Edunes Online Education
Numericals on Thermodynamics:
1. Mass enters an open system with one inlet and one exit at a constant rate of 50 kg/min. At the exit, the mass flow rate is 60 kg/min. If the system initially contains 1000 kg of working fluid, determine the time when the system mass becomes 500 kg.
2. Mass leaves an open system with a mass flow rate of \( c \cdot m \), where c is a constant and m is the system mass. If the mass of the system at t = 0 is \( m_0 \), derive an expression for the mass of the system at time t.
3. Water enters a vertical cylindrical tank of cross-sectional area 0.01 \( m^2 \) at a constant mass flow rate of 5 kg/s. It leaves the tank through an exit near the base with a mass flow rate given by the formula 0.2h kg/s, where h is the instantaneous height in m. If the tank is empty initially, develop an expression for the liquid height h as a function of time t. Assume density of water to remain constant at 1000 \( kg/m^3 \).
4. A conical tank of base diameter D and height H is suspended in an inverted position to hold water. A leak at the apex of the cone causes water to leave with a mass flow rate of \( c \sqrt{h} \), where c is a constant and h is the height of the water level from the leak at the bottom.
- (a) Determine the rate of change of height h.
- (b) Express h as a function of time t and other known constants, \( \rho \) (constant density of water), D, H, and c if the tank was completely full at t=0.
5. Steam enters a mixing chamber at 100 kPa, 20 m/s, with a specific volume of 0.4 \( m^3/kg \). Liquid water at 100 kPa and \( 25 ^\circ C \) enters the chamber through a separate duct with a flow rate of 50 kg/s and a velocity of 5 m/s. If liquid water leaves the chamber at 100 kPa and \( 43 ^\circ C \) with a volumetric flow rate of 3.357 \( m^3/min \) and a velocity of 5.58 m/s, determine the port areas at the inlets and exit. Assume liquid water density to be 1000 \( kg/m^3 \) and steady state operation.
6. Air is pumped into and withdrawn from a 10 \( m^3 \) rigid tank as shown in the accompanying figure. The inlet and exit conditions are as follows. Inlet: \( v_1 \, = \, 2 \, m^3/kg \), \( V_1 \) = 10 m/s, \( A_1 \) = 0.01 \( m^2 \); Exit: \( v_2 \) = 5 \( m^3/kg \), \( V_2 \) = 5m/s, \( A_2 \) = 0.015 \( m^2 \). Assuming the tank to be uniform at all time with the specific volume and pressure related through pv = 9.0 \( kPa.m^3 \), determine the rate of change of pressure in the tank.
7. A gas flows steadily through a circular duct of varying cross-section area with a mass flow rate of 10 kg/s. The inlet and exit conditions are as follows. Inlet: \( V_1 \) = 400 m/s, \( A_1 \) = 179.36 \( cm^2 \); Exit: \( V_2 \) = 584 m/s, \( v_2 \) = 1.1827 m/kg.
- (a) Determine the exit area.
- (b) Do you find the increase in velocity of the gas accompanied by an increase in flow area counter intuitive? Why?
8. Steam enters a turbine with a mass flow rate of 10 kg/s at 10 MPa, \( 600 ^\circ C \), 30 m/s, it exits the turbine at 45 kPa, 30 m/s with a quality of 0.9. Assuming steady-state operation, determine
- (a) the inlet area, and
- (b) the exit area.
Mass enters an open system with one inlet and one exit at a constant rate of \( 50 \text{ kg/min} \). At the exit, the mass flow rate is \( 60 \text{ kg/min} \). If the system initially contains \( 1000 \text{ kg} \) of working fluid, determine the time when the system mass becomes $500 \text{ kg}$.
1. Identify the Given Data:
Inlet mass flow rate ($\dot{m}_{in}$): $50 \text{ kg/min}$
Exit mass flow rate ($\dot{m}_{out}$): $60 \text{ kg/min}$
Initial mass in the system ($m_{initial}$): $1000 \text{ kg}$
Final mass in the system ($m_{final}$): $500 \text{ kg}$
2. Determine the Rate of Change of Mass:
The continuity equation for an open system is:
$\dfrac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out}$
Substituting the given values:
$\dfrac{dm}{dt} = 50 \text{ kg/min} - 60 \text{ kg/min} = -10 \text{ kg/min}$
This means the system is losing mass at a constant rate of $10 \text{ kg}$ every minute.
