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Naddik [55]
3 years ago
12

Air is compressed steadily from 100kPa and 20oC to 1MPa by an adiabatic compressor. If the mass flow rate of the air is 1kg/s an

d the compressor efficiency is 80%, determine a) the temperature of air at the exit and b) the required power input of the compressor.
Engineering
1 answer:
igomit [66]3 years ago
6 0

Answer:

(a). 575 kJ/kg.

(b). 290kw.

Explanation:

We have the following set of information or parameters from the question above;

Pressure(1) = 100kPa, Pressure (2) = 1MPa, temperature(1) = b1= 12°C = 285K = 285kJ/kg, efficiency = 80% and the mass flow rate of the air = 1kg/s.

At a temperature of 12°C, we have the value from steam table; gx= 1.2, thus gx22 = 1.2 × (1000/100) = 12.

We have that the value for b12 = 517.

Therefore, the value for h2a can be calculated as;

80/100 = (517 - 285)/ (tp at exist) - 285.

0.8 = 232/ (tp at exist) - 285.

232 = 0.8 × (tp at exist) - 285).

232 = 0.8 (tp at exist) - 228 .

(tp at exist) = 575.

Therefore, the temperature 575 kJ/kg.

Thus, the required power input of the compressor = 1kg/s × ( 575 - 285) = 290kw.

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A circuit contains a 40 ohm resistor and a 60 ohms resistor connected in parallel. If you test this circuit with a DMM you shoul
lorasvet [3.4K]
<h2>Answer:</h2>

24Ω

<h2>Explanation:</h2>

When resistors are connected in parallel, the reciprocal of their combined resistance, when read with a DMM (Digital Multimeter - for measuring various properties of a circuit or circuit element such as resistance...) is the sum of the reciprocals of their individual resistances.

For example if two resistors of resistances R₁ and R₂ are connected together in parallel, the reciprocal of their combined resistance Rₓ is given by;

\frac{1}{R_x} = \frac{1}{R_1} + \frac{1}{R_2}

Solving for Rₓ gives;

R_{x} = \frac{R_1 * R_2}{R_1 + R_2}          ------------------(i)

From the question;

Let

R₁ = resistance of first resistor = 40Ω

R₂ =  resistance of second resistor = 60Ω

Now,

To get their combined or total resistance, Rₓ, substitute these values into equation (i) as follows;  

R_{x} = \frac{40 * 60}{40 + 60}

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R_{x} = 24 Ω

Therefore, the total resistance is 24Ω

4 0
3 years ago
A wooden cylinder (0 02 x 0 02 x 0 1m) floats vertically in water with one-third of ts length immersed. a)-Determine the density
Anuta_ua [19.1K]

Answer:

a)- the density of wood is 333.33 Kg/m³

b)-unstable condition

c)-unstable condition

Explanation:

Given data

wooden cylinder = 0.02m  x 0.02m x 0.1m

floating = 1/3 × Length

to find out

density of wood,  is it stable condition and wood would float stably in alcohol with density 700 kg/m3

Solution

First we find out density of wood

we know density of water is 1000 kg/m³

and we know wood float 1/3 of length so fraction of density will be

density of wood/ density of water = 1/3

density of wood = 1/3 ×  density of water

density of wood  = 1/3 × 1000 = 333.33 Kg/m³

Now in 2nd part we know for stable condition in partially submerged of body the metacentric height is greater than the centroid of body

we check these condition now,

metacentric height (GM)= I/v  

I = ( 0.02 × 0.02³ / 12 )  

v = ( 0.02 × 0.02 × 0.1 )

(GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

and we know centroid of body (BM) =  0.05 - 0.033 = 0.017

we know height is 0.1m so G act at 0.05 and B act at (0.1 × 0.3 ) = 0.033

we can see that now metacentric height is less than centroid of body so our body is unstable condition

Now in 3rd part we use alcohol so we calculate ratio of density of wood and density of alcohol i.e. = 333.33 / 700 = 0.48

so now our new G will be 0.05 and B will be (0.1 × 0.48 ) = 0.048

metacentric height (GM)= I/v =  ( 0.02 × 0.02³ / 12 ) / ( 0.02 × 0.02 × 0.1 ) = 0.000333

centroid of body (BM) =  0.05 - 0.048 = 0.002

we can see that now metacentric height is less than centroid of body so our body is unstable condition

5 0
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vitfil [10]

Answer:

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Explanation:

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Final temperature= 20.3 ◦C

The final temperature of stream will be 20.3 ◦C. Thechange is very small so the minnows will be able to handle this temperature.

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