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Chapter 4: Problem 42

In a steam power plant, steam flows steadily from the boiler to the turbine through a \(0.25 \mathrm{~m}\) diameter pipe. Steam conditions at the boiler end are found to be \(P=4.5 \mathrm{Mpa}\), \(T=400^{\circ} \mathrm{C}, h=3204.7 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.06475 \mathrm{~m}^{3} / \mathrm{kg}\). At the turbine end, steam conditions are found to be \(P=4 \mathrm{MPa}\), \(T=393^{\circ} \mathrm{C}, h=3196.4 \mathrm{~kJ} / \mathrm{kg}\) and \(v=0.073 \mathrm{~m}^{3} / \mathrm{kg}\). There is a heat loss of \(6 \mathrm{~kJ} / \mathrm{kg}\) from the pipeline. Compute the steam flow rate.

Short Answer

Expert verified
The steam flow rate is \(0.758 \mathrm{~kg / s}\).

Step by step solution

01

Identify known variables

At the boiler end:- Pressure, \(P_1 = 4.5 \mathrm{~MPa}\)- Temperature, \(T_1 = 400^{\circ}\mathrm{C}\)- Specific enthalpy, \(h_1 = 3204.7 \mathrm{~kJ} / \mathrm{kg}\)- Specific volume, \(v_1 = 0.06475 \mathrm{~m}^{3} / \mathrm{kg}\)At the turbine end:- Pressure, \(P_2 = 4 \mathrm{~MPa}\)- Temperature, \(T_2 = 393^{\circ}\mathrm{C}\)- Specific enthalpy, \(h_2 = 3196.4 \mathrm{~kJ} / \mathrm{kg}\)- Specific volume, \(v_2 = 0.073 \mathrm{~m}^{3} / \mathrm{kg}\)Heat loss in the pipeline, \(q_{loss} = 6 \mathrm{~kJ} / \mathrm{kg}\)
02

Apply the steady-flow energy equation

The steady-flow energy equation can be written as\[ h_1 + \frac{v_1^2}{2} + gz_1 = h_2 + \frac{v_2^2}{2} + gz_2 + q_{loss} \ \text{Since the elevation change and kinetic energy change are negligible:} \ h_1 = h_2 + q_{loss} \]Rearranging for the mass flow rate, we get:\[ \dot{m} = \frac{\text{Area} * v_1}{v_{1}} \]
03

Calculate the cross-sectional area of the pipe

The diameter of the pipe is given as \(d = 0.25 \mathrm{~m}\). The cross-sectional area, \(A\), of the pipe can be calculated using:\[ A = \frac{\pi d^2}{4} = \frac{\pi (0.25)^2}{4} = 0.0491 \mathrm{~m}^2 \]
04

Find the specific volume change and mass flow rate

Using the specific volume change from state 1 to state 2:\[ v_{1} \rightarrow v_{2} \]Substitute into the mass flow rate equation:\[ \dot{m} = \frac{A}{v_{1}} (h_1 - h_2 - q_{loss}) = \frac{0.0491 \mathrm{~m}^2}{0.06475 \mathrm{~m}^3 / \mathrm{kg}} (3204.7 - 3196.4 - 6) \]Calculating,\[ \dot{m} = 0.758 \mathrm{~kg / s} \]

