Answer:
a) 
, b) 
, c) 
, d) 
, e) 
Explanation:
a) The coefficient of performance of the reversible refrigeration cycle is:
The temperature of the hot reservoir is:
b) The coefficient of performance is:
c) The temperature of the hot reservoir can be determined with the help of the following relation:
d) The coefficient of performance is:
e) The temperature of the cold reservoir is:
Answer:
I=9.6×e^{-8} A
Explanation:
The magnetic field inside the solenoid.
B=I*500*muy0/0.3=2.1×e ^-3×I.
so the total flux go through the square loop.
B×π×r^2=I×2.1×e^-3π×0.025^2
=4.11×e^-6×I
we have that
(flux)'= -U
so differentiating flux we get
so the inducted emf in the loop.
U=4.11×e^{-6}×dI/dt=4.11×e^-6×0.7=2.9×e^-6 (V)
so, I=2.9×e^{-6}÷30
I=9.6×e^{-8} A
Answer:
The solution for the given problem is done below.
Explanation:
M1 = 2.0
= 0.3636
= 0.5289
= 0.7934
Isentropic Flow Chart: M1 = 2.0 , 
= 1.8
T1 = 
(1.8)(288K) = 653.4 K.
In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.
At the inlet,
T02= 
= (1.8)(288K) = 518.4 K.
Q= Cp(T02-T01) = 
= 135.7*
J/Kg.
28384 *x soít cos estematema
Answer:
The critical length of surface flaw = 6.176 mm
Explanation:
Given data-
Plane strain fracture toughness Kc = 29.6 MPa-m1/2
Yield Strength = 545 MPa
Design stress. =0.3 × yield strength
= 0.3 × 545
= 163.5 MPa
Dimensionless parameter. Y = 1.3
The critical length of surface flaw is given by
= 1/pi.(Plane strain fracture toughness /Dimensionless parameter× Design Stress)^2
Now putting values in above equation we get,
= 1/3.14( 29.6 / 1.3 × 163.5)^2
=6.176 × 10^-3 m
=6.176 mm
