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3 years ago

6

Consider a space shuttle weighing 100 kN. It is travelling at 310 m/s for 30 minutes. At the same time, it descends 2200 m. Cons

ider a drag of 12 kN on the shuttle. Calculate the work done by the shuttle engine.

1 answer:

mixas84 [53]3 years ago

6 0

Answer:

work done = 48.88 × $10^{9}$ J

Explanation:

given data

mass = 100 kN

velocity = 310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time - force ×velocity ×time - mass ×descends ) 10³ J

work done = ( 100 × 310 × 1800 - 12×310 ×1800 - 100 × 2200 ) 10³ J

work done = 48.88 × $10^{9}$ J

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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

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Answer:

a

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

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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

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

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

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