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Andrej [43]
3 years ago
15

Nitrogen enters a steady-flow heat exchanger at 150 kPa, 10°C, and 100 m/s, and it receives heat as it flows through it. Nitroge

n leaves the heat exchanger at 100 kPa with a velocity of 200 m/s.
a) Determine the Mach number of the nitrogen at the inlet and the amount of heat received if the exit Mach number is 0.51.
Engineering
1 answer:
diamong [38]3 years ago
5 0

Answer:

A) Mach number at Inlet;M1 = 0.292

B) amount of heat received;Q_in = 105.59 KJ/kg

Explanation:

A) The formula for the speed of nitrogen at the inlet is;

C1 = √kRT1

Where;

k is a constant = 1.4

R is the gas constant of nitrogen R = 296.8 J/kg.K

T1 is the temperature at inlet = 10°C = 10 + 273 K = 283K

Thus;

C1 = √(1.4 * 296.8 * 283)

C1 = 342.9 m/s

Formula for mach number at inlet is;

M1 = V1/C1

We are given V1 = 100 m/s

Thus; M1 = 100/342.9

M1 = 0.292

B) We are given mach number at exit; M2 = 0.51

Also; V2 = 200 m/s

Thus;M2 = V2/C2

C2 = V2/M2 = 200/0.51

C2 = 392.16 m/s

Now, we know that;

C2 = √kRT2

So, 392.16 = √(1.4 * 296.8 * T2)

392.16² = 415.52*T2

T2 = 392.16²/415.52

T2 = 370.11 K

Now formula for steady flow balance equation is;

Q_in = =cp(T2 -T1) + [(v2)² - (v1)²]/2

Where Q_in is the amount of heat energy received and cp is the specific heat of nitrogen,= 1.040 KJ/kg.K

Plugging in the relevant values;

Q_in = 1.04(370.11 -283) + [(200)² - (100)²]/2](1KJ/Kg/1000m²/s²)

Q_in = 90.59 + 15

Q_in = 105.59 KJ/Kg

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Let A→=(150iˆ+270jˆ) mm , B→=(300iˆ−450jˆ) mm , and C→=(−100iˆ−250jˆ) mm . Find scalars r and s, if possible, such that R→=rA→+s
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Answer: r = 0.8081; s = -0.07071

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A = (150i + 270j) mm

B = (300i - 450j) mm

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R = rA + sB + C = 0i + 0j

R = r(150i + 270j) + s(300i - 450j) + (-100i - 250j) = 0i + 0j

R = (150r + 300s - 100)i + (270r - 450s - 250)j = 0i + 0j

Equating the i and j components;

150r + 300s - 100 = 0

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5 0
3 years ago
A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16
abruzzese [7]

Answer:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The net work per cycle is 845.88 kJ/kg

The power developed in horsepower ≈ 45374 hP

Explanation:

1) The three possible assumptions are

a) All processes are reversible internally

b) Air, which is the working fluid circulates continuously in a closed loop

cycle

c) The process of combustion is depicted as a heat addition process

2) The diagrams are attached

5) The dimension of the cylinder bore diameter = 3.7 in. = 0.09398 m

Stroke length = 3.4 in. = 0.08636 m.

The volume of the cylinder v₁= 0.08636 ×(0.09398²)/4 = 5.99×10⁻⁴ m³

The clearance volume = 16% of cylinder volume = 0.16×5.99×10⁻⁴ m³

The clearance volume, v₂  = 9.59 × 10⁻⁵ m³

p₁ = 14.5 lbf/in.² = 99973.981 Pa

T₁ = 60 F = 288.706 K

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{v_{1}}{v_{2}}  \right )^{K-1}

Otto cycle T-S diagram

T₂ = 288.706*6.25^{0.393} = 592.984 K

The maximum temperature = T₃ = 5200 R = 2888.89 K

\dfrac{T_{3}}{T_{4}} = \left (\dfrac{v_{4}}{v_{3}}  \right )^{K-1}

T₄ = 2888.89 / 6.25^{0.393} = 1406.5 K

Work done, W = c_v×(T₃ - T₂) - c_v×(T₄ - T₁)

0.718×(2888.89  - 592.984) - 0.718×(1406.5 - 288.706) = 845.88 kJ/kg

The power developed in an Otto cycle = W×Cycle per second

= 845.88 × 2400 / 60  = 33,835.377 kW = 45373.99 ≈ 45374 hP.

8 0
3 years ago
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