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natima [27]
3 years ago
10

A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of hea

t transfer to the river water. Will the actual heat transfer rate be higher or lower than this value
Engineering
2 answers:
Serhud [2]3 years ago
5 0

Answer:

<em>A The heat transfer to the river would be 900 MW</em>

<em>B. The actual heat transfer would be lower than the theoretical heat transfer.</em>

Explanation:

Part A

For a steam power plant, the process is continuous and the law of thermodynamic is conserved. The steam power plant consists of the cold and hot reservoir and the rate of heat transfer in the steam power plant can be gotten with the expression below.

W = Q_{H} - Q_{c} ......................1

where W is the work output of the steam power plant

Q_{H}  is the heat transfer at the hot reservoir

Q_{c} is the heat transfer at the cold reservoir( heat transfer to the nearby river)

The efficiency of the steam power plant would be used to obtain the heat transfer at the hot reservoir (Q_{H}). The efficiency can be obtained with equation 2;

e = \frac{W}{Q_{H} }

e is the thermal efficiency of the steam power plant = 40 % = 0.4

W is the work output of the steam power plant = 600 MW

Q_{H}  is the heat transfer at the hot reservoir

Rearranging equation 2 to make  Q_{H}  the subject formula we have;

Q_{H}  = W/e

substituting the values we have;

Q_{H} = 600 MW/ 0.4

Q_{H} = 1500 MW

The heat transfer at the hot reservoir is 1500 MW and putting it into equation 1 to find the heat transfer to the river;

Q_{c} = Q_{H} - W

Q_{c} = 1500 MW - 600 MW

Q_{c} = 900 MW

Therefore the heat transfer to the river would be 900 MW

Part B

The actual heat transfer would be lower than the theoretical heat transfer. Losses dues to friction, head losses and heat loss by evaporation to the surroundings account for the lower value.

umka21 [38]3 years ago
3 0

Answer:

As thermal efficiency is of 40% hence rate of heat transfer is 40% of 600 MW

Rate of heat transfer=(40/100)*600 MW

Rate of heat transfer=240 MW and it is lower than 600 MW

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Write a statement that calls the recursive method backwardsAlphabet() with parameter startingLetter.
Tcecarenko [31]

Recursion refers to the act of calling a function itself. With the use of this strategy, complex problems can be reduced to more manageable, simpler ones. Recursion might be a little challenging to comprehend. The best method to figure out how it works is to experiment with it.

<h3>How to write a programme by recursive method ?</h3>

The process of making a function call itself is known as recursion. With the use of this strategy, complex problems can be reduced to more manageable, simpler ones. Recursion might be a little challenging to comprehend. Experimenting with it is the most effective way to learn how it functions.

public class Recursive Calls {

public static void backwards Alphabet(char currLetter) {

if (currLetter == 'a') {

System.out.println(currLetter);

}

else {

System.out.print(currLetter + " ");

backwards Alphabet(--currLetter);

}

return;

}

public static void main (String [] args) {

char starting Letter = '-';

starting Letter = 'z';

// Your solution goes here

backwards Alphabet(starting Letter);

return;

}

}

To learn more about recursive method refer to :

brainly.com/question/24167967

#SPJ4

6 0
1 year 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
Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C (40F). Of a joint spa
DENIUS [597]

Answer:

41.5° C

Explanation:

Given data :

1025 steel

Temperature = 4°C

allowed joint space = 5.4 mm

length of rails = 11.9 m

<u>Determine the highest possible temperature </u>

coefficient of thermal expansion ( ∝ ) = 12.1 * 10^-6 /°C

Applying thermal strain ( Δl / l )  = ∝ * ΔT

                                    ( 5.4 * 10^-3 / 11.9 )  = 12.1 * 10^-6 * ( T2 - 4 )

∴  ( T2 - 4 ) =  ( 5.4 * 10^-3 / 11.9 ) / 12.1 * 10^-6

hence : T2 = 41.5°C

8 0
3 years ago
- You have a bin wrench turning a 1/2 13 UNC bolt. You overcome 1200 lbs of resistance when you
andrew11 [14]

Fr

is my guess but yeah

6 0
3 years ago
Suppose that you and two classmates are discussing the design of a roller coaster. One classmate says that each summit must be l
Kitty [74]

Answer:

The second classmate is right.

Explanation:

The height of first summit provides the potential energy it will use to climb the following ones.

Ep = m * g * h

Where

m: mass

g: acceleration of gravity

h: height

When the train goes downwards the potential energy is converted into kinetic energy (manifested as speed) and when it climbs it consumes its kinetical energy. As long as no summit is taller than the first the train should have enough energy to climb them.

Also it must be noted that friction also consumes energy, and if the track is too lomg all the energy might be consumed by it.

8 0
3 years ago
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