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sveta [45]
3 years ago
15

Steam undergoes an isentropic compression in an insulated piston–cylinder assembly from an initial state where T1 5 1208C, p1 5

1 bar to a final state where the pressure p2 5 100 bar. Determine the final temperature, in 8C, and the work, in kJ per kg of steam.
Engineering
1 answer:
balu736 [363]3 years ago
3 0

Answer:

temperature T2 = 826.9°C

\frac{W}{m} = -1142.7 kJ/kg

Explanation:

given data

initial state temperature = 120°C

final state pressure p1 = 1 bar

pressure p2 =  100 bar

solution

we use here super heated water table A6 that is

specific internal energy u1 = 2537.3 kJ/kg

specific entropy s1 = 7.4668 kJ/kg.K

and

here specific entropy stage 1 = stage 2

so for specific entropy and pressure 100 bar

specific internal energy u2 = 3680 kJ/kg

and temperature T2 = 826.9°C

so here now we get specific work of steam is

ΔU = -W     ...........1

m ( u1 - u2) = W

\frac{W}{m} = u1 - u2

\frac{W}{m} = 2537.3 - 3680

\frac{W}{m} = -1142.7 kJ/kg

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A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the
kramer

Answer:

Maximum shear stress is;

τ_max = 1427.12 psi

Explanation:

We are given;

Power = 2 HP = 2 × 746 Watts = 1492 W

Angular speed;ω = 450 rev/min = 450 × 2π/60 rad/s = 47.124 rad/s

Diameter;d = 1 in

We know that; power = shear stress × angular speed

So,

P = τω

τ = P/ω

τ = 1492/47.124

τ = 31.66 N.m

Converting this to lb.in, we have;

τ = 280.2146 lb.in

Maximum shear stress is given by the formula;

τ_max = (τ•d/2)/J

J is polar moment of inertia given by the formula; J = πd⁴/32

So,

τ_max = (τ•d/2)/(πd⁴/32)

This reduces to;

τ_max = (16τ)/(πd³)

Plugging in values;

τ_max = (16 × 280.2146)/((π×1³)

τ_max = 1427.12 psi

7 0
3 years ago
Give me uses of a grinding machine in agriculture.
Tju [1.3M]

Answer:

The grinding machine is used for roughing and finishing flat, cylindrical, and conical surfaces; finishing internal cylinders or bores; forming and sharpening cutting tools; snagging or removing rough projections from castings and stampings; and cleaning, polishing, and buffing surfaces.

4 0
3 years ago
g Consider a thin opaque, horizontal plate with an electrical heater on its backside. The front end is exposed to ambient air th
xxTIMURxx [149]

Answer:

The electrical power is 96.5 W/m^2

Explanation:

The energy balance is:

Ein-Eout=0

qe+\alpha sGs+\alpha skyGsky-EEb(Ts)-qc=0

if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }

\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }

if Gl≈El(l,5800)

\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt= 2*5800=11600 um-K, at this value, F=0.941

\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77

The hemispherical emissivity is equal to:

E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))

lt=2*333=666 K, at this value, F=0

E=0+(1-0.7)(1)=0.3

The hemispherical absorptivity is equal to:

qe=EoTs^{4}+h(Ts-T\alpha  )-\alpha sGs-\alpha oTsky^{4}=(0.3*5.67x10^{-8}*333^{4})+10(60-20)-(0.77-600)-(0.3*5.67x10^{-8}*233^{4})=96.5 W/m^{2}

3 0
3 years ago
8. Block A shown in the figure below weighs 2000 N. The chord attached to A passes over a
Kobotan [32]

Answer:

Read the passage. Then, answer the questions about the metaphor in boldface in the text.

Lately, I've been so overwhelmed with school and sports. There was a time when I enjoyed going to classes and going to practice every afternoon. Now, everything is piling up and wearing me down. Thankfully, I get to see you every day. You are truly the sunshine of my life. Thank you for making me laugh when I'm feeling down.

What is the context of the passage?

What is being compared in the metaphor?

What is the meaning of the metaphor?

Explanation:

4 0
2 years ago
A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s
Scorpion4ik [409]

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

3 0
3 years ago
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