3. Calculate the Total Change in Mass:
$\Delta m = m_{final} - m_{initial}$
$\Delta m = 500 \text{ kg} - 1000 \text{ kg} = -500 \text{ kg}$
4. Solve for Time ($t$):
Since the flow rates are constant, the time can be found using the formula:
$\Delta m = \left( \dfrac{dm}{dt} \right) \times t$
Rearranging for $t$:
$t = \dfrac{\Delta m}{\dfrac{dm}{dt}}$
$t = \dfrac{-500 \text{ kg}}{-10 \text{ kg/min}}$
$t = 50 \text{ minutes}$
Answer:
The system mass will become $500 \text{ kg}$ after $50 \text{ minutes}$.
Solution:
1. Conservation of Mass Principle:
For an open system, the rate of change of mass within the control volume is equal to the net mass flow rate across the boundary:
$\dfrac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out}$
2. Identify Given Parameters:
Inlet mass flow rate ($\dot{m}_{in}$): $0$ (since mass only leaves the system)
Exit mass flow rate ($\dot{m}_{out}$): $c \cdot m$
Initial condition: At $t = 0$, $m = m_0$
3. Set up the Differential Equation:
Substituting the values into the continuity equation:
$\dfrac{dm}{dt} = 0 - (c \cdot m)$
$\dfrac{dm}{dt} = -c \cdot m$
4. Solve using Separation of Variables:
Rearrange the equation to group $m$ terms and $t$ terms:
$\dfrac{dm}{m} = -c \, dt$
5. Integrate Both Sides:
Integrate from the initial state ($t=0, m=m_0$) to an arbitrary state at time $t$:
$\int_{m_0}^{m} \dfrac{1}{m} \, dm = \int_{0}^{t} (-c) \, dt$
Calculating the integrals:
$[\ln(m)]_{m_0}^{m} = [-ct]_{0}^{t}$
$\ln(m) - \ln(m_0) = -ct$
6. Simplify the Expression:
Using logarithmic properties ($\ln a - \ln b = \ln \dfrac{a}{b}$):
$\ln\left(\dfrac{m}{m_0}\right) = -ct$
To solve for $m$, take the exponential ($e$) of both sides:
$\dfrac{m}{m_0} = e^{-ct}$
$m(t) = m_0 e^{-ct}$
Final Expression:
The mass of the system at any time $t$ is given by:
$m(t) = m_0 e^{-ct}$
This expression shows that the mass in the system decreases exponentially over time.
Solution:
1. Conservation of Mass:
The rate of change of mass in the tank is the difference between the inlet and outlet mass flow rates:
$\frac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out}$
2. Express Mass ($m$) in terms of Height ($h$):
For a cylindrical tank: $m = \rho \cdot V = \rho \cdot A \cdot h$
Since $\rho$ and $A$ are constant, $\frac{dm}{dt} = \rho A \frac{dh}{dt}$.
3. Substitute Given Values:
$\dot{m}_{in} = 5$
$\dot{m}_{out} = 0.2h$
$\rho = 1000$
$A = 0.01$
The equation becomes:
$(1000 \cdot 0.01) \frac{dh}{dt} = 5 - 0.2h$
$10 \frac{dh}{dt} = 5 - 0.2h$
4. Separate Variables:
Divide both sides by $10$:
$\frac{dh}{dt} = 0.5 - 0.02h$
$\frac{dh}{0.5 - 0.02h} = dt$
5. Integrate:
Integrate from $t = 0$ (where $h = 0$ as the tank is empty) to time $t$:
$\int_{0}^{h} \frac{dh}{0.5 - 0.02h} = \int_{0}^{t} dt$
Using the integral rule $\int \frac{1}{a+bx} dx = \frac{1}{b}\ln(a+bx)$:
$\left[ \frac{\ln(0.5 - 0.02h)}{-0.02} \right]_{0}^{h} = t$
$\ln(0.5 - 0.02h) - \ln(0.5) = -0.02t$
6. Solve for $h$:
$\ln\left(\frac{0.5 - 0.02h}{0.5}\right) = -0.02t$
$\frac{0.5 - 0.02h}{0.5} = e^{-0.02t}$
$0.5 - 0.02h = 0.5e^{-0.02t}$
$0.02h = 0.5(1 - e^{-0.02t})$
$h(t) = 25(1 - e^{-0.02t})$
Final Expression:
The liquid height $h$ as a function of time $t$ is:
$h(t) = 25(1 - e^{-0.02t})$
Question 5:
Steam enters a mixing chamber at $100\text{ kPa}$, $20\text{ m/s}$, with a specific volume of $0.4\text{ m}^3/\text{kg}$. Liquid water at $100\text{ kPa}$ and $25^\circ\text{C}$ enters through a separate duct at $50\text{ kg/s}$ and $5\text{ m/s}$. Liquid water leaves at $100\text{ kPa}$ and $43^\circ\text{C}$ with a volumetric flow rate of $3.357\text{ m}^3/\text{min}$ and a velocity of $5.58\text{ m/s}$. Determine the port areas at the inlets and exit. (Assume $\rho_{liquid} = 1000\text{ kg/m}^3$ and steady state).
