Distance, since distance represents how far something has travelled, which would be in our case 2.5m.
Answer:
The upper limit on the flow rate = 39.46 ft³/hr
Explanation:
Using Ergun Equation to calculate the pressure drop across packed bed;
we have:
![\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cdelta%20P%7D%7BL%7D%3D%20%5Cfrac%7B150%20%5Cmu_oU%281-%20%5Cepsilon%20%29%5E2%7D%7Bd%5E2p%20%5Cepsilon%5E3%7D%20%2B%20%5Cfrac%7B1.75%20%5Crho%20U%5E2%281-%5Cepsilon%29%7D%7Bdp%20%5Cepsilon%5E3%7D)
where;
L = length of the bed
= viscosity
U = superficial velocity
= void fraction
dp = equivalent spherical diameter of bed material (m)
= liquid density (kg/m³)
However, since U ∝ Q and all parameters are constant ; we can write our equation to be :
ΔP = AQ + BQ²
where;
ΔP = pressure drop
Q = flow rate
Given that:
9.6 = A12 + B12²
Then
12A + 144B = 9.6 -------------- equation (1)
24A + 576B = 24.1 --------------- equation (2)
Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So
288 B = 4.9
B = 0.017014
From equation (1)
12A + 144B = 9.6
12A + 144(0.017014) = 9.6
12 A = 9.6 - 144(0.017014)
![A = \frac{9.6 -144(0.017014}{12}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B9.6%20-144%280.017014%7D%7B12%7D)
A = 0.5958
Thus;
ΔP = AQ + BQ²
Given that ΔP = 50 psi
Then
50 = 0.5958 Q + 0.017014 Q²
Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;
Q² + 35.02Q - 2938.8 = 0
Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;
Q = 39.46 ft³/hr
Answer:
C
Explanation:
V=1/p
By means of cross multiplication so by that we will have pv=1 which also implies p1v1=p2v2 coz boyles law states that the volume of a given mass of gas is inversely proportional to pressure provided that the temperature in kelvin remains constant
Answer:
See explanation below
Explanation:
The equation to use for this is the following:
dU = q + w
As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.
Therefore the change in the energy would be:
dU = -2.59x10^4 - 6.46^4
dU = -9.05x10^4 kJ
Answer:
Asthenosphere
Explanation:
The asthenosphere is a part of the upper mantle just below the lithosphere that is involved in plate tectonic movement and isostatic adjustments.