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netineya [11]
3 years ago
7

A 1.9-mm-diameter tube is inserted into an unknown liquid whose density is 960 kg/m3, and it is observed that the liquid rises 5

mm in the tube, making a contact angle of 15°. Determine the surface tension of the liquid.
Engineering
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

a 9mm dimeter tube is inserted into un known whose density is 960kg/m3 and it is obsered that the liquid rises 5mm inthe tube,making a contact angle of15. determin the surface tention of the fluid.

Explanation:

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A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t
gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

7 0
3 years ago
Consider the diffusion of water vapor through a polypropylene (PP) sheet 1 mm thick. The pressures of H2O at the two faces are 3
Neko [114]

Answer:

\boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

Explanation:

Diffusion flux of a gas, J is given by

J=P_m\frac {\triangle P}{\triangle x} where P_m is permeability coefficient, \triangle P is pressure difference and x is thickness of membrane.

The pressure difference will be 10,000 Pa- 3000 Pa= 7000 Pa

At 298 K, the permeability coefficient of water vapour through polypropylene sheet is 38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)

Since the thickness of sheet is given as 1mm= 0.1 cm then

J=38\times 10^{-13}(cm^{3}. STP)(cm)/(cm^{2}.s.Pa)\times \frac {7000 pa}{0.1cm}=0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}

Therefore, the diffusion flux is \boxed{0.000000266 \frac {cm^{3}.STP}{cm^{2}.s}}

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