Chapter 9, Problem 9.101P

To determine

(a)

The value of mass flow for isothermal flow.

Answer to Problem 9.101P

The value of mass flow for isothermal flow is 6.731 kg/h

Explanation of Solution

Given Information:

The value of pressure at entrance, p₁=102000 Pa

The value of pressure at entrance, p₂=100000 Pa

The value of friction coefficient f=0.028

The diameter of the tube, D = 0.01 m

The length of the tube, L = 3.0 m

Concept Used:

$\begin{array}{l}\left(\frac{\dot{m}}{A}\right)=\frac{p_1{}^2-p_2{}^2}{RT\left[\frac{fL}{D}+2\ln \frac{p_1}{p_2}\right]}\\ \dot{m}=massflowrateinkg/s\end{array}$

Calculation:

$\begin{array}{l}{\left(\frac{\dot{m}}{A}\right)}^2=\frac{{(102000)}^2-{(100000)}^2}{(287)(293)\left[\frac{(0.028)3}{0.01}+2\ln \frac{102}{100}\right]}=\frac{404000000}{(84091)(8.4+0.0396)}=569.26\\ \left(\frac{\dot{m}}{A}\right)=23.86\\ \dot{m}=23.86\times A\\ =23.86\times \left(\frac{\pi }{4}{(0.01)}^2\right)=0.001874kg/s=6.731kg/h.\end{array}$

Conclusion:

The value of mass flow for isothermal flow is 6.731 kg/h.

To determine

(b)

The value of mass flow for adiabatic flow.

Answer to Problem 9.101P

The value of mass flow for adiabatic flow is 6.79 kg/h

Explanation of Solution

Given Information:

The value of temperature at entrance, T₀ = 293 K

The value of pressure at entrance, p₁ = 102000 Pa

The diameter of tube,D=0.01 m

The value of k for adiabatic flow = 1.4

Concept Used:

$\begin{array}{l}{p}_1={\rho }_{1}R{T}_1\\ {\rho }_1=densityofairatentrance\\ Thevelocityofairatentrance,a_0=\sqrt{kR{T}_0}\\ fromequation9.75:\\ \frac{V_1}{V_2}=\frac{p_2}{p_1}\\ fromequation9.74:\\ V_1{}^2=\frac{a_0{}^2(1-{(\frac{V_1}{V_2})}^2)}{kfL/D+(k+1)\ln (\frac{V_2}{V_1})}\\ \dot{m}={\rho }_1(\pi /4\times D^2)(V_1)\end{array}$

Calculation:

$\begin{array}{l}{p}_1=R{T}_1\\ {\rho }_1=\frac{p_1}{RT}=\frac{102000}{287(293)}=1.2129kg/m^3\\ Thevelocityofairatentrance,a_0=\sqrt{1.4\times 287\times 293}=\sqrt{11772.4}=343.11m/s\\ fromequation9.75:\\ \frac{V_1}{V_2}=\frac{100000}{102000}=0.9803\\ fromequation9.74:\\ V_1{}^2=\frac{{(343.11)}^2(1-{(0.9803)}^2)}{(1.4)(0.028\times 3/0.01)+(1.4+1)\ln (1.02)}=\frac{117724.47\times 0.039}{(1.4\times 8.4)+(2.4\times 0.0198)}=\frac{4591.25}{11.76+0.04752}=388.84\\ V_1=19.72m/s\\ \dot{m}=1.219(\pi /4\times {0.01}^2)(19.72)\\ \dot{m}=0.001887kg/s=6.79kg/h\end{array}$

Conclusion:

The value of mass flow for adiabatic flow is 6.79 kg/h.

To determine

(c)

The value of mass flow for incompressible flow.

Answer to Problem 9.101P

The value of mass flow for incompressible flow is 6.81 kg/h

Explanation of Solution

Given Information:

The value of temperature at entrance, T₀ = 293 K

The value of pressure at entrance, p₁ = 102000 Pa

The value of pressure at exit, p₂ = 10000 Pa

The change in pressure = p₁-p₂ = 2000 Pa

The diameter of tube, D=0.01 m

Concept Used:

$\begin{array}{l}{p}_1={\rho }_{1}R{T}_1\\ {\rho }_1=densityofairatentrance\\ Thevelocityofairatentrance,V_1{}^2=\frac{\Delta p}{(fLD)({\rho }_1/2)}\\ {\dot{m}}_{incompressible}={\rho }_1(\pi /4\times D^2)(V_1)\end{array}$

Calculation:

$\begin{array}{l}{\rho }_1=\frac{p_1}{RT}=\frac{102000}{287(293)}=1.2129kg/m^3\\ {\rho }_1=densityofairatentrance\\ Thevelocityofairatentrance,V_1{}^2=\frac{2000}{(0.028\times 3\times 0.01)(1.2129/2)}=\frac{2000}{5.12}=390.625\\ V_1=19.76m/s\\ {\dot{m}}_{incompressible}=(1.219((\pi /4\times {0.01}^2)(19.76)=0.001892kg/s=6.81kg/h\end{array}$

Conclusion:

The value of mass flow for incompressible flow is 6.81 kg/h.