Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:

It is necessary to get the Reynold's number:

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

The overall heat transfer coefficient:

Here

Substituting values:

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Answer: Attached below is the missing diagram
answer :
A) 1) Wr > WI, 2) Qc' > Qc
B) 1) QH' > QH, 2) Qc' > Qc
Explanation:
л = w / QH = 1 - Qc / QH and QH = w + Qc
<u>A) each cycle receives same amount of energy by heat transfer</u>
<u>(</u> Given that ; Л1 = 1/3 ЛR )
<em>1) develops greater bet work </em>
WR develops greater work ( i.e. Wr > WI )
<em>2) discharges greater energy by heat transfer</em>
Qc' > Qc
solution attached below
<u>B) If Each cycle develops the same net work </u>
<em>1) Receives greater net energy by heat transfer from hot reservoir</em>
QH' > QH ( solution is attached below )
<em>2) discharges greater energy by heat transfer to the cold reservoir</em>
Qc' > Qc
solution attached below
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