Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the
Answer:
Maximum and mean velocity will be equal i.e 0.5 m/s
Explanation:
given data:
viscosity = 0.001 m^2/s
Thickness of thin film = 2 mm = 0.002 m
Neglecting the body weight hence no shear stress
Mean velocity is given V
V = 0.5 m/s
Maximum and mean velocity will be equal i.e 0.5 m/s
Answer:
a) 1.3 cycles
b) iv. 1.1 cycles.
Explanation:
a) h = 0.97
T = 1
m = 10
(1 - h = 0.03)
Tave = 0.97 + 0.03*10 = 1 - 27 = 1.3 cycles
b) Tave = 0.99 + 0.01*10
= 1.09 cycles
Therefore, The answer is iv. 1.1 cycles.
Answer:
Power consume by compressor=113,726.87 KW
Explanation:
Given:
Actually compressor is an open system, so here we will use first law of thermodynamics for open system .
We know that first law of thermodynamics for steady flow
We know thatand we take the air as ideal gas.
System is in steady state then mass flow rate in =mass flow rate out
Mass flow rate=
So mass flow rate = ,
=1.23×200×2 Kg/s
=541.17 Kg/s
,
=80.07 m/s
Enthalpy of ideal gas h=
So
Now by putting the values
Here Q=0 because heat transfer is zero here.
W= -210.15 KJ/kg
So power consume by compressor=541.17×210.15
=113,726.87 KW