Chapter 3, Problem 3.182P

To determine

The initial rate of temperature rise of the air in the tank.

Answer to Problem 3.182P

The initial rate of temperature rise of the air in the tank is $3.2\text{K}/\text{s}$.

Explanation of Solution

Given information:

The tank is insulated. The initial temperature inside the tank is $20°\text{C}$ and initial pressure is $200\text{kPa}$ . The gas is ideal with initial mass flow rate in the tank is $0.013\text{kg}/\text{s}$.

Write the expression for the unsteady flow energy equation for the control volume of the tank.

$\begin{array}{c}\left[\begin{array}{l}{Q}_{\dot{}}+W_{\dot{}}+m_{\dot{}}\left(u+\frac{p}{\rho }+\frac{V^2}{2}+gz\right)-Q_{\dot{}}-\\ W_{\dot{}}-m_{\dot{}}\left(u^{\prime }+\frac{p^{\prime }}{\rho }+\frac{{V^{\prime }}^2}{2}+g{z}^{\prime }\right)\end{array}\right]=\frac{d}{dt}\left({\displaystyle \int e\rho dv}\right)\\ \left[\begin{array}{c}\left(Q_{\dot{}}-Q_{\dot{}}\right)+\left(W_{\dot{}}-W_{\dot{}}\right)+\\ m_{\dot{}}\left(u+\frac{p}{\rho }+\frac{V^2}{2}+gz\right)-m_{\dot{}}\left(u^{\prime }+\frac{p^{\prime }}{\rho }+\frac{{V^{\prime }}^2}{2}+g{z}^{\prime }\right)\end{array}\right]=\frac{d}{dt}\left({\displaystyle \int e\rho dv}\right)\end{array}$ ...(I)

Here, the rate of the heat inlet to the control volume is $Q_{\dot{}}$, the rate of work inlet to the control volume is $W_{\dot{}}$, mass flow rate into the system is $m_{\dot{}}$, the rate of heat outlet to the control volume is $Q_{\dot{}}$, the rate of work outlet to the control volume is $W_{\dot{}}$, mass flow rate outside the system is $m_{\dot{}}$, the energy stored in the control volume is $e$, the density is $\rho$, the volumetric change is $dv$, the specific internal energy at inlet is $u$, the specific internal energy at outlet is $u^{\prime }$, the pressure at the inlet is $p$, the pressure at the outlet is $p^{\prime }$, the velocity at the inlet is $V$, the velocity at the outlet is $V^{\prime }$, the datum height at inlet is $z$ and the datum height at the outlet is $z^{\prime }$.

Since there is no energy interactions with the control volume so,

$\begin{array}{c}\left(Q_{\dot{}}-Q_{\dot{}}\right)=0\\ \left(W_{\dot{}}-W_{\dot{}}\right)=0\end{array}$

Since there is no mass leaving from the tank so,

$m_{\dot{}}=0$

Since the velocity at inlet and the datum are negligible.

$\begin{array}{c}V=0\\ z=0\end{array}$

Substitute $0$ for $\left(Q_{\dot{}}-Q_{\dot{}}\right)$, $0$ for $m_{\dot{}}$, $0$ for $V$, $0$ for $z$ and $0$ for $\left(W_{\dot{}}-W_{\dot{}}\right)$ in Equation (I).

$\begin{array}{l}\left[\begin{array}{c}0+0+\\ m_{\dot{}}\left(u+\frac{p}{\rho }+\frac{0^2}{2}+g\times 0\right)-0\times \left(u^{\prime }+\frac{p^{\prime }}{\rho }+\frac{{V^{\prime }}^2}{2}+g{z}^{\prime }\right)\end{array}\right]=\frac{d}{dt}\left({\displaystyle \int e\rho dv}\right)\\ m_{\dot{}}\left(u+\frac{p}{\rho }\right)-\frac{d}{dt}\left({\displaystyle \int e\rho dv}\right)=0\\ m_{\dot{}}\left(u+\frac{p}{\rho }\right)-\frac{d}{dt}\left(e\rho v\right)=0\end{array}$ ...(II)

Write the expression for the energy stored by the control volume.

$e=C_{v}T$

Here, the specific heat at constant volume is $C_v$ and the temperature is $T$.

