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Gekata [30.6K]
3 years ago
10

Saturated water vapor undergoes a throttling process from 1bar to a 0.35bar. What is the change in temperature for this process?

O -4.2C O -11.3C CA-17.7C O No change in temperature for a throttling process
Engineering
1 answer:
mamaluj [8]3 years ago
4 0

Answer:

-25.63°C.

Explanation:

We know that throttling is a constant enthalpy process

      h_1=h_2

From steal table

We know that if we know only one property in side the dome then we will find the other property by using steam property table.

  Temperature at saturation pressure 1 bar is 99.63°C and  Temperature at saturation pressure 0.35 bar is about 74°C .

So from above we can say that change in temperature is -25.63°C.

But there is no any option for that .

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If i see i dead army person will i be asked questions and if i see the person who did it
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3 years ago
Read 2 more answers
Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati
snow_tiger [21]

Answer:

D=0.41m

Explanation:

From the question we are told that:

Discharge rate V_r=0.35 m3/s

Distance d=4km

Elevation of the pumping station h_p= 140 m

Elevation of the Exit point h_e= 150 m

Generally the Steady Flow Energy Equation SFEE is mathematically given by

h_p=h_e+h

With

P_1-P_2

And

V_1=V-2

Therefore

h=140-150

h=10

Generally h is give as

h=\frac{0.5LV^2}{2gD}

h=\frac{8Q^2fL}{\pi^2 gD^5}

Therefore

10=\frac{8Q^2fL}{\pi^2 gD^5}

D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}

D=0.41m

8 0
3 years ago
The uniform slender rod has a mass m.
Nikolay [14]

Answer: 3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

6 0
3 years ago
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