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Unformatted text preview: Physics 7B W10 Quiz 1 Consider a fluid (density ρ) filled syringe (a tube with cross sectional area At leading to a much narrower hollow needle with cross sectional area An). Imagine that you hold the syringe horizontally in air and press the plunger to establish a steady
state flow with fluid leaving the end of the needle with velocity v. a) Suppose there is no friction in the syringe tube or needle. Comparing the fluid gauge pressure inside the body of the syringe and inside the needle, which is greater? Explain. Consider the energy density equation (Bernoulli’s equation). The ρgΔy term is zero as there is no height difference; the Epump term is zero because there is no pump between the body of the syringe and the needle; the IR term is zero because there is no resistance. Therefore, we have only ΔP+(1/2)ρΔ(v2)=0. We know from continuity that the velocity of the fluid must be higher in the needle because it has a smaller cross section, therefore the pressure must be lower. b) Now suppose there is friction in the needle. Comparing the pressure in the body of the syringe and the end of the needle, which is greater? If there is not enough information to answer this question, then state so. Explain your answer. By adding resistance, we can no longer ignore the IR term in the energy density equation, however the only way to increase the pressure in the needle would be for the needle to have a negative resistance, which is not possible. So the pressure in the needle will still be lower. c) Assume again that there is no friction. Write an expression to show how one could determine the force which must be applied to the plunger to establish this steady state stream of fluid. Since we are assuming no friction, we can use ΔP+ ΔP+(1/2)ρΔ(v2)=0. The end of the needle is open to atmospheric pressure (or gauge pressure of zero). The velocity of the fluid in the body of the syringe is approximately zero, therefore Δv is the v that was given for the fluid exiting the needle. The change in pressure is then (1/2)ρΔ(v2), and the force is (1/2)ρΔ(v2)As ...
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 Winter '08
 Taylor
 Physics

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