Answer:
The discharge of the stream at this location is 40 cubic meters per second.
Explanation:
The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:
Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)
Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)
Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)
<u>Volume Flow Rate = 40 cubic meters per second</u>
Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>
This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.
Answer:
The correct answers are It is the resistance of an object to changes in its motion, and It is a force
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
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
Through a dam... hope this helps:)
