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lisov135 [29]
3 years ago
11

A gas expands in a piston-cylinder assembly from p1 = 8 bar, V1 = 0.02 m3 to p2 = 2 bar. The relation between pressure and volum

e during the gas expansion process is pV1.2 = constant. The mass of the gas is 0.25 kg. If the specific internal energy of the gas decreases by 55 kJ/kg during the process, determine the heat transfer, in kJ. Kinetic and potential energy effects are negligible.
Engineering
1 answer:
Charra [1.4K]3 years ago
7 0

Answer:

The heat transfer is 29.75 kJ

Explanation:

The process is a polytropic expansion process

General polytropic expansion process is given by PV^n = constant

Comparing PV^n = constant with PV^1.2 = constant

n = 1.2

(V2/V1)^n = P1/P2

(V2/0.02)^1.2 = 8/2

V2/0.02 = 4^(1/1.2)

V2 = 0.02 × 3.2 = 0.064 m^3

W = (P2V2 - P1V1)/1-n

P1 = 8 bar = 8×100 = 800 kPa

P2 = 2 bar = 2×100 = 200 kPa

V1 = 0.02 m^3

V2 = 0.064 m^3

1 - n = 1 - 1.2 = -0.2

W = (200×0.064 - 800×0.02)/-0.2 = -3.2/-0.2 = 16 kJ

∆U = 55 kJ/kg × 0.25 kg = 13.75 kJ

Heat transfer (Q) = ∆U + W = 13.75 + 16 = 29.75 kJ

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The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet
Crank

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, I_s = 2 A

power supplied, P_s = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A

The number of turns on the input side is calculated as;

\frac{N_p}{N_s} = \frac{I_s}{I_p}  \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns

4 0
3 years ago
Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the
Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

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3 years ago
The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

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s_max = 0.8394m

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Answer:

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