Answer: 337J/kg
Explanation:
(a) For air, take k = 1.40, R = 287 J/kg ⋅ K, and cp = 1005 J/kg ⋅K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity:+ = = + = +2 2 2p 21 1 1c T V constant 1005( 260) (75) 1005(207) V or .2 2 2
Ans2mV 335 s≈2 1 p 2 1 2 1 207 273 30Meanwhile, s s c ln(T /T ) R ln( p /p ) 1005 ln 287 ln ,260 273 140+ − = − = − ÷ ÷ +or s 2 − s1 = −105 + 442 ≈ 337 J/kg ⋅ K Ans.(a)
(b) For argon, take k = 1.67, R = 208 J/kg ⋅ K, and cp = 518 J/kg ⋅ K. Repeat part (a):2 2 2p 21 1 1c T V 518(260) (75) 518(207) V , solve .2 2 2 Ans+ = + = + 2mV 246 1 207 273 30s s 518 l n 208 ln 54 320 . (b)260 273 140
Answer:
Q = 8.845 DEGREE
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
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
putting these value in volume equation and solve for Discharge
Q = 8.845 DEGREE
Answer:
Height of oil is 7.06 meters.
Explanation:
The situation is shown in the attached figure
The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by
Applying the given values we get
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