Vector is perpendicular to x axis or i component.
Hence i component is 0
j component is 63.5
Which of the following is true of an adiabatic process?
2 answers:
Adiabatic process is that process in which heat of a closed system will remain conserved or we can say that system can not exchange its thermal energy from its surrounding
So here from 1st law of thermodynamics we can say
since we know that
so we will have
so we have
so here correct answers are
<em>A. -ΔU = W </em>
<em>B. Heat is constant. </em>
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Answer:
time spent = 0.2276
Explanation:
given data
distance = 135 mi
usual speed = 65 mph
today speed = 73 mph
solution
we get here time that is express as
time = ...................1
usual time = = 2.0769 h
today time = = 1.8493 h
so we get here time spent as
time spent = 2.0769 h - 1.8493 h
time spent = 0.2276
Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W