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Key Concepts

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

Steady-Flow Energy Equation
The steady-flow energy equation is crucial for understanding how energy is conserved in a control volume. It states that the energy at the inlet of a system plus any energy added to the system is equal to the energy at the outlet plus any energy lost. In this problem, it simplifies to:\[ h_1 = h_2 + q_{loss} \]Here, \( h_1 \) and \( h_2 \) are the specific enthalpies at the boiler and turbine ends, and \( q_{loss} \) represents the heat loss per kilogram. This equation helps in understanding the energy changes that the steam undergoes as it travels through the system.
Specific Enthalpy
Specific enthalpy is the total heat content per unit mass of a substance. It includes both internal energy and the product of the pressure and volume of the fluid. In this exercise, specific enthalpies at the boiler (\( h_1 = 3204.7 \text{ kJ/kg} \)) and at the turbine (\( h_2 = 3196.4 \text{ kJ/kg} \)) are given. The difference in specific enthalpy, along with the heat loss, helps determine the mass flow rate. Large values indicate significant energy content, making it essential to track changes in enthalpy for calculating energy transformations.
Heat Loss in Pipeline
Heat loss in a pipeline indicates how much energy is lost to the surroundings as steam flows through the pipe. In this case, the heat loss is given as 6 kJ/kg. It is subtracted from the steam's specific enthalpy from the boiler to the turbine. This loss affects the total energy available for the turbine's operation. The steady-flow energy equation incorporates this loss to show how it impacts the steam's energy state from entry to exit.
Mass Flow Rate
The mass flow rate is a measure of the mass of steam flowing per unit time, usually expressed in kg/s. To calculate this for the given problem, we need the cross-sectional area of the pipe and the specific volume of the steam. With the diameter of the pipe given as 0.25 m, the area is found using:\[ A = \frac{\pi d^2}{4} = 0.0491 \text{ m}^2 \]The mass flow rate equation is:\[ \dot{m} = \frac{A}{v_1} \]Plugging in the known values: \[ \dot{m} = \frac{0.0491 \text{ m}^2}{0.06475 \text{ m}^3/\text{kg}} = 0.758 \text{ kg/s} \]This result tells us how much steam is flowing through the system, crucial for balancing energy inputs and outputs in the system.

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Most popular questions from this chapter

A \(0.5 \mathrm{~m}^{3}\) tank initially contains air at \(300 \mathrm{kPa}, 350 \mathrm{~K}\). Air slowly escapes from the tank until the pressure drops to \(100 \mathrm{kPa}\). The air that remains in the tank undergoes a process described by \(p v^{1.3}=\) constant. For a control volume enclosing the tank, determine the heat transfer, in kJ. Assume ideal gas behavior with constant specific heats.

A \(0.75 \mathrm{~m}^{3}\) capacity tank contains a two-phase liquid-vapor mixture at \(240^{\circ} \mathrm{C}\) and a quality of \(0.6 .\) Heat is transferred to the mixture and saturated vapor from the tank are withdrawn at \(240^{\circ} \mathrm{C}\) to maintain the constant pressure inside the tank by means of a pressure regulating value. This process continues until the tank has only saturated vapor left in it. Neglect the effect of kinetic and potential energies. Determine the amount of heat transfer in \(\mathrm{kJ}\).

The intake to a hydraulic turbine installed in a flood control dam is located at an elevation of \(10 \mathrm{~m}\) above the turbine exit. Water enters at \(20^{\circ} \mathrm{C}\) with negligible velocity and exits from the turbine at \(10 \mathrm{~m} / \mathrm{s}\). The water passes through the turbine with no significant changes in temperature or pressure between the inlet and exit, and heat transfer is negligible. The acceleration of gravity is constant at \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). If the power output at steady state is \(500 \mathrm{~kW}\), what is the mass flow rate of water, in \(\mathrm{kg} / \mathrm{s}\) ?

A water heater operating under steady flow conditions receives water at the rate of \(5 \mathrm{~kg} / \mathrm{s}\) at \(80^{\circ} \mathrm{C}\) temperature with specific enthalpy of \(320.5 \mathrm{~kJ} / \mathrm{kg}\). Water is heated by mixing steam at temperature \(100.5^{\circ} \mathrm{C}\) and specific enthalpy of 2650 \(\mathrm{kJ} / \mathrm{kg}\). The mixture of water and steam leaves the heater in the form of liquid water at temperature \(100^{\circ} \mathrm{C}\) with specific enthalpy of \(421 \mathrm{~kJ} / \mathrm{kg}\). Calculate the required steam flow rate to the heater per hour.

A pump steadily delivers water through a hose terminated by a nozzle. The exit of the nozzle has a diameter of \(2.5 \mathrm{~cm}\) and is located \(4 \mathrm{~m}\) above the pump inlet pipe, which has a diameter of \(5.0 \mathrm{~cm}\). The pressure is equal to 1 bar at both the inlet and the exit, and the temperature is constant at \(20^{\circ} \mathrm{C}\). The magnitude of the power input required by the pump is \(8.6 \mathrm{~kW}\), and the acceleration of gravity is \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the mass flow rate delivered by the pump, in \(\mathrm{kg} / \mathrm{s}\).
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