Solution:
1. Define State Points:
Inlet 1 (Steam): $P_1 = 100\text{ kPa}$, $V_1 = 20\text{ m/s}$, $v_1 = 0.4\text{ m}^3/\text{kg}$
Inlet 2 (Liquid Water): $\dot{m}_2 = 50\text{ kg/s}$, $V_2 = 5\text{ m/s}$, $\rho_2 = 1000\text{ kg/m}^3$
Exit 3 (Liquid Mixture): $\dot{Q}_3 = 3.357\text{ m}^3/\text{min}$, $V_3 = 5.58\text{ m/s}$, $\rho_3 = 1000\text{ kg/m}^3$
2. Calculate Exit Mass Flow Rate ($\dot{m}_3$):
First, convert the volumetric flow rate to $\text{m}^3/\text{s}$:
$\dot{Q}_3 = \dfrac{3.357}{60} = 0.05595\text{ m}^3/\text{s}$
Now, find the mass flow rate:
$\dot{m}_3 = \rho_3 \cdot \dot{Q}_3 = 1000 \cdot 0.05595 = 55.95\text{ kg/s}$
3. Find Steam Inlet Mass Flow Rate ($\dot{m}_1$):
Using the Conservation of Mass (Steady State):
$\dot{m}_1 + \dot{m}_2 = \dot{m}_3$
$\dot{m}_1 + 50 = 55.95 \implies \dot{m}_1 = 5.95\text{ kg/s}$
4. Calculate Port Areas ($A = \dfrac{\dot{m} \cdot v}{V}$ or $A = \dfrac{\dot{Q}}{V}$):
Area at Inlet 1 (Steam):
- $A_1 = \dfrac{\dot{m}_1 \cdot v_1}{V_1} = \dfrac{5.95 \cdot 0.4}{20}$
- $A_1 = 0.119\text{ m}^2$
Area at Inlet 2 (Liquid Water):
- Specific volume $v_2 = \dfrac{1}{\rho} = \dfrac{1}{1000} = 0.001\text{ m}^3/\text{kg}$
- $A_2 = \dfrac{\dot{m}_2 \cdot v_2}{V_2} = \dfrac{50 \cdot 0.001}{5}$
- $A_2 = 0.01\text{ m}^2$
Area at Exit 3:
- $A_3 = \dfrac{\dot{Q}_3}{V_3} = \dfrac{0.05595}{5.58}$
- $A_3 = 0.01002\text{ m}^2$ (approximately $0.01\text{ m}^2$)
Final Results:
| Port | Area ($\text{m}^2$) |
|---|---|
| Inlet 1 (Steam) | $0.119$ |
| Inlet 2 (Water) | $0.01$ |
| Exit 3 (Mixture) | $0.01$ |
Question 6:
Air is pumped into and withdrawn from a $10\text{ m}^3$ rigid tank. The conditions are:
Inlet: $v_1 = 2\text{ m}^3/\text{kg}$, $V_1 = 10\text{ m/s}$, $A_1 = 0.01\text{ m}^2$
Exit: $v_2 = 5\text{ m}^3/\text{kg}$, $V_2 = 5\text{ m/s}$, $A_2 = 0.015\text{ m}^2$
Assuming the tank is uniform at all times with $pv = 9.0\text{ kPa}\cdot\text{m}^3$, determine the rate of change of pressure in the tank.
