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Nataly_w [17]
3 years ago
11

Calculate the electroosmotic velocity of an aqueous solution through a glass capillary 5 cm long with a 0.5 mm internal diameter

under a potential difference of 100 V. The zeta potential at the capillary wall is 80 mV
Engineering
1 answer:
natita [175]3 years ago
8 0

Answer:

Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

Explanation:

We have given applied voltage v = 100 volt

Length of capillary L = 5 mm = 0.005 m

Zeta potential of the capillary surface \xi =80mV=0.08volt

Dielectric constant of glass is between 5 to 10 here we are taking dielectric constant as \epsilon =10

Viscosity of glass is \eta =10^8

Electroosmotic velocity is given as v_{eo}=\frac{\epsilon \xi }{\eta }\times \frac{v}{L}

v_{eo}=\frac{10\times 0.08 }{10^8 }\times \frac{100}{0.005}=1.6\times 10^{-4}m/sec

So Electroosmotic velocity will be equal to 1.6\times 10^{-4}m/sec

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h_{1} = h_{f} @ 410kPa = 273.01kJ/kg\\v_{1} = v_{f} @ 410kPa = 0.001842 m^{3}/kgw_{p,in} =  \frac{v_{1}(P_{2}-P_{1})   }{n_{p} } \\\\= \frac{0.01842(3250-410)}{0.9} \\\\ =5.81kJ/kg\\h_{2} =h_{1} + w_{p,in}\\          = 273.01+5.81\\           = 278.82 kJ/kg\\\\w_{T,out} = m^{.}  (h_{3} -h_{4} )\\=(305.6)(761.54-689.74)\\=305.6(71.8)\\=21,942kW\\\\

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c. The thermal efficiency of the cycle  n_{th}  =\frac{W^{.} _{net} }{Q^{._{in} } } \\\\= \frac{20,166}{162,656} \\=0.124\\=12.4%

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