Air flows through a 0.25-m-diameter duct. At the inlet the velocity is 300 m/s, and the stagnation temperature is 90°C. If the M
ach number at the exit is 0.3, determine the direction and the rate of heat transfer. For the same conditions at the inlet, determine the amount of heat that must be transferred to the system if the flow is to be sonic at the exit of the duct.
1 answer:
Answer:
a. 318.2k
b. 45.2kj
Explanation:
Heat transfer rate to an object is equal to the thermal conductivity of the material the object is made from, multiplied by the surface area in contact, multiplied by the difference in temperature between the two objects, divided by the thickness of the material.
See attachment for detailed analysis
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<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:
A) energy loss E = pgQtH
Where p = density in kg/m3
g = gravity acceleration in m/s2
Q = flow rate in m3/s
t = time taken for flow in sec
H = height of flow in m
B) power required to run pump;
P = pgQH
Explanation:
Detailed explanation and calculation is shown in the image below
