Reynold's Number and Poiseuille's Equation Questions in English

(C) The Reynolds number $(R_{e})$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
It is defined as the ratio of inertial forces to viscous forces.
The formula for the Reynolds number is given by:
$R_{e} = \frac{\rho v d}{\eta}$
Where:
$\rho$ = density of the fluid
$v$ = velocity of the fluid
$d$ = diameter of the pipe
$\eta$ = coefficient of viscosity of the fluid
Thus,the correct option is $C$.

(B) The Reynold's number $(R_e)$ is given by the formula:
$R_e = \frac{\rho v D}{\eta}$
where $\rho$ is the density of the liquid, $v$ is the velocity, $D$ is the diameter of the tube, and $\eta$ is the coefficient of viscosity.
Given:
Density $\rho = 10^3 \, kg/m^3$
Velocity $v = 2 \, m/s$
Radius $r = 2 \, cm = 0.02 \, m$, so diameter $D = 2r = 0.04 \, m$
Coefficient of viscosity $\eta = 8 \times 10^{-2} \, \text{decapoise} = 0.08 \, Pa \cdot s$
Substituting the values:
$R_e = \frac{10^3 \times 2 \times 0.04}{8 \times 10^{-2}}$
$R_e = \frac{1000 \times 0.08}{0.08}$
$R_e = 1000$

(D) According to Poiseuille's equation,the rate of flow $Q$ through a capillary tube is given by $Q = \frac{\pi P r^4}{8 \eta L}$.
From this,we can see that $Q \propto P \times r^4$.
Let the initial state be $Q_1 = Q$,$P_1 = P$,and $r_1 = a$.
Let the final state be $Q_2$,$P_2 = 4P$,and $r_2 = \frac{a}{4}$.
Using the proportionality $Q \propto P \times r^4$,we have $\frac{Q_2}{Q_1} = \frac{P_2}{P_1} \times \left( \frac{r_2}{r_1} \right)^4$.
Substituting the values: $\frac{Q_2}{Q} = \frac{4P}{P} \times \left( \frac{a/4}{a} \right)^4$.
$\frac{Q_2}{Q} = 4 \times \left( \frac{1}{4} \right)^4 = 4 \times \frac{1}{256} = \frac{1}{64}$.
Therefore,$Q_2 = \frac{Q}{64}$.

(A) According to Poiseuille's equation,the rate of flow $Q$ is given by $Q = \frac{\pi P r^4}{8 \eta L}$.
Since the capillaries are connected in series,the rate of flow $Q$ is the same for all three capillaries.
Therefore,the pressure difference $\Delta P$ across a capillary is given by $\Delta P = Q \cdot \frac{8 \eta L}{\pi r^4} \propto \frac{L}{r^4}$.
Let $P_1, P_2, P_3$ be the pressure differences across the first,second,and third capillaries respectively.
Given $P_1 = P$,$L_1 = L, r_1 = r$.
For the second capillary: $L_2 = L/2, r_2 = r/2$.
$P_2 = P_1 \cdot \frac{L_2}{L_1} \cdot \left(\frac{r_1}{r_2}\right)^4 = P \cdot \frac{L/2}{L} \cdot \left(\frac{r}{r/2}\right)^4 = P \cdot \frac{1}{2} \cdot (2)^4 = P \cdot \frac{16}{2} = 8P$.
For the third capillary: $L_3 = L/3, r_3 = r/3$.
$P_3 = P_1 \cdot \frac{L_3}{L_1} \cdot \left(\frac{r_1}{r_3}\right)^4 = P \cdot \frac{L/3}{L} \cdot \left(\frac{r}{r/3}\right)^4 = P \cdot \frac{1}{3} \cdot (3)^4 = P \cdot \frac{81}{3} = 27P$.
Thus,the pressure difference across the second capillary is $8P$.

