One of the most common passenger vehicles in the North American market is the Fo

Question

One of the most common passenger vehicles in the North American market is the Ford F-150 truck. One engine option for this vehicle is a 4-stroke, 8-cylinder gasoline engine (V8 5.0L Vi-TCT, video: http://www.ford.com/trucks/f150/features/Feature12/#page=Feature12).  In this exercise , we will develop an ideal thermodynamic model for this engine.

Cylinders  8

Cylinder head Aluminum

Block material Aluminum

Bore/Stroke 3.63"/3.65"

Compression Ratio 10.5:1

Horesepower/rmp 360@5,500

A)Using the information above determine: i) the total displacement volume of the engine (L), ii) the total volume at BDC (L), the mass of air- the

Explanation / Answer

A)

i)

Displcement volume of each cylinder = pi/4 *bore^2 *stroke = 3.14/4*(3.63^2)*3.65 = 37.755 in^3

Dispalcement volume of 8 cylinders, V_d = 8*37.755 = 302.04 in^3 = 0.1748 ft^3 = 4.95 litre

ii)

Compresion ratio = V_bdc / V_tdc

= V_bdc / (V_bdc - V_d)

10.5 = V_bdc / (V_bdc - 4.95)

V_bdc = 5.471 L

iii)

For air, gas constant R = 287 J/kg-K

At BDC, P1*V1 = m*R*T1

100*10^3 *(5.471*10^-3) = m*287*300

m = 0.0063543 kg = 6.354 grams

B)

i)

Mass of fuel = Mass of air / (Air-to-Fuel ratio)

= 6.354/25

= 0.254 grams

ii)

Heat input per cycle = mass of fuel*calorific value

= (0.254*10^-3)*33

= 0.008387 MJ

= 8.387 kJ

C)

i)

For compression stroke,

P2 / P1 = (V1 / V2)^gamma

P2/100 = 10.5^1.4

P2 = 2689.5 kPa

T2 / T1 =(P2/P1)^((gamma-1)/gamma)

T2 / 300 = (2689.5/100)^((1.4-1)/1.4)

T2 = 768.4 K

For process 2-3:

Heat input = m_air*Cv_air*(T3 - T2)

8.387 = (6.354*10^-3)*0.717*(T3 - 768.4)

T3 = 2609.3 K

For process 3-4:

T3 / T4 = (V4 / V3)^(gamma-1)

2609.3 / T4 = (10.5)^(1.4-1)

T4 = 1018.7 K

ii)

Specific Work done in process 1-2:

W1-2 = R(T2 - T1)/(1-gamma)

W1-2 = 287*(768.4 - 300) / (1 - 1.4)

W1-2 = -336077 J/kg

W1-2 = -336.077 kJ/kg

W2-3 = 0 (constant volume process)

W3-4 = R*(T4 - T3)/(1 - gamma)

W3-4 = 287*(1018.7 - 2609.3) / (1 - 1.4)

W3-4 = 1141255.5 J/kg

W3-4 = 1141.255 kJ/kg

W4-1 = 0 (constant volume process)

Net specific work done per cycle = -336.077 + 1141.255 = 805.18 kJ/kg

iii)

Net Work done per cycle = m_air*specific work

= (6.354*10^-3)*805.18

= 5.116 kJ

Thermal efficiency = Net work / heat input

= 5.116 / 8.387

= 0.61 or 61 %

D)

Work done per cycle = 5.116 kJ

2 revs = 1 cycle

5500 revs = 5500/2 cycle = 2750 cycles

Time taken per cycle = 60 sec / 2750 = 0.02181 sec

Power = Work done / time taken

= 5.116 / 0.02181

= 234.5 kW

= 314.9 hp

Manufacturer specified power (360 hp) is higher than ideal power calculated (314.9 hp) which cannot be true.

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