Question

Refrigerant-134a enters a diffuser steadily as saturated vapor at 600 kPa with a velocity of 160 m/s, and it leaves at 700 kPa and 40∘C. The refrigerant is gaining heat at a rate of 2 kJ/s as it passes through the diffuser. If the exit area is 80 percent greater than the inlet area, determine (a) the exit velocity and (b) the mass flow rate of the refrigerant.

Answer 1

Answer:

a) V_2 = 82.1 m/s

b) m = 0.298 Kg/s

Explanation:

from A-11 to A-13 we have the following data

P_1 = 600 kpa

V_1 = 0.033925 m^3/kg

h_1 = 262.52 kJ/kg

P_2 = 700 kpa

V_2 = 0.0313 m^3/kg

T_2 = 40°C = 313K

h_2 = 278.66 kJ/kg

Now, from the conversation of mass,

A_2*V_2/u_2 = A_1*V_1/u_1

V_2 = A_1/A_2*u_2/u_1*V_1

V_2 = A_1/1.8*A_1 * 0.0313 /0.033925*160

V_2 = 82.1 m/s

now from the energy balance equation

E_in = E_out

Q_in + m(h_1 + V_1^2/2) =  m(h_2 + V_2^2/2)

m = 0.298 Kg/s

Answer 2

Answer:

A) V2 = 82.1 m/s

B) m' = 0.298 kg/s

Explanation:

A) From the tables attached,

At P1 = 600 KPa, we have specific volume and specific enthalpy as;

v1 = 0.033925 m³/kg

And h1 = 262.52 Kj/kg or 262520 J/kg

Also at P2 = 700 KPa, and T2 = 40°C = 313K, we have specific volume and specific enthalpy as ;

v2 = 0.03133 m³/kg and h2 = 278.66 Kj/kg or 278660 J/kg

From conservation of mass, we know that;

A1V1/v1 = A2V2/v2

So, making V2 the subject, we have;

(A1V1v2)/(v1A2) = V2

Now, from the question, the exit area (A2) is 80% greater than the inlet Area (A1). Thus, A2 = A1 + 0.8A1 = 1.8A1

So,

(A1V1v2)/(v1 x 1.8 xA1) = V2

So, A1 will cancel out to give;

(V1v2)/(v1 x 1.8) = V2

Plugging in relevant values;

(160 x 0.0313)/(0.033925 x 1.8) = 82.1 m/s

B) From energy balance, we know that;

E'in = E'out

Thus,

Q'in + m'(h1 + (v1²/2)) = m'(h2 + (v2²/2))

From the question, Q'in = 2kj = 2000J

Thus;

2000 + m'(262520 + (160²/2)) = m'(278660 + (82.1²/2))

So,

2000 + 275320m' = 282030 m'

m'(282030 - 275320) = 2000

m' = 2000/6710 = 0.298 kg/s