Write the expression for the enthalpy.

$h=u+pv$

Here, the enthalpy is $h$.

Write $h$ for $u+pv$ and $C_{v}T$ for $e$ in Equation (II).

$m_{\dot{}}\left(h\right)-\frac{d}{dt}\left(\rho v{C}_{v}T\right)=0$ ...(III)

Write the expression for the specific enthalpy.

$h=C_{p}T$

Here, the specific heat at constant pressure is $C_p$.

Substitute $C_{p}T$ for $h$ in Equation (III).

$\begin{array}{l}{m}_{\dot{}}\left(C_{p}T\right)-\frac{d}{dt}\left(\rho v{C}_{v}T\right)=0\\ m_{\dot{}}{C}_{p}T-\frac{d}{dt}\left(\rho v{C}_{v}T\right)=0\\ m_{\dot{}}{C}_{p}T-\rho v{C}_v\frac{dT}{dt}-\left(v{C}_{v}T\right)\frac{d\rho }{dt}=0\end{array}$ ...(IV)

Write the expression for the mass flow rate at inlet with respect to time.

$\begin{array}{c}{m}_{\dot{}}=\frac{d}{dt}\left(\rho v\right)\\ =v\frac{d\rho }{dt}\end{array}$

Substitute $m_{\dot{}}$ for $v\frac{d\rho }{dt}$ in Equation (IV).

$\begin{array}{l}{m}_{\dot{}}{C}_{p}T-\rho v{C}_v\frac{dT}{dt}-m_{\dot{}}\left(C_{v}T\right)=0\\ \rho v{C}_v\frac{dT}{dt}=m_{\dot{}}\left(C_p-C_v\right)T\\ \frac{dT}{dt}=\frac{m_{\dot{}}\left(C_p-C_v\right)T}{\rho v{C}_v}\end{array}$ ...(V)

Write the expression for the ideal gas equation.

$\begin{array}{l}p=\rho RT\\ \rho =\frac{p}{RT}\end{array}$

Substitute $\frac{p}{RT}$ for $\rho$ in Equation (V).

$\frac{dT}{dt}=\frac{m_{\dot{}}\left(C_p-C_v\right)T}{\left(\frac{p}{RT}\right)v{C}_v}$ ...(VI)

Calculation:

Convert the temperature from $°\text{C}$ to $°\text{K}$.

$\begin{array}{c}T=20°\text{C}\\ =\left(20+273\right)\text{K}\\ =\text{293}\text{K}\end{array}$

Substitute $0.013\text{kg}/\text{s}$ for $m_{\dot{}}$, $200\text{L}$ for $v$, $293\text{K}$ for $T$, $200\text{kPa}$ for $p$, $287\text{J}/\text{kg}\cdot \text{K}$ for $R$, $1005\text{J}/\text{kg}\cdot \text{K}$ for $C_p$ and $718\text{J}/\text{kg}\cdot \text{K}$ for $C_v$ in Equation (VI).

$\begin{array}{c}\frac{dT}{dt}=\frac{\left(0.013\text{kg}/\text{s}\right)\left\{\left(1005\text{J}/\text{kg}\cdot \text{K}\right)-\left(718\text{J}/\text{kg}\cdot \text{K}\right)\right\}\left(293\text{K}\right)}{\left(\frac{\left(200\text{kPa}\right)}{\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(\text{293}\text{K}\right)}\right)\left(200\text{L}\right)\left(718\text{J}/\text{kg}\cdot \text{K}\right)}\\ =\frac{\left(0.013\text{kg}/\text{s}\right)\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(293\text{K}\right)}{\left(\frac{\left(200\text{kPa}\right)\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{\left(287\text{J}/\text{kg}\cdot \text{K}\right)\left(\text{293}\text{K}\right)}\right)\left(200\text{L}\right)\left(\frac{1{\text{m}}^3}{1\times {10}^3\text{L}}\right)\left(718\text{J}/\text{kg}\cdot \text{K}\right)}\\ =\frac{1093.183}{341.53}\text{K}/\text{s}\\ =3.2\text{K}/\text{s}\end{array}$

Conclusion:

The initial rate of temperature rise of the air in the tank is $3.2\text{K}/\text{s}$.