Answer:
Q = 8.845 DEGREE
Step-by-step explanation:
given data:
combine Mass for 6 cylinder (M) =15 Kg/hr
mass of each cylinder (m) = 15/6 = 2.5 Kg/hr = 0.000694 Kg/ sec
Engine speed (N)= 1500rpm
Diameter of one nozzle hole ( d) = 200 micrometer = 0.0002 m
Discharge Coefficient (Cd) = 0.75
Pressure difference = 100 MPa
Density of fuel = 800 kg/m^3
velocity of fuel is
![v = cd\sqrt{(2*P)/(p)}](https://img.qammunity.org/2020/formulas/engineering/college/q11pk1kkjylv4r6x8vee20jcfrh66m7cpx.png)
![v = 0.75 \sqrt{(2* 100* 10^6)/(800)} = 375 m/sec](https://img.qammunity.org/2020/formulas/engineering/college/fhp1mlczh7de4cm2tq2terrbt3nxnt09og.png)
injected fuel volume (V) =Area of given Orifices × Fuel velocity × time of single injection × no of injection/sec
we know that p = m/ V
So
![V = (0.000694)/(800) =8.68*10^(-7) m3/sec](https://img.qammunity.org/2020/formulas/engineering/college/asfa1him33daon8mbdot53pdr8u1p2w7nx.png)
putting these value in volume equation and solve for Discharge
![8.68* 10^(-7) = (((3.14)/(4))* 6*( .0002* .0002) * 375 * ((Q)/(360)) * (30)/(750) * ((750)/(60))](https://img.qammunity.org/2020/formulas/engineering/college/2aw94x26a7sj380iq6r9j46mz43cczmimm.png)
Q = 8.845 DEGREE