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2 years ago

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Select the correct answer. Which existing technology did engineers use to enhance the speed of propeller-driven airplanes

1 answer:

Musya8 [376]2 years ago

7 0

metallurgy:

Explanation:

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Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago

A single-cylinder pump feeds a boiler through a delivery

Studentka2010 [4]

Answer:

Net discharge per hour will be 3.5325 $m^3/hr$

Explanation:

We have given internal diameter d = 25 mm

Time = 1 hour = 3600 sec

So radius $r=\frac{d}{2}=\frac{25}{2}=12.5mm=12.5\times 10^{-3}m$

We know that area is given by

$A=\pi r^2=3.14\times (12.5\times 10^{-3})^2=490.625\times 10^{-6}m^2$

We know that discharge is given by $Q=AV$ , here A is area and V is velocity

So $Q=AV=490.625\times 10^{-6}\times 2=981.25\times 10^{-6}m^3/sec$

So net discharge in 1 hour = $981.25\times 10^{-6}m^3/sec\times 3600=3.5325m^3/hour$

8 0

3 years ago

What is the function of the rotor and the stator in an AC motor

sweet-ann [11.9K]

The rotor is situated inside the stator and is mounted on the AC motor's shaft. It is the rotating part of the AC motor. And while we know this, the major function of the rotor and the stator is helping the motor shaft rotate.

7 0

3 years ago

The football team has a total of 20 jerseys. There are 6

nadezda [96]

20% of the jerseys are medium sized

3 0

3 years ago

The rate of heat transfer between a certain electric motor and its surroundings varies with time as , Q with dot on top equals n

tekilochka [14]

Answer:

the required expression is $E_2 - E_1$ = 4[ 1 - $e^{(-0.05t)}$ ]

Explanation:

Given the data in the question;

Q = -0.2[ 1 - $e^(-0.05t)$ ]

ω = 100 rad/s

Torque T = 18 N-m

Electric power input = 2.0 kW

now, form the first law of thermodynamics;

dE/dt = dQ/dt + dw/dt = Q' + w'

dE/dt = Q' + w' ------ let this be equation 1

w' is the net power on the system

w' = $w_{elect$ - $w_{shaft$

$w_{shaft$ = T × ω

we substitute

$w_{shaft$ = 18 × 100

$w_{shaft$ = 1800 W

$w_{shaft$ = 1.8 kW

so

w' = $w_{elect$ - $w_{shaft$

w' = 2.0 kW - 1.8 kW

w' = 0.2 kW

hence, from equation 1, dE/dt = Q' + w'

we substitute

dE/dt = -0.2[ 1 - $e^{(-0.05t)$ ] + 0.2

dE/dt = -0.2 + 0.2 $e^{(-0.05t)$ ] + 0.2

dE/dt = 0.2 $e^{(-0.05t)$

Now, the change in total energy, increment E, as a function of time;

ΔE = $\int\limits^t_0}\frac{dE}{dt} . dt$

ΔE = $\int\limits^t_0} 0.2e^{(-0.05t)} dt$

ΔE = $\int\limits^t_0} \frac{0.2}{-0.05} [e^{(-0.05t)}]^t_0$

$E_2 - E_1$ = 4[ 1 - $e^{(-0.05t)}$ ]

Therefore, the required expression is $E_2 - E_1$ = 4[ 1 - $e^{(-0.05t)}$ ]

5 0

2 years ago