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Rainbow [258]
3 years ago
6

. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298

K and 20 bars to 200 bars. Compute the heat removal and work required.

Engineering
1 answer:
andriy [413]3 years ago
5 0

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

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7 0
2 years ago
simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The
Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

q(x)=\frac{x}{L} q_{0}

σ = 120 MPa = 120*10⁶ Pa

L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\

We can see the pic shown in order to understand the question.

We apply

∑MB = 0  (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6   (↑)

Then, we apply

v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x

Then, we can get the maximum bending moment as follows

M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\  x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m

then we get  

M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}

We get the inertia as follows

I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

y=\frac{h}{2} =\frac{0.3m}{2}=0.15m

If M = Mmax, we have

(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4}   }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}

8 0
3 years ago
9. Imagine that you're performing measurements on a circuit with a multimeter. You measure a total circuit
ratelena [41]

Answer:

C

Explanation:

the total resistance is equal to the total potential difference divided by the Current

3 0
2 years ago
Read 2 more answers
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
3 years ago
What type of intersection is this?
mote1985 [20]
Diverging Diamond Interchange
6 0
3 years ago
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