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KatRina [158]
3 years ago
10

A mass of 1.3 kg of air at 120 kPa and 24°C is contained in a gas-tight, frictionless piston–cylinder device. The air is now com

pressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the magnitude of work input during this process.
Engineering
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

The magnitude of work input is 178.79 kJ

Explanation:

The process is an isothermal compression process because the temperature inside the cylinder is constant.

W = nRT ln(P1/P2)

n is the number of moles of air in the cylinder = mass/MW = 1.3/29 = 0.045 kgmol

R is gas constant = 8.314 kJ/kgmol.K

T is temperature of air = 24 °C = 24+273 = 297 K

P1 is intial pressure of air = 120 kPa

P2 is final pressure of air = 600 kPa

W = 0.045×8.314×297 ln(120/600) = 111.11661 ln 0.2 = 111.11661 × -1.609 = -178.79 kJ

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Steel balls 10 mm in diameter are annealed by heating to 1150 k and then slowly cooling to 450 k in an air environment for which
Darina [25.2K]

Answer:

The answer is below

Explanation:

Given that:

ρ = 800 kg/m3, c = 600 J/kg-K,  k = 40 W/m-K, Initial temperature = Ti = 1150,

Environment temperature = T = 450 K, Final temperature = T∞ = 325 K

Diameter = 10 mm = 0.01 m, A = 6

The estimated time for cooling process (t) is given as:

t=\frac{\rho dc}{hA} ln\frac{T_i-T_\infty}{T-T_infty}=\frac{7800*0.01*600}{25*6} ln\frac{1150-325}{450-325}\\  \\t=589\ s\\t=0.1635\ h

The estimated cooling time is 589 s

4 0
3 years ago
Cimmaan08 for you! Thank you again :)
fomenos

Answer:

Your welcome!!! I hope that helped!!!

Explanation:

5 0
2 years ago
Read 2 more answers
Which of the following impulse responses correspond(s) to stable LTI systems? (a) h1(t) = e-(1-2j)ru(t) (b) h2(t) = e-r cos(2t)u
Illusion [34]

Answer:

LTI system is stable if Impulse response is finite.

so the correct answer is "b"

(b) h2(t) = e-r cos(2t)u(t)

6 0
4 years ago
To purchase a new car, you borrow $20,000. The bank offers a 6-year loan at an interest rate of 3.25% compounded annually. If yo
Natalka [10]

Answer:

SPCA factor

Single payment compound amount factor.

Total amount pay A = $24,230.95 (Approx)

Interest paid = $4,230.95  (Approx)

Explanation:

Given:

P = $20,000

n = 6 year

r = 3.25%

Find:

Total amount pay A

Computation:

A=p(1+r)ⁿ

A=20,000[1+3.5%]⁶

A=20,000[(1.0325)⁶]

Total amount pay A = $24,230.95 (Approx)

Interest paid = $24,230.95 - 20,000

Interest paid = $4,230.95  (Approx)

6 0
3 years ago
Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newton
SashulF [63]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude q_{1},q_{2} the magnitude of force between them is given by

F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}

where

k_{e} is constant

r is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}

Applying value of the constant we get

F_{1}=\frac{62.208}{0.7^{2}}

Thus F_{2}=126.955Newtons

5 0
3 years ago
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