Numerical Assignment On Thermodynamics, Mass And Heat Transfer

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Numericals on Thermodynamics:

Numerical Assignment 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 Air is pumped into and withdrawn from a 10 \( m^3 \) rigid tank as shown in the accompanying figure.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.

To solve Question 1 from the provided text, we can use the principle of conservation of mass for an open system.

Question 1:
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 $500 \text{ kg}$.

Solution:

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 $50 \text{ minutes}$.

Question 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 $t$.

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.

Question 3: Water enters a vertical cylindrical tank of cross-sectional area $0.01\text{ m}^2$ at a constant mass flow rate of $5\text{ kg/s}$. It leaves the tank through an exit near the base with a mass flow rate given by $0.2h\text{ kg/s}$, 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 $\rho = 1000\text{ kg/m}^3$).

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).