Given:
Temperature of water, =
=273 +(-6) =267 K
Temperature surrounding refrigerator, =
=273 + 21 =294 K
Specific heat given for water, = 4.19 KJ/kg/K
Specific heat given for ice, = 2.1 KJ/kg/K
Latent heat of fusion, = 335KJ/kg
Solution:
Coefficient of Performance (COP) for refrigerator is given by:
Max =
= = 9.89
Coefficient of Performance (COP) for heat pump is given by:
Max =
= 10.89
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%
This question is incomplete, the complete question as well as the missing diagram is uploaded below;
Consider a mixing tank with a volume of 4 m³. Glycerin flows into a mixing tank through pipe A with an average velocity of 6 m/s, and oil flow into the tank through pipe B at 3 m/s. Determine the average density of the mixture that flows out through the pipe at C. Assume uniform mixing of the fluids occurs within the 4 m³ tank.
Take = 880 kg/m³ and
= 1260 kg/m³
Answer:
the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
Explanation:
Given that;
Inlet velocity of Glycerin, = 6 m/s
Inlet velocity of oil, = 3 m/s
Density velocity of glycerin, = 1260 kg/m³
Density velocity of glycerin, Take = 880 kg/m³
Volume of tank V = 4 m
from the diagram;
Diameter of glycerin pipe, = 100 mm = 0.1 m
Diameter of oil pipe, = 80 mm = 0.08 m
Diameter of outlet pipe = 120 mm = 0.12 m
Now, Appling the discharge flow equation;
π/4 × ()²
+ π/4 × (
)²
= π/4 × (
)²
we substitute
π/4 × (0.1 )² × 6 + π/4 × (0.08 )² × 3 = π/4 × (0.12)²
0.04712 + 0.0150796 = 0.0113097
0.0621996 = 0.0113097
= 0.0621996 / 0.0113097
= 5.5 m/s
Now we apply the mass flow rate condition
so we substitute
1260 × π/4 × (0.1 )² × 6 + 880 × π/4 × (0.08 )² × 3 = p × π/4 × (0.12)² × 5.5
1260 × 0.04712 + 880 × 0.0150796 = p × 0.06220335
59.3712 + 13.27 = 0.06220335p
72.6412 = 0.06220335p
p = 72.6412 / 0.06220335
p = 1167.8 kg/m³
Therefore, the average density of the mixture that flows out through the pipe at C is 1167.8 kg/m³
