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Chapter 4: Problem 111

A rigid tank whose volume is \(2 \mathrm{~m}^{3}\), initially containing air at 1 bar, \(295 \mathrm{~K}\), is connected by a valve to a large vessel holding air at \(6 \mathrm{bar}, 295 \mathrm{~K}\). The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of \(350 \mathrm{~K}\). Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in \(\mathrm{kJ}\).

Short Answer

Expert verified
The heat transfer is 1894.76 kJ.

Step by step solution

01

- Determine the initial mass of air in the tank

Use the ideal gas law to find the initial mass: \[ PV = nRT \] Given: \(P_1 = 1 \text{ bar} = 100 \text{ kPa}, V = 2 \text{ m}^3, T_1 = 295 \text{ K}, R = 0.287 \text{ kJ/kg.K} \) \[ m_1 = \frac{P_1 V}{R T_1} \]
02

- Solve for initial mass

Substitute the known values into the equation: \[ m_1 = \frac{(100 \text{ kPa}) (2 \text{ m}^3)}{0.287 \text{ kJ/kg.K}} (295 \text{ K}) = \frac{200}{84.565} 鈮� 2.366 \text{ kg} \]
03

- Determine the final mass of air in the tank

Use the ideal gas law again for the final state: Given: \(P_2 = 6 \text{ bar} = 600 \text{ kPa}, T_2 = 350 \text{ K} \) \[ m_2 = \frac{P_2 V}{R T_2} \]
04

- Solve for final mass

Substitute the known values into the equation: \[ m_2 = \frac{(600 \text{ kPa}) (2 \text{ m}^3)}{0.287 \text{ kJ/kg.K}} (350 \text{ K}) = \frac{1200}{100.45} \approx 11.95 \text{ kg} \]
05

- Apply the First Law of Thermodynamics

The First Law of Thermodynamics for a closed system (tank is not closed, but for simplification treat changes in mass): \[ \text{Q - W} = \text{螖 U} \] When work is zero (W=0), then: \[ Q = \text{螖 U} = \text{m}_2 \text{u}_2 - \text{m}_1 \text{u}_1 \] Specific internal energy change: \[ \text{螖U} = \text{m}_2 c_v (T_2 - T_1) + c_v T_1 (\text{m}_2 - \text{m}_1) \]
06

- Solve for heat transfer, assuming Ideal Gas Model

Using air specific heat at constant volume ( \( c_v = 0.718 \text{ kJ/kg.K} \)): \[ Q = 11.95 \text{ kg} (0.718 \text{ kJ/kg.K}) (350 \text{ K} - 295 \text{ K}) + 0.718 (295) (11.95 - 2.366) = 331.28 + 1563.48 = 1894.76 \text{ kJ} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is crucial for understanding how gases behave under various conditions of pressure, volume, and temperature. The equation is expressed as \(PV=nRT\) where:
  • \(P\) stands for pressure
  • \(V\) is the volume
  • \(n\) is the moles of gas
  • \(R\) is the universal gas constant
  • \(T\) is the absolute temperature
In the given problem, this law helps us find the initial and final masses of the air in the tank using given values for pressure, volume, and temperature. We rearrange to \(m = \frac{PV}{RT}\), given that \(PV=nRT\), and \(n = \frac{m}{M}\) where \(M\) is the molar mass. This simplification works because we are dealing with a specific type of gas鈥攁ir.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy in a closed system is constant. Energy can neither be created nor destroyed, just converted from one form to another. The mathematical form is expressed as \(Q - W = \text{螖U}\). In simpler terms:
  • \(Q\): Heat added to the system
  • \(W\): Work done by the system
  • \(螖U\): Change in internal energy
For the tank problem, we neglect work (\(W=0\)), making the equation simply \(Q = 螖U\). This allows us to easily compute the heat transfer, knowing that the internal energy change depends on mass and temperature change of the air.
Heat Transfer
Heat transfer dictates how much thermal energy is moved from one body to another. Here, calculating the heat transfer between the tank contents and the surroundings involves understanding changes in internal energy. When the air in the tank goes from the initial state to its final state, the internal energy change is tied to temperature and can be calculated with specific internal energy: \(Q = m_2 c_v (T_2 - T_1) + c_v T_1 (m_2 - m_1)\). This essentially means we look at how the internal energy shifts based on the different temperatures and masses before and after.
Specific Heat Capacity
Specific Heat Capacity (\(c_v\) for constant volume) is the quantity of heat needed to increase the temperature of a unit mass by one degree at constant volume. Air's specific heat at constant volume is about \(c_v = 0.718 \text{ kJ/kg.K}\), a constant used in our heat transfer calculations. It's essential because it helps us understand how much energy is needed to change the system's internal temperature.
Mass Calculation
Mass Calculation using the Ideal Gas Law is a primary step in understanding how systems change. Initially, we calculate the mass in the tank using: \(m_1 = \frac{P_1 V}{R T_1}\). After increasing pressure and temperature, the new mass is calculated similarly: \(m_2 = \frac{P_2 V}{R T_2}\). These formulas show how gases distributed in fixed volumes respond to shifts in other variables (pressure and temperature), thus providing a foundation for further thermodynamic analysis.