(A) According to Poiseuille's equation,the volume of liquid flowing per second $(Q)$ in a capillary tube is given by:
$Q = \frac{V}{t} = \frac{\pi r^4 \Delta P}{8 \eta L}$
Since mass $m = V \cdot d$,the mass of liquid flowing per second is:
$\frac{m}{t} = \frac{\pi r^4 \Delta P}{8 \eta L} \cdot d$
For two liquids with equal mass $m$ flowing through identical tubes under the same pressure difference $\Delta P$:
$\frac{m}{t_1} = \frac{\pi r^4 \Delta P}{8 \eta_1 L} d_1 \implies \eta_1 = \frac{\pi r^4 \Delta P \cdot d_1 t_1}{8 L m} \dots (I)$
$\frac{m}{t_2} = \frac{\pi r^4 \Delta P}{8 \eta_2 L} d_2 \implies \eta_2 = \frac{\pi r^4 \Delta P \cdot d_2 t_2}{8 L m} \dots (II)$
Dividing equation $(I)$ by $(II)$:
$\frac{\eta_1}{\eta_2} = \frac{d_1 t_1}{d_2 t_2}$

(C) According to Poiseuille's law,the rate of flow of liquid through a capillary tube is given by:
$V = \frac{\pi P r^4}{8 \eta \ell}$
where $P$ is the pressure difference,$r$ is the radius,$\eta$ is the coefficient of viscosity,and $\ell$ is the length of the tube.
Given that $P$ and $\eta$ remain constant,the rate of flow $V$ is proportional to $\frac{r^4}{\ell}$.
Therefore,the ratio of the new rate of flow $V^{\prime}$ to the original rate of flow $V$ is:
$\frac{V^{\prime}}{V} = \frac{(2r)^4}{2\ell} \times \frac{\ell}{r^4}$
$\frac{V^{\prime}}{V} = \frac{16r^4}{2\ell} \times \frac{\ell}{r^4} = \frac{16}{2} = 8$
Thus,$V^{\prime} = 8V$.

(B) The Reynolds number $(Re)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
It is defined by the formula: $Re = \frac{\rho v d}{\eta}$.
Where:
$\rho$ = density of the fluid
$v$ = velocity of the fluid
$d$ = diameter of the pipe
$\eta$ = coefficient of viscosity of the fluid.
Comparing this with the given options, the correct expression is $\frac{\rho v d}{\eta}$.

(B) The formula for the critical velocity $(v_c)$ of a fluid flowing through a pipe is given by Reynolds number $(R_e)$:
$R_e = \frac{\rho v_c d}{\eta}$
Rearranging for critical velocity $(v_c)$:
$v_c = \frac{R_e \eta}{\rho d}$
From this expression,it is clear that the critical velocity $(v_c)$ is directly proportional to the coefficient of viscosity $(\eta)$ and inversely proportional to the density $(\rho)$ and the diameter $(d)$ of the tube.
Therefore,the correct statement is that the critical velocity is directly proportional to $\eta$.

(C) The Reynolds number $(Re)$ is a dimensionless quantity used to predict the flow regime of a fluid.
For flow through a pipe:
$1$. If $Re < 2000$,the flow is streamline (laminar).
$2$. If $2000 < Re < 3000$,the flow is in a transition state.
$3$. If $Re > 3000$,the flow is turbulent.
Given that the Reynolds number is $900$,which is less than $2000$,the flow of the liquid is streamline.

(C) The Reynolds number $(R_e)$ is a dimensionless quantity used to predict the flow pattern of a fluid in a pipe. It represents the ratio of inertial forces to viscous forces.
Mathematically,it is defined as:
$R_e = \frac{\text{Inertial force}}{\text{Viscous force}}$
For a liquid of density $\rho$ flowing with velocity $V$ through a tube of diameter $d$ with coefficient of viscosity $\eta$,the formula is:
$R_e = \frac{\rho V d}{\eta}$
Thus,the correct option is $C$.

(B) The Reynolds number $(R_{n})$ for the flow of a fluid through a pipe is given by the formula:
$R_{n} = \frac{v_{c} \rho d}{\eta}$
Where:
$v_{c}$ is the critical velocity,
$\rho$ is the density of the fluid $(10^3 \,kg/m^3)$,
$d$ is the diameter of the tube $(2 \times r = 2 \times 1 \,cm = 2 \times 10^{-2} \,m)$,
$\eta$ is the coefficient of viscosity $(1 \times 10^{-3} \,Ns/m^2)$,
$R_{n}$ is the critical Reynolds number $(2500)$.
Rearranging the formula to solve for $v_{c}$:
$v_{c} = \frac{R_{n} \eta}{\rho d}$
Substituting the given values:
$v_{c} = \frac{2500 \times 10^{-3}}{10^3 \times 2 \times 10^{-2}}$
$v_{c} = \frac{2.5}{20} = 0.125 \,m/s$
Thus, the speed of flow required to convert the streamline flow into turbulent flow is $0.125 \,m/s$.

