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EleoNora [17]
3 years ago
8

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

as a thermal efficiency equal to one third the thermal efficiency of R.
a. If each cycle receives the same amout of energy by heat transfer from the hot reservor, detrmine which cycle:
1. develops greater net work.
2. discharges greater energy by heat transfer to the cold reservoir

b. If each cycle develops the same net work, determine which cycle:
1. receives greater net energy by heat transfer from the hot reservoir
2. discharges greater energy by heat transfer to the cold reservoir.
Engineering
1 answer:
anygoal [31]3 years ago
4 0

Answer: Attached below is the missing diagram

answer :

A)   1) Wr > WI,     2) Qc' > Qc

B)   1) QH' > QH,   2) Qc' > Qc

Explanation:

  л = w / QH = 1 - Qc / QH  and  QH = w + Qc

<u>A) each cycle receives same amount of energy by heat transfer</u>

<u>(</u> Given that ; Л1 = 1/3 ЛR )

<em>1) develops greater bet work </em>

WR develops greater work ( i.e. Wr > WI )

<em>2) discharges greater energy by heat transfer</em>

 Qc' > Qc

solution attached below

<u>B) If Each cycle develops the same net work </u>

<em>1) Receives greater net energy by heat transfer from hot reservoir</em>

QH' > QH   ( solution is attached below )

<em>2) discharges greater energy  by heat transfer to the cold reservoir</em>

Qc' > Qc

solution attached below

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(0.845)(2610 W)t = (6.03 kg)(2.26 x 10⁶ J/kg)

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A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
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Three tool materials (high-speed steel, cemented carbide, and ceramic) are to be compared for the same turning operation on a ba
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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\%

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