Question

Can you please do a part a

4. EXCEL Problem - each student is to create and submit a spreadsheet to solve this problem The pressure drop across a short constriction in a pipe is a function of fluid density, fluid velocity, and the pipe and constriction diameters, D and d, respectively. Experimental data collected in a pipe with diameter D = 0.2 ft, fluid density of 1.94 slugs/ft?, and velocity of 3 ft/s are given in the table d (ft) AP (lbf/ft) 0.06 493.8 0.08 156.2 0.10 64.0 0 .15 126 a) Determine the appropriate dimensionless terms b) Plot the dimensionless data and fit a trendline to develop a dimensionless empirical equation for AP. c) What are the limits of application for this equation?

Answer 1

Solution :

$\textup{Given Data : -}$

$\rho=1.94\, \, slug/ft^3$

.

$\textup{Part a) To determine the dimensionless terms :}$

$\textup{Using FLT system, dimensions of each parameters :}$

$\Delta P\Rightarrow FL^{-2}$

$\rho\Rightarrow FL^{-4}T^2$

$V\Rightarrow LT^{-1}$

DL

DL

$\textup{Applying Buckingham pi - theorem}:$

$\textup{There are 5 parameters and 3 dimensiones (F,L and T)}.$

$\textup{Considering repeating variable as : }\rho,\, \, V,\, \, \textup{and }D$

$\textup{No of }\pi\textup{ terms : }5-3=2$

$\bullet\, \, \, \pi_1\, \, \textup{term : }\pi_1=\Delta P \rho^aV^bD^c$

$\therefore F^0L^0T^0=(FL^{-2})(F^aL^{-4a}T^{2a})(L^bT^{-b})(L^c)$

$\textup{Now, equating exponents : }$

$\therefore F:0=1+a$

$\therefore L:0=-2-4a+b+c$

T:0 = 2a - b

$\textup{On solving we get, }a=-1,\, \, b=-2,\, \, \, c=0$

$\therefore \pi_1=\Delta P \rho^{-1}V^{-2}D^0=\frac{\Delta P}{\rho V^2}$

.

$\bullet\, \, \, \pi_2\, \, \textup{term : }\pi_2=d \rho^aV^bD^c$

$\therefore F^0L^0T^0=(L)(F^aL^{-4a}T^{2a})(L^bT^{-b})(L^c)$

$\therefore F:0=a$

$\therefore L:0=1-4a+b+c$

T:0 = 2a - b

$\textup{On solving we get, }a=0,\, \, b=0,\, \, \, c=-1$

$\therefore \pi_2=d \rho^0V^0D^{-1}=\frac{d}{D}$

$\textup{we can also transform it as : }\pi_2=\frac{D}{d}$

.

$\textup{Thus, }\pi_1=\phi(\pi_2)\Rightarrow\frac{\Delta P}{\rho V^2}=\phi \left ( \frac{D}{d} \right )$

${\color{DarkBlue} \therefore \frac{\Delta P}{\rho V^2}=\phi \left ( \frac{D}{d} \right )}$

$\textup{For given data :}$

π2 D/d 3.333 | 2.5 I 2 I 1.333 ΔΡ/pv2 28.282 8.9462 3.6655 0.72165

.

$\textup{Part b) To develop equation of dimensionless form :}$

$\textup{Let us assume, }\pi_1=a\pi_2 ^b$

$\therefore \log\pi_1=\log a+b \log\pi_2$

$\textup{Plotting of log values in excel we get, }$

Log Pi_2 3.333 | 28.282 2.5 8.9462 2 3.6655 | 1.333 0.7217 Log value value of Pi_2 of Pi_1 0.52283531 1.45151012 0.39794001 0.9516386 0.30103 0.56413322 0.12483015 -0.14167338|

PLOT y = 4.0028x - 0.6412 Log value of Pl_1 Log value of Pi 2
$\textup{(Note : Since the log plot is linear the above assumption was correct.)}$

$\textup{Comparing the equation we get from excel : }$

$\therefore \log a=-0.6412\Rightarrow a=0.22846\, \, \, \, \textup{ and }b=4.0028$

$\therefore \pi_1=0.22846\pi_2 ^{4.0028}$

${\color{DarkBlue} \therefore \frac{\Delta P}{\rho V^2}=0.22846 \left ( \frac{D}{d} \right )^{4.0028}}$

.

$\textup{Part c) Limitation of the equation :}$

$\textup{This equation is applicable over the range of data :}$

${\color{DarkBlue} 1.333\leq \frac{D}{d}\leq 3.333}$