Respuesta :
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
The approach that does not permit the transfer of substances in and out of the environment is called a closed system. The difference in the internal power is referred to as ΔU and heat as ΔQ.
The average temperature of the air is 10.74°C.
This can be estimated as:
Given,
- Time (t) = 45 minutes
- Initial Temperature (T 1) = 7°C
- Fan work W = 120 W
- Radiator Volume (V 1) = 15 L
- Initial Steam Pressure (P 1) = 200 kPa
- Initial Steam Temperature (Ts 1) = 200°C
- Final Steam Pressure (P 2) = 100 kPa
The radiator is the closed system and the energy is given as:
[tex]\text{-Qout} = \Delta \text{U}[/tex]
[tex]\text{-Qout} = \text{m(u2 - u1)}[/tex]
For calculating u2:
- At P1 and Ts1
P1 = 200 kPa, Ts1 = 200°C, u1 = 2654.6 KJ/Kg, v1 = 1.08049 m³/Kg
- At P2 = 100 kPa, v1 = v2
vf = 0.001043 m³, vg = 1.6941 m³/kg, uf = 417.4 kj/kg, ufg = 2088.2 kj/kg
[tex]\text{x2} = \dfrac{(\text{v2 - vf)}} {(\text{vg - vf)}}[/tex]
[tex]\text{x2} = \dfrac{(1.08049 - 0.001043)}{(1.6941 - 0.001043)}[/tex]
[tex]\text{x2} = \dfrac{1.079447}{1.693057}[/tex]
x2 = 0.63757
[tex]\text{u2} = \text{uf + x2 (ufg)}[/tex]
[tex]\text{u2} = 417.4 + 0.63757 \times 2088.2[/tex]
u2 = 417.4 + 1331.374
u2 = 1748.774
For calculating m:
[tex]\text{m} = \dfrac{\text{V1}}{\text{v1}}[/tex]
[tex]\text{m }= \dfrac{15 \text{L}}{1.08049}[/tex]
m = 0.01388 Kg
Hence, [tex]\text{-Qout} = \text{m(u2 - u1)}[/tex]
= 0.01388 (1748.774 - 2654.6)
[tex]= 0.01388 \times -905.826[/tex]
= -12.572
Q out = 12.572 KJ
The dimensions of the room is given as: 3×4×5
Volume = 3×4×5
[tex]\text{Mass} = P (\dfrac{V}{RT})[/tex]
R = Gas Constant = 0.287
T = T 1 = 7°C = 273+ 7 = 280 K
P = P2 = 100 kPa
[tex]\text{Mass} =\dfrac{ (100 \times 72)}{(0.287 \times 280)}[/tex]
Mass = 89.597 Kg
The work is given by:
Work = Fan work × Time
Work = 120 × 45 minutes
Work = 120 × 45 × 60 seconds
Work = 324000 J = 324 KJ
When the air in the room is considered as a system then:
∆Energy = E(in) - E(out)
E(in) = Q(out) + W(fan,in)
E(out) = W(out)
∆Energy = ∆U
∆U = Q(out) + W(fan,in) - W(out)
∆U + W(out) = 12.572 + 324
∆U + W(out) = 336.572
∆U + W(out) = ∆H
This is given by [tex]\text{mc (T2 - T1)}[/tex]
Where,
- m = 89.597 kg
- c = 1.005
- T 1 = 7°C
[tex]336.572 = 89.597 \times 1.005 (\text{T2} - 7)[/tex]
336.572 = 90.044985 (T2 - 7)
T2 - 7 = 3.74
T2 = 3.74 + 7
T2 = 10.74°C
Therefore, the average temperature is 10.74°C.
To learn more about closed system follow the link:
https://brainly.com/question/15839384
