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PtichkaEL [24]
3 years ago
11

Air enters the compressor of an ideal cold air-standard Brayton cycle at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The co

mpressor pressure ratio is 7, and the turbine inlet temperature is 1200 K. For constant specific heats with k = 1.4 and Cp = 1.005 kJ/kg, calculate the percent thermal efficiency (enter a number only)
Engineering
1 answer:
ICE Princess25 [194]3 years ago
7 0

Answer:

The thermal efficiency of cycle is 42.6%.

Explanation:

Given that

T_1=300 K

P_1=100KPa

mass flow rate = 6 kg/s

Compression ratio = 7

Turbine inlet temperature = 1200 K

γ=1.4

We know that thermal efficiency of Brayton cycle given as

\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}

Now by putting the values

\eta=1-\dfrac{1}{r_p^{\frac{\gamma-1}{\gamma}}}

\eta=1-\dfrac{1}{7^{\frac{1.4-1}{1.4}}}

η=0.426

So the thermal efficiency of cycle is 42.6%.

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A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

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3 years ago
What is considered the greatest engineering achievement of the 20th century?
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6 0
2 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

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3 years ago
Major processing methods for fiberglass composited include which of the following? Mark all that apply) a)- Open Mold b)- Closed
Novay_Z [31]

Answer:

it is f all of the above

Explanation:

let me know if im right

im not positive if im right but i should be right

4 0
3 years ago
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