ENGINEERING COLLEGE

Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.

Answers

Answer 1

Answer:

Answer:

The power input, in kW is -86.396 kW

Explanation:

Given;

initial pressure, P₁ = 1.05 bar

final pressure, P₂ = 12 bar

initial temperature, T₁ = 300 K

final temperature, T₂ = 400 K

Heat transfer, Q = 6.5 kW

volumetric flow rate, V = 39 m³/min = 0.65 m³/s

mass of air, m = 28.97 kg/mol

gas constant, R = 8.314 kJ/mol.k

R' = R/m

R' = 8.314 /28.97 = 0.28699 kJ/kg.K

Step 1:

Determine the specific volume:

p₁v₁ = RT₁

$v_1 = (R'T_1)/(p_1) = ((0.28699.(kJ)/(kg.K) )(300 k))/((1.05 bar *\ (10^5 N/m^2)/(1 bar) *(1kJ)/(1000N.m) )) \n\nv_1 = 0.81997 \ m^3/kg$

Step 2:

determine the mass flow rate; m' = V / v₁

mass flow rate, m' = 0.65 / 0.81997

mass flow rate, m' = 0.7927 kg/s

Step 3:

using steam table, we determine enthalpy change;

h₁ at T₁ = 300.19 kJ /kg

h₂ at T₂ = 400.98 kJ/kg

Δh = h₂ - h₁

Δh = 400.98 - 300.19

Δh = 100.79 kJ/kg

step 4:

determine work input;

W = Q - mΔh

Where;

Q is heat transfer = - 6.5 kW, because heat is lost to surrounding

W = (-6.5) - (0.7927 x 100.79)

W = -6.5 -79.896

W = -86.396 kW

Therefore, the power input, in kW is -86.396 kW

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In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length of cracks is 62 m and the total internal volume is 210 m3 . Due to the wind, 9.4 x 10-5 kg/s of air enters per meter of crack and exits up a chimney. Assume air temperature is the same inside and out and air density is constant at 1.186 kg/m3 . If windows and doors are not opened or closed, estimate the time required for one complete air change in the building.

Answers

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 * 10^(-5) kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4* 10^(-5) * 62 = 582.8* 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8* 10{-5}}{1.156} = 5.04* 10^(-3) m^3/s$

Time period $= (210)/(5.04* 10^(-3)) = 41654.08 s$

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating

Answers

Answer:

The answer is below

Explanation:

A 4-pole, 3-phase induction motor operates from a supply whose frequency is 60 Hz. calculate: 1- the speed at which the magnetic field of the stator is rotating. 2- the speed of the rotor when the slip is 0.05. 3- the frequency of the rotor currents when the slip is 0.04. 4- the frequency of the rotor currents at standstill.

Given that:

number of poles (p) = 4, frequency (f) = 60 Hz

1) The synchronous speed of the motor is the speed at which the magnetic field of the stator is rotating. It is given as:

$n_s=(120f)/(p)=(120*60)/(4)=1800\ rpm$

2) The slip (s) = 0.05

The speed of the motor (n) is the speed of the rotor, it is given as:

$n=n_s-sn_s\n\nn=1800-0.05(1800)\n\nn=1800-90\n\nn=1710\ rpm$

3) s = 0.04

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\n\nf_r=0.04*60\n\nf_r=2.4\ Hz$

4) At standstill, the motor speed is zero hence the slip = 1:

$s=(n_s-n)/(n_s)\n \nn=0\n\ns=(n_s-0)/(n_s)\n\ns=1$

The rotor frequency is the product of the supply frequency and slip it is given as:

$f_r=sf\n\nf_r=1*60\n\nf_r=60\ Hz$

What is the De Broglie wavelength of an electron under 150 V acceleration?

Answers

Answer:

0.1 nm

Explanation:

Potential difference of the electron = 150 V

Mass of electron $m=9.1* 10^(-31)kg$

Charge on electron $1.6* 10^(-19)C$

Plank's constant $h=6.67* 10^(-34)$

If the velocity of the electron is v

Then according to energy conservation $eV =(1)/(2)mv^2$

$v=\sqrt{(2eV)/(m)}=\sqrt{(2* 1.6* 10^(-`19)* 150)/(9.1* 10^(-31))}=7.2627* 10^(6)m/sec$

According to De Broglie $\lambda =(h)/(mv)=(6.67* 10^(-34))/(9.1* 10^(-31)* 7.2627* 10^(6))=0.1nm$

One piece of evidence that supports the Theory of Plate Tectonics is the discovery of what in both South America and Africa? The ancient atmosphere in both places was identical. The rates of weathering of rock are similar. Fossil remains of the same land-dwelling animal. Plants on both continents have similar flowers.

Answers

Answer: Fossil remains of the same land-dwelling animal.

Explanation: Fossil remains which were found to belong to same land dwelling animals, in South America and Africa was used as evidence to help support the theory of Tectonics plates, what this theory simply means is that the whole continents of earth were once fused together until a tectonic plate caused it’s division. Since same remains were found in Africa and South America this shows that both continents were once fused together.

Answer:

Fossil remains of the same land-dwelling animal

Explanation:

Fossil remains tell us where the animals once lived and how by the movement of plate spearated their remaind that was burried thousands of years ago.

If new research showed that the standard heights for crest vertical curve design were H1=3.0 ft and H2=1.5 ft, respectively, by what percent would the minimum length of curvature increase or decrease for a crest curve with a stopping sight distance of 500 ft and grades of +3% and -2%, respectively.

Answers

Answer:

A = 5

S<L, L = 714.89ft

S>L, L = 650.29ft

L = 115.85ft

Percentage min. Length of curvature = 6.2 %

Explanation: see explanation at the attached file

At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answers

Answer and Explanation:

The computation is shown below

Total average cost + total variable cost = total cost

Let number of output be x

So,

Total fixed average cost = x × $30

Total variable cost = x × $15

Total cost = $2,500

Therefore,

$20 × x + $30 × x = $2,500

50 × x = $2,500

x = 50

Now the total variable cost is

= 50 × $20

= $1,000

And, the fixed cost is

= 50 × $30

= $1,500

Answer:

12 4

Explanation:

because the production average is variable

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