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ivolga24 [154]
3 years ago
5

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp

/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Engineering
1 answer:
cricket20 [7]3 years ago
6 0

Answer:

Given that;

Jello there, see explanstion for step by step solving.

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

Explanation:

A horizontal channel of height H has two fluids of different viscosities and densities flowing because of a pressure gradient dp/dx1. Find the velocity profiles of two fluids if the height of the flat interface is ha.

See attachment for more clearity

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Independent auto lots usually have <u>higher</u> finance rates than dealerships

<u>Explanation:</u>

The finance rates that are charged by the dealers are lower than the finance charges that are charged by the independent auto. In case if you are getting financed through dealerships, you can also negotiate with them to charge finance rates and lower the charges of the finance.

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The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
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Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

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