Answer: 337J/kg
Explanation:
(a) For air, take k = 1.40, R = 287 J/kg ⋅ K, and cp = 1005 J/kg ⋅K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity:+ = = + = +2 2 2p 21 1 1c T V constant 1005( 260) (75) 1005(207) V or .2 2 2
Ans2mV 335 s≈2 1 p 2 1 2 1 207 273 30Meanwhile, s s c ln(T /T ) R ln( p /p ) 1005 ln 287 ln ,260 273 140+ − = − = − ÷ ÷ +or s 2 − s1 = −105 + 442 ≈ 337 J/kg ⋅ K Ans.(a)
(b) For argon, take k = 1.67, R = 208 J/kg ⋅ K, and cp = 518 J/kg ⋅ K. Repeat part (a):2 2 2p 21 1 1c T V 518(260) (75) 518(207) V , solve .2 2 2 Ans+ = + = + 2mV 246 1 207 273 30s s 518 l n 208 ln 54 320 . (b)260 273 140
Answer:
1.25 cm/day
Explanation:
An air thickness , (l) = 0.15 cm
Air Temperature =
Mass Diffusion coefficient (D) =
If the air pressure
We are to determine how fast will the water level drop in a day.
From the property of air at T = 20° C
from saturated water properties.
The mass flow of can be calculated as:
where:
R(constant) = 8.314 kJ/mol.K
Since 1 mole = 18 cm ³ of water
will be:
Again:
Converting the above value to cm/day: we have:
= 1.25 cm/day
∴ the rate at which the water level drop in a day = 1.25 cm/day
Answer:
Explanation:
Normal strain :
It is the ratio of deformation to the original dimension.Normal strain occurs due to longitudinal force.
Normal strain = ΔL/L
Shear strain:
It is the ratio of deformation to perpendicular dimension.Shear strain occurs due to tangential force.
From the second diagram:
if angle Φ is small then we can write
So
This is the shear strain equation.
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