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Rama09 [41]
3 years ago
8

g Calculate a better value for the convection coefficient using resources from heat transfer. Assume forced convection of the ai

r inside the chimney moving at 4 ft/s. Assume convection on the outside of the chimney is due to natural convection. Use the temperature plots or probe of the inside and outside surfaces to calculate the bulk fluid temperature and obtain accurate material properties for air based on these temperatures. Hint: Your new results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.
Engineering
1 answer:
ollegr [7]3 years ago
4 0

Answer:results for h should on the same order of magnitude of the value I provided. If it isn't, check your units. Update to you calculated values for h and resolve your model.

Explanation:

You might be interested in
The C language allows developers to pass functions as parameters to other functions. Provide a declaration for a function named
MariettaO [177]

Answer:

answer is given below

Explanation:

 Declaration for garnet  

  • void garnet(char,int (*f)(int));
  • void - garnet return nothing

garnet accepts two parameters

  1. 'Char'
  2. function

Function as a parameter

  • int (*f) (int))

and here

first int refers return type of function

  • (*f)  function as pointer variable

second int

  • arguments that accept by inner function- function as a parameter

so here Garnet is a function that accepts two parameters. One parameter that contains the character and another parameter that takes an argument (here I am taking an integer) and returns an integer. Function The function returns nothing. Zero is mentioned here.

5 0
4 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
What is the duty of a choreographer?
s2008m [1.1K]

Answer:

A Choreographer designs and directs the dance or stylized movement in musical productions, working closely with the director and musical director. A choreographer works with dancers to interpret and develop ideas and transform them into the finished performance.

Explanation:

3 0
3 years ago
Read 2 more answers
On a date when the earth was 147.4x106 km from the sun, a spacecraft parked in a 200 km altitude circular earth orbit was launch
rewona [7]

Answer:

ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

Explanation:

Distance of earth from sun = R_{2} = 147.4 \times 106 Km

Spacecraft perihelion = R_{2} = 120\times106Km

gravitational parameters are now given as

\mu_{sun} = 132.7\times 10^{9}

\mu_{earth} = 398600

radius of earth  = 6378 Km

Heliocentric spacecraft velocity at earth sphere of influence =

   V_{D}^{v} =\sqrt{2\mu_{sun}} \sqrt{\frac{R_{2} }{R_{1}(R_{1} +R_{2} ) } }

V_{D}^{v} =28.43\frac{km}{s}

Heliocentric velocity of earth = v_{earth} = 30.06\frac{km}{sec}

V_{infinity}= v_{earth}-V_{D}^{v} =30.06-28.43=1.57g\frac{km}{s}

assume

r_{p} =r_{earth} +r_{altitude} =6378 + 200 = 6578Km

Geometric spacecraft velocity of spacecraft at perigee of departure hyperbola

v_{p}=\sqrt{v^{2} _{infinity}+\frac{2\mu_{earth} }{r_{p} } } = 11.12\frac{km}{s}

geometric space craft velocity in its circular parking orbit

v_{c}=\sqrt{\frac{\mu_{earth} }{r_{p} } }  = 7.784 \frac{km}{s}

              ΔV =  =v_{p} -v_{c} = 3.337 \frac{km}{s}

7 0
4 years ago
What are the main causes of injuries when using forklifts?
Julli [10]

Answer:

1, 3, 4, 5

Explanation:

just did it

4 0
3 years ago
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