A student determines the density, solubility, and boiling point of two liquids, Liquid 1 and Liquid 2. Then he stirs the two liq
1 answer:
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Answer:
The algebraic equation is:
Explanation:
Given information:
mb = book's mass
vb = tangential speed
R = radius of the path
Question: Derive an algebraic equation for the vertical force, Fv = ?
To derive the equation, we need to draw a force diagram for this case, please, see the attached diagram. As you can see, there are three types of forces acting on the system. Two up and one of the weight acting down. Therefore, the algebraic equation is as follows:
The variables were defined above and g is the gravity.
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0
This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u /><u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u /><u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
Answer:
part a : <em>The dry unit weight is 0.0616 </em><em />
part b : <em>The void ratio is 0.77</em>
part c : <em>Degree of Saturation is 0.43</em>
part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>
Explanation:
Part a
Dry Unit Weight
The dry unit weight is given as
Here
is the dry unit weight which is to be calculated
- γ is the bulk unit weight given as
- w is the moisture content in percentage, given as 12%
Substituting values
<em>The dry unit weight is 0.0616 </em><em />
Part b
Void Ratio
The void ratio is given as
Here
- e is the void ratio which is to be calculated
is the water unit weight which is 62.4
or 0.04
- G is the specific gravity which is given as 2.72
Substituting values
<em>The void ratio is 0.77</em>
Part c
Degree of Saturation
Degree of Saturation is given as
Here
- e is the void ratio which is calculated in part b
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Substituting values
<em>Degree of Saturation is 0.43</em>
Part d
Additional Water needed
For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as
Here
is the zero air unit weight which is to be calculated
- G is the specific gravity which is given as 2.72
- w is the moisture content in percentage, given as 12% or 0.12 in fraction
Now as the volume is known, the the overall weight is given as
As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.