Solution:
1. Conservation of Mass:
For the control volume (the tank):
$\dfrac{dm}{dt} = \dot{m}_{in} - \dot{m}_{out}$
2. Calculate Mass Flow Rates:
Inlet ($\dot{m}_{in}$):
$\dot{m}_1 = \dfrac{A_1 V_1}{v_1} = \dfrac{0.01 \cdot 10}{2} = 0.05\text{ kg/s}$
Exit ($\dot{m}_{out}$):
$\dot{m}_2 = \dfrac{A_2 V_2}{v_2} = \dfrac{0.015 \cdot 5}{5} = 0.015\text{ kg/s}$
3. Determine Rate of Change of Mass ($\dfrac{dm}{dt}$):
$\dfrac{dm}{dt} = 0.05 - 0.015 = 0.035\text{ kg/s}$
4. Relate Mass to Pressure:
The total mass in the tank is $m = \dfrac{\text{Volume}}{\text{specific volume}} = \dfrac{V}{v}$.
From the given relation $pv = 9.0$, we have $v = \dfrac{9.0}{p}$.
Substituting $v$ into the mass equation:
$m = \dfrac{V}{\left(\dfrac{9.0}{p}\right)} = \dfrac{V \cdot p}{9.0}$
Since the tank is rigid, Volume ($V = 10\text{ m}^3$) is constant:
$m = \dfrac{10}{9.0}p = \dfrac{10}{9}p$
5. Calculate Rate of Change of Pressure ($\dfrac{dp}{dt}$):
Differentiate the mass equation with respect to time:
$\dfrac{dm}{dt} = \dfrac{10}{9} \dfrac{dp}{dt}$
Substitute the value of $\dfrac{dm}{dt}$ calculated in step 3:
$0.035 = \dfrac{10}{9} \dfrac{dp}{dt}$
$\dfrac{dp}{dt} = \dfrac{0.035 \cdot 9}{10}$
$\dfrac{dp}{dt} = 0.0315\text{ kPa/s}$
Final Result:
The rate of change of pressure in the tank is $0.0315\text{ kPa/s}$. Since the value is positive, the pressure in the tank is increasing.
Question 7:
A gas flows steadily through a circular duct of varying cross-section area with a mass flow rate of $10 \text{ kg/s}$. The inlet and exit conditions are as follows:
Inlet: $V_1 = 400 \text{ m/s}$, $A_1 = 179.36 \text{ cm}^2$
Exit: $V_2 = 584 \text{ m/s}$, $v_2 = 1.1827 \text{ m}^3/\text{kg}$
- (a) Determine the exit area.
- (b) Do you find the increase in velocity of the gas accompanied by an increase in flow area counter-intuitive? Why?
Solution:
Part (a): Determine the exit area
1. Identify Given Data:
Mass flow rate ($\dot{m}$): $10 \text{ kg/s}$ (Constant for steady flow)
Exit Velocity ($V_2$): $584 \text{ m/s}$
Exit Specific Volume ($v_2$): $1.1827 \text{ m}^3/\text{kg}$
2. Use the Mass Flow Rate Equation:
The formula for mass flow rate is:
$\dot{m} = \frac{A_2 V_2}{v_2}$
3. Rearrange to solve for Exit Area ($A_2$):
$A_2 = \frac{\dot{m} \cdot v_2}{V_2}$
4. Substitute the values:
$A_2 = \frac{10 \cdot 1.1827}{584}$
$A_2 = \frac{11.827}{584}$
$A_2 \approx 0.02025 \text{ m}^2$
5. Convert to $\text{cm}^2$ (for comparison with inlet):
$A_2 = 0.02025 \times 10,000 = 202.5 \text{ cm}^2$
Part (b): Discussion on Intuition
Is it counter-intuitive?
At first glance, yes. Usually, we expect a fluid to speed up when the duct narrows (like a nozzle). Here, the velocity increased ($400 \rightarrow 584 \text{ m/s}$), but the area also increased ($179.36 \rightarrow 202.5 \text{ cm}^2$).
Why does this happen?
This occurs because the fluid is a compressible gas.
In incompressible flow (like water), $A_1V_1 = A_2V_2$. If $V$ goes up, $A$ must go down.
In compressible flow, the density ($\rho$) or specific volume ($v$) changes significantly.
In this specific case, the gas expanded so much (its specific volume increased) that the duct had to widen just to let the higher-volume flow pass through, even though it was moving faster. This is typical of supersonic flow in a divergent section or high-magnitude expansion.
Final Results:
Exit Area ($A_2$): $202.5 \text{ cm}^2$
Reasoning: The increase in area despite increased velocity is due to the significant increase in the gas's specific volume (expansion).
