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Chapter 4: Problem 60

Steam at a pressure of \(0.10\) bar and a quality of \(94.2 \%\) enters a shell- and-tube heat exchanger where it condenses on the outside of tubes through which cooling water flows, exiting as saturated liquid at \(0.10\) bar. The mass flow rate of the condensing steam is \(4.2 \times 10^{5} \mathrm{~kg} / \mathrm{h}\). Cooling water enters the tubes at \(20^{\circ} \mathrm{C}\) and exits at \(40^{\circ} \mathrm{C}\) with negligible change in pressure. Neglecting stray heat transfer and ignoring kinetic and potential energy effects, determine the mass flow rate of the cooling water, in \(\mathrm{kg} / \mathrm{h}\), for steady-state operation.

Short Answer

Expert verified
11.04 脳 10^6 kg/h

Step by step solution

01

Identify Given Data

First, identify the given data. - Pressure of steam: 0.10 bar - Quality of steam: 94.2% - Mass flow rate of condensing steam: 4.2 脳 10^5 kg/h - Cooling water inlet temperature: 20掳C - Cooling water outlet temperature: 40掳C
02

Find Enthalpy Values

Next, use steam tables to find the specific enthalpies for the steam at specific conditions. Since the steam exits as saturated liquid at 0.10 bar: - Enthalpy of saturated liquid (hf): approximately 191.81 kJ/kg - Initial enthalpy (mixed quality): \[ h_{initial} = h_{f} + x \times h_{fg} \ = 191.81 + 0.942 \times (2392.8) \ = 2387.98 \text{ kJ/kg} \]
03

Calculate Heat Released by Steam

To calculate the heat released during condensation: Use the formula: \[ Q = \text{mass flow rate of steam} \times (h_{initial} - h_{final}) \ = 4.2 \times 10^5 \times (2387.98 - 191.81) \text{ kJ} \] So, \[ Q = 4.2 \times 10^5 \times 2196.17 \text{ kJ/h} \ = 922.39 \times 10^6 \text{ kJ/h} \]
04

Calculate Mass Flow Rate of Cooling Water

Use energy balance for the cooling water to find its mass flow rate. \[ Q = \text{mass flow rate of water} \times c_p \times \theta \ \text{where } c_p \text { (specific heat capacity of water) } = 4.18 \text{ kJ/kg掳C} \ \theta \text{ (temperature change of water) } = 40 - 20 \text{ 掳C} \ = \text{mass flow rate of water} \times 4.18 \times 20 \ \text{mass flow rate of water} = \frac{922.39 \times 10^6}{4.18 \times 20} \] So, \[ \text{mass flow rate of water} = 11.04 \times 10^6 \text{ kg/h} \]
05

Conclusion

The mass flow rate of the cooling water required for steady-state operation is 11.04 脳 10^6 kg/h.

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Key Concepts

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

Thermodynamics
In the context of a heat exchanger, thermodynamics deals with the principles governing the transfer of energy between fluids. When steam condenses in a heat exchanger, it releases latent heat, which is absorbed by the cooling water flowing through the tubes. Understanding thermodynamics helps to analyze how heat energy is transferred and conserved within the system. In this problem, the first law of thermodynamics, which is the principle of energy conservation, is particularly important. This principle states that energy cannot be created or destroyed but only transferred from one form to another. In a heat exchanger, this means that the heat lost by the steam must be equal to the heat gained by the water, leading to the determination of the mass flow rate of cooling water needed to absorb the condensing steam's heat energy.
Enthalpy Calculation
Enthalpy is a thermodynamic property that represents the total heat content of a system. It is a crucial concept in this exercise because we need to calculate the heat released by the steam as it condenses. For steam with a given quality (percentage of vapor) entering the heat exchanger, we use steam tables to determine specific enthalpy values. Specifically, the initial enthalpy is calculated using the formula: \[ h_{initial} = h_{f} + x \times h_{fg} \]Where:
  • \( h_{f} \) is the enthalpy of the saturated liquid at the given pressure,
  • \( x \) is the quality of the steam, and
  • \( h_{fg}\) is the enthalpy of vaporization.
By finding these values, we can then determine the heat (\( Q \)) released during condensation by multiplying the mass flow rate of the steam with the difference in initial and final enthalpy. This precise calculation is vital to ensuring the energy balance in the system.
Energy Balance
Energy balance is an essential concept in thermodynamics, particularly for closed systems like heat exchangers where steady-state conditions are assumed. It involves accounting for all forms of energy entering and leaving the system to ensure none is lost or created in the process. In this heat exchanger problem, the energy released by the condensing steam must equal the energy absorbed by the cooling water, thus maintaining a perfect energy balance. By using an energy balance equation and the specific heat capacity of water, we can solve for the mass flow rate of the cooling water needed to absorb the released heat. The basic formula used in this context is: \( Q = \text{mass flow rate of water} \times c_p \times \theta \), where
  • \( Q \) = heat absorbed (kJ),
  • \( c_p \) = specific heat capacity of water (kJ/kg掳C),
  • \( \theta \) = temperature change of the water (掳C).
This equation ensures that we account for all heat exchanged and allows us to calculate the necessary parameters for system equilibrium.

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Most popular questions from this chapter

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.

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 \(\mathrm{mm}\).

Liquid water at \(20^{\circ} \mathrm{C}\) enters a pump though an inlet pipe having a diameter of \(160 \mathrm{~mm}\). The pump operates at steady state and supplies water to two exit pipes having diameters of \(80 \mathrm{~mm}\) and \(120 \mathrm{~mm}\), respectively. The velocity of the water exiting the \(80 \mathrm{~mm}\) pipe is \(0.5 \mathrm{~m} / \mathrm{s}\). At the exit of the \(120 \mathrm{~mm}\) pipe the velocity is \(0.1 \mathrm{~m} / \mathrm{s}\). The temperature of the water in each exit pipe is \(22^{\circ} \mathrm{C}\). Determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), in the inlet pipe and each of the exit pipes, and (b) the volumetric flow rate at the inlet, in \(\mathrm{m}^{3} / \mathrm{s}\).

Water is drawn steadily by a pump at a volumetric flow rate of \(0.8 \mathrm{~m}^{3} / \mathrm{min}\) through a pipe with a \(10 \mathrm{~cm}\) diameter inlet. The water is delivered to a tank placed at a height of \(10 \mathrm{~m}\) above the pipe inlet. The water exits through a nozzle having a diameter of \(2.5 \mathrm{~cm}\). Water enters at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure but there is no significant change in pressure and temperature at exit. Six percent of the power input to the pump is dissipated in the form of heat into the surroundings. Take the value of acceleration due to gravity \((g)\) equal to \(9.8 \mathrm{~m} / \mathrm{s}^{2}\). Determine (a) the velocity of the water at the inlet and exit, and (b) the power required by the pump.

Forced-air warming systems involving inflatable thermal blankets are commonly used to prevent subnormal body temperature (hypothermia) during and following surgery. A. heater and blower provide a stream of warmed air to the blanket. While the air temperature leaving the heater/blower is monitored by a temperature sensor, the temperature of the air providing warming to patients can vary widely, causing in some instances overheating and localized burning of patients. The object of this project is to develop costeffective modifications of existing thermal blankets that would control the air temperature and eliminate injurious "hot spots." The modifications must conform to standards governing the safety of systems involving heating in medical applications. Summarize your conclusions in a memorandum.
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