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Dmitriy789 [7]
3 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]3 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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aleksley [76]

Answer:

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Explanation:

The natural air change per hour is given by the formula:

Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume

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Therefore,

0.4 = (60 min)(Volume Flow)/(19456 ft³)

Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)

Volume Flow =  (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s

Now, we find heat loss coefficient:

Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air

Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)

<u>Uair = 0.0749 KW/k = 74.9 W/k</u>

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3 years ago
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3 years ago
If the total energy change of an system during a process is 15.5 kJ, its change in kinetic energy is -3.5 kJ, and its potential
drek231 [11]

Answer:

The change in specific internal energy is 3.5 kj.

Explanation:

Step1

Given:

Total change in energy is 15.5 kj.

Change in kinetic energy is –3.5 kj.

Change in potential energy is 0 kj.

Mass is 5.4 kg.

Step2

Calculation:

Change in internal energy is calculated as follows:

\bigtriangleup E=\bigtriangleup KE+\bigtriangleup PE+\bigtriangleup U15.5=-3.5+0+\bigtriangleup U

\bigtriangleup U=19 kj.

Step3

Specific internal energy is calculated as follows:

\bigtriangleup u=\frac{\bigtriangleup U}{m}

\bigtriangleup u=\frac{19}{5.4}

\bigtriangleup u=3.5 kj/kg.

Thus, the change in specific internal energy is 3.5 kj/kg.

7 0
4 years ago
If you have a hole diameter of 0.250 with a tolerance of ±0.005, what are the limits of the hole size?
sleet_krkn [62]

Answer:

The limits of the hole size are;

The maximum limit of the hole diameter 0.255

The minimum limit of the hole diameter = 0.245

Explanation:

Tolerance is a standardized form of language that can be used to define the intended 'tightness' or 'clearance' degree between mating parts in a mechanical assembly process and in metal joining processes such as welding and brazing processes

In tolerancing, the size used in the description of a part is known as the nominal size while allowable variation of the nominal size that will still allow the part to function properly is known as the tolerance

A tolerance given in the form ±P is known as bilateral tolerancing, with the value being added to or subtracted from the nominal size to get the maximum and minimum allowable limits of the dimensions of the nominal size

Therefore;

The given nominal dimension of the hole diameter = 0.250

The bilateral tolerance of the dimension, = ±0.005

Therefore;

The maximum limit of the diameter of the hole = 0.250 + 0.005 = 0.255

The minimum limit of the diameter of the hole = 0.250 - 0.005 = 0.245

7 0
3 years ago
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