Answer:
a) a = - 0.0524 m/s²
b) t = 6.17 s
Explanation:
Given
τ = 0.18*τ₀ / x
τ₀ = 3.6 m/s
Determine:
a) The acceleration of the air at x= 2m.
Knowing
a = dτ / dt
Multiplying both the numerator and denominator by dx
a = (dτ / dx) (dx / dt)
Substituting τ for dx / dt
a = τ*(dτ / dx)
⇒ a = τ*(-0.18*τ₀ / x²) = (0.18*τ₀ / x)(-0.18*τ₀ / x²)
⇒ a = - 0.18²*τ₀² / x³ = - 0.18²*(3.6)² / x³
⇒ a = - 0.4199 / x³
If x = 2m
⇒ a = - 0.4199 / (2)³
⇒ a = - 0.0524 m/s²
b) The time required for the air to flow from x=1 to x= 3m.
If
τ = dx / dt = 0.18*τ₀ / x
⇒ dt = (x / 0.18*τ₀) dx
⇒ t = (1 / 0.18*τ₀) ∫x dx
⇒ t = (1 / 0.18*τ₀)*(1 / 2)*x²
then
t = (1 / (0.18*3.6))*(1 / 2)*((3)² - (1)²)
⇒ t = 6.17 s
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Answer:
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Explanation:
Answer:
Option B
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Explanation:
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Answer:
T=833.8 °C
Explanation:
Given that
m= 2 kg
T₁=200 °C
time ,t= 10 min = 600 s
Work input = 1 KW
Work input = 1 x 600 KJ=600 KJ
Heat input = 0.5 KW
Q= 05 x 600 = 300 KJ
Gas is ideal gas.
We know that for ideal gas internal energy change given as
ΔU= m Cv ΔT
For air Cv= 0.71 KJ/kgK
From first law of thermodynamics
Q = ΔU +W
Heat input taken as positive and work in put taken as negative.
300 KJ = - 600 KJ + ΔU
ΔU = 900 KJ
ΔU= m Cv ΔT
900 KJ = 2 x 0.71 x (T- 200 )
T=833.8 °C
So the final temperature is T=833.8 °C