A difficult question that involves rearranging a complicated formula, then applying unit conversions to variable values, then evaluating the formula for the selected value. The variable values are randomised.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "In the design of orifice plate flowmeters, the volumetric flowrate, \\(Q\\) (m\\(^3\\)s\\(^{-1}\\)), is given by \\[Q=C_dA_o\\sqrt{\\frac{2gh}{1-A_o^2/A_p^2}}\\] where \\(C_d\\) is a dimensionless discharge coefficient, \\(h\\) (m) is the head difference across the orifice plate, \\(A_o\\) (m\\(^2\\)) is the area of the orifice and \\(A_p\\) (m\\(^2\\)) is the area of the pipe.
", "advice": "$\\begin{align}Q&=C_dA_o\\sqrt{\\frac{2gh}{1-A_o^2/A_p^2}}\\\\ \\frac{Q}{C_dA_o}&=\\sqrt{\\frac{2gh}{1-A_o^2/A_p^2}}\\\\ \\left(\\frac{Q}{C_dA_o}\\right)^2&=\\frac{2gh}{1-A_o^2/A_p^2}\\\\ \\left(\\frac{C_dA_o}{Q}\\right)^2&=\\frac{1-A_o^2/A_p^2}{2gh}\\\\ \\left(\\frac{C_d}{Q}\\right)^2A_o^2&=\\frac{1}{2gh}-\\frac{A_o^2}{2ghA_p^2}\\\\ \\left(\\frac{C_d}{Q}\\right)^2A_o^2 + \\frac{A_o^2}{2ghA_p^2} &=\\frac{1}{2gh}\\\\ \\left[\\frac{C_d^2}{Q^2} + \\frac{1}{2ghA_p^2}\\right]A_o^2 &=\\frac{1}{2gh}\\\\ \\left[\\frac{C_d^22ghA_p^2+Q^2}{Q^22ghA_p^2}\\right]A_o^2 &=\\frac{1}{2gh}\\\\ A_o^2&=\\frac{1}{2gh}\\times\\frac{Q^22ghA_p^2}{C_d^22ghA_p^2+Q^2}\\\\A_o&=\\frac{QA_p}{\\sqrt{C_d^22ghA_p^2+Q^2}}\\end{align}
\n$\\sqrt{\\var{Ao_eval_on_pi}}$
\nTo begin with, we need to work out the values of each of the variables in the correct units for the equation.
\n$Q = \\var{Q}$ cm$^3$s$^{-1}$ = $\\var{Q}\\div 1000000$ m$^3$s$^{-1}= \\var{Q_eval}$ m$^3$s$^{-1}$
\nPipe diameter = $\\var{A_p}$ cm = $\\var{A_p} \\div 100$ m = $\\var{2*Apr_eval}$ m
\nPipe radius = Pipe diameter $\\div 2 = \\var{2*Apr_eval}\\div 2 = \\var{Apr_eval}$ m
\nA$_p = \\pi\\times$ pipe radius$^2 = \\pi\\times \\var{Apr_eval}^2 = \\var{Ap_eval}$ m$^2$
\nC_d = $0.8$
\ng = $9.81$ ms$^{-2}$
\nh = $\\var{h} $ mm = $\\var{h}\\div 1000$ m = $\\var{h_eval}$ m
\nSo:
\n\\[ \\begin{align}A_o&=\\frac{QA_p}{\\sqrt{C_d^22ghA_p^2+Q^2}}\\\\
&= \\frac{\\var{Q_eval}\\times\\var{dpformat(Ap_eval,5,\"scientific\")}}{\\sqrt{(\\var{C_d})^2\\times 2\\times 9.81 \\times \\var{h_eval}\\times (\\var{dpformat(Ap_eval,5,\"scientific\")})^2+(\\var{Q_eval})^2}} \\\\
&= \\frac{\\var{numerator}}{\\sqrt{\\var{den_under_root_term1}+\\var{dpformat(Q_eval*Q_eval,5,\"scientific\")}}} \\\\
&= \\frac{\\var{numerator}}{\\sqrt{\\var{den_under_root}}}\\\\
&= \\frac{\\var{numerator}}{\\var{denominator}}\\\\
&=\\var{Ao_eval}\\text{ m}\\\\
\\text{And } A_o &= \\pi \\times \\text{ radius}^2\\\\
\\var{Ao_eval} & = \\pi\\times \\text{ radius}^2\\\\
\\text{radius}^2 & =\\frac{\\var{Ao_eval}}{\\pi} \\\\
& = \\var{Ao_eval_on_pi}\\\\
\\text{radius}&=\\sqrt{\\var{Ao_eval_on_pi}}\\\\
\\text{radius}&=\\var{root_Ao_eval_on_pi} \\text{ m}\\\\
& =\\var{root_Ao_eval_on_pi}\\times 100 \\text{ cm}\\\\
& =\\var{root_Ao_eval_on_pi*100} \\text{ cm}\\\\
\\text{Since diameter }&=\\text{ radius }\\times 2\\\\
\\text{diameter } &= \\var{root_Ao_eval_on_pi*100}\\times 2 \\text{ cm}\\\\
\\text{diameter } &= \\var{precround(root_Ao_eval_on_pi*100*2,3)} \\text{ cm}\\\\
\\end{align}\\]
used in the advice. The value inside the denominator square root for area Ao
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\\(A_o=\\)[[0]]
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\n[Hint: be very careful with units.]
\norifice diameter\\(=\\)[[0]] cm
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