- #1
ApeXaviour
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Hi. This is a two part question, I'm pretty sure about the first bit but in the second part my answer seems a bit weird. Not the order of magnitude I was expecting. Just wondering can anyone confirm or deny my suspician? Point me in more correct direction if I am indeed wrong. This isn't for anything, its just a problem sheet given out and this last question has been bugging me :)
(i)Water flows down a broad channel at an angle \theta to the horizontal. \theta is sufficiently small that the flow is laminar. By balancing the pressure from the gradient with the shear force due to viscosity (i.e. constant velocity flow) find an expression for the velocity of the surface of the water. Recall that the velocity adjacent to the channel bed is zero.
(ii)If the Liffey[A river in Dublin
] is 5m deep, the surface water velocity is 0.3m/s and the viscosity of the water is 10^-3 kg/ms, find the height of the river surface between heuston stn and talbot bridge, a distance of 2km.
This is what I've got so far:
u is the velocity
\rho density of the water
z is the height off the river bed (water depth when at surface)
\tau is the shear stress
\eta the viscosity
w l arbitrary width and lengthThe shear stress between planes in a liquid is: \tau = \eta \frac{du}{dz}
So the total force per unit area is
\frac{F_\tau}{A} = w l \eta \frac{du}{dz}
Which must be equal to the gravitational force per unit area for steady flow:
\frac{F_g}{A} = m g Sin(\theta) = w l z \rho g Sin(\theta)
so:
w l \eta \frac{du}{dz}= w l z \rho g Sin(\theta)
\int_0^u \,du = \int_0^z \frac{z \rho g Sin(\theta)}{\eta} \,dz
And hence the surface velocity is given by:
u = \frac{z^{2} \rho g Sin(\theta)}{2 \eta}
So when I put in the figures I get
Sin(\theta) = \frac{2 u \eta}{z^{2} \rho g} = 2.4 \times 10^{-9} \cong \theta
Then from triginometric relations, H the height change is given by:
(2 \times 10^{3})(Tan(\theta)) \cong (\theta)(2 \times 10^{3}) = 4.8 \times 10^{-6}m Which to me seems - and maybe my intuition is out of whack - just a little ridiculous.
Thanks
-Dec
(i)Water flows down a broad channel at an angle \theta to the horizontal. \theta is sufficiently small that the flow is laminar. By balancing the pressure from the gradient with the shear force due to viscosity (i.e. constant velocity flow) find an expression for the velocity of the surface of the water. Recall that the velocity adjacent to the channel bed is zero.
(ii)If the Liffey[A river in Dublin
] is 5m deep, the surface water velocity is 0.3m/s and the viscosity of the water is 10^-3 kg/ms, find the height of the river surface between heuston stn and talbot bridge, a distance of 2km.This is what I've got so far:
u is the velocity
\rho density of the water
z is the height off the river bed (water depth when at surface)
\tau is the shear stress
\eta the viscosity
w l arbitrary width and lengthThe shear stress between planes in a liquid is: \tau = \eta \frac{du}{dz}
So the total force per unit area is
\frac{F_\tau}{A} = w l \eta \frac{du}{dz}
Which must be equal to the gravitational force per unit area for steady flow:
\frac{F_g}{A} = m g Sin(\theta) = w l z \rho g Sin(\theta)
so:
w l \eta \frac{du}{dz}= w l z \rho g Sin(\theta)
\int_0^u \,du = \int_0^z \frac{z \rho g Sin(\theta)}{\eta} \,dz
And hence the surface velocity is given by:
u = \frac{z^{2} \rho g Sin(\theta)}{2 \eta}
So when I put in the figures I get
Sin(\theta) = \frac{2 u \eta}{z^{2} \rho g} = 2.4 \times 10^{-9} \cong \theta
Then from triginometric relations, H the height change is given by:
(2 \times 10^{3})(Tan(\theta)) \cong (\theta)(2 \times 10^{3}) = 4.8 \times 10^{-6}m Which to me seems - and maybe my intuition is out of whack - just a little ridiculous.
Thanks
-Dec
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