Basics of Fluid Mechanics, 2014a

1.6. FLUID PROPERTIES 23
straight. The new density is
ρ 2 =
ρ 1
(1 + α Δ T ) 3
} {{ }
thermal expansion
(1.IX.a)
The more accurate calculations require looking into the steam tables. As estimated
value of the density using Young’s modulus and V 2 ∝ (L 2 ) 38 .
ρ 2 ∝ 1
(L 2 ) 3 =⇒ ρ 2 ∼ =
m
( (L 1 1 − ΔP
E
)) 3
(1.IX.b)
It can be noticed that ρ 1
∼ = 3 m/L1 and thus
ρ 1
(1 + αΔ T ) 3 = ρ 1
(
1 − ΔP ) 3
(1.IX.c)
E
The change is then
Thus the final pressure is
1+αΔ T =1− ΔP
E
P 2 = P 1 − EαΔ T
(1.IX.d)
(1.IX.e)
In this case, what happen when the value of P 1 − EαΔ T becomes negative or very
very small? The basic assumption falls and the water evaporates.
If the expansion of the water is taken into account then the change (increase) of
water volume has to be taken into account. The tank volume was calculated earlier
and since the claim of “strong” steel the volume of the tank is only effected by the
temperature.
∣
V 2 ∣∣∣tank
=(1+α ΔT ) 3
(1.IX.f)
V 1
The volume of the water undergoes also a change and is a function of the temperature
and pressure. The water pressure at the end of the process is unknown but the volume
is known. Thus, the density at end is also known
ρ 2 =
m w
T 2 | tank
(1.IX.g)
The pressure is a function volume and the temperature P = P (v, T) thus
dP =
∼β v
( { }} {
∂P
∂v
)
dv +
∼E
({ }} {
∂P
∂T
)
dT
(1.IX.h)
8 This leads E (L 2 − L 1 )=ΔPL 1 . Thus, L 2 = L 1 (1 − ΔP/E)