Question

The radiator of a steam heating system has a volume of 20 L and is filled with superheated water vapor at 200 kPa and 200°C. At this moment both the inlet and the exit valves to the radiator are closed. After a while it is observed that the temperature of the steam drops to 80°C as a result of heat transfer to the room air, which is at 21°C. Assuming the surroundings to be at 0°C, determine (a) the amount of heat transfer to the room and (b) the maximum amount of heat that can be 462 EXERGY supplied to the room if this heat from the radiator is supplied to a heat engine that is driving a heat pump. Assume the heat engine operates between the radiator and the surroundings.

Answer 1

Answer:

a = 30.1 kj

b = 115 kj

Explanation:

To determine the mass we use the formula m = V/v1

v1 =1.08m3/kg, and V = 20L

m = 20/1000 × 1.08 = 0.0185kg

Next we determine the initial specific internal energy, u1.

Using softwares and appropriate values of T1 and p1, we get

u1 = 2650kj/kg.

After this we determine the final specific internal energy, u2 using the formula u2 = uf + x2 × ufg

Therefore we need to find x2 first.

x2 = u2 - uf/ug - uf

x2 = 1.08 - 0.001029/3.4053 -0.001029

x2 = 0.3180

But u2 = uf + x2× uf=334.97 + 0.3180×2146.6 = 1017.59 kj/kg

Now heat transfer Q= DU

Q = m x (u1 - u2)

Q = 0.0185(2650-1017.59

Q = 30.1 kj

Calculating the b part of the question we use the formula

W = m( u1-u2) - m. To. (s1 - s2)

Where s1 = 7.510kj/kgk

And s2 = 3.150 kj/kgk

We need to convert To and Ta to k values by adding 273 to 0 and 21 respectively.

Putting the values into the formula, we get W = 30.1 - 0.0185 × 273 (7.510-3.150)

W = 8.179kj

Finally maximum heat transfer

Qm = W/1 - to/ta

Qm = 8.179/1 - 273/294

Qm = 115kj