

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m_a_m_a [10]

Answer:

Explanation:

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("Joan", "unlisted", 12.7) :: Nil

type listOfTuples = List[(String, String, Double)]

def find_salary(name: String) = {

def search(t: listOfTuples): Double = t match {

case (name_, _, salary) :: t if name == name_ => salary

case _ :: t => search(t)

case Nil =>

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}

search(db)

}

def select(pred: (String, String, Double) => Boolean) = {

def search(found: listOfTuples): listOfTuples = found match {

case (p1, p2, p3) :: t if pred(p1, p2, p3) => (p1, p2, p3) :: search(t)

case (p1, p2, p3) :: t if !pred(p1, p2, p3) => search(t)

case Nil => Nil

case _ => throw new Exception("Invalid Argument in select function")

}

search(db)

}

println("Searching the salary of 'Joan' at db: " + find_salary("Joan"))

println("")

val predicate = (_:String, _:String, salary:Double) => (salary < 100.0)

println("All employees that match with predicate 'salary < 100.0': ")

println("\t" + select(predicate) + "\n")

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3 years ago

28. What is the value of a resistor in a series circuit if you measure 0.5 amps flowing through it and 15 volts

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3 years ago

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2 years ago

Read 2 more answers

What is the measurement below?

Bess [88]

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3 years ago

Steam at 4 MPa and 350°C is expanded in an adiabatic turbine to 125kPa. What is the isentropic efficiency (percent) of this turb

guajiro [1.7K]

Answer:

$\eta_{turbine} = 0.603 = 60.3\%$

Explanation:

First, we will find actual properties at given inlet and outlet states by the use of steam tables:

AT INLET:

At 4MPa and 350°C, from the superheated table:

h₁ = 3093.3 KJ/kg

s₁ = 6.5843 KJ/kg.K

AT OUTLET:

At P₂ = 125 KPa and steam is saturated in vapor state:

h₂ = $h_{g\ at\ 125KPa}$ = 2684.9 KJ/kg

Now, for the isentropic enthalpy, we have:

P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K

Since s₂ is less than $s_g$ and greater than $s_f$ at 125 KPa. Therefore, the steam is in a saturated mixture state. So:

$x = \frac{s_2-s_f}{s_{fg}} \\\\x = \frac{6.5843\ KJ/kg.K - 1.3741\ KJ/kg.K}{5.91\ KJ/kg.K}\\\\x = 0.88$

Now, we will find $h_{2s}$ (enthalpy at the outlet for the isentropic process):

$h_{2s} = h_{f\ at\ 125KPa}+xh_{fg\ at\ 125KPa}\\\\h_{2s} = 444.36\ KJ/kg + (0.88)(2240.6\ KJ/kg)\\h_{2s} = 2416.088\ KJ/kg$

Now, the isentropic efficiency of the turbine can be given as follows:

$\eta_{turbine} = \frac{h_1-h_2}{h_1-h_{2s}}\\\\\eta_{turbine} = \frac{3093.3\ KJ/kg-2684.9\ KJ/kg}{3093.3\ KJ/kg-2416.088\ KJ/kg}\\\\\eta_{turbine} = \frac{408.4\ KJ/kg}{677.212\ KJ/kg}\\\\\eta_{turbine} = 0.603 = 60.3\%$

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3 years ago