Question

Air enters a 34 kW electrical heater at a rate of 0.8 kg/s with negligible velocity and a temperature of 60 °C. The air is discharged at a height 50 m above the inlet at a temper-ature of 200 °C and a velocity of 50 m / s. What is the work done in the heater?

Answer 1

Answer:

79 kW.

Explanation:

The equation for enthalpy is:

H2 = H1 + Q - L

Enthalpy is defined as:

H = G*(Cv*T + p*v)

This is specific volume.

The gas state equation is:

p*v = R*T (with specific volume)

The specific gas constant for air is:

287 K/(kg*K)

Then:

T1 = 60 + 273 = 333 K

T2 = 200 + 273 = 473 K

p1*v1 = 287 * 333 = 95.6 kJ/kg

p2*v2 = 287 * 473 = 135.7 kJ/kg

The Cv for air is:

Cv = 720 J/(kg*K)

So the enthalpies are:

H1 = 0.8*(0.72 * 333 + 95.6) = 268 kW

H2 = 0.8*(0.72 * 473 + 135.7) = 381 kW

Ang the heat is:

Q = 34 kW

Then:

H2 = H1 + Q - L

381 = 268 + 34 - L

L = 268 + 34 - 381 = -79 kW

This is the work from the point of view of the air, that's why it is negative.

From the point of view of the machine it is positive.