A small single-cylinder, two-stroke cycle SI engine operates at 8000 RPM with a volumetric efficiency of Tlv = 0.85. The engine is square (bore = stroke) and has a displacement of 6.28 em 3. The fuel-air ratio FA = 0.067. Calculate:
(a) Average piston speed. [m/sec]
(b) Flow rate of air into engine. [kg/see]
(c) Flow rate of fuel into engine. [kg/see]
(d) Fuel input for one cycle. [kg/cycle]

a) V_avg = 5.33 m/s

b) flow(m_a) =  0.00084 kg/s

c) flow(m_f) = 5.63*10^-5 kg/s  

d) m_f = 4.2225*10^-7 kg / cycle        

Explanation:

Given:

- The rotation of crankshaft N = 8000 RPM

- The total volume displaced V_d = 6.28 cm^3

- The volumetric efficiency n_v = 0.85

- The Fuel to air ratio FA = 0.067

- Stroke length = Bore size

Find:

(a) Average piston speed. [m/sec]

(b) Flow rate of air into engine. [kg/see]

(c) Flow rate of fuel into engine. [kg/see]

(d) Fuel input for one cycle. [kg/cycle]

Solution:

- To compute the average piston speed we need the Stroke length of the engine. We will used the amount of volume displaced to calculate it:

                                   V_d = A_b * S

                                   V_d = pi*d^2*S / 4

Since, we know that stroke length and bore dia are equal hence S = d:

                                   V_d = pi*S^3 / 4

- Input values:

                                   6.28 = pi*S^3 / 4

                                     S = cuberoot ( 7.9959)

                                     S = 2.0 cm

- The average piston speed can now be calculated from the formula:

                                    V_avg = S*N / 30

                                    V_avg = 0.02*8000 / 30

                                    V_avg = 5.33 m/s

- We can use the relation of volumetric efficiency to calculate the flow rate of air into the engine flow(m_a):

                                     n_v = flow(m_a) / p_a*V_d*N

Where, p_a is the density of air at atmospheric conditions, using ideal gas law we can calculate:

                                     p_a = P_atm / R*T_amb

Where. P_atm = 101 KPa , R = 0.287 KJ/kgK , T_amb = 25 + 273 = 298 K

                                     p_a = 101 / 0.287*298 = 1.181 kg/m^3

Hence,

                                    flow(m_a) =  p_a*V_d*N*n_v

- Plug values in:

                                   flow(m_a) =  1.181*6.28*10^-6*8000/60*.85

                                   flow(m_a) =  0.00084 kg/s

- We will use the Fuel to Air ratio to compute the flow rate of fuel in a engine flow(m_f):

                                 flow(m_f) = FA*flow(m_a)

- Plug in the values:

                                 flow(m_f) = 0.067*0.00084

                                  flow(m_f) = 5.63*10^-5 kg/s                  

- The fuel in put per cycle can be calculated as follows:

                                  m_f = flow(m_f)*60 / N

- plug in values:

                                  m_f = 5.63*10^-5*60 / 8000

                                  m_f = 4.2225*10^-7 kg / cycle        

false

a hangover caused by drinking a lethal dosage of alcohol

answer:

(b)

explanation: