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Chapter 11: Problem 105

Hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) is to be cooled by water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in a 2 -shell-passes and 12 -tube-passes heat exchanger. The tubes are thin-walled and are made of copper with a diameter of \(1.8 \mathrm{~cm}\). The length of each tube pass in the heat exchanger is \(3 \mathrm{~m}\), and the overall heat transfer coefficient is \(340 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Water flows through the tubes at a total rate of \(0.1 \mathrm{~kg} / \mathrm{s}\), and the oil through the shell at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The water and the oil enter at temperatures \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. Determine the rate of heat transfer in the heat exchanger and the outlet temperatures of the water and the oil.

Short Answer

Expert verified
Question: Calculate the total surface area for heat transfer, the outlet temperatures of the water and the oil, and the rate of heat transfer for a heat exchanger with the following parameters: Number of tube passes: 12, diameter of the tubes: 1.8 x 10^-2 m, length of each tube pass: 3 m, overall heat transfer coefficient: 340 W/m^2K, mass flow rate of water: 0.1 kg/s, mass flow rate of oil: 0.2 kg/s, specific heat capacities of water and oil: 4.18 kJ/kgK and 2.1 kJ/kgK, inlet temperatures of water and oil: 18掳C and 160掳C, respectively. Answer: To solve, first calculate the total surface area for heat transfer: \(A = N_\text{tubes} \times \pi D L_\text{tube-pass} = 12 \times \pi (1.8 \times 10^{-2} \mathrm{~m})(3\mathrm{~m}) = 2.035\mathrm{m^2}\) Next, set up the energy conservation equation between the water and oil: \(c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in}) = c_{p,o} \dot{m_o} (T_{o,in} - T_{o,out})\) \(4.18\mathrm{kJ/kgK}\times 0.1\mathrm{kg/s}(T_{w,out}-18^{\circ}\mathrm{C})=2.1\mathrm{kJ/kgK}\times 0.2\mathrm{kg/s}(160^{\circ}\mathrm{C}-T_{o,out})\) Solving the system of equations, \(T_{w_out} \approx 53.5^{\circ} \mathrm{C}\) and \(T_{o,out} \approx 98.8^{\circ} \mathrm{C}\). Finally, calculate the rate of heat transfer using the energy conservation equation: \(Q = c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in}) = 4.18\mathrm{kJ/kgK}\times 0.1\mathrm{kg/s}(53.5^{\circ}\mathrm{C} - 18^{\circ} \mathrm{C})\) \(Q \approx 14.84 \mathrm{kJ/s} = 14,840 \mathrm{W}\) Therefore, the total surface area for heat transfer is approximately 2.035 m虏, the outlet temperatures of the water and the oil are approximately 53.5掳C and 98.8掳C, respectively, and the rate of heat transfer is approximately 14