Which lists the order of Energy Career Pathways from the source to the customer?

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago

A debugging process where you, the programmer, pretend you are a computer and step through each statement while recording the va

Ipatiy [6.2K]

Answer:

hand tracing

Explanation:

as a programmer when we pretend computer in the debugging process by the step of each statement in recording

then there value of variable is hand tracing because as The hand tracking feature is the use of hands as an input method

so while recording value of each variable each step is hand tracing

5 0

3 years ago

How does the "E" in STEM work with the other letters

KIM [24]

Answer:

if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society

4 0

3 years ago

A vertical curve is designed for 55 mi h and has an intial grade of +2.5

Dafna1 [17]

Vertical curves are the curves which provide a vehicle to negotiate elevation rate change at a slow rate.

<u>Explanation:</u>

A vertical curve gives a progress between two slanted roadways, permitting a vehicle to arrange the height rate change at a slow rate as opposed to a sharp cut.

In any event four distinct criteria for building up lengths of list vertical bends are perceived somewhat. These are (1) front light sight separation, (2) rider comfort, (3) waste control, and (4) a dependable guideline for outward presentation.

8 0

3 years ago

Air enters a 28-cm diameter pipe steadily at 200 kPaand 208C with a velocity of 5 m/s. Air is heated as it flows, and leaves the

Alina [70]

Answer:

(a) $\dot V_1$ = 0.308 m³/s

(b) $\dot m$ = 0.732 kg/m³

(c) v₂ = 5.94 m/s.

Explanation:

(a) The volume flow rate is given by the cross sectional area of the pipe × Velocity of flow of air

Diameter of pipe = 28 cm = 0.28 m

The cross sectional area, A, of the pipe = 0.28²/4×π = 0.0616 m²

Volume flow rate = 5 × 0.0616 = 0.308 m³/s

$\dot V_1$ = 0.308 m³/s

(b) From the general gas equation, we have;

p₁v₁ = RT₁ which gives;

p₁/ρ₁ = RT₁

ρ₁ = p₁/(RT₁)

Where:

ρ₁ = Density of the air

p₁ = 200 kPa

T₁ = 20 C =

R = 0.287 kPa·m³/(kg·K)

ρ₁ = 200/(0.287 ×293.15) = 2.377 kg/m³

The mass flow rate = Volume flow rate × Density

The mass flow rate, $\dot m$ = 2.377×0.308 = 0.732 kg/m³

$\dot m$ = 0.732 kg/m³

(c) The density at exit, ρ₂, is found from the the universal gas equation as follows;

ρ₂ = p₂/(RT₂)

Where:

p₂ = Pressure at exit = 180 kPa

T₂ = Exit temperature = 40°C = 273.15 + 40 = 313.15 K

∴ ρ₂ = 180/(0.287×313.15) = 2.003 kg/m³

$\dot m$ = ρ₂× $\dot V_2$

$\dot V_2$ = $\dot m$ /ρ₂ = 0.732/2.003 = 0.366 m³/s

$\dot V_2$ = v₂ × A

v₂ = $\dot V_2$ /A = 0.366/0.0616 = 5.94 m/s.

v₂ = 5.94 m/s.

5 0

3 years ago