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Dmitriy789 [7]
2 years ago
12

An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown

liquid is 1.5 m and the depth of the oil (specific weight = 8.5 kN/m3) floating on top is 5.0 m. A pressure gage connected to the bottom of the tank reads 65 kPa. What is the specific gravity of the unknown liquid?
Engineering
1 answer:
barxatty [35]2 years ago
4 0

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>  

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nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

firstly we calculate the ; power, angular velocity and torque of the tube

power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

Torque (T) = power / angular velocity = 100000 / 213.71 = 467.92 N.m/s

next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

    39.06 * 10^-8 - 12.402 * 10^-8 =26.66 *10^-8

c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

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2 years ago
A 50-kg iron block and 20-kg copper block, both initially at 80 oC, are dropped into a large lake at 15 oC. Thermal equilibrium
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An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening
Viktor [21]

Given:

Pressure, P_{1} = 1300 kPa

Temperature,  T_{1} = 500^{\circ}

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

velocity, v = 40 m/s

A = 1m^{2}

Solution:

For air propertiess at

P_{1} = 1300 kPa

T_{1} = 500^{\circ}

h_{1} = 793kJ/K

v_{1} = 0.172\frac{m^{3}}{kg}

and also at

P_{2} = 100 kPa

T_{2} = 127^{\circ}  

h_{2} = 401 KJ/K

v_{2} =  1.15\frac{m^{3}}{kg}

a) Mass flow rate is given by:

m' = \frac{Av}{v_{1}}

Now,

m = \frac{0.2\times 40}{0.172} = 46.51 kg/s

b) for the power produced by turbine, P = m'(h_{1} - h_{2})

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5 0
2 years ago
A conical enlargement in a vertical pipeline is 5 ft long and enlarges the pipe diameter from 12 in. to 24 in. diameter. Calcula
makkiz [27]

Answer:

F_y = 151319.01N = 15.132 KN

Explanation:

From the linear momentum equation theory, since flow is steady, the y components would be;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

We are given;

Length; L = 5ft = 1.52.

Initial diameter;d1 = 12in = 0.3m

Exit diameter; d2 = 24 in = 0.6m

Volume flow rate of water; Q2 = 10 ft³/s = 0.28 m³/s

Initial pressure;p1 = 30 psi = 206843 pa

Thus,

initial Area;A1 = π•d1²/4 = π•0.3²/4 = 0.07 m²

Exit area;A2 = π•d2²/4 = π•0.6²/4 = 0.28m²

Now, we know that volume flow rate of water is given by; Q = A•V

Thus,

At exit, Q2 = A2•V2

So, 0.28 = 0.28•V2

So,V2 = 1 m/s

When flow is incompressible, we often say that ;

Initial mass flow rate = exit mass flow rate.

Thus,

ρ1 = ρ2 = 1000 kg/m³

Density of water is 1000 kg/m³

And A1•V1 = A2•V2

So, V1 = A2•V2/A1

So, V1 = 0.28 x 1/0.07

V1 = 4 m/s

So, from initial equation of y components;

-V1•ρ1•V1•A1 + V2•ρ2•V2•A2 = P1•A1 - P2•A2 - F_y

Where F_y is vertical force of enlargement pressure and P2 = 0

Thus, making F_y the subject;

F_y = P1•A1 + V1•ρ1•V1•A1 - V2•ρ2•V2•A2

Plugging in the relevant values to get;

F_y = (206843 x 0.07) + (1² x 1000 x 0.07) - (4² x 1000 x 0.28)

F_y = 151319.01N = 15.132 KN

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