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8090 [49]
3 years ago
9

The temperature in a pressure cooker is 130 degree C while the water is boiling. Determine the pressure inside the cooker.

Engineering
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer:

The pressure inside the cooker is 1.0804 atm.

Explanation:

Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.

<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>

So, Standard conditions:

T₁ = 373.15 K

P₁ = 1 atm

Given ,

The water boils at Temperature =  130 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So, the temperature, T₂ = (130 + 273.15) K = 403.15 K

To find pressure inside the cooker (P₂) :

<u>Applying Amontons's Law as:</u>

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

So,

P_2=\frac {P_1}{T_1} \times {T_2}

P_2=\frac {1 atm}{373.15 K} \times {403.15 K}

P_2=1.0804 atm

<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>

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b.

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M[0][0] = 0,M[1][1] = 0,M[2][2] = 0,M[3][3] = 0,M[4][4] = 0,M[5][5] = 0,

M[1][2] = M[1][1]+M[2][2]+p[0]*p[1]*p[2] = 0+0+5*10*3 = 150

M[2][3] = M[3][3]+M[2][2]+p[1]*p[2]*p[3] = 0+0+10*3*12 = 360

M[3][4] = M[3][3]+M[4][4]+p[2]*p[3]*p[4] = 0+0+3*12*5 = 180

M[4][5] = M[4][4]+M[5][5]+p[3]*p[4]*p[5] = 0+0+12*5*50 = 3000

M[1][3] = min{M[1][1]+M[2][3]+p[0]*p[1]*p[3] , M[1][2]+M[3][3]+p[0]*p[2]*p[3]}

= {0 + 360 + 600 , 150+0+180} = {960,330} = 330

M[2][4] = min{M[2][2]+M[3][4]+p[1]*p[2]*p[4] , M[2][3]+M[4][4]+p[1]*p[3]*p[4]}

= {0 + 180 + 150 , 360+0+600} = {960,330} = 330

M[3][5] = min{M[3][3]+M[4][5]+p[2]*p[3]*p[5] , M[3][4]+M[5][5]+p[2]*p[4]*p[5]}

= {0 + 3000 + 1800 , 180+0+750} = {4800,930} = 930

M[1][4] = min{M[1][1] + M[2][4] +p[0]*p[1]*p[4] ,M[1][2] + M[3][4] +p[0]*p[2]*p[4] ,

M[1][3] + M[4][4] +p[0]*p[3]*p[4]}

{0+330+250 , 150+180+75 , 330+0+300} = 405

M[2][5] = min{M[2][2] + M[3][5] +p[1]*p[2]*p[5] ,M[2][3] + M[4][5] +p[1]*p[3]*p[5] ,

M[2][4] + M[5][5] +p[1]*p[4]*p[5]}

{0+930+1500 , 360+3000+6000,330+0+2500} = 2430

M[1][5] = min{M[1][1] +M[2][5]+p[0]*p[1]*p[5] , M[1][2] +M[3][5]+p[0]*p[2]*p[5],

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{0+2430+2500 , 150+930+750 , 330+3000+3000 , 405+0+1250} = 1655

(a)

MemoizedCutRod(p, n)

r: array(0..n) := (0 => 0, others =>MinInt)

return MemoizedCutRodAux(p, n, r)

MemoizedCutRodAux(p, n, r)

if r(n) = 0 and then n /= 0 then -- check if need to calculate a new solution

q: int := MinInt

for i in 1 .. n loop

q := max(q, p(i) + MemoizedCutRodAux(p, n-i, r))

end loop

end if

r(n) := q

end if

return r(n)

8 0
3 years ago
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype
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Explanation:

Given

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Properties of sea water

\nu _{sea}=1.26\times 10^{-5} ft/s

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8 0
4 years ago
Calculate total hole mobility if the hole mobility due to lattice scattering is 50 cm2 /Vsec and the hole mobility due to ionize
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Answer:

The total hole mobility is 41.67 cm²/V s

Explanation:

Data given by the exercise:

hole mobility due to lattice scattering = ul = 50 cm²/V s

hole mobility due to ionized impurity = ui = 250 cm²/V s

The total mobility is equal:

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5 0
3 years ago
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Answer:

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4 years ago
Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa m . It has bee
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in this question we are asked to determine if an aircraft will fracture for a given fracture toughness.

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solving for Y we have;

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Y = 2.50

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From our results we can say that fracture will not occur since Kc (32.2 MPa√m) is less than KIc (35  MPa√m) of the material.

cheers i hope this helps!!!!

8 0
3 years ago
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