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cluponka [151]
3 years ago
9

In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temp

erature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.
Engineering
1 answer:
Fantom [35]3 years ago
8 0

Answer:

i did not known answer but anobody help you

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How high is the tallest building in the world?
nekit [7.7K]

Answer:

The Burj Khalifa in Dubai is the tallest building and its height is 2,716 feet.

Explanation:

6 0
3 years ago
You have a 12 volt power source running through a circuit that has 3kΩ of resistance, how many amps (in mA) can flow through the
siniylev [52]

Answer:

4mA

Explanation:

For this problem, we will simply apply Ohm's law:

V = IR

V/R = I

I = V / R

I = 12 volt / 3kΩ

I = 4mA

Hence, the current in the circuit is 4mA.

Cheers.

5 0
3 years ago
Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t
kodGreya [7K]

Answer:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

h = 20 \frac{W}{m^2 K} represent the heat transfer coefficient.

p represent the perimeter for this case and would be given by:

p = 2*0.05m +2*0.001m= 0.102m

k = 200 \frac{W}{m C} represent the thermal conductivity

w = 5cm =0.05 m represent the width

h = 1mm =0.001m represent the thickness

A= wh= 0.05m *0.001m = 0.00005 m^2

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}

Where m is a coefficient given by:

m = \sqrt{\frac{hp}{kA}}=\sqrt{\frac{20 W/m^2 C 0.102 m}{200 W/ mC 0.00005 m^2}}= 14.28 m^{-1}

The value of x for this case represent the distance x =5 cm =0.05m

T_b =130 C represent the base temperature

T_{\infty}= 20 represent the temperature of the sorroundings or the ambient.

If we replace we have this:

\frac{T-20}{130-20}= e^{-14.28*0.05}

And if we solve for T we got:

T= 20 + 110e^{-14.28*0.05} = 73.86 C

The answer for this case would be T = 73.86 C at 5cm from the base of the fin.

3 0
3 years ago
the increase of current when 15 V is applied to 10000ohm rheostat which is adjusted to 1000ohm value​
Anastasy [175]
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm

Required:
•). The magnitude of current= ?

•••••••••••••SOLUTION•••••••••••••

We can find the relation ship between current, voltage and resistance with the help of Ohms law.

According to ohms law;

V= IR.

Rearranging the above equation;

I= V/ R

Putt the values in the above equation; we get

I= 15V/ 1000ohm

I = 0.015 A( ampere)

••••••••••••••• CONCLUSION•••••••

The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.
4 0
3 years ago
Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and
ra1l [238]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data :

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

Determine final temperature for an Isentropic  process

T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}  ----------- ( 1 )

T2 = 303 ( \frac{450}{90})^{(1.667- 1)/1.667} =  576.9K

Work done in a piston-cylinder device can be calculated using this formula

w_{in} = c_{v} ( T2 - T1 )    ------- ( 2 )

where : cv = 3.1156 kJ/kg.k  for helium gas

             T2 = 576.9K ,    T1 = 303 K

substitute given values Back to equation 2

w_{in}  = 853.4 kJ/kg

work done in a steady flow compressor can be calculated using this

w_{in} = c_{p} ( T2 - T1 )

where : cp ( constant pressure of helium gas )  = 5.1926 kJ/kg.K

             T2 = 576.9 k , T1 = 303 K

substitute values back to equation 3

w_{in} = 1422.3 kJ / kg

4 0
3 years ago
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