Numerical Assignment On Thermodynamics, Mass And Heat Transfer

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 Air is pumped into and withdrawn from a 10 \( m^3 \) rigid tank as shown in the accompanying figure.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

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 $50 \text{ minutes}$.