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4 years ago

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A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as

an air-standard Otto cycle with a pressure of 14.7 lbf/in. 2 , a temperature of 80 8 F, and a volume of 0.0196 ft 3 at the beginning of compression. The compression ratio is 10, and maximum pressure in the cycle is 1080 lbf/in.Write possible Assumptions no less than three assumptionsDetermine, using a cold air-standard analysis with k 5 1.4, the power developed by the engine, in horsepower, and the mean effective pressure, in lbf/in.

1 answer:

Ulleksa [173]4 years ago

4 0

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 = 101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ = 254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP = 9196.41/(0.00055501 - 0.000055501) = 18410899.5 kPa

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The power cost savings for the 8 inches pipe offsets the increased cost for the pipe therefore to save costs the 8-inch pipe should be used

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The given parameters in the question are;

The distance of the river from the the site, d = 2,500 ft.

The planned flow rate = 600 gal/min

The diameter of the pipe, d = 6-in.

The pipe material = Steel

The cost of pumping = 3 cents per kilowatt-hour

The Bernoulli's equation is presented as follows;

$\dfrac{P_a}{\rho} + g\cdot Z_a + \dfrac{V^2_a}{2} + \eta\cdot W_p = \dfrac{P_b}{\rho} + g\cdot Z_b + \dfrac{V^2_b}{2} +h_f +W_m$

${P_a} = {P_b} = Atmospheric \ pressure$

$Z_a = Z_b$

Vₐ - 0 m/s (The river is taken as an infinite source)

$W_m$ = 0

The head loss in 6 inches steel pipe of at flow rate of 600 gal/min = 1.19 psi/100 ft

Therefore; $h_f$ = 1.19 × 2500/100 = 29.75 psi

$\eta\cdot W_p = \dfrac{V^2_b}{2} +h_f$

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(6 in.)²/4) = 6.80829479 ft./s

$V_b$ ≈ 6.81 ft./s

The pressure of the pump = P = 62.4 lb/ft³× (6.81 ft./s)²/2 + 29.75 psi ≈ 30.06 psi

The power of the pump = Q·P ≈ 30.06 psi × 600 gal/min = 7,845.50835 W

The power consumed per hour = 7,845.50835 × 60 × 60 W

The cost = 28,243.8301 kW × 3 = $847.31 per hour

Annual cost = $847.31 × 8766 = $7,427,519.46

Pipe cost = $15/ft × 2,500 ft = $37,500

Annual charges = 20% × Installed cost = 0.2 × $37,500 = $7,500

Total cost = $37,500 + $7,500 + $7,427,519.46 = $7,475,519.46

For the 8-in pipe, we have;

$V_b$ = Q/ $A_b$ = 600 gal/min/(π·(8 in.)²/4) = 3.83 ft./s

$h_f$ = 1.17 ft/100 feet

Total head loss = (2,500 ft/100 ft) × 1.17 ft. = 29.25 ft.

$h_p = \dfrac{V^2_b}{2 \cdot g} +h_f$

∴ $h_p$ = (3.83 ft./s)²/(2 × 32.1740 ft/s²) + 29.25 ft. ≈ 29.5 ft.

The power of the pump = ρ·g·h × Q

Power of pump = 62.4 lb/ft³ × 32.1740 ft/s² × 29.5 ft.× 600 gal/min = 3,363.8047 W

The cost power consumed per annum = 3,363.8047 W × 60 × 60 × 3 × 8766 = $3,184,608.1

The Cost of the pipe = $20/ft × 2,500 ft. = $50,000

The total cost plus charges = $3,184,608.1 + $50,000 + $10,000 = $3,244,608.1

Therefore it is more affordable to use the 8-in pipe which provides substantial savings in power costs

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