Question 1 |
For a steady incompressible laminar flow between two infinite parallel stationary plates, the shear stress variation is
linear with zero value at the plates | |
linear with zero value at the center | |
quadratic with zero value at the Plates | |
quadratic with zero value at the centre |
Question 1 Explanation:
Shear stress variation between parallel stationary plates.
Question 2 |
With reference to a standard Cartesian (x,y) plane, the parabolic velocity distribution profile of fully developed laminar flow in x-direction between two parallel, stationary and identical plates that are separated by distance, h, is given by the expression
u=-\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x}\left [ 1-4(\frac{y}{h})^{2} \right ]
In this equation, the y=0 axis lies equidistant between the plates at a distance h/2 from the two plates, p is the pressure variable and \mu is the dynamic viscosity term. The maximum and average velocities are, respectively
u=-\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x}\left [ 1-4(\frac{y}{h})^{2} \right ]
In this equation, the y=0 axis lies equidistant between the plates at a distance h/2 from the two plates, p is the pressure variable and \mu is the dynamic viscosity term. The maximum and average velocities are, respectively
u_{max}=-\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x} and u_{average}=\frac{2}{3}u_{max} | |
u_{max}=\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x} and u_{average}=\frac{2}{3}u_{max} | |
u_{max}=-\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x} and u_{average}=\frac{3}{8}u_{max} | |
u_{max}=\frac{h^{2}}{8\mu }\frac{\mathrm{d} p}{\mathrm{d} x} and u_{average}=\frac{3}{8}u_{max} |
Question 2 Explanation:
Velocily expression for a laminar flow between two parallel plates is
U=-\frac{n^{2}}{8 u}\left(\frac{d p}{d x}\right)\left[1-4\left(\frac{y}{n}\right)^{2}\right]
End condition
U=U_{max}\text{ at }y=0
\Rightarrow U_{max} = -\frac{h^{2}}{8\mu}\left(\frac{dp}{dx} \right )
Discharges dQ = Area x Velocily
\Rightarrow \Delta 0=\left[-\frac{\hbar^{2}}{8 \mu}\left(\frac{d \rho}{d x}\right)\left(1-4\left(\frac{y}{n}\right)^{2}\right)\right](d y \times 1)
\begin{aligned} &\Rightarrow Q=-\frac{h^{2}}{8 \mu}\left(\frac{d p}{d x}\right) \int_{-h / 2}^{h / 2}\left(1-\frac{4 y^{2}}{h^{2}}\right) d y\\ &=-\frac{h^{2}}{8 \mu}\left(\frac{d p}{d x}\right)\left[y-\frac{4 y^{3}}{3 h^{2}}\right]_{-n / 2}^{h / 2}\\ &=-\frac{h^{2}}{8 \mu}\left(\frac{d p}{d x}\right)\left[\left\{\frac{h}{2}-\left(-\frac{h}{2}\right)\right\}-\left\{\frac{4}{3 h^{2}}\left(\frac{h^{3}}{8}-\left(-\frac{h^{3}}{8}\right)\right)\right\}\right]\\ &=-\frac{h^{3}}{12 \mu}\left(\frac{d p}{d x}\right)\\ &Q=A V\\ &-\frac{h^{3}}{12 \mu}\left(\frac{d p}{d x}\right)=(h \times 1) \times U_{a v g}\\ &\Rightarrow \quad U_{a v g}=-\frac{h^{2}}{12 \mu}\left(\frac{d p}{d x}\right) \end{aligned}
Question 3 |
Flow rate of a fluid (density = 1000 kg/m^{3}) in a small diameter tube is 800 mm^{3}/s. The length and the diameter of the tube are 2 m and 0.5 mm, respectively. The pressure drop in 2 m length is equal to 2.0 MPa. The viscosity of the fluid is
0.025N.s/m^{2} | |
0.012N.s/m^{2} | |
0.00192N.s/m^{2} | |
0.00102N.s/m^{2} |
Question 3 Explanation:
Applying Hagen-Poiseuille equation, the drop in pressure is given by,
\begin{aligned} p_{1}-p_{2} &=\frac{32 \mu V L}{D^{2}} \\ \text{But}\quad Q &=A V \\ \Rightarrow\quad V &=\frac{Q}{A} \\ &=\frac{800 \times 4}{\pi \times(0.5)^{2}} \\ &=4074.37 \mathrm{mm} / \mathrm{s} \simeq 4.07 \mathrm{m} / \mathrm{s} \\ \therefore \quad 2 \times 10^{6} &=\frac{32 \times \mu \times 4.07 \times 2}{\left(0.5 \times 10^{-3}\right)^{2}} \\ \Rightarrow\quad \mu &=1.92 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} \\ &=0.00192 \mathrm{Ns} / \mathrm{m}^{2} \end{aligned}
\begin{aligned} p_{1}-p_{2} &=\frac{32 \mu V L}{D^{2}} \\ \text{But}\quad Q &=A V \\ \Rightarrow\quad V &=\frac{Q}{A} \\ &=\frac{800 \times 4}{\pi \times(0.5)^{2}} \\ &=4074.37 \mathrm{mm} / \mathrm{s} \simeq 4.07 \mathrm{m} / \mathrm{s} \\ \therefore \quad 2 \times 10^{6} &=\frac{32 \times \mu \times 4.07 \times 2}{\left(0.5 \times 10^{-3}\right)^{2}} \\ \Rightarrow\quad \mu &=1.92 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} \\ &=0.00192 \mathrm{Ns} / \mathrm{m}^{2} \end{aligned}
Question 4 |
An upward flow of oil (mass density 800 kg/m^3, dynamic viscosity 0.8 kg/m-s)
takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown
in the figure. The pressures at sections 1 and 2 are measured as p_1 = 435 kN/m^2
and p_2 = 200 kN/m^2.
