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ikadub [295]
4 years ago
9

A rigid container container contain 3kg of air.the initial pressure = 0.5MPa and final pressure = 1.2MPa, the initial temp. =75

degree celcius. heat absorb =195KJ. assume Cv =0.714KJ/Kg, find final temperature, change in U from 1-2 and work done from 1-2.
Engineering
1 answer:
Stolb23 [73]4 years ago
3 0

Answer:

T_2=562.2 C

ΔU =195 KJ

W=0 KJ

Explanation:

Given that

P_1=0.5 MPa

P_2=1.2 MPa

T_1=75 C

Mass of air m=3 kg

heat absorb by air =195 KJ

Cv=0.741 KJ/kgK

If we assume that air is as ideal gas so

\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}

\dfrac{0.5}{273+75}=\dfrac{1.2}{T_2}

T_2=562.2 C

Given that container is rigid it means that volume of system is constant so W=0

From first law of thermodynamics

Q=ΔU + W

⇒Q= ΔU                  (W=0)

So ΔU =195 KJ

And work done will be zero because because it is a constant volume process.

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victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

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32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

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\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
4 years ago
A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
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Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

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5 0
2 years ago
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5 is the correct one to choose for this
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3 years ago
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Does a food market have any rooms in particular? Also whats units?
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A box-shaped aquarium has horizontal dimensions 0.5 m by 1 m, and depth 0.5 m, and is filled two-thirds of the way to the surfac
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Answer:

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From the above data given will  be

p = 1000 kg/m^3, h = 2/3 * 0.5 = 0.33 m , a =2g , g = 9.81

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Py =  -Ph ( g - 2g ) = Phg ------ ( 3 )

Py = 1000 * 0.33 * 9.81

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