Business, 15.02.2020 02:57 ryleighgilmore1

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.

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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 minutesInitial Temperature (T 1) = 7°CFan work W = 120 WRadiator Volume (V 1) = 15 LInitial Steam Pressure (P 1) = 200 kPaInitial Steam Temperature (Ts 1) = 200°CFinal Steam Pressure (P 2) = 100 kPa

The radiator is the closed system and the energy is given as:

$\text{-Qout} = \Delta \text{U}$

$\text{-Qout} = \text{m(u2 - u1)}$

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

$\text{x2} = \dfrac{(\text{v2 - vf)}} {(\text{vg - vf)}}$

$\text{x2} = \dfrac{(1.08049 - 0.001043)}{(1.6941 - 0.001043)}$

$\text{x2} = \dfrac{1.079447}{1.693057}$

x2 = 0.63757

$\text{u2} = \text{uf + x2 (ufg)}$

$\text{u2} = 417.4 + 0.63757 \times 2088.2$

u2 = 417.4 + 1331.374

u2 = 1748.774

For calculating m:

$\text{m} = \dfrac{\text{V1}}{\text{v1}}$

$\text{m }= \dfrac{15 \text{L}}{1.08049}$

m = 0.01388 Kg

Hence, $\text{-Qout} = \text{m(u2 - u1)}$

= 0.01388 (1748.774 - 2654.6)

$= 0.01388 \times -905.826$

= -12.572

Q out = 12.572 KJ

The dimensions of the room is given as: 3×4×5

Volume = 3×4×5

$\text{Mass} = P (\dfrac{V}{RT})$

R = Gas Constant = 0.287

T = T 1 = 7°C = 273+ 7 = 280 K

P = P2 = 100 kPa

$\text{Mass} =\dfrac{ (100 \times 72)}{(0.287 \times 280)}$

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 $\text{mc (T2 - T1)}$

Where,

m = 89.597 kgc = 1.005T 1 = 7°C

$336.572 = 89.597 \times 1.005 (\text{T2} - 7)$

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.

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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

Answer from: krystalhurst97

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

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