Physics, 10.03.2020 07:16 manywonders

⦁ A gas-turbine power plant operating on an ideal Brayton cycle has a pressure ratio of 8. The gas temperature is 300 K at the compressor inlet and 1400 K at the turbine inlet. Utilizing the air-standard assumptions, determine (a) the gas temperature at the exits of the compressor and the turbine, (b) the back work ratio, and (c) the thermal efficiency.

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Answer from: jessejames48

The value of temperature at compressor outlet = 543.43 K

The temperature at turbine outlet = 773.04 K

Back work ratio = 0.388

The value of efficiency of the Ideal Brayton cycle =0.448

Explanation:

Pressure ratio ( $r_{p}$ )= 8 And specific heat ratio ( $\gamma$ ) = 1.4

Compressor inlet temperature ( $T_{1}$ ) = 300 K

Turbine inlet temperature ( $T_{3}$ ) = 1400 K

(a). Temperature ratio inside the compressor is given by

$\frac{T_{2} }{T_{1} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{1}$ & $T_{2}$ represents the compressor inlet and outlet temperature.

Put all the values in the given formula we get

⇒ $\frac{T_{2} }{300}$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $\frac{T_{2} }{300}$ = 1.811

⇒ $T_{2}$ = 543.43 K

This is the value of temperature at compressor outlet.

The temperature ratio inside the turbine is given by

$\frac{T_{3} }{T_{4} }$ = $r_{p}\frac{\gamma - 1}{\gamma}$

Where $T_{3}$ & $T_{4}$ represents the turbine inlet & outlet temperatures.

Put all the values in the given formula we get

⇒ $\frac{1400}{T_{4} }$ = $8^{\frac{1.4 - 1}{1.4} }$

⇒ $T_{4}$ = 773.05 K

This is the temperature at turbine outlet.

(b). The work done by the turbine is given by the formula ( $W_{T}$ ) = $C_{p}$ ( $T_{3}$ - $T_{4}$ )

Put all the values in the above formula we get $W_{T}$ = 1.005 × (1400 - 773.05)

$W_{T}$ = 630.08 $\frac{KJ}{Kg}$

And work done inside the pump is given by $W_{c}$ = $C_{p}$ ( $T_{2}$ - $T_{1}$ )

Put all the values in the above formula we get $W_{c}$ = 1.005 × (543.43 - 300)

$W_{c}$ = 244.65 $\frac{KJ}{Kg}$

Back work ratio is given by $r_{b}$ = $\frac{W_{C} }{W_{}T}$

Put the values of $W_{c}$ & $W_{T}$ in above formula we get

⇒ $r_{b}$ = $\frac{244.65}{630.08}$

⇒ $r_{b}$ = 0.388

This is the value of back work ratio.

(C). Thermal Efficiency is given by E = 1 - $\frac{1}{r_{p} ^{\frac{\gamma - 1}{\gamma} } }$

⇒ E = 1 - $\frac{1}{8 ^{\frac{1.4 - 1}{1.4} } }$

⇒ E = 0.448

This is the value of efficiency of the Ideal Brayton cycle.

Answer from: Quest

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6 $\frac{m}{s^{2} }$

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

Answer from: Quest

answer///isobars;

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