Explanation:
a converging nozzle has an exit area of 0.001 m2. Air enters the nozzle with negligible velocity at a pressure of 1 MPa and a temperature of 360 K. For isentropic flow of an ideal gas with k = 1.4 and the gas constant R = Ru/MW = 287 J/kg-K, determine the mass flow rate in kg/s and the exit Mach number for back pressures
100% (3 ratings)
A_2 = 0.001 m^2 P_1 = 1 MPa, T_1 = 360 k P_2 = 500 kpa p^gamma - 1/gamma proportional T (1000/500)^1.4 - 1/1.4 = (360/T_2) 2^4/14 = 360/T_2 T_2
Answer:
Explanation:
Given that:
The Inside pressure (p) = 1402 kPa
= 1.402 × 10³ Pa
Force (F) = 13 kN
= 13 × 10³ N
Thickness (t) = 18 mm
= 18 × 10⁻³ m
Radius (r) = 306 mm
= 306 × 10⁻³ m
Suppose we choose the tensile stress to be (+ve) and the compressive stress to be (-ve)
Then;
the state of the plane stress can be expressed as follows:

Since d = 2r
Then:







When we take a look at the surface of the circular cylinder parabolic variation, the shear stress is zero.
Thus;

Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Answer:
heat transfer rate is -15.71 kW
Explanation:
given data
Initial pressure = 4 bar
Final pressure = 12 bar
volumetric flow rate = 4 m³ / min
work input to the compressor = 60 kJ per kg
solution
we use here super hated table for 4 bar and 20 degree temperature and 12 bar and 80 degree is
h1 = 262.96 kJ/kg
v1 = 0.05397 m³/kg
h2 = 310.24 kJ/kg
and here mass balance equation will be
m1 = m2
and mass flow equation is express as
m1 =
.......................1
m1 =
m1 = 1.2353 kg/s
and here energy balance equation is express as
0 = Qcv - Wcv + m × [ ( h1-h2) +
+ g (z1-z2) ] ....................2
so here Qcv will be
Qcv = m × [
] ......................3
put here value and we get
Qcv = 1.2353 × [ {-60}+ (310.24-262.96) ]
Qcv = -15.7130 kW
so here heat transfer rate is -15.71 kW