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(I) Engine oil (assume SAE $10,$ Table 3$)$ passes through afine 1.80 -mm-diameter tube that is 8.6 $\mathrm{cm}$ long. What pressuredifference is needed to maintain a flow rate of 6.2 $\mathrm{mL} / \mathrm{min}$ ?

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6900$P a$

Physics 101 Mechanics

Chapter 13

Fluids

Fluid Mechanics

University of Michigan - Ann Arbor

University of Sheffield

McMaster University

Lectures

03:45

In physics, a fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases, plasmas and, to some extent, plastic solids.

09:49

A fluid is a substance that continually deforms (flows) under an applied shear stress. Fluids are a subset of the phases of matter and include liquids, gases and plasmas. Fluids display properties such as flow, pressure, and tension, which can be described with a fluid model. For example, liquids form a surface which exerts a force on other objects in contact with it, and is the basis for the forces of capillarity and cohesion. Fluids are a continuum (or "continuous" in some sense) which means that they cannot be strictly separated into separate pieces. However, there are theoretical limits to the divisibility of fluids. Fluids are in contrast to solids, which are able to sustain a shear stress with no tendency to continue deforming.

02:40

(I) Engine oil (assume SAE…

05:56

Engine oil (assume SAE 10,…

01:33

(II) Engine oil (assume SA…

03:56

SAE $30 \mathrm{W}$ oil at…

02:02

A fish tank has dimensions…

we're going to use a crossbow yays equation in order to find the pressure difference. This is given by equation 13 12 and so we can then say that cue the volumetric flow rate would be equal in two pi. Times are to the fourth power times three. Change in pressure, divided by eight times the viscosity times eight times new the viscosity times l and so we can solve For the change in pressure, this would be equal to eight times the volumetric floor eight times the viscosity knew at times the length l divided by pi are to the fourth power and so we can solve. This would be eight multiplied by 6.2 times 10 to the negative third leaders per minute. We're going to multiply this by one minute for every 60 seconds and then multiply this by 10 to the negative third meters cubed for every leader. And so we can then multiply this by the viscosity 0.2 pascal seconds multiplied bye. 8.6 times 10 to the negative second meters. This would all be divided by pie multiplied by 0.9 times 10 to the negative third meters quantity to the fourth and we'll put the pressure. You're here. Piece of two minus piece of one ends up equaling approximately 6900. Ta Scouse! This would be our final answer. That is the end of the solution. Thank you for watching.

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