## User: Snehal Bhise

Snehal Bhise •

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#### Posts by Snehal Bhise

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... Data:-
$u=1.5 poise=0.15 N-s/m^2$
$S_{oil}=0.9,L=20m$
$D=3cm=3\times 10^{-2}m$
$P_A=200kPa=200\times 10^3N/m^2$
$P_B=500kPa=500\times 10^3N/m^2$
To find:-
(a) Direction of flow=?
(b)Rate of flow=?
Solution:-
(a)Direction of flow:
Total energy at point A
$T.E_A=\frac{P_A}{S_{oil}.9}+Z_A$
...

written 5 weeks ago by
Snehal Bhise •

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... An oil of dynamic viscosity 1.5 poise and relative density 0.9 flows through a 3 cm diameter vertical pipe. Two pressure guages are fixed 20 m apart. The guage A fixed at the top records 200kPa and the guage B fixed at the bottom records 500 kPa. Find the direction of flow and rate of flow. ...

written 5 weeks ago by
Snehal Bhise •

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... Data:-
Upstream conditions:
$v_1=660m/s, P_1=110kPa=100\times 10^3 N/m^2$
$k=1.4, T_1=-20^\circ C=-20+273=253 K, R=287 J/kgK$
To find:-
$P_2=?,V_2=?,T_2=?$
Solution:-
![enter image description here][1]
Sonic speed at upstream
$c_1=\sqrt{K.R.T_1}=\sqrt{1.9\times 287\times 253}=318.83 m/s$
U ...

written 5 weeks ago by
Snehal Bhise •

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... Calculate pressure, velocity and temperature just downstream of the shock wave. [Take ratio of specific heat k=1.4 and gas constant R=287 J/(kg.k)]. ...

written 5 weeks ago by
Snehal Bhise •

**0**• updated 18 days ago by Yashbeer ♦♦**150**0

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... Mach Number:-
It is defined as the ratio of actual velocity v to the sonic velocity c
$\therefore M=\frac{v}{c}$
when
$M\gt 1$.......Flow is supersonic
$M\lt 1$.......Flow is subsonic
$M=1$.......Flow is sonic
Stagnation pressure:-
The stagnation pressure $P_0$ is related to mach number and ...

written 5 weeks ago by
Snehal Bhise •

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... **Topic** Compressible Fluid Flow
**Marks:** 5M
**Difficulty:** Medium ...

written 5 weeks ago by
Snehal Bhise •

**0**• updated 18 days ago by Yashbeer ♦♦**150**0

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... Data:-
$P_1=60kPa=60\times 10^3 Pa,P_2=78.8\times 10^3 Pa$
$T_1=27^\circ=300 K,V_1=486 m/s,A_1=0.02 m^2$
To find:-
$(i) M_2=?$
$(ii) \text{Type of nozzle=?}$
Solution:-
At section (1) since velocity
$C_1=\sqrt{\gamma.R.T_1}=\sqrt{1.4\times 287\times 300}=347.188 m/sec$
Mach no. at section ( ...

written 5 weeks ago by
Snehal Bhise •

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... At section 2, further downstream, the pressure is 78.8 kPa(abs). Assuming isotropic flow. Calculate the mach no. at section 2. Also identify the type of nozzle. ...

written 5 weeks ago by
Snehal Bhise •

**0**• updated 18 days ago by Yashbeer ♦♦**150**0

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Answer:
A: Module 5-Boundary Layer Flows

... To find:-
$\rho *=?$ $\theta=?$
Solution:-
(1) Displacement thickness
$\rho *=\int _0^\rho (1-\frac{u}{U})dy$
$=\int _0^\rho (1-sin (\frac{\pi y}{2\rho}))dy$
$=[y-[\frac{(\frac{\pi y}{2\rho})}{\frac{\pi}{2\rho}}]]_0^\rho$
$=[y+\frac{2\rho}{\pi}cos(\frac{\pi y}{2\rho})]_0^\rho$
$=[(\rho + \fr ...

written 5 weeks ago by
Snehal Bhise •

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... The velocity profile within a laminar boundary layer over a fiat plate is given by
$\frac{u}{U}=sin (\frac{\pi y}{2\rho})$
where, 'U' is the mean stream velocity and $'\rho'$ is the boundary layer thickness. Determine
(1) Displaced thickness
(2) Momentum thickness ...

written 5 weeks ago by
Snehal Bhise •

**0**#### Latest awards to Snehal Bhise

Centurion
5 weeks ago,
created 100 posts.

Rising Star
5 weeks ago,
created 50 posts within first three months of joining.