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3 years ago

5

find the volume of the pond with the following dimension length 40m breadth 10m height 1.2m depth 0.9m express in both meters an

d feet

1 answer:

Anna35 [415]3 years ago

8 0

Answer:

The volume for this is 29.7

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

4 years ago

marusya05 [52]

Answer:

disable yahoo from activating.

Explanation:

either force quit it or add chrome to your user bar at the bottom of the screen (if ure on a computer) if yahoo is on ur bar make sure to force quit it by right clicking and clicking "force quit" and it should stop

5 0

3 years ago

Conclude from the scenario below which type of documentation Holly should use, and explain why this would be the best choice. Ho

NARA [144]

Answer:

Okay

Explanation:

7 0

3 years ago

A keyboard, monitor and mouse are all considered as computer hardware and a program is known as a software.

Iteru [2.4K]

Answer:

This should be true because program is a software and mouse and keyboard are specific computer hardware to make your exsperience with a computer more efficent and faster.

5 0

3 years ago

Suppose you wanted to convert an AC voltage to DC. Your AC voltage source has Vrms= 60V. If you use a full wave rectifier direct

vovangra [49]

Answer:

$V_{dc}=84.15\ V$

Explanation:

Given that

Vrms= 60 V

Vf= 0.7 V

We know that peak value of AC voltage given as

$V_{o}=\sqrt{2}\ V_{rms}$

Now by putting the values

$V_{o}=60\sqrt{2}\ V$

The output voltage of the DC current given as

$V_{dc}=V_{o}-V_f$

$V_{dc}=60\sqrt{2}-0.7\ V$

$V_{dc}=84.15\ V$

Therefore output voltage of the DC current is 84.15 V.

7 0

3 years ago