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3 years ago

3. Determine the most unfavorable arrangement of the crane loads and

Engineering

1 answer:

Gnoma [55]3 years ago

3 0

Answer:

There are 350cm in 3.5m.

Explanation:

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Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$ ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0

3 years ago

How much does 1 gallon of water weigh in pound given that the density of water is 1gram/ cm3

MAXImum [283]

Explanation:

There are 8.35 pounds in a gallon of water. Water weighs 1 gram per cubic centimeter or 1 000 kilogram per cubic meter, i.e. density of water is equal to 1 000 kg/m³; at 25°C (77°F or 298.15K) at standard atmospheric pressure.

6 0

3 years ago

An angle grinder is best suited for use with which material?.

Kaylis [27]

Answer:

Angle grinders are used mostly for copper, iron, steel, lead, and other metals.

Explanation:

I hope it helps! Have a great day!

Lilac~

4 0

2 years ago

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where

Harrizon [31]

Answer:

a. 430.944 pascal

b. 0.0625psi

c. 1.73008inH20

Explanation:

The pressure rise Ap associated with wind hitting a win- dow of a building can be estimated using the formula Ap-p(12/2), where p is density of air and V is the speed of the wind. Apply the grid method to calculate pressure rise for P-1.2 kg/m and V-60 mph. a. Express your answer in pascals. b. Express your answer in pounds-force per square inch (psi). c. Express your answer in inches of water column

checking the dimensional consistency

Dp= $\frac{kg}{m^3} (\frac{m^2}{s^2} \\\frac{x}{y} \\\\\frac{kg}{ms^2} \\which s the same as\\ \frac{N}{m^2}$

$Dp=∈(\frac{v^2}{2} )$

convert 1 mile to meter

1mile=1609m

1h=3600s

60mile/h=26.8m/s

slotting intpo the relation

$Dp=1.2kg/m^3(\frac{26.8^2}{2} )$

430.944kg/(ms^2)

which is the same as 430.944N/m^2

expressing in pascal.

We know that

1 pascal=1 N/m^2

430.944 pascal

2. 1 pascal=0.000145psi

answer=0.0625psi

3.1 pascal=0.00401inH20

answer=430.944*0.0040146

1.73008inH20

5 0

3 years ago

1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un

ElenaW [278]

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

Bearing life: $L_1 = L , L_2 = 2L$

Catalog rating: $C_1 = C , C_2 = ? ,$

From given equation bearing life equation,

$F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)$

we Dividing eqn (2) with (1)

$\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C$

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

$R_1 = 0.9 , R_2 = 0.99$

Now calculating life adjustment factor for both value of reliability from Weibull parametres

$a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}$

$= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968$

Similarly

$a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215$

Now calculating bearing life for each value

$L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L$

Now using given ball bearing life equation and dividing each other similar to previous problem

$\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C$

Catalog rating increased by factor of 0.61

6 0

3 years ago