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3 years ago

What is the biggest expectation when engineers test out designs?

2 answers:

6 0

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

Gnom [1K]3 years ago

3 0

Answer: Designs need to be feasible.

Explanation: don’t know, guessed got it right

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(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

A) would the 1040 steel be a possible candidate for this application

Yield strength of 1040 steel < stress ( in order to be a possible candidate )

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

3 years ago

Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2

motikmotik

Answer:

The answer is " $112.97 \ \frac{ft}{s}$ "

Explanation:

Air flowing into the $p_1 = 20 \ \frac{lbf}{in^2}$

Flow rate of the mass $m = 230.556 \frac{lbm}{s}$

inlet temperature $T_1 = 700^{\circ} F$

Pipeline $A= 5 \times 4 \ ft$

Its air is modelled as an ideal gas Apply the ideum gas rule to the air to calcule the basic volume v:

$\to \bar{R} = 1545 \ ft \frac{lbf}{lbmol ^{\circ} R}\\\\ \to M= 28.97 \frac{lb}{\bmol}\\\\ \to pv=RT \\\\\to v= \frac{\frac{\bar{R}}{M}T}{p}$

$= \frac{\frac{1545}{28.97}(70^{\circ}F+459.67)}{20} \times \frac{1}{144}\\\\=9.8 \frac{ft3}{lb}$

$V= \frac{mv}{A}$

$= \frac{230.556 \frac{lbm}{s} \times 9.8 \frac{ft^3}{lb}}{5 \times 4 \ ft^2}\\\\= 112.97 \frac{ft}{s}$

8 0

3 years ago

A brake caliper is considered a suspension item. True False

user100 [1]

True

Suspension is the system of tires, tire air, springs, shock absorbers and linkages that connects a vehicle to its wheels and allows relative motion between the two.[1] Suspension systems must support both road holding/handling and ride quality

5 0

3 years ago

Which material has the highest cp value?

Nataly [62]

Answer: B. Water

Explanation:

6 0

2 years ago

64A geothermal pump is used to pump brine whose density is 1050 kg/m3at a rate of 0.3 m3/s from a depth of 200 m. For a pump eff

grin007 [14]

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

5 0

3 years ago