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2 years ago

7

Given the inherent costs of regulation it is safe to say that there is always a negative economic impact associated with regulat

ions.
True
False

1 answer:

Alexus [3.1K]2 years ago

6 0

the answer is true. <u> </u>

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Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor i

Temka [501]

Answer:

Explanation:

Find the temperature at exit of compressor

$T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k$

Find the work done by the compressor

$\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg$

Find the actual workdone by the compressor

$\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg$

Find the temperature at exit of the turbine

$T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k$

Find the actual workdone by the turbine

$1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg$

Find the temperature of the regeneration

$\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k$

Find the heat supplied

$Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg$

Find the thermal efficiency

$n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4$

60.4%

Find the mass flow rate

$m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42$

Find the actual workdone by the compressor

$\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg$

Find the actual workdone by the turbine

$\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg$

Find the temperature of the compressor exit

$\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k$

Find the temperature at the turbine exit

$4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k$

Find the temperature of regeneration

$\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k$

6 0

3 years ago

Read 2 more answers

What is the power system?

Inessa [10]

Answer: An electric power system is a network of electrical components deployed to supply, transfer, and use electric power.

Explanation:

4 0

4 years ago

What is the built-in pollution control system in an incinerator called

Kobotan [32]

Explanation:

hbyndbnn☝️

7 0

3 years ago

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = $\frac{wet mass - dry mass }{wet mass}$ = $\frac{0.525 - 0.49}{0.525}$ = 0.07 or 7%

water content = $\frac{wet mass - dry mass}{wet mass}$ = 7%

8 0

3 years ago

Read 2 more answers

Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

$T_0 = T+\frac{V^2}{2c_p}$

Where

T = Static temperature

V = Velocity of Fluid

$c_p =$ Specific Heat

Re-arrange to find the static temperature we have that

$T = T_0 - \frac{V^2}{2c_p}$

$T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})$

$T = 672.88K$

Now the pressure of helium by using the Adiabatic pressure temperature is

$P = P_0 (\frac{T}{T_0})^{k/(k-1)}$

Where,

$P_0$ = Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

$P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}$

$P = 0.399Mpa$

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0

3 years ago