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2 years ago

13

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

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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A flat, circular hydrostatic air bearing has an outer diameter of 160 mm and a 5-mm-deep recess from the 50-mm diameter to the b

OLEGan [10]

Answer:

W=15510.06

Air film thickness =1.8255×10^-4m

Torque=8.2l where l is length of bearing

Explanation:

Please find attached file for complete answer solution and explanation.

7 0

3 years ago

Pineapple contains about 20 wt% solid and the balance water. To make pineapple jam, crush

Allisa [31]

The system is exactly, specified problem and the values of all the unknows will be determined unkowns will have a unique value.

<u>Explanation:</u>

given informtion:

pineapple:

water = 80 % = 0.8

solids = 20% = 0.2

pineapple : sugar = 1:5

jam:

water = 30% = 0.3

solids and sugar = 70%

basis: 1 kg pineapple

therefore, sugar = 5 kg

the process flow sheet is as:

As per the figures shown in the attached file, DOF = 0, thus the system is exactly, specified problem and the values of all the unknows will be determined unkowns will have a unique value.

8 0

3 years ago

A bridge foundation has a nominal load capacity of 1700kN. If the resistance factor for this foundation is 0.65, what is the ult

Alekssandra [29.7K]

Answer:

The appropriate answer will be "2353.8 K". A further solution is given below.

Explanation:

The given values are:

Ultimate capacity,

= 1700kN

Resistance factor,

F.S = 0.65

According to the the LRFD, the allowable bearing pressure will be:

⇒ $Pallo w = \phi_c\times \frac{Pn}{F.S}$

On substituting the values, we get

⇒ $=0.9\times \frac{1700}{0.65}$

⇒ $=2353.8 \ K$

According to the the ASD, the allowable bearing pressure will be:

⇒ $Pallow=\frac{Pn}{\Omega \times F.S}$

On substituting the values, we get

⇒ $=\frac{1700}{1.67\times 0.65}$

⇒ $= \frac{1700}{1.0855}$

⇒ $=1845.35 \ K$

5 0

3 years ago

An engineer must design a rectangular box that has a volume of 9 m3 and that has a bottom whose length is twice its width. What

Jet001 [13]

Answer:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

Explanation:

Given

$Volume = 9m^3$

Represent the height as h, the length as l and the width as w.

From the question:

$Length = 2 * Width$

$l = 2w$

Volume of a box is calculated as:

$V = l*w*h$

This gives:

$V = 2w *w*h$

$V = 2w^2h$

Substitute 9 for V

$9 = 2w^2h$

Make h the subject:

$h = \frac{9}{2w^2}$

The surface area is calculated as:

$A = 2(lw + lh + hw)$

Recall that: $l = 2w$

$A = 2(2w*w + 2w*h + hw)$

$A = 2(2w^2 + 2wh + hw)$

$A = 2(2w^2 + 3wh)$

$A = 4w^2 + 6wh$

Recall that: $h = \frac{9}{2w^2}$

So:

$A = 4w^2 + 6w * \frac{9}{2w^2}$

$A = 4w^2 + 6* \frac{9}{2w}$

$A = 4w^2 + \frac{6* 9}{2w}$

$A = 4w^2 + \frac{3* 9}{w}$

$A = 4w^2 + \frac{27}{w}$

To minimize the surface area, we have to differentiate with respect to w

$A' = 8w - 27w^{-2}$

Set A' to 0

$0 = 8w - 27w^{-2}$

Add $27w^{-2}$ to both sides

$27w^{-2} = 8w$

Multiply both sides by $w^2$

$27w^{-2}*w^2 = 8w*w^2$

$27 = 8w^3$

Make $w^3$ the subject

$w^3 = \frac{27}{8}$

Solve for w

$w = \sqrt[3]{\frac{27}{8}}$

$w = \frac{3}{2}$

Recall that : $h = \frac{9}{2w^2}$ and $l = 2w$

$h = \frac{9}{2 * \frac{3}{2}^2}$

$h = \frac{9}{2 * \frac{9}{4}}$

$h = \frac{9}{\frac{9}{2}}$

$h = 9/\frac{9}{2}$

$h = 9*\frac{2}{9}$

$h= 2$

$l = 2w$

$l = 2 * \frac{3}{2}$

$l = 3$

Hence, the dimension that minimizes the surface area is:

$Length =3$ $Height = 2$ and $Width = \frac{3}{2}$

6 0

3 years ago

An ideal Carnot engine extracts of heat from a high-temperature reservoir at 1200 during each cycle, and rejects heat to a low-t

Brut [27]

Answer:

efficiency of the engine is 0.4167

Explanation:

given data

T high = 1200 K

T low = 700 K

solution

we get here efficiency that is express as

efficiency = $1 - \frac{T\ low}{T \ high }$ .........................1

put here value of both temperature we get efficiency

efficiency = 1 - $\frac{700}{1200}$

efficiency = 1 - 0.58333

efficiency = 0.4167

so efficiency of the engine is 0.4167

3 0

3 years ago