Answer:
Explanation:
from steam tables , at 250 kPa, and at
T₁ = 80⁰C ⇒ h₁ = 335.02 kJ/kg
T₂ = 20⁰C⇒ h₂ = 83.915 kJ/kg
T₃ = 42⁰C ⇒ h₃ = 175.90 kJ/kg
we know
according to energy balance equation
A point charge of 8.0 C is placed 0.050 m from a rod that has an electric field of 1.5 × 103 N/C. What is the electric potential
2 answers:
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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:
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 = 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
In this problem we have the electric field intensity E:
E = 6.5 × newtons/coulomb
We have the magnitude of the load:
q = 6.4 × coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × )(6.5 ×
)(1.2 ×
)
PE = 5.0 x joules
None of the options shown is correct.