+{"orderNumber":1,"title":"Friction on a plate","skill":0.6666666666666666,"guidance":" This question is to refresh yourself using control volumes, which were used in ME1 and will be used often in ME2.","durationLowerBound":15,"durationUpperBound":20,"masterContent":"A fluid with density $\\rho = 800\\,\\mathrm{kg/m}^3$, flows at $U_0 = 3\\,\\mathrm{m/s}$ over a flat plate of length $L = 1\\,\\mathrm{m}$ and width $W = 1\\,\\mathrm{m}$. At the trailing edge the boundary-layer thickness is $\\delta = 25\\,\\mathrm{mm}$. Assume the velocity profile at the trailing edge to be linear (in the image shown), and the flow to be two-dimensional. \n\n{ width=60% }","publish":true,"displayFinalAnswer":true,"displayStructuredTutorial":true,"displayWorkedSolution":true,"displayChatbot":true,"parts":[{"orderNumber":0,"content":"Compute the mass flow rate across the top surface of the control volume (noted ''ab'' in the figure).","answerContent":"$$\n\\dot{m}= \\boxed{30 \\space \\mathrm{kg/s} \\>}\n$$","responseAreas":[{"orderNumber":0,"contentAfter":"","preResponseText":"$ \\dot{m}= $","postResponseText":"","inputSymbols":[],"displayInputSymbols":false,"includeInPdf":false,"saveAllowed":false,"evaluationFunctionName":"comparePhysicalQuantities","livePreview":false,"gradeParams":{"rtol":0.05,"strict_syntax":false},"separateFeedback":true,"commonFeedbackColor":"#C4CDD5","correctFeedbackColor":"#22C55E","correctFeedbackPrefix":"Correct","incorrectFeedbackColor":"#ff5630","incorrectFeedbackPrefix":"Incorrect","tests":[{"id":"673a7103-7930-4b11-8a62-98732b27bb5e","payload":"30 kgs-1","expectedResponse":{"isCorrect":false}}],"cases":[{"id":"092d2819-a2b5-481c-b0f9-dbfecfb176fa","answer":"30 kgs-1","feedback":"To enter a negative exponent, put the exponent in parentheses (e.g. ```kg*s^(-1)``` ). The expression ```kgs-1``` is interpreted differently and is incorrect.. ","isCorrect":false,"params":null}],"response":{"responseInput":{"responseType":"NUMERIC_UNITS","answer":"30 kg/s","config":null}}}],"workedSolution":{"content":"From mass conservation:\n\n---\n\n$\\dot{m}_{ad}=\\dot{m}_{ab}+\\dot{m}_{bc}$\n\n---\n\nThis becomes:\n\n---\n\n$\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{U_0\\space \\mathrm{d}y}=\\dot{m}_{ab}+\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{u(y)\\space\\mathrm{d}y}$\n\n---\n\nSince we know that $u(y)$ is linear, we can express it in terms of $y$ and $U_0$:\n\n---\n\n$$\nu(y)=U_0\\frac{y}{\\delta},\n$$\n\n---\n\nTherefore:\n\n$$\n\\begin{aligned}\\dot{m}_{ab}&=\\rho W U_0\\left(\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{1-\\frac{y}{\\delta}\\space\\mathrm{d}y} \\right)\\\\\\dot{m}_{ab}&=\\rho W U_0\\frac{\\delta}{2},\\\\\\dot{m}_{ab}&=(800)(1)(3)\\frac{25\\times 10^{-3}}{2}.\\end{aligned}\n$$\n\n---\n\n$$\n\\dot{m}= \\boxed{30 \\space \\mathrm{kg/s} \\>}\n$$","children":[]}},{"orderNumber":1,"content":"Determine the drag force on the plate.","answerContent":"$$\nF_{\\mathrm{plate}}= \\boxed{30\\space \\mathrm{N}\\space}\n$$","responseAreas":[{"orderNumber":0,"contentAfter":"","preResponseText":"$F_D =$","postResponseText":"","inputSymbols":[],"displayInputSymbols":false,"includeInPdf":false,"saveAllowed":false,"evaluationFunctionName":"comparePhysicalQuantities","livePreview":false,"gradeParams":{"rtol":0.05,"strict_syntax":false},"separateFeedback":true,"commonFeedbackColor":"#C4CDD5","correctFeedbackColor":"#22C55E","correctFeedbackPrefix":"Correct","incorrectFeedbackColor":"#ff5630","incorrectFeedbackPrefix":"Incorrect","tests":[],"cases":[],"response":{"responseInput":{"responseType":"NUMERIC_UNITS","answer":"30 N","config":null}}}],"workedSolution":{"content":"From momentum conservation:\n\n***\n\n***\n\n$$\nF_{\\small \\mathrm{fluid}}=M_{ab}+M_{bc}-M_{ad}\n$$\n\n***\n\nSince we are calculating the horizontal resultant force, we need to consider the horizontal momentum in the section \"ab\"; this means that we ignore any vertical velocities. Since the velocity of the fluid at $y=\\delta$ (i.e. at \"ab\") is always $U_0$, we can say:\n\n***\n\n$$\nM_{ab}=U_0\\dot{m}_{ab},\n$$\n\n***\n\nWe then also convert the other momentum flowrate terms into their mathematical forms, and proceed to find the force of the plate on the fluid, as shown below:\n\n***\n\n$$\nF_{\\mathrm{fluid}}=M_{ab}+\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{u(y)^2\\space\\mathrm{dy}}-\\rho W\\Large{\\int}_{\\small 0}^{\\small \\delta}\\normalsize{U_0^2\\space \\mathrm{dy}}\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=U_0\\dot{m}_{ab}+\\rho WU_0^2\\left(\\LARGE{\\int}_{\\small 0}^{\\small \\delta}\\normalsize \\frac{y^2}{\\delta^2}-1\\space\\mathrm{dy} \\right)\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=(3)(30)+(800)(1)(3)^2\\left[ -\\frac{2}{3}(25\\times 10^{-3})\\right]\n$$\n\n***\n\n$$\nF_{\\mathrm{fluid}}=-30 \\space \\mathrm{N}\n$$\n\n***\n\nHowever, the question asks us to find the drag force **of the fluid on the plate**, hence:\n\n***\n\n$$\nF_{\\mathrm{plate}}=-F_{\\mathrm{fluid}}\n$$\n\n***\n\n$$\nF_{\\mathrm{plate}}= \\boxed{30\\space \\mathrm{N}\\space}\n$$\n","children":[]}}]}
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