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3. Cassava paste is pumped through a tube. Determine the ratio $v_{max}/v_{mean}$ for this fluid, given that its rheological behavior may be characterized by the Bingham model. Note: The Bingham model is often written in simple form as: $\tau = \tau_0 + \mu_p \dot{\gamma}$ However, in order to work out a fluid flow solution, it may be written in tensor form as: $\tau_{ij} = -\mu_p + \sqrt{\frac{1}{2} \sum_{i=1}^{3} \sum_{j=1}^{3} \left[ \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right] \left[ \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right]} \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right)$ Solve the fluid flow equation for steady-fully developed flow with constant pressure gradient to determine the velocity profile, then determine $v_{max}$ and $v_{mean}$ 

          3. Cassava paste is pumped through a tube. Determine the ratio $v_{max}/v_{mean}$ for this fluid, given that its rheological behavior may be characterized by the Bingham model.
Note: The Bingham model is often written in simple form as:
$\tau = \tau_0 + \mu_p \dot{\gamma}$
However, in order to work out a fluid flow solution, it may be written in tensor form as:
$\tau_{ij} = -\mu_p + \sqrt{\frac{1}{2} \sum_{i=1}^{3} \sum_{j=1}^{3} \left[ \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right] \left[ \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right]} \left( \frac{\partial v_i}{\partial x_j} + \frac{\partial v_j}{\partial x_i} \right)$
Solve the fluid flow equation for steady-fully developed flow with constant pressure gradient to determine the velocity profile, then determine $v_{max}$ and $v_{mean}$
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3. Cassava paste is pumped through a tube. Determine the ratio vmax/vmean for this fluid, given that its rheological behavior may be characterized by the Bingham model. Note: The Bingham model is often written in simple form as: τ = τ0 + γ̇ However, in order to work out a fluid flow solution, it may be written in tensor form as: τij = - + √((1)/(2)∑i=1^3∑j=1^3[ (∂ vi)/(∂ xj) + (∂ vj)/(∂ xi)] [ (∂ vi)/(∂ xj) + (∂ vj)/(∂ xi)])( (∂ vi)/(∂ xj) + (∂ vj)/(∂ xi)) Solve the fluid flow equation for steady-fully developed flow with constant pressure gradient to determine the velocity profile, then determine vmax and vmean

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University Physics with Modern Physics
Hugh D. Young 14th Edition
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Cassava paste is pumped through a tube. Determine the ratio (v_(max))/(v_(mean )) for this fluid, given that its rheological behavior may be characterized by the Bingham model. Note: The Bingham model is often written in simple form as: au = au _(0)+mu _(p)gamma ^(˙) However, in order to work out a fluid flow solution, it may be written in tensor form as: au _(ij)=-(mu _(p)+( au _(0))/(|sqrt((1)/(2)sum_(i=1)^3 sum_(j=1)^3 [(delv_(i))/(delx_(j))+(delv_(j))/(delx_(i))][(delv_(j))/(delx_(i))+(delv_(i))/(delx_(j))])|))((delv_(i))/(delx_(j))+(delv_(j))/(delx_(i))) Solve the fluid flow equation for steady-fully developed flow with constant pressure gradient to determine the velocity profile, then determine v_(max ) and v_(mean ) Cassava paste is pumped through a tube. Determine the ratio vmax/vmean for this fluid given that its rheological behavior may be characterized by the Bingham model. Note: The Bingham model is often written in simple form as: T=To+pY However,in order to work out a fluid flow solution,it may be written in tensor form as: To Ov ov Ov ax Ov Ox Ox Ox Solve the fluid flow equation for steady-fully developed flow with constant pressure gradient to determine the velocity profile, then determine vmax and vmean
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Transcript

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00:01 In this question we are given the velocity profile for laminar flow in a smooth circular tube which is u of r is equal to 2 times v average multiplied by 1 minus small r over capital r whole square this is given the value of capital r is 5 millimeters which is 0 .000 5 meter so this is 0 .005 meter which is the inner radius of the tube and we are given the length of the tool l as 2 meters and the question says that if heavy weight oil flows laminarily through the tube at a rate of 1 milliliter per second that is we are given the volumetric flow rate v dot volumetric flowers which is represented by cube that is 1 milliliter per second so this is equal to 10 to the power minus 6 meter cube per second and we had given the density row of oil that is 888 kg per meter cube and the dynamic viscosity is also given that is 0 .9 kg per meter second.
01:35 So first of all we see the diagram.
01:40 So this is the diagram.
01:42 Now this dynamic viscosity can be equal to 0 .9 newton per meter square second.
01:52 Now first of all we find umax.
01:56 So umax is given by two times.
02:02 Average so here if we see so area of the since it is a circular tube so area of cross -section is a circle and area of circle is pi r square so we have area is equal to pi multiplied by radius is 0 .005 meter and it's squared so from here we get area is equal to 7 .8 5 4 into 7 .854 multiplied by 10 to the par minus 5 meters square.
02:41 So this is the area.
02:43 Now we know that the volumatic flow rate is nothing but area multiplied is the average volume.
02:51 So from here we know that volumatic flow rate is 10 to the per minus 6.
02:56 We know area that is here it is 7 .854 minus 6.
03:00 We know area that is here it is 7 .8554 multiply with 10 to the plus minus 5 multiply we average so from here we average we get as 0 .01273 meter per second so u max will be 2 multiplied as 0 .01273 meter per second u max is equal to 0 .025 4 .6 meter per second.
03:41 Now if we calculate the pressure change in the fluid due to viscous effects.
03:48 Here the pressure change the pressure change in the fluid in the fluid due to viscous effect is so we know that umax is nothing but 1 over 4 mu that is the mu that is the dynamic viscosity multiplied by minus dp over dx that is the change in pressure multiplied by capital r square so here we know all our values the umax is 0 .0 546 is equal to 1 over 4 multiplied by 0 .9 multiply minus dp over d x multiplied 0 .005 whole square so from here we get dp over d x as minus dp over dx as 3666 .6 .6 .24 nuten per meter square or we can write dp over dx as minus 3666 .24 newton per meter square this is negative because the pressure is decreasing now we are required to tell which pressure is higher p1 or p2 so if we see so minus dp over d x here minus if we write minus dp over d x is 3666 .24 so so dp what does dp represent? dp is the change in pressure.
05:56 So we can write it as p2 minus p1 over dx is equal to 3666 .24...
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