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3 years ago

The volume of microbial culture is observed to increase according to the formula

Engineering

1 answer:

saw5 [17]3 years ago

3 0

The expression of V(m³)=e^(t(s)) to make V in in³ and t in minutes is;

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

We are given that;

Volume of microbial culture is observed to increase according to the formula;

V = e^(t)

where;

t is in seconds

V is in m³

We want to now express V in in³ and t in minutes.

Now, from conversions;

1 m³ = 61024 in³

Also; 1 second = 1/60 minutes

according to formula for exponential decay, we know that;

V = ae^(bt)

Thus, we have;

61024V = ae^(¹/₆₀b(t(h))

V(in³) = (¹/₆₁₀₂₄)a $e^{\frac{1}{60}bt(h)$

Read more about subject of formula at; brainly.com/question/790938

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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

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

A construction crew lifts approximately 400 lb. of material several times during a day from a flatbed truck to a 25 ft. rooftop.

Irina18 [472]

Answer:

2ib

Explanation:

if you divide 10 divided by 2 it gives you 5 and then subtract it by 2.2 = 2.8

there goes your answer.

5 0

2 years ago

Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin

Vladimir79 [104]

Answer:

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 = 1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 = 1.013 μm

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 = 0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.243 μm

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 = 0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 = 0.608 μm

6 0

3 years ago

1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con

QveST [7]

Answer:

Explanation: Here it is: 67 Hope that helps! :)

5 0

3 years ago

A steam power plant operates on an ideal reheat- regenerative Rankine cycle and has a net power output of 80 MW. Steam enters th

trasher [3.6K]

Answer:

flow(m) = 54.45 kg/s

thermal efficiency u = 44.48%

Explanation:

Given:

- P_1 = P_8 = 10 KPa

- P_2 = P_3 = P_6 = P_7 = 800 KPa

- P_4 = P_5 = 10,000 KPa

- T_5 = 550 C

- T_7 = 500 C

- Power Output P = 80 MW

Find:

- The mass flow rate of steam through the boiler

- The thermal efficiency of the cycle.

Solution:

State 1:

P_1 = 10 KPa , saturated liquid

h_1 = 192 KJ/kg

v_1 = 0.00101 m^3 / kg

State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

h_2 = 192 + 0.00101*(790) = 192.80 KJ/kg

State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

v_3 = 0.00111 m^3 / kg

State 4:

P_4 = 10,000 KPa , constant volume process work done:

h_4 = h_3 + v_3 * ( P_4 - P_3)

h_4 = 721 + 0.00111*(9200) = 731.21 KJ/kg

State 5:

P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

- Fraction of steam y = flow(m_6 / m_3).

- Use energy balance of steam bleed and cold feed-water:

E_6 + E_2 = E_3

flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

y*h_6 + (1-y)*h_3 = h_3

y*2810 + (1-y)*192.8 = 721

Compute y: y = 0.2018

- Heat produced by the boiler q_b:

q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

Compute q_b: q_b = 3303.58 KJ/ kg

-Heat dissipated by the condenser q_c:

q_c = (1-y)*(h_8 - h_1)

q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c: q_c = 1834.26 KJ/ kg

- Net power output w_net:

w_net = q_b - q_c

w_net = 3303.58 - 1834.26

w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

flow(m) = P / w_net

compute flow(m) flow(m) = 80,000 /1469.32 = 54.45 kg/s

- Thermal efficiency u:

u = 1 - (q_c / q_b)

u = 1 - (1834.26/3303.58)

u = 44.48 %

5 0

3 years ago