Question

A liquid refrigerant (sg=1,2) is flowing at a weight flow rate of 20,9 N/h. Refrigerant flashes into a vapor and its specific weight become 11,5 N/m^3. If the weigt flow rate remains at 20,9 N/h, compute the volume flow rate[10^-4 m^3/sec]. Express results in 10^-4 m^3/sec.

Answer 1

Answer:

Explanation:

volume of 20.9 N

= 20.9 / 11.5 m³

= 1.8174 m³

In one hour 1.8174 m³ flows

in one second volume flowing = 1.8174 / 60 x 60

= 5 x 10⁻⁴ m³

Rate of volume flow = 5 x 10⁻⁴ m³ / s .

Answer 2

The volume flow rate of the liquid refrigerant in terms of 10¯⁴ m³/s is;

V' = 5.0483 × 10¯⁴ m³/s

We are given;

Weight flow rate; W' = 20.9 N/h

Specific weight; γ = 11.5 N/m³

Thus;

Volume flow rate is;

V' = W'/γ

V' = 20.9/11.5

V' = 1.8174 m³/h

Now, we want to express the volume flow rate in m³/s

Thus;

1.8174 m³/h = 1.8174/(60 × 60) m³/s

>> 0.00050483 m³/s

Expressing it in terms of 10¯⁴ m³/s gives us;

V' = 5.0483 × 10¯⁴ m³/s

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