Chapter 3, Problem 3.28P

To determine

i.

The mass flow rate.

Answer to Problem 3.28P

The mass flow rate is 0.1174kg/s

Explanation of Solution

Given:

Diameter of the duct = 2 cm

At section 1:

Temperature = 345 K

Pressure = 1.1 MPa

At section 2:

Temperature = 298 K

Density = 1.34 kg/m³

Concept Used:

$\begin{array}{l}Speedofair=\sqrt{\gamma RT}\\ where,\\ \gamma ofair=1.4\\ R=287J/K\end{array}$

$Machnumber=\frac{Speedofairinthetube}{Idealspeedofair}$

$Massflowrate=\rho *Areaofcross\sec tion*velocity$

Calculation:

From Table A.4, for argon gas $R=287J/Kand\gamma =1.4$

Let us calculate the speed of air in section 2.

$\begin{array}{l}Speedofair=\sqrt{\gamma RT}\\ Speedofair=\sqrt{1.4*298*287}\\ Speedofair=346m/s\end{array}$

$Machnumber=\frac{Speedofairinthetube}{Idealspeedofair}$

$\begin{array}{l}Speedofairinthetube=Machnumber*Idealspeedofair\\ Speedofairinthetube=0.9*346\\ Speedofairinthetube=311.4\end{array}$

$\begin{array}{l}Massflowrate=\rho *Areaofcross\sec tion*velocity\\ FromtableA.4\\ \rho =11.8/9.81kg/m^3\\ \rho =1.20kg/m^3\\ Massflowrate=1.20*\frac{\pi *{0.02}^2}{4}*311.4\\ Massflowrate=0.1174kg/s\end{array}$

Conclusion:

The mass flow rate is 0.1174kg/s.

To determine

ii.

The pressure in section 2.

Answer to Problem 3.28P

The pressure in section 2 is 0.1146 MPa

Explanation of Solution

Given:

Diameter of the duct = 2 cm

At section 1:

Temperature = 345 K

Pressure = 1.1 MPa

At section 2:

Temperature = 298 K

Density = 1.34 kg/m³

Concept Used:

$P=\rho RT$

Calculation:

$\begin{array}{l}{P}_2={\rho }_{2}R{T}_2\\ P_2=1.34*287*298\\ P_2=0.1146MPa\end{array}$

Conclusion:

The final pressure is 0.1146 MPa.

To determine

iii.

The velocity in section 2.

Answer to Problem 3.28P

The velocity in section 2 is 311.4 m/s

Explanation of Solution

Given:

Diameter of the duct = 2 cm

At section 1:

Temperature = 345 K

Pressure = 1.1 MPa

At section 2:

Temperature = 298 K

Density = 1.34 kg/m³

Concept Used:

$\begin{array}{l}Speedofair=\sqrt{\gamma RT}\\ where,\\ \gamma ofair=1.4\\ R=287J/K\end{array}$

$Machnumber=\frac{Speedofairinthetube}{Idealspeedofair}$

Calculation:

From Table A.4, for argon gas $R=287m^2/s^2-Kand\gamma =1.4$

Let us calculate the speed of air in section 2

$\begin{array}{l}Speedofair=\sqrt{\gamma RT}\\ Speedofair=\sqrt{1.4*298*287}\\ Speedofair=346m/s\end{array}$

$Machnumber=\frac{Speedofairinthetube}{Idealspeedofair}$

$\begin{array}{l}Speedofairinthetube=Machnumber*Idealspeedofair\\ Speedofairinthetube=0.9*346\\ Speedofairinthetube=311.4m/s\end{array}$

Conclusion:

The speed of air in section 2 is 311.4 m/s.

To determine

iv.

The change of entropy.

Answer to Problem 3.28P

The change in entropy is 544 J/K

Explanation of Solution

Given:

Diameter of the duct = 2 cm

At section 1:

Temperature = 345 K

Pressure = 1.1 MPa

At section 2:

Temperature = 298 K

Density = 1.34 kg/m³

Concept Used:

$S_2-S_1=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_2}{P_1}\right)$

Calculation:

$\begin{array}{l}{C}_p-C_v=R\\ C_p-C_v=287J/K\\ \frac{C_p}{C_v}=\gamma \\ \frac{C_p}{C_v}=1.4\\ Fromtheabovetworelationsweget{C}_p=717.5m^2/(s^2-K)\\ S_2-S_1=C_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{P_2}{P_1}\right)\\ S_2-S_1=717.5*\ln \left(\frac{298}{345}\right)-287\ln \left(\frac{0.1146MPa}{1.1MPa}\right)\\ S_2-S_1=544J/K\end{array}$

Conclusion:

The change in entropy is 544 J/K.

To determine

v.

The change in entropy.

Explanation of Solution

When air moves from section 1 to section 2 which is 67 meters below section 1, it experiences a change in entropy of 544 J/K. This is because of the fact that there is a considerable amount of friction action on surface of the tube. This friction generates the above quantity of entropy. We can also see the effect of friction on the air by looking at the change in density of air from section1 to section 2.