Question

A well insulated 3-m X 4-m X 6-m room initially at 7 degrees C is heated by the radiator of a steam heating system. The radiator has a volume of 15 L and is filled with superheated vapor at 200 kPa and 200 degrees C. At this moment both the inlet and the exit valves to the radiator are closed. A 120-W fan is used to distribute the air to the room. The pressure of the steam is observed to drop to 100 kPa after 45 min as a result of heat transfer to the room. Assuming constant specific heats for air at room temperature, determine the average termperature of air in 45 min. Assume the air pressure in the room remains constant at 100 kPa.

Answer 1

Answer:

The average temperature of air in 45 minutes is 10.74°C

Explanation:

Given

Time t = 45 minutes

Initial Air Temperature, T1 = 7°C

Fan work W = 120W

Radiator Volume = V1 = 15L

Initial Steam Pressure = P1 = 200kPa

Initial Steam Temperature = Ts1 = 200°C

Final Steam Pressure = P2 = 100kPa

The radiator is our system.

We take the system as a close system.

The energy balance is given as follows

-Qout = ∆U

-Qout = m(u2 - u1)

Calculating u2

From steam table (see attachment below)

At P1 and Ts1

P1 = 200kPa, Ts1 = 200°C

u1 = 2654.6KJ/Kg

v1 = 1.08049 m³/Kg

At P2 = 100kPa,  v1 = v2

vf = 0.001043m³,   vg = 1.6941m³/kg    uf = 417.4kj/kg      ufg = 2088.2kj/kg

x2 = (v2 - vf)/(vg - vf)

x2 =  (1.08049 - 0.001043)/(1.6941 - 0.001043)

x2 = 1.079447/1.693057

x2 = 0.63757

u2 = uf + x2ufg

u2 = 417.4 + 0.63757 * 2088.2

u2 = 417.4 + 1331.374

u2 = 1748.774

Solving for m

m = V1/v1

m = 15L/1.08049

m = 0.015/1.08049

m = 0.01388Kg

So, -Qout = m(u2 - u1) becomes

0.01388(1748.774 - 2654.6)

= 0.01388 * -905.826

= -12.572

-Qout = -12.572

So, Qout = 12.572KJ

The dimensions of the room is 3 * 4 * 5.

The mass and volume of air in the room is given by

Volume = 3 * 4 * 6 = 72m³

Mass = P * V/RT

Where R = Gas Constant = 0.287

T = T1 = 7°C = 273+ 7 = 280K

P = P2 = 100kPa

Mass = (100 * 72)/(0.287 * 280)

Mass = 7200/80.36

Mass = 89.597Kg

The amount of work is defined by

Work = Fan work W * Time t

Work = 120 * 45 minutes

Work = 120 * 45 * 60seconds

Work = 324000J

Work (fan, in) = 324KJ

Taking the air in the room as the system

The energy balance is as follows

∆Energy = E(in) - E(out)

E(in) = Q(out) + W(fan,in)

E(out) = W(out)

∆Energy = ∆U

So, we have

∆U = Q(out) + W(fan,in) - W(out)

∆U + W(out) = Q(out) + W(fan,in)

∆U + W(out)  = 12.572 + 324

∆U + W(out) = 336.572

∆U + W(out) = ∆H (Change in heat energy)

This is given by mc(T2 - T1)

Where m = 89.597kg

c = specific heat of air at constant pressure = 1.005

T1 = 7°C

Substituting in these values, we have

336.572 = 89.597 *1.005 (T2 - 7)

=> 336.572 = 90.044985(T2 - 7)

T2 - 7 = 3.74

T2 = 3.74 + 7

T2 = 10.74°C