ENGINEERING COLLEGE

A refrigeration system was checked for leaks. The system temperature and surroundings were 75°F when the system was charged with nitrogen to 100 psig. The temperature then dropped to 50°F. What should the pressure be if no nitrogen has escaped?A) 9 psig
B) 94 psig
C) 100 psig
D) 90 psig

Answers

Answer 1

Answer:

The temperature and pressure of an ideal gas are directly proportional

The pressure of the system should be in the range B) 94 psig

The given refrigerator parameters are;

The temperature of the system and the surrounding, T₁ = 75 °F = 237.0389 K

The pressure to which the system was charged with nitrogen, P₁ = 100 psig

The temperature to which the system dropped, T₂ = 50 °F = 283.15 K

The required parameter;

The pressure, P₂, of the system at 50°F

Method:

The relationship between pressure and temperature is given by Gay-Lussac's law as follows;

At constant volume, the pressure of a given mass of gas is directly proportional to its temperature in Kelvin

Mathematically, we have;

$(P_1)/(T_1) = \mathbf{(P_2)/(T_2)}$

Plugging in the values of the variables gives;

$\mathbf{(100 \ psig)/(297.0389)} = (P_2)/(283.15)$

Therefore;

$P_2 = \mathbf{283.15 \, ^(\circ)F * (100 \ psig)/(297.0389\ ^(\circ)F) \approx 95.3 \, ^(\circ)F}$

The closest option to the above pressure is option B) 94 psig

Learn more about Gay-Lussac's law here;

brainly.com/question/16302261

Answer 2

Answer: B 94 psig because the temperature dropped too

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Answers

Answer: true

Explanation:

Propane (C3H8) burns completely with 150% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. The combustion products leave at 1 atm. Determine the mole fraction of water in the products, in lbmol(water)/lbmol(products).

Answers

Answer:

y=0.12 lbmol(water)/lbmol(products)

Explanation:

First we find the humidity of air. Using humidity tables and the temperature we find that is 0.01 g water/L air.

Now we set the equation assuming dry air:

$C_(3)H_(8)+7.5(O_(2)+3.76N_(2)) \longrightarrow 3CO_(2)+4H_(2)O+2.5O_(2)+28.2N_(2)$

With this equation we have almost all moles that exit the reactor, we are just missing the initial moles of water due to the humidity. So we proceed to calculate it with the ideal gas law:

PV=nRT

Vair=867.7L

With the volume and the fraction of water, we can calculate the mass of water:

0.01 * 867.7=8.677 g of water

Now we calculate the moles of water:

8.677 g / 18 g/mol = 0.48 moles of water

Now we can calculate the total moles of water in the exit of the reactor:

0.48 + 4 = 4.48 moles of water

And finally we just need to sum all moles at the exit of the reactor and divide:

3 moles of CO2 + 2.5 moles of O2+ 4.48 moles of H2O + 28.2 moles of N2

And we have 38.18 moles in total, then:

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The structure of a house is such that it loses heat at a rate of 5400 kJ/h per degree Cdifference between the indoors and outdoors. A heat pump that requires a power input of 6 kW isused to maintain this house at 21 C. Determine the lowest outdoor temperature for which the heatpump can meet the heating requirements of this house

Answers

Answer: Tl = - 13.3°C

the lowest outdoor temperature is - 13.3°C

Explanation:

Given that;

Temperature of Th = 21°C = 21 + 273 = 294 K

the rate at which heat lost is Qh = 5400 kJ/h°C

the power input to heat pump Wnet = 6 kw

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Answers

Answer:

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Explanation:

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