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2 years ago

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A student proposes a complex design for a steam power plant with a high efficiency. The power plant has several turbines, pumps,

and feedwater heaters. Steam enters the first turbine at T1 (the highest temperature of the cycle) and saturated liquid exits the condenser at T7 (the lowest temperature of the cycle). The rate of heat transfer to the boiler (the only energy input to the system)is Qb. Determine the maximum possible efficiency and power output for this complex steam power plant design.

1 answer:

lozanna [386]2 years ago

6 0

Answer:

Hello your question lacks some values here are the values

T1 = 500⁰c, T7 = 70⁰c, Qb = 240000 kj/s

answer : A) 56%

B) 134400 kw ≈ 134.4 Mw

Explanation:

Given values

T1 (tmax) = 500⁰c = 773 k

T7(tmin) = 70⁰c = 343 k

Qb = 240000 kj/s

A) Determine the maximum possible efficiency

$n_{max}$ = 1 - $\frac{tmin}{tmax}$ * 100

= 1 - ( 343 / 773 )

= 1 - 0.44 = 0.5562 * 100 ≈ 56%

B) Determine the power output for this complex steam power plant design

$p_{out}$ = Qb * max efficiency

= 240000 kj/s * 56%

= 240000 * 0.56 = 134400 kw ≈ 134.4 Mw

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A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos

Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

$T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }$

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\$

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

$T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\$

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0

2 years ago

What does a manufacturing engineer do?

Lisa [10]

Answer: Manufacturing engineers may be responsible for solving production problems, conducting cost-benefit assessments, or designing and manufacturing goods and systems using computer-aided design software. Plant engineers and process engineers are two terms used by professionals in this industry. Manufacturing engineers are in charge of new and existing production lines' technical management, maintenance, and development. Employers are looking for people that are commercially aware and have good technical and analytical skills. They are in charge of keeping production costs low while preserving the product or service's quality, and they have considerable project expertise and insight.

Explanation: See above.

I hope this helps.

6 0

2 years ago

Read 2 more answers

1. Sewage-treatment plant, a large concrete tank initially contains 440,000 liters liquid and 10,000 kg fine suspended solids. T

Elenna [48]

Answer:

Concentration = 10.33 kg/m³

Explanation:

We are given;

Mass of solids; 10,000 kg

Volume; V = 440,000 L = 440 m³

Rate at which water is pumped out = 40,000 liter/h

Thus, at the end of 5 hours we amount of water that has been replaced with fresh water is = 40,000 liter/h x 5 hours = 200,000 L = 200 m³

Now, since the tank is perfectly mixed, therefore we can calculate a ratio of fresh water to sewage water as;

200m³/440m³ = 5/11

Thus, the amount left will be calculated by multiplying that ratio by the amount of solids;

Thus,

Amount left; = 10000 x (5/11) = 4545 kg

The concentration would be calculated by:

Concentration = amount left/initial volume

Thus,

Concentration = 4545/440 = 10.3 kg/m³

8 0

3 years ago

Use Routh's stability criterion to determine how many roots with positive real parts the following equations have:

Pavlova-9 [17]

Answer:

a) no roots not in LHP

b) 2 roots not in LHP

c) 2 roots not in the LHP

d) 2 roots not in the LHP

e) 2 roots not in LHP

Explanation:

$a) s^4 + 8s^3 + 32s^2 + 80s + 100 = 0\\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:32\:\:\:\:\:\:100\\s^3:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\\s^2:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:100\\s^1:\:\:\:80-\frac{800}{22} =43.6\\s^0:\:\:\:100$

No roots not in the LHP

$b) s^5 + 10s^4 + 30s^3 + 80s^2+344s + 480 =0 \\\\s^5:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:30\:\:\:\:\:\:344\\s^4:\:\:\:10\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:80\:\:\:\:\:\:480\\s^3:\:\:\:22\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:296\\s^2:\:\:\:-545\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:480\\s^1:\:\:\:490\\s^0:\:\:\:480$

2 roots not in the LHP

$c) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^3:\:\:\:2\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-2\\s^2:\:\:\:8\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:8\\s^1:\:\:\:-4\\s^0:\:\:\:8$

There are roots in the RHP (not all coefficients are greater than 0).

2 roots not in the LHP

$d) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^3:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:20\\s^2:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:78\\s^1:\:\:\:-58\\s^0:\:\:\:78$

There are two sign changes in the first column of the Routh array.

2 roots not in the LHP

$e) s^4 + 2s^3 + 7s^2 -2s + 8 = 0 \\\\s^4:\:\:\:1\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^3:\:\:\:4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:12\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: new \:\:row \\s^2:\:\:\:3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:25\\s^1:\:\:\:12-\frac{100}{3}=-21.3 \\s^0:\:\:\:25$

2 roots not in LHP

check:

$a (s) = 0$ ⇒

$s^2 = -3 \limits^+_- 4j = 5e^{j(\pi \limits^+_- 0.92)}\\\\s = \sqrt5 e^{j( \frac{\pi}{2} \limits^+_- 0.46)+n\pi j},\:\:\:\:\: n= 0, 1\\$

3 0

3 years ago

A ramjet operates by taking in air at the inlet, providing fuel for combustion, and exhausting the hot air through the exit. Th

UNO [17]

Answer:

15300 N

Explanation:

$\rho_i$ = Density of air at inlet

$\dfrac{m}{t}$ = Mass flow rate = 60 kg/s

$v_i$ = Inlet velocity = 225 m/s

$\rho_o$ = Density of gas at outlet = $0.25\ \text{kg/m}^3$

$A_i$ = Inlet area

$A_o$ = Outlet area = $0.5\ \text{m}^2$

Since mass flow rate is the same in the inlet and outlet we have

$\rho_iv_iA_i=\rho_ov_oA_o\\\Rightarrow v_o=\dfrac{\dfrac{m}{t}}{\rho_oA_o}\\\Rightarrow v_o=\dfrac{60}{0.25\times 0.5}\\\Rightarrow v_o=480\ \text{m/s}$

Thrust is given by

$F=\dfrac{m}{t}(v_o-v_i)\\\Rightarrow F=60\times (480-225)\\\Rightarrow F=15300\ \text{N}$

The thrust generated is 15300 N.

8 0

3 years ago