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
where R is hydraulic radius
S = bed slope
P is perimeter
solving for y
solving for y value by using iteration method ,we get
y ≈ 2.5
<u>Answer</u>:
a. r(t) = 6.40 cos (ωt + 38.66°) units
b. r(t) = 6.40 cos (ωt - 38.66°) units
c. r(t) = 6.40 cos (ωt - 38.66°) units
d. r(t) = 6.40 cos (ωt + 38.66°) units
<u>Explanation</u>:
To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:
(i) Convert the phasor to polar form. The polar form is written as;
r∠Ф
Where;
r = magnitude of the phasor =
Ф = direction = tan⁻¹ ()
(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:
r(t) = r cos (ωt + Φ)
Where;
ω = angular frequency of the sinusoid
Φ = phase angle of the sinusoid
(a) 5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
5 + j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
(b) 5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
5 - j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(c) -5 + j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (-0.8)
Φ = -38.66°
-5 + j4 = 6.40∠-38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt - 38.66°)
(d) -5 - j4
<em>(i) convert to polar form</em>
r =
r =
r =
r = 6.40
Φ = tan⁻¹ ()
Φ = tan⁻¹ (0.8)
Φ = 38.66°
-5 - j4 = 6.40∠38.66°
(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>
r(t) = 6.40 cos (ωt + 38.66°)
Answer:
T = 167 ° C
Explanation:
To solve the question we have the following known variables
Type of surface = plane wall ,
Thermal conductivity k = 25.0 W/m·K,
Thickness L = 0.1 m,
Heat generation rate q' = 0.300 MW/m³,
Heat transfer coefficient hc = 400 W/m² ·K,
Ambient temperature T∞ = 32.0 °C
We are to determine the maximum temperature in the wall
Assumptions for the calculation are as follows
Therefore by the one dimensional conduction equation we have
During steady state
= 0 which gives
From which we have
Considering the boundary condition at x =0 where there is no heat loss
= 0 also at the other end of the plane wall we have
hc (T - T∞) at point x = L
Integrating the equation we have
from which C₁ is evaluated from the first boundary condition thus
0 = from which C₁ = 0
From the second integration we have
From which we can solve for C₂ by substituting the T and the first derivative into the second boundary condition s follows
→ C₂ =
T(x) = and T(x) = T∞ +
∴ Tmax → when x = 0 = T∞ +
Substituting the values we get
T = 167 ° C