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Most popular questions from this chapter

Refrigerant \(134 \mathrm{a}\) exits a heat exchanger through \(0.75\)-in.diameter tubing with a mass flow rate of \(0.9 \mathrm{lb} / \mathrm{s}\). The temperature and quality of the refrigerant are \(-15^{\circ} \mathrm{F}\) and \(0.05\), respectively. Determine the velocity of the refrigerant, in \(\mathrm{m} / \mathrm{s}\).

Refrigerant \(134 a\) enters an insulated compressor operating at steady state as saturated vapor at \(-20^{\circ} \mathrm{C}\) with a mass flow rate of \(1.2 \mathrm{~kg} / \mathrm{s}\). Refrigerant exits at 7 bar, \(70^{\circ} \mathrm{C}\). Changes in kinetic and potential energy from inlet to exit can be ignored. Determine (a) the volumetric flow rates at the inlet and exit, each in \(\mathrm{m}^{3} / \mathrm{s}\), and (b) the power input to the compressor, in \(\mathrm{kW}\).

Liquid water flows isothermally at \(20^{\circ} \mathrm{C}\) through a oneinlet, one-exit duct operating at steady state. The duct's inlet and exit diameters are \(0.02 \mathrm{~m}\) and \(0.04 \mathrm{~m}\), respectively. At the inlet, the velocity is \(40 \mathrm{~m} / \mathrm{s}\) and pressure is 1 bar. At the exit, determine the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), and velocity, in \(\mathrm{m} / \mathrm{s}\).

Steam enters a turbine operating at steady state at \(700^{\circ} \mathrm{F}\) and \(450 \mathrm{lbf} / \mathrm{in}^{2}\) and leaves as a saturated vapor at \(1.2 \mathrm{lbf} / \mathrm{in} .^{2}\) The turbine develops \(12,000 \mathrm{hp}\), and heat transfer from the turbine to the surroundings occurs at a rate of \(2 \times 10^{6} \mathrm{Btu} / \mathrm{h}\). Neglecting kinetic and potential energy changes from inlet to exit, determine the volumetric flow rate of the steam at the inlet, in \(\mathrm{ft}^{3} / \mathrm{s}\).

A tank of volume \(1.2 \mathrm{~m}^{3}\) initially contains steam at \(8 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\). Steam is withdrawn slowly from the tank until the pressure drops to \(p\). Heat transfer to the tank contents maintains the temperature constant at \(400^{\circ} \mathrm{C}\). Neglecting all kinetic and potential energy effects and assuming specific enthalpy of the exiting steam is linear with the mass in the tank. (a) determine the heat transfer, in \(\mathrm{kJ}\), if \(p=2 \mathrm{MPa}\). (b) plot the heat transfer, in \(\mathrm{kJ}\), versus \(p\) ranging from \(0.5\) to \(8 \mathrm{MPa}\).
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