Consider a low speed open circuit subsonic wind tunnel with an inlet to throat area ratio of 12. The tunnel is turned on and the pressure difference between the inlet (the settling chamber) and the test section is read as a height difference of 10 cm on a U-tube mercury manometer. (The density of liquid mercury is 1.36 x 10-4 kg/m3 ). Calculate the velocity of the air in the test section.

velocity = 147.57 m/s

Explanation:

given data

inlet to throat area ratio = 12

height difference Δh = 10 cm = 0.1 m

density of liquid mercury = 1.36 × kg/m³

solution

we get here weight of mercury that is express as

weight of mercury = density of liquid mercury ×  g    .................1

weight of mercury = 1.36 × × 9.8

weight of mercury = 1.33 ×  N/m²  

and

area ratio is

 = 12

so velocity of air in the test section will be

velocity =      .......................1

put here value and we get

velocity =    

velocity = 147.57 m/s

98.13kJ

Explanation:

Given that;

The rigid 10-L vessel initially contains a mixture of liquid water & vapor at

We are to calculate the heat transfer required during the process by obtaining our data from the steam tables.

In order to do that, let start with our Energy Balance

So, Energy Balance for closed rigid tank system is given as:

Since the K.E and P.E are insignificant;

∴ K.E = P.E = 0

Where;

m = mass flow rate of the mixture

= corresponding change in the internal energy at state point 2 and 1

However, since we are informed that the vessel is rigid, then there is no work done in the system, then W turn out to be equal to zero .i,e

W = 0

we have our above equation re-written as:

We were told to obtain our data from the steam table, so were going to do just that

∴  At inlet temperature , the given quality of mixture of liquid water and vapor = 123% = 0.123

Using the equation:

where;

= specific volume at state 1

= specific volume of the liquid

= specific volume of the liquid vapor mixture

The above data from the steam table is given as;

 = 0.001043 m³/kg

= 1.6720 m³/kg

so; we have

At   = 100°C and ;

the following steam data from the tables were still obtained for the internal energy; which is given as:

Internal Energy at the state 1

where;

specific internal energy of the liquid    = 419.06 kJ/kg

The specific internal energy of the liquid vapor mixture = 2087.0 kJ/kg

∴ since ;

 = 419.06 + (0.123 × 2087.0)

 = 675.761 kJ/kg

As the tank is rigid, so as the volume which is kept constant:

Now, let take a look at when from the data in the steam tables

Specific volume of the liquid = 0.00113 m³/kg

specific volume of the liquid vapor mixture = 0.19384 m³/kg

The quality of the mixture at the final state 2 can be determined  by using the  equation shown below:

    = 1.0657

From our usual steam table; we still obtained data for the Internal Energy when

Specific internal energy of the liquid = 761.92 kJ/kg

Specific internal energy of the liquid vapor mixture = 1820.88 kJ/kg

Calculating the internal energy at finsl state point 2 ; we have:

= 761.92 + (1.0657 × 1820.88)

= 761.92 + 1940.511816

= 2702.431816

≅ 2702.43 kJ/kg

Furthermore, let us calculate the mass in the system; we have:

where V₁ = the volume 10 - L given by the system and v₁ = specific volume at state 1 as 0.2065

V₁ = the volume 10 - L = 10  × ( 0.001 m³/L)

v₁ = 0.2065

∴

mass (m) =  

= 0.04842 kg

Now, we gotten all we nee do calculate for the heat transfer that is required during the process:

Therefore, the heat transfer that is required during the process = 98.13 kJ

There you have it!, I hope this really helps alot!