## Snehal Bhise

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3 weeks ago
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#### Posts by Snehal Bhise

<prev • 102 results • page 1 of 11 • next >
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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 6 weeks ago by Snehal Bhise0
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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 6 weeks ago by Snehal Bhise0
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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 6 weeks ago by Snehal Bhise0
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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 6 weeks ago by Snehal Bhise0 • updated 21 days ago by Yashbeer ♦♦ 150
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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 6 weeks ago by Snehal Bhise0
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... **Topic** Compressible Fluid Flow **Marks:** 5M **Difficulty:** Medium ...
written 6 weeks ago by Snehal Bhise0 • updated 21 days ago by Yashbeer ♦♦ 150
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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 6 weeks ago by Snehal Bhise0
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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 6 weeks ago by Snehal Bhise0 • updated 21 days ago by Yashbeer ♦♦ 150
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... To find:- $\rho *=?$ $\theta=?$ Solution:- (1) Displacement thickness $\rho *=\int _0^\rho (1-\frac{u}{U})dy$ $=\int _0^\rho (1-sin (\frac{\pi y}{2\rho}))dy$ $=[y-[\frac{(\frac{\pi y}{2\rho})}{\frac{\pi}{2\rho}}]]_0^\rho$ $=[y+\frac{2\rho}{\pi}cos(\frac{\pi y}{2\rho})]_0^\rho$ $=[(\rho + \fr ... written 6 weeks ago by Snehal Bhise0 1 answer 74 views 1 answer ... The velocity profile within a laminar boundary layer over a fiat plate is given by$\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 ...