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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.

Short Answer

Expert verified
The required steam flow rate is 811.08 kg/hr.

Step by step solution

01

Define the given values

Start by identifying and noting down all the given quantities: - Mass flow rate of water, \(\frac{dm_{w}}{dt} = 5 \text{ kg/s}\) - Initial temperature of water, \(T_{w} = 80^\text{o} C\) - Specific enthalpy of incoming water, \(h_{w} = 320.5 \text{ kJ/kg}\) - Temperature of steam, \(T_{s} = 100.5^\text{o} C\) - Specific enthalpy of steam, \(h_{s} = 2650 \text{ kJ/kg}\) - Final temperature of the mixture, \(T_{f} = 100^\text{o} C\) - Specific enthalpy of mixture, \(h_{f} = 421 \text{ kJ/kg}\)
02

Establish energy balance

Using the principle of conservation of energy for the steady flow heater, the energy entering should equal the energy leaving: \(\frac{dm_{w}}{dt} \times h_{w} + \frac{dm_{s}}{dt} \times h_{s} = ( \frac{dm_{w}}{dt} + \frac{dm_{s}}{dt} ) \times h_{f} \)
03

Substitute known values

Now substituting our known values into the energy balance equation: \(5 \times 320.5 + \frac{dm_{s}}{dt} \times 2650 = (5 + \frac{dm_{s}}{dt}) \times 421 \)
04

Simplify the equation

Next, we solve for the unknown steam flow rate. Start by simplifying the equation: \[1602.5 + 2650 \frac{dm_{s}}{dt} = 2105 + 421 \frac{dm_{s}}{dt}\]Combine like terms gives: \[2650 \frac{dm_{s}}{dt} - 421 \frac{dm_{s}}{dt} = 2105 - 1602.5\]Further simplifying: \[2229 \frac{dm_{s}}{dt} = 502.5\]
05

Solve for steam flow rate

Now divide both sides by 2229: \( \frac{dm_{s}}{dt} = \frac{502.5}{2229} = 0.2253 \text{ kg/s} \)
06

Convert to hourly rate

Finally, convert the steam flow rate from per second to per hour: \(\frac{dm_{s}}{dt} \times 3600 = 0.2253 \times 3600 = 811.08 \text{ kg/hr}\)

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Key Concepts

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

Steady Flow Conditions
In thermodynamics, steady flow conditions refer to a process where the flow parameters, such as mass flow rate, pressure, and temperature, remain constant over time. This does not mean that the system itself is not evolving, but rather that at any given point, the properties of the flow do not change. This simplification allows us to make predictions and calculations more consistent and easier to manage.

In the context of the water heater problem, the steady flow condition implies that the flow rates and temperatures of the water and steam entering and exiting the heater remain steady. This assumption is critical because it allows us to apply the conservation of energy principles straightforwardly, without accounting for fluctuating conditions.
  • **Mass flow rate and other properties are consistent over time.**
  • **Allows for straightforward application of conservation laws.**
  • **Simplifies calculations and predictions.**
This condition sets a stable basis to apply an energy balance which is integral to solving the problem.
Energy Balance
Energy balance is a principle in thermodynamics that states that the total energy entering a system must equal the total energy leaving the system, under steady flow conditions. This principle is derived from the first law of thermodynamics and is fundamental in solving energy transfer problems.

In this exercise, we use the energy balance to calculate the steam flow rate required to heat the water. The energy brought by the incoming water and the steam must equal the energy of the mixture leaving the heater.
Using the given parameters:
  • **Mass flow rate of water,** \(\frac{dm_{w}}{dt}\rho_h=5\text{kg/s}\)
  • **Specific enthalpy of water,** \(h_{w}=320.5\text{kJ/kg}\)
  • **Specific enthalpy of steam,** \(h_{s} = 2650\text{kJ/kg}\)
  • **Specific enthalpy of mixture,** \(h_{f}=421\text{kJ/kg}\)
The energy entering via the water and steam can be expressed as: \(\frac{dm_{w}}{dt} \times h_{w} + \frac{dm_{s}}{dt} \times h_{s} \) must equal the energy leaving the system: \(\frac{dm_{w}}{dt} + \frac{dm_{s}}{dt}) \times h_{f} \)
This balance helps determine the required steam flow rate to ensure the system remains in energy equilibrium.
Enthalpy
Enthalpy is a measure of the total energy of a thermodynamic system. It includes internal energy and the product of pressure and volume. In simpler terms, enthalpy encompasses the heat content of a system, making it a crucial parameter in thermal calculations.

In this problem, enthalpy plays a key role in understanding how energy is transferred during the heating process. The specific enthalpy represents the energy per unit mass. For each substance within the heater, the specific enthalpy allows us to quantify the energy content per kilogram.
  • Incoming water has a specific enthalpy of \(320.5 \text{kJ/kg}\).
  • Steam has a specific enthalpy of \(2650 \text{kJ/kg}\).
  • The leaving mixture has a specific enthalpy of \(421 \text{kJ/kg}\).
This information enables us to set up an energy balance equation incorporating these specific enthalpies and flow rates to find the required steam flow rate.

Understanding enthalpy allows students to grasp how energy is stored and transferred in terms of heat, making it easier to solve related problems.

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Most popular questions from this chapter

Air enters a compressor operating at steady state with a pressure of 1 bar, a temperature of \(20^{\circ} \mathrm{C}\), and a volumetric flow rate of \(0.25 \mathrm{~m}^{3} / \mathrm{s}\). The air velocity in the exit pipe is 210 \(\mathrm{m} / \mathrm{s}\) and the exit pressure is \(1 \mathrm{MPa}\). If each unit mass of air passing from inlet to exit undergoes a process described by \(p v^{1.34}=\) constant, determine the exit temperature, in \({ }^{\circ} \mathrm{C}\), and the diameter of the exit pipe, in \(\mathrm{mm}\).

Steam enters a well-insulated turbine operating at steady state at \(5 \mathrm{MPa}\) with a specific enthalpy of \(3000 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(9 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.1 \mathrm{MPa}\), specific enthalpy is \(2540 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(12 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in \(\mathrm{kW}\).

A \(0.5-\mathrm{m}^{3}\) tank contains ammonia, initially at \(-40^{\circ} \mathrm{C}, 8\) bar. A leak develops, and refrigerant flows out of the tank at a constant mass flow rate of \(0.04 \mathrm{~kg} / \mathrm{s}\). The process occurs slowly enough that heat transfer from the surroundings maintains a constant temperature in the tank. Determine the time, in s, at which half of the mass has leaked out, and the pressure in the tank at that time, in bar.

Steam enters a horizontal pipe operating at steady state with a specific enthalpy of \(3200 \mathrm{~kJ} / \mathrm{kg}\) and a mass flow rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). At the exit, the specific enthalpy is \(1900 \mathrm{~kJ} / \mathrm{kg}\). If there is no significant change in kinetic energy from inlet to exit, determine the rate of heat transfer between the pipe and its surroundings, in \(\mathrm{kW}\).

Air enters a one-inlet, one-exit control volume at 8 bar, \(600 \mathrm{~K}\), and \(40 \mathrm{~m} / \mathrm{s}\) through a flow area of \(20 \mathrm{~cm}^{2}\). At the exit, the pressure is 2 bar, the temperature is \(400 \mathrm{~K}\), and the velocity is \(350 \mathrm{~m} / \mathrm{s}\). The air behaves as an ideal gas. For steady-state operation, determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\). (b) the exit flow area, in \(\mathrm{cm}^{2}\).

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