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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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A gas turbine operates with a regenerator and two stages of reheating and intercooling. Air enters this engine at 14 psia and 60
Rzqust [24]

Answer:

flow(m) = 7.941 lbm/s

Q_in = 90.5184 Btu/lbm

Q_out = 56.01856 Btu/lbm

Explanation:

Given:

- T_1 = 60 F = 520 R

- T_6 = 940 = 1400 R

- Heat ratio for air k = 1.4

- Compression ratio r = 3

- W_net,out = 1000 hp

Find:

mass flow rate of the air

rates of heat addition and rejection

Solution:

- Using ideal gas relation compute T_2, T_4, T_10:

                     T_2 = T_1 * r^(k-1/k)

                     T_2 = T_4 = T_10 = 520*3^(.4/1.4) = 711.744 R

- Using ideal gas relation compute T_7, T_5, T_9:

                     T_7 = T_6 * r^(-k-1/k)

                     T_7 = T_5 = T_9 = 1400*3^(-.4/1.4) = 1022.84 R

- The mass flow rate is obtained by:

                     flow(m) = W_net,out / 2*c_p*(1400-1022.84-711.744+520)

                     flow(m) = 1000*.7068 / 2*0.24*(1400-1022.84-711.744+520)

                     flow(m) = 7.941 lbm/s

- The heat input is as follows:

                     Q_in = c_p*(T_6 - T_5)

                     Q_in = 0.24*(1400 - 1022.84)

                     Q_in = 90.5184 Btu/lbm

- The heat output is as follows:

                     Q_out = c_p*(T_10 - T_1)

                     Q_out = 0.24*(711.744 - 520)

                    Q_out = 56.01856 Btu/lbm

                                           

                     

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Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss
klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

ΔU = Q + W ⇒ Q = ΔU -W

Q = 79 - (-63) = 142 kJ

Therefore, the gas gained heat by an amount of 142 kJ.

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3 years ago
Starting with the column in Problem 1, perform enough additional calculations to determine the effects of increasing fc from 500
Hoochie [10]

Answer:

Explanation:

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