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
