A cooling process at constant pressure involves the removal of 3.5 kW from the s

Question

A cooling process at constant pressure involves the removal of 3.5 kW from the system. Compute the mass flow rate required and the change in entropy of the following substances: (a) air decreasing in temperature from 0oC to -20oC; (b) R12 condensing at -20oC; (c) ammonia condensing at -20oC. A cooling process at constant pressure involves the removal of 3.5 kW from the system. Compute the mass flow rate required and the change in entropy of the following substances: (a) air decreasing in temperature from 0oC to -20oC; (b) R12 condensing at -20oC; (c) ammonia condensing at -20oC.

Explanation / Answer

Here a cooling process at constant pressure involves the removal of 3.5 kW from the system.

so( latent heat / sensible heat) q = 3.5 kW = 3009.5 kilocalorie /hour

Now for mass flow rate for (a) air decreasing in temperature from 0oC to -20oC;

Here T1 = 0oC and T2 = 20oC

Now q = mCp (T1-T2)

       m = q / [Cp * (T1 -T2) ]

      m = 3009.5 (kilocalorie /hour) / [ 0.2403 kcal/(kg oC) * (0+20)oC ]

         (here Cp = specific heat of air = 0.2403 kcal/(kg oC)

   m = 627 kg/hr = mass flow rate of air required

(b) R12 condensing at -20oC

for condensation

q = m*lemda ( where lemda = latent heat of vapurization of R12 =39.71 Kcal /Kg)

so m = q / lemda

    m = 3009.5 (kilocalorie /hour) / 39.71 (Kcal /Kg)

   m = 75.79 kg /hr = mass flow rate of R12 required

(c) ammonia condensing at -20oC.

or condensation

q = m*lemda ( where lemda = latent heat of vapurization of NH3 =327.32 Kcal /Kg)

so m = q / lemda

    m = 3009.5 (kilocalorie /hour) / 327.32 (Kcal /Kg)

   m = 9.20 kg /hr = mass flow rate of NH3 required

       

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