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scoundrel [369]
4 years ago
7

It is proposed to use water instead of refrigerant-134a as the working fluid in air-conditioning applications where the minimum

temperature never falls below the freezing point. Can water be used as the working fluid in air-conditioning applications?
Engineering
1 answer:
slega [8]4 years ago
5 0

Answer:

No, water can't be used.

Explanation:

No, water cannot be used as the working fluid in air-conditioning applications.

This is because, if we assume the water is maintained at 10°C in the evaporator, the evaporator pressure will now be the saturation pressure that corresponds to this pressure, which in this case would be 1.2 kPa.

So we can conclude that for the refrigerants in the evaporator the temperature of a saturated pressure would be very low and so it's not practical to maintain it with water

Thus, it's is not practical to design refrigeration or air conditioning devices with water as the working fluid because it will involve extremely low pressures.

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The temperature of a gas is related to its absolute pressure and specific volume through T = 3.488pv, where T, p, and v are expr
tangare [24]

Answer:

0.425kPA \approx 0.43kPa

Explanation:

In order to solve the problem we must resort to Taylor's approximations in which it is possible to obtain an approximation through a polynomial function.

For the particular case we proceed to make a linear approach.

Our values are defined as,

T = 3.488pv \rightarrowRelation of temperature, pressure and volume

T_1 = 350K \rightarrow Initial Temperature

v_1 = 1m^3/Kg \rightarrow Specific Volume

T_2 = 355K \rightarrow Final Temperature

v_2 = 1.01m^3/kh \rightarrow Final Specific Volume

The previous equation can be expressed as function of pressure, i.e,

P = \frac{T}{3.488v}

We can differentiate the expression in function of temperature and the specific volume, then

Temperature:

\frac{dP}{dT} = \frac{1}{3.488v}

Volume

\frac{dP}{dv} = -\frac{T}{3.488v^2}

PART A) Then the total change of the pressure is given by,

\Delta P = \frac{dP}{dT} + \frac{dP}{dv}

\Delta P =  \frac{1}{3.488v}(T_2-T_1)-\frac{T}{3.488v^2}(v_2-v_1)

Replacing the values given, we have

\Delta P =  \frac{1}{3.488(1.01)}(355-350)-\frac{355}{3.488(1.01)^2}(1.01-1)

\Delta P = 0.43kPa

PART B) Now we can calculate the exact change in pressure through the general equation, that is

\Delta P = \frac{1}{3.488}(\frac{T_2}{v_2}-\frac{T_1}{v_1})

Replacing the values we have:

\Delta P =  \frac{1}{3.488}(\frac{355}{1.01}-\frac{350}{1})

\Delta P = 0.425kPA

We can conclude that the approximation by Taylor's theorem is close to the value calculated by the general expression.

7 0
4 years ago
6.
zhenek [66]
The answer is b i did the same thing and i got it right
5 0
3 years ago
Read 2 more answers
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