The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 178C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air.

A) The specific humidity of the air with the given parameters is;

w1 = 0.00967

B) The relative humidity of the air with the given parameters is;

Φ1 = 0.459

C) The enthalpy of the air in KJ/kg dry air with the given parameters is;

h1 = 49.75 KJ/Kg

Correct question is;

The dry- and wet-bulb temperatures of atmospheric air at 95 kPa are 25 and 17 °C, respectively. Determine (a) the specific humidity, (b) the relative humidity, and (c) the enthalpy of the air, in kJ/kg dry air.

We are given;

Atmospheric Pressure;P = 95 KPa

Dry temperature;T1 = 25 °C

Wet temperature;T2 = 17°C

A) From table A-4 attached and at temperature of 17°C and by interpolation, we have a saturation pressure of P_g2 = 1.938 kpa

First of all, we will calculate the specific humidity from the given pressure and saturation pressure with the formula;

w2 = (0.622 × P_g2)/(P - P_g2)

w2 = (0.622 × 1.938)/(95 - 1.938)

w2 = 0.013

Now, let's calculate specific humidity with the enthalpies at 17 °C and by interpolation. We have specific enthalpies from table A-4 as;

h_fg2 = 2460 KJ/Kg

h_g1 = 2546.5 KJ/Kg

h_f2 = 71.36 KJ/Kg

The formula for the specific humidity under these conditions is;

w1 = (c_p(T2 - T1) + w2•h_fg2)/(h_g1 - h_f2)

c_p of air has a value of 1.005 KJ/Kg.°C

Thus;

w1 = (1.005(17 - 25) + 0.013*2460)/(2546.5 - 71.36)

w1 = 0.00967

B) The relative humidity is determined from the equation;

Φ1 = (w1*p)/(0.622 + w1)p_g1

From table A-4 attached and at temperature of 25 °C, we have a saturation pressure of P_g1 = 3.1698 KPa

Φ1 = (0.00967*95)/(0.622 + 0.00967)3.1698

Φ1 = 0.459

C) For the enthalpy of air, h1 we will use the formula;

h1 = (c_p × T1) + (w1 × h_g1)

h1 = (1.005 × 25) + (0.00967 × 2546.5)

h1 = 49.75 KJ/Kg

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water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2