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blondinia [14]
3 years ago
10

Determine the maximum height (in inches) that a lift pump can raise water (0.9971 g/ml) from a well at normal atmospheric pressu

re. Also present the cancellation of units to arrive at the final unit? Pls help. with solution​
Engineering
1 answer:
solniwko [45]3 years ago
7 0

Answer:Height = 30 inches​

Mass of water / Volume of water x Gravity constant for Earth x Density of Water. This cancels out to give us mass/volume, which is then cancelling mass, giving us the final unit of length/area (in this case inches).​

Explanation:

Height = 30 inches.

Units cancelled:

30/30 x 9.8 x 0.9971 = 30/39.97

The final units are in inches!​

2. Determine the weight (in pounds) that an ice cube at 32°F will displace when placed in water at 75°F.

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = Weight displaced by ice cube in pounds​

Weight of ice cube / Volume of water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 0.5/22.67

The final units are in pounds!​

3. Determine the mass (in grams) of an ice cube at 32°F when placed in water at 75°F (assume no change in volume).

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = Mass of ice cube in grams

Mass of Ice Cube / Volume of Water x Gravity constant for Earth x Density of Ice = 0.5/75 x 32 x 0.92 = 15/22.67

The final units are in grams!​

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Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed
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Answer:

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Explanation:

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A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:
Yakvenalex [24]

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You set a sensor to transmit the wind speed on top of a mountain. Which of these languages are you most likely to have used?
kifflom [539]

The programming language that is most likely used to transmit the wind speed is: B. SQL.

<h3>What is SQL?</h3>

SQL is an acronym for structured query language and it can be defined as a domain-specific programming language that is designed and developed for the management of various data that are saved in a relational or structured database.

This ultimately implies that, a structured query language (SQL) can be used to communicate with a database in accordance with the American National Standards Institute (ANSI) standards.

In conclusion, the programming language that is most likely used to transmit the wind speed is SQL.

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8 0
2 years ago
It is appropriate to use the following yield or failure criterion for ductile materials (a) Maximum shear stress or Tresca crite
Nataly [62]

Answer:

(b)Distortion energy theory.

Explanation:

The best suitable theory for ductile material:

       (1)Maximum shear stress theory (Guest and Tresca theory)

It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.

      (2)Maximum distortion energy theory(Von Mises henkey's        theory)

It states that maximum shear train energy per unit volume at any point  is equal to strain energy per unit volume under the state of uni axial stress condition.

But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.

6 0
3 years ago
Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
Tpy6a [65]

Answer:

Among all three tools, the ceramic tool is taking the least time for the production of a batch, however, machining from the HSS tool is taking the highest time.

Explanation:

The optimum cutting speed for the minimum cost

V_{opt}= \frac{C}{\left[\left(T_c+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]^n}\;\cdots(i)

Where,

C,n = Taylor equation parameters

T_h =Tool changing time in minutes

C_e=Cost per grinding per edge

C_m= Machine and operator cost per minute

On comparing with the Taylor equation VT^n=C,

Tool life,

T= \left[ \left(T_t+\frac{C_e}{C_m}\right)\left(\frac{1}{n}-1\right)\right]}\;\cdots(ii)

Given that,  

Cost of operator and machine time=\$40/hr=\$0.667/min

Batch setting time = 2 hr

Part handling time: T_h=2.5 min

Part diameter: D=73 mm =73\times 10^{-3} m

Part length: l=250 mm=250\times 10^{-3} m

Feed: f=0.30 mm/rev= 0.3\times 10^{-3} m/rev

Depth of cut: d=3.5 mm

For the HSS tool:

Tool cost is $20 and it can be ground and reground 15 times and the grinding= $2/grind.

So, C_e= \$20/15+2=\$3.33/edge

Tool changing time, T_t=3 min.

C= 80 m/min

n=0.130

(a) From equation (i), cutting speed for the minimum cost:

V_{opt}= \frac {80}{\left[ \left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]^{0.13}}

\Rightarrow 47.7 m/min

(b) From equation (ii), the tool life,

T=\left(3+\frac{3.33}{0.667}\right)\left(\frac{1}{0.13}-1\right)\right]}

\Rightarrow T=53.4 min

(c) Cycle time: T_c=T_h+T_m+\frac{T_t}{n_p}

where,

T_m= Machining time for one part

n_p= Number of pieces cut in one tool life

T_m= \frac{l}{fN} min, where N=\frac{V_{opt}}{\pi D} is the rpm of the spindle.

\Rightarrow T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 47.7}=4.01 min/pc

So, the number of parts produced in one tool life

n_p=\frac {T}{T_m}

\Rightarrow n_p=\frac {53.4}{4.01}=13.3

Round it to the lower integer

\Rightarrow n_p=13

So, the cycle time

T_c=2.5+4.01+\frac{3}{13}=6.74 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times6.74+\frac{3.33}{13}=\$4.75/pc

(e) Total time to complete the batch= Sum of setup time and production time for one batch

=2\times60+ {50\times 6.74}{50}=457 min=7.62 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times4.01}{457}=0.4387=43.87\%

Now, for the cemented carbide tool:

Cost per edge,

C_e= \$8/6=\$1.33/edge

Tool changing time, T_t=1min

C= 650 m/min

n=0.30

(a) Cutting speed for the minimum cost:

V_{opt}= \frac {650}{\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]^{0.3}}=363m/min [from(i)]

(b) Tool life,

T=\left[ \left(1+\frac{1.33}{0.667}\right)\left(\frac{1}{0.3}-1\right)\right]=7min [from(ii)]

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

T_m= \frac{\pi D l}{fV_{opt}}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 363}=0.53min/pc

n_p=\frac {7}{0.53}=13.2

\Rightarrow n_p=13 [ nearest lower integer]

So, the cycle time

T_c=2.5+0.53+\frac{1}{13}=3.11 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times3.11+\frac{1.33}{13}=\$2.18/pc

(e) Total time to complete the batch=2\times60+ {50\times 3.11}{50}=275.5 min=4.59 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.53}{275.5}=0.0962=9.62\%

Similarly, for the ceramic tool:

C_e= \$10/6=\$1.67/edge

T_t-1min

C= 3500 m/min

n=0.6

(a) Cutting speed:

V_{opt}= \frac {3500}{\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]^{0.6}}

\Rightarrow V_{opt}=2105 m/min

(b) Tool life,

T=\left[ \left(1+\frac{1.67}{0.667}\right)\left(\frac{1}{0.6}-1\right)\right]=2.33 min

(c) Cycle time:

T_c=T_h+T_m+\frac{T_t}{n_p}

\Rightarrow T_m=\frac{\pi \times 73 \times 250\times 10^{-6}}{0.3\times 10^{-3}\times 2105}=0.091 min/pc

n_p=\frac {2.33}{0.091}=25.6

\Rightarrow n_p=25 pc/tool\; life

So,

T_c=2.5+0.091+\frac{1}{25}=2.63 min/pc

(d) Cost per production unit:

C_c= C_mT_c+\frac{C_e}{n_p}

\Rightarrow C_c=0.667\times2.63+\frac{1.67}{25}=$1.82/pc

(e) Total time to complete the batch

=2\times60+ {50\times 2.63}=251.5 min=4.19 hr.

(f) The proportion of time spent actually cutting metal

=\frac{50\times0.091}{251.5}=0.0181=1.81\%

3 0
4 years ago
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