Explanation:
Note: Refer the diagram below
Obtaining data from property tables
State 1:
State 2:
State 3:
State 4:
Throttling process
(a)
Magnitude of compressor power input
(b)
Refrigerator capacity
(c)
Cop:
System grounding on a power system means electrically connecting the __?__ of every wye-connected transformer or generator to ea
1 answer:
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Given:
diameter of sphere, d = 6 inches
radius of sphere, r = = 3 inches
density, = 493 lbm/
S.G = 1.0027
g = 9.8 m/ = 386.22 inch/
Solution:
Using the formula for terminal velocity,
=
(1)
where,
V = volume of sphere
= drag coefficient
Now,
Surface area of sphere, A =
Volume of sphere, V =
Using the above formulae in eqn (1):
=
=
=
Therefore, terminal velcity is given by:
=
inch/sec
Answer:
the highest rate of heat transfer allowed is 0.9306 kW
Explanation:
Given the data in the question;
Volume = 4L = 0.004 m³
V = V
= 0.002 m³
Using Table ( saturated water - pressure table);
at pressure p = 175 kPa;
v = 0.001057 m³/kg
v = 1.0037 m³/kg
u = 486.82 kJ/kg
u 2524.5 kJ/kg
h = 2700.2 kJ/kg
So the initial mass of the water;
m₁ = V/v
+ V
/v
we substitute
m₁ = 0.002/0.001057 + 0.002/1.0037
m₁ = 1.89414 kg
Now, the final mass will be;
m₂ = V/v
m₂ = 0.004 / 1.0037
m₂ = 0.003985 kg
Now, mass leaving the pressure cooker is;
m = m₁ - m₂
m = 1.89414 - 0.003985
m = 1.890155 kg
so, Initial internal energy will be;
U₁ = mu
+ m
u
U₁ = (V/v
)u
+ (V
/v
)u
we substitute
U₁ = (0.002/0.001057)(486.82) + (0.002/1.0037)(2524.5)
U₁ = 921.135288 + 5.030387
U₁ = 926.165675 kJ
Now, using Energy balance;
E - E
= ΔE
QΔt - mh
= m₂u₂ - U₁
QΔt - mh
= m₂u
- U₁
given that time = 75 min = 75 × 60s = 4500 sec
so we substitute
Q(4500) - ( 1.890155 × 2700.2 ) = ( 0.003985 × 2524.5 ) - 926.165675
Q(4500) - 5103.7965 = 10.06013 - 926.165675
Q(4500) = 10.06013 - 926.165675 + 5103.7965
Q(4500) = 4187.690955
Q = 4187.690955 / 4500
Q = 0.9306 kW
Therefore, the highest rate of heat transfer allowed is 0.9306 kW
Answer:
Explanation:
Given that,
The area of glass =
The thickness of the glass
The area of the styrofoam
The thickness of the styrofoam
The thermal conductivity of the glass
The thermal conductivity of the styrofoam
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
The heat loss due to conduction in the wall is
The net heat loss of the wall and the window is
The percentage of heat lost by the window is