$(A)$ The Reynolds number $(Re)$ is a dimensionless quantity that represents the ratio of inertial forces to viscous forces in a fluid flow.
It is used to predict the flow regime of a fluid.
For a flow in a pipe, if the Reynolds number is less than $2000$, the flow is considered to be streamline or laminar.
If the Reynolds number is between $2000$ and $4000$, the flow is in a transition state.
If the Reynolds number is greater than $4000$, the flow is turbulent.
Among the given options, the condition for streamline flow is satisfied by the range less than $1000$.

(B) According to Poiseuille's law,the rate of steady volume flow $V$ through a capillary tube is given by $V = \frac{\pi p r^4}{8 \eta l}$.
This can be rewritten as the pressure difference $p = V \left( \frac{8 \eta l}{\pi r^4} \right) = V R_H$,where $R_H = \frac{8 \eta l}{\pi r^4}$ is the hydraulic resistance.
For the first tube,$R_1 = \frac{8 \eta l}{\pi r^4}$.
For the second tube,$R_2 = \frac{8 \eta l}{\pi (r/2)^4} = \frac{8 \eta l}{\pi r^4 / 16} = 16 R_1$.
In a series combination,the total pressure difference $p$ is the sum of pressure differences across each tube: $p = p_1 + p_2 = V' R_1 + V' R_2$,where $V'$ is the new rate of flow.
Since $p = V R_1$,we have $V R_1 = V' (R_1 + 16 R_1) = V' (17 R_1)$.
Therefore,$V = 17 V'$,which gives $V' = \frac{V}{17}$.

(A) Given: Diameter $d = 8 \text{ cm}$,so radius $r = 4 \text{ cm} = 4 \times 10^{-2} \text{ m}$.
Length $l = 3140 \text{ m}$.
Rate of flow $Q = 2 \times 10^{-3} \text{ m}^3/\text{s}$.
Viscosity $\eta = 10^{-3} \text{ SI units}$.
According to Poiseuille's equation for laminar flow in a pipe:
$Q = \frac{\pi P r^4}{8 \eta l}$
Rearranging for pressure $P$:
$P = \frac{Q(8 \eta l)}{\pi r^4}$
Substituting the values:
$P = \frac{2 \times 10^{-3} \times 8 \times 10^{-3} \times 3140}{3.14 \times (4 \times 10^{-2})^4}$
$P = \frac{16 \times 3140 \times 10^{-6}}{3.14 \times 256 \times 10^{-8}}$
$P = \frac{3140 \times 10^2}{3.14 \times 16}$
$P = \frac{1000 \times 10^2}{16} = \frac{10^5}{16} = 6.25 \times 10^3 \text{ N/m}^2$.

(B) According to Poiseuille's equation,the resistance to fluid flow $R$ is given by $R = \frac{8 \eta l}{\pi r^4}$,where $\eta$ is the viscosity,$l$ is the length,and $r$ is the radius of the tube.
Since the length $l$ and viscosity $\eta$ are the same for both tubes,we have $R \propto \frac{1}{r^4}$.
Let $r_1 = 4 \ mm$ and $r_2 = 2 \ mm$. The ratio of resistances is $\frac{R_1}{R_2} = \left(\frac{r_2}{r_1}\right)^4 = \left(\frac{2}{4}\right)^4 = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$.
Thus,$R_2 = 16 R_1$. Let $R_1 = R$,then $R_2 = 16 R$.
Since the tubes are connected in series,the same volume flow rate $Q$ passes through both. The pressure difference across each tube is $\Delta P = Q \times R$.
For the first tube,$\Delta P_1 = Q \times R_1 = Q \times R$.
For the second tube,$\Delta P_2 = Q \times R_2 = Q \times 16 R$.
The total pressure difference is $P = \Delta P_1 + \Delta P_2 = Q R + 16 Q R = 17 Q R$.
Therefore,$Q R = \frac{P}{17}$.
The pressure difference across the first tube is $\Delta P_1 = Q R = \frac{P}{17}$.

