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2 years ago

What is the physical mechanism that causes the friction factor to be higher in turbulent flow?

Engineering

1 answer:

Montano1993 [528]2 years ago

3 0

Answer:

The standard enthalpy of combustion of ethene gas, , is at . Given the following enthalpies of formation, calculate for .

Explanation:

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3 years ago

An aluminum metal rod is heated to 300oC and, upon equilibration at this temperature, it features a diameter of 25 mm. If a tens

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3 years ago

Read 2 more answers

There are three options for heating a particular house: a. Gas: $1.33/therm where 1 therm=105,500 kJ b. Electric Resistance: $0.

sergejj [24]

Answer:

Option ‘a’ is the cheapest for this house.

Explanation:

Cheapest method of heating must have least cost per kj of energy. So, convert all the energy in the same unit (say kj) and take select the cheapest method to heat the house.

Given:

Three methods are given to heat a particular house are as follows:

Method (a)

Through Gas, this gives energy of amount $1.33/therm.

Method (b)

Through electric resistance, this gives energy of amount $0.12/KWh.

Method (c)

Through oil, this gives energy of amount $2.30/gallon.

Calculation:

Step1

Change therm to kj in method ‘a’ as follows:

$C_{1}=\frac{\$ 1.33}{therm}\times(\frac{1therm}{105500kj})$

$C_{1}=1.2606\times10^{-5}$ $/kj.

Step2

Change kWh to kj in method ‘b’ as follows:

$C_{2}=\frac{\$ 0.12}{kWh}\times(\frac{1 kWh }{3600kj})$

$C_{2}=3.334\times10^{-5}$ $/kj.

Step3

Change kWh to kj in method ‘c’ as follows:

$C_{3}=\frac{\$ 2.30}{gallon}\times(\frac{1 gallon }{138500kj})$

$C_{3}=1.66\times10^{-5}$ $/kj.

Thus, the method ‘a’ has least cost as compare to method b and c.

So, option ‘a’ is the cheapest for this house.

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3 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

$=0.3324 \ m^2$

$F=\rho g A h$

$=7200 \times 9.81 \times 0.3324 \times 0.3$

= 7043.42 N

3 0

3 years ago