Chapter 9, Problem 9.1P

To determine

(a)

The velocity V₂ and change in entropy (s₂− s₁) at the exit of the duct for air.

Answer to Problem 9.1P

At the exit of the duct for air.

The velocity $V_2=334.835\text{m/s}$ and change in entropy is $s_2-s_1=336.848\text{J/kgK}$

Explanation of Solution

Given information:

At section 1,

Pressure p₁ = 140 kPa

Temperature T₁ = 260° C

Velocity V₁ = 75 m/s

Downstream, at section 2, the values are

Pressure p₂ = 30 kPa

Temperature T₂ = 207° C

${\gamma }_{air}=1.4$

Gas constant for air R = 287 J/kg-K

The specific heat at constant pressure c_p = 1005 J/kg-K

Calculation:

For calculating the downstream velocity, the adiabatic steady-flow energy equation can be used

$c_p{T}_1+\frac{1}{2}{V}_1{}^2=c_p{T}_2+\frac{1}{2}{V}_2{}^2$

$1005\times 260+\frac{1}{2}{\left(75\right)}^2=1005\times 207+\frac{1}{2}{V}_2{}^2$

$V_2=334.835\text{m/s}$

Now, for calculating the entropy change

$s_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$

$s_2-s_1=1005\ln \left(\frac{207+273}{260+273}\right)-287\ln \left(\frac{30}{140}\right)$

$s_2-s_1=336.848\text{J/kgK}$

Conclusion:

Thus, at the exit of the duct for air, the velocity $V_2=334.835\text{m/s}$ and change in entropy is $s_2-s_1=336.848\text{J/kgK}$.

To determine

(b)

The velocity V₂ and change in entropy (s₂− s₁) at the exit of the duct for argon.

Answer to Problem 9.1P

At the exit of the duct for argon, the velocity V₂ = 246.034 m/s and change in entropy is $s_2-s_1=266.1596\text{J/kgK}$

Explanation of Solution

Given information:

At section 1,

Pressure p₁ = 140 kPa

Temperature T₁ = 260° C

Velocity V₁ = 75 m/s

Downstream, at section 2, the values are

Pressure p₂ = 30 kPa

Temperature T₂ = 207° C

${\gamma }_{air}=1.67$

Gas constant for air R = 208 J/kg-K

The specific heat at constant pressure c_p = 518 J/kg-K

Calculation:

For calculating the downstream velocity, the adiabatic steady-flow energy equation can be used

For Argon,

let us take

k = 1.67

The gas constant for Argon R = 208 J/kg.K

Specific heat at constant pressure c_p = 518 J/kg.K

$c_p{T}_1+\frac{1}{2}{V}_1{}^2=c_p{T}_2+\frac{1}{2}{V}_2{}^2$

$518\times 260+\frac{1}{2}{\left(75\right)}^2=518\times 207+\frac{1}{2}{V}_2{}^2$

$V_2=246.034\text{m/s}$

Now, for calculating the entropy change:

$s_2-s_1=c_p\ln \left(\frac{T_2}{T_1}\right)-R\ln \left(\frac{p_2}{p_1}\right)$

$s_2-s_1=518\ln \left(\frac{207+273}{260+273}\right)-208\ln \left(\frac{30}{140}\right)$

$s_2-s_1=266.1596\text{J/kgK}$

Conclusion:

Thus, At the exit of the duct for argon, the velocity V₂ = 246.034 m/s and change in entropy is $s_2-s_1=266.1596\text{J/kgK}\text{.}$.