(A) To determine the nature of the flow, we calculate the Reynolds number $(R_e)$.
The formula for the Reynolds number is $R_e = \frac{\rho v D}{\eta}$, where $\rho$ is the density, $v$ is the velocity, $D$ is the diameter of the pipe, and $\eta$ is the coefficient of viscosity.
Given:
Density $\rho = 10^3 \,kg \,m^{-3}$
Velocity $v = 20 \,cm \,s^{-1} = 0.2 \,m \,s^{-1}$
Radius $r = 2 \,cm = 0.02 \,m$, so diameter $D = 2r = 0.04 \,m$
Viscosity $\eta = 10^{-3} \,kg \,m^{-1} \,s^{-1}$
Substituting these values:
$R_e = \frac{10^3 \times 0.2 \times 0.04}{10^{-3}}$
$R_e = \frac{10^3 \times 0.008}{10^{-3}} = 8 \times 10^3 = 8000$
Since the Reynolds number $R_e > 2000$, the flow is turbulent.

(C) Given: Diameter $D = 1.5 \ cm = 1.5 \times 10^{-2} \ m$,Flow rate $Q = 7.5 \times 10^{-5} \ m^3 \ s^{-1}$,Viscosity $\eta = 10^{-3} \ Pa \cdot s$,Density of water $\rho = 10^3 \ kg \cdot m^{-3}$.
Flow rate $Q = A \cdot v = \frac{\pi}{4} D^2 v$,so velocity $v = \frac{4Q}{\pi D^2}$.
Reynolds number $R_e = \frac{\rho v D}{\eta} = \frac{\rho (\frac{4Q}{\pi D^2}) D}{\eta} = \frac{4 \rho Q}{\pi \eta D}$.
Substituting the values: $R_e = \frac{4 \times 10^3 \times 7.5 \times 10^{-5}}{3.14 \times 10^{-3} \times 1.5 \times 10^{-2}}$.
$R_e = \frac{0.3}{4.71 \times 10^{-5}} \approx 6369.4$.
Since $R_e > 4000$,the flow is turbulent. Among the given options,the flow is turbulent with a Reynolds number greater than $6000$.

(B) The Reynolds number $(R_e)$ is a dimensionless quantity used to predict flow patterns in different fluid flow situations.
For flow through a pipe,the generally accepted criteria are:
$1$. If $R_e < 2000$,the flow is laminar.
$2$. If $2000 < R_e < 3000$,the flow is unsteady or in transition.
$3$. If $R_e > 3000$,the flow is turbulent.
Option $B$ states that for $1000 < R_e < 2000$,the flow is steady. While flow is laminar in this range,the classification of 'steady' is not the standard terminology used to define this specific range in fluid mechanics,and option $C$ is also technically incorrect based on the standard threshold of $3000$. However,in many simplified textbook contexts,$R_e > 2000$ is often cited as the threshold for turbulence. Comparing the options,option $B$ is the most inaccurate description of flow regimes.

(B) Given: Diameter of tap,$D = 1.25 \text{ cm} = 1.25 \times 10^{-2} \text{ m}$.
Density of water,$\rho = 10^3 \text{ kg/m}^3$.
Coefficient of viscosity,$\eta = 10^{-3} \text{ Pa-s}$.
Volume flow rate,$Q = 3 \text{ L/min} = \frac{3 \times 10^{-3} \text{ m}^3}{60 \text{ s}} = 5 \times 10^{-5} \text{ m}^3/\text{s}$.
Using the formula for velocity $v = \frac{Q}{A} = \frac{4Q}{\pi D^2}$.
Reynold's number $R_e$ is given by $R_e = \frac{\rho v D}{\eta} = \frac{4 \rho Q}{\pi D \eta}$.
Substituting the values:
$R_e = \frac{4 \times 10^3 \times 5 \times 10^{-5}}{3.14159 \times 1.25 \times 10^{-2} \times 10^{-3}} \approx 5093$.
Since $R_e > 3000$,the flow is turbulent.

(A) pure number which determines the type of flow of a liquid through a pipe is known as Reynold's number $(R_e)$.
$(i)$ If $R_e < 2100$,the flow is laminar.
(ii) If $2100 < R_e < 4000$,the flow is unsteady or transitional.
(iii) If $R_e > 4000$,the flow is turbulent.

(C) According to Poiseuille's law,the rate of flow of a liquid $(Q)$ through a capillary tube is given by the formula:
$Q = \frac{V}{t} = \frac{\pi p R^4}{8 \eta L}$
Where:
$V$ is the volume of the liquid,
$t$ is the time of flow,
$p$ is the pressure difference,
$R$ is the radius of the tube,
$\eta$ is the coefficient of viscosity,
$L$ is the length of the tube.
Rearranging the formula to solve for the volume $V$:
$V = \frac{\pi p R^4 t}{8 \eta L}$
Since $\frac{\pi}{8}$ is a dimensionless constant,the volume $V$ is proportional to the remaining terms:
$V \propto \frac{p R^4 t}{\eta L}$
Therefore,the correct option is $C$.

