=
=
3
=
=
Temperature has a srong effect on viscosity May depend on the rate of shear strain Assumptions often used in fluid mechanics-
− − ∆∆ ∀ − → → − − ∴ ∆∆ − −−−∆ − − ∆ ∴ ∆ ∆ =
-----------------------------------------------------------
*viscosity is constant (Newtonian fluid) *viscosity is 0 (ideal fluid, inviscid fluid, flow is frictionless)
--------------------------------------------------------------
=
=
(conservation of energy)
1
+
=
2
2 1 1
+
1
=
2
+
1 2
2
2
+
2
Further common assumptions ONLY FOR SV 1
3
1
+
2
=
1
=0
--------------------------------------------------------------------------------------------------------------------------Pascals principle ----------------------------------------------------------- ‘if an external pressure is applied to a confined fluid, Ideal Gas equation
=×
=×
the pressure at every point within the fluid increases by that amount’
4°
=
eg Hydraulic Lift 1 = 2 1
=
.
°
×
=
= . = . = = . . = . . . . . .
=
2
Can be used to obtain mechanical advantage
----------------------------------------------------------Pressure vs depth (incompressible (incompressible fluids)
=
=
2
=
1
2
=
=
= 8.31 8.3145 45
1
1
-------------------------------------------------------------Real Gas equation
2
=
1
=
1
Work done is the same by which the surface A 2 rises is smaller than the change in the height of surface with area A
Z= compressibility compressibility & is dimensionless dimensionless
= -------------------------------------------------------------- -------------------------------------------------------------Root-mean-square atomic velocity Buoyancy Pressure increases with depth. So the pressure at the bottom of a floating object is greater than on top. Thus the water exerts a net upward force on the object. This is the boyant force.
=
Pressure vs depth (compressible fluids)
=
>
T= Temperature Kelvins m= mass ( + ) Archimedes’ Principa Principal M= Molar mass of gas = The boyant force on an object immersed in fluid is ----------------------------------------------------------------------------------------------------------------------- equal to the weight of fluid displaced by that object. STP = = For pressure of fluid in container with lid open. P=101.325 kPa T=273.15K 22.414L Assume fluid is incompressible. -------------------------------------------------------------Pressure on the top surface = = 1.01 1.0132 325 5 × 105 2 =
+
=0 =0
′ ′
∆ ∆ − =
1
2
=
∴ =
+
-----------------------------------------------------------
&
+
=
− ∆ ≡ − ∆∀∀ ∆∀∀
----------------------------------------------------------.
=
=
− − Δ − =
1
Force on the top surface 1
=
=
1
2
Pressure on the bottom surface 2
=
2
Force on then bottom surface 2
=
=
2
2
F B is the net force exerted by the fluid on the submerged object
=
2
1
=
2
1
=
=
=
----------------------------------------------------------- --------------------------------------------------------------
− − Δ Δ Δ Δ ΔΔ Δ Δ =
=
=
=
=
Mark Riley
(conservation of mass) 1
=
2
=
=
(
)
(
)
=0
For multiple inputs & outputs
=
--------------------------------------------------------------
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