If the flow is reversed, keeping the same discharge and the pressure at section 1 is maintained as 435 kN/m^{2}, the pressure at section 2 is equal to
If the flow is reversed, keeping the same discharge and the pressure at section 1 is maintained as 435 kN/m^{2}, the pressure at section 2 is equal to
488kN/m^{2} | |
549kN/m^{2} | |
586kN/m^{2} | |
614kN/m^{2} |
Question 4 Explanation:
When the flow is reversed, then
\begin{aligned} \frac{\rho_{1}}{\rho g}+\frac{32\mu VL}{\rho g D^{2}}&=\frac{p_{2}}{\rho g}+5 \sin45^{\circ}\\ \Rightarrow \frac{435\times 10^{3}}{800\times 9.81}&+\frac{32\times 0.8 \times 16.19 \times 5}{800\times 9.81\times (0.1)^{2}}\\ &=\frac{\rho_{2}}{800\times 9.81}+\frac{5}{\sqrt{2}}\\ \Rightarrow p_{2} &= 614.48kH/m^{2} = 614kN/m^{2} \end{aligned}
\begin{aligned} \frac{\rho_{1}}{\rho g}+\frac{32\mu VL}{\rho g D^{2}}&=\frac{p_{2}}{\rho g}+5 \sin45^{\circ}\\ \Rightarrow \frac{435\times 10^{3}}{800\times 9.81}&+\frac{32\times 0.8 \times 16.19 \times 5}{800\times 9.81\times (0.1)^{2}}\\ &=\frac{\rho_{2}}{800\times 9.81}+\frac{5}{\sqrt{2}}\\ \Rightarrow p_{2} &= 614.48kH/m^{2} = 614kN/m^{2} \end{aligned}
Question 5 |
An upward flow of oil (mass density 800 kg/m^3, dynamic viscosity 0.8 kg/m-s)
takes place under laminar conditions in an inclined pipe of 0.1 m diameter as shown
in the figure. The pressures at sections 1 and 2 are measured as p_1 = 435 kN/m^2
and p_2 = 200 kN/m^2.
The discharge in the pipe is equal to
The discharge in the pipe is equal to
0.1m^{3}/s | |
0.127m^{3}/s | |
0.144m^{3}/s | |
0.161m^{3}/s |
Question 5 Explanation:
Applying Bornouth's equation between (1) and (2)
\begin{aligned} \frac{\rho_{1}}{\rho g}&=\frac{\rho_{2}}{\rho g}+\sin 45^{\circ}+\frac{32\mu VL}{\rho gD^{2}}\\ \Rightarrow \frac{430\times 10^{3}200\times 10^{3}}{800\times }&-\frac{5}{\sqrt{2}}\\ &=\frac{32\times 0.8\times 5}{800\times 9.81 \times (0.1)^{2}}\\ \Rightarrow V&=16.19 m/s\\ \therefore Q=AV\\ &= \frac{\pi \times (0.1)^{}}{4} \times 16.19\\ &=0.127m^{3}/s \end{aligned}
\begin{aligned} \frac{\rho_{1}}{\rho g}&=\frac{\rho_{2}}{\rho g}+\sin 45^{\circ}+\frac{32\mu VL}{\rho gD^{2}}\\ \Rightarrow \frac{430\times 10^{3}200\times 10^{3}}{800\times }&-\frac{5}{\sqrt{2}}\\ &=\frac{32\times 0.8\times 5}{800\times 9.81 \times (0.1)^{2}}\\ \Rightarrow V&=16.19 m/s\\ \therefore Q=AV\\ &= \frac{\pi \times (0.1)^{}}{4} \times 16.19\\ &=0.127m^{3}/s \end{aligned}
There are 5 questions to complete.