$(A)$ The Reynolds number $(Re)$ is a dimensionless quantity used to predict the flow regime of a fluid.
For flow in a pipe, the criteria are generally defined as follows:
$1$. If $Re < 2000$, the flow is streamlined or laminar.
$2$. If $2000 < Re < 3000$, the flow is in a transition state.
$3$. If $Re > 3000$, the flow is turbulent.
Comparing the given options with the criteria for streamlined flow $(Re < 2000)$:
- Option $A$: $900 < 2000$ (Streamlined)
- Option $B$: $2100$ (Transition)
- Option $C$: $2900$ (Transition)
- Option $D$: $4000$ (Turbulent)
Therefore, the correct option is $A$.

(D) According to Poiseuille's law,the volume flow rate $Q$ is given by:
$Q = \frac{\pi \Delta P r^4}{8 L \eta}$
where $r$ is the radius,$\eta$ is the viscosity,and $\Delta P$ is the pressure difference.
Since the flow rate $Q$ does not depend on the density of blood,the change in density does not affect the flow rate.
Taking the logarithmic differentiation of the formula:
$\frac{\Delta Q}{Q} = 4 \frac{\Delta r}{r} - \frac{\Delta \eta}{\eta}$
Given that the radius $r$ increases by $1 \%$ $(\frac{\Delta r}{r} = 0.01)$ and the viscosity $\eta$ increases by $1 \%$ $(\frac{\Delta \eta}{\eta} = 0.01)$:
$\frac{\Delta Q}{Q} = 4(0.01) - 0.01 = 0.04 - 0.01 = 0.03$
Therefore,the percentage change in the flow rate is $0.03 \times 100 = 3 \%$.

(A) The Reynolds number for a fluid flow is given by $R_e = \frac{2 v \rho r}{\eta}$,where $v$ is the velocity of flow,$\rho$ is the density of the fluid,$r$ is the radius of the tube,and $\eta$ is the viscosity of the fluid.
From the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A = \pi r^2$. Thus,$\pi r_1^2 v_1 = \pi r_2^2 v_2$,which implies $\frac{v_1}{v_2} = \frac{r_2^2}{r_1^2}$.
The ratio of Reynolds numbers is $\frac{R_1}{R_2} = \frac{v_1 r_1}{v_2 r_2} = \left(\frac{r_2^2}{r_1^2}\right) \times \left(\frac{r_1}{r_2}\right) = \frac{r_2}{r_1}$.
Given $r = r_0 e^{-\alpha x}$,we have $r_1 = r_0 e^{-\alpha x_1}$ and $r_2 = r_0 e^{-\alpha (x_1 + 3)}$.
Substituting these values,$\frac{R_1}{R_2} = \frac{r_0 e^{-\alpha (x_1 + 3)}}{r_0 e^{-\alpha x_1}} = e^{-\alpha (x_1 + 3) + \alpha x_1} = e^{-3 \alpha}$.
Given $\alpha = \frac{1}{3} \text{ m}^{-1}$,we get $\frac{R_1}{R_2} = e^{-3(1/3)} = e^{-1} = \frac{1}{e}$.

(D) Given: Diameter of pipe $d = 5 \,mm = 5 \times 10^{-3} \,m$. Density of gasoline $\rho = 720 \,kg/m^3$. Viscosity of gasoline $\eta = 6 \times 10^{-3} \,Poise = 6 \times 10^{-4} \,Pa \cdot s$. The critical Reynolds number for pipe flow is $R_e = 2000$. The formula for critical velocity $v_c$ is $v_c = \frac{R_e \cdot \eta}{\rho \cdot d}$. Substituting the values: $v_c = \frac{2000 \times 6 \times 10^{-4}}{720 \times 5 \times 10^{-3}} = \frac{1.2}{3.6} = \frac{1}{3} \approx 0.33 \,m/s$. Thus,the flow becomes turbulent when the velocity is $ > 0.33 \,m/s$.

(C) The velocity of a fluid,such as water,flowing through a tube,below which the flow remains streamline (or laminar) and above which it becomes turbulent,is defined as the critical velocity.