Answer:
W=2 MW
Explanation:
Given that
COP= 2.5
Heat extracted from 85°C
Qa= 5 MW
Lets heat supplied at 150°C = Qr
The power input to heat pump = W
From first law of thermodynamics
Qr= Qa+ W
We know that COP of heat pump given as
W=2 MW
For Carnot heat pump
2.5 T₂ - 895= T₂
T₂=596.66 K
T₂=323.6 °C
Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality =
exit quality =
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Answer:
Explanation:
Reynolds number:
Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is low then flow is called laminar flow .
Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.
For plate can be given as
Where ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.
Flow on plate is a external flow .The values of Reynolds number for different flow given as
Answer:
a) 280MPa
b) -100MPa
c) -0.35
d) 380 MPa
Explanation:
GIVEN DATA:
mean stress
stress amplitude
a)
--------------1
-----------2
solving 1 and 2 equation we get
b)
c)
stress ratio
d)magnitude of stress range
= 280 -(-100) = 380 MPa