Answer:
Final Pressure in the tank after letting 0.5 kg out of the tank is 504.6 kPa
Explanation:
Diesel cycle:
All diesel engine work on diesel cycle .In diesel cycle there are four process .These processes are as follows
1. Adiabatic reversible compression
2.Heat addition at constant pressure
3.Adiabatic reversible expansion
4.Constant volume heat rejection
In general compression ratio in diesel engine is high as compare to petrol engine.But the efficiency of diesel cycle is less as compare to petrol cycle for same compression ratio.
Applications of diesel cycle:
Generally diesel cycle used for heavy vehicle or equipment because heavy vehicle or equipment is required high initial torque.So this cycle have lots of applications such as in industrial machining,in trucks,power plant,in mining ,in defense or military,large motors ,compressor and pump etc.
Answer:
speed of air is 200 m/s
Explanation:
given data
length dl = 10 m
diameter dp = 2 m
maximum allowable speed Vp = 10 m/s
scale = 1/20th
to find out
speed of air in the tunnel to achieve dynamic similarity
solution
we know in dynamic similarity ratio of force at corresponding point in model are equal and different dimensionless no is use for dynamic similarity
so
Rem = Rep
Re is modal law
so
and
VpDp = VmDm
so
so
10 ×
Vm = 200 m/s
so speed of air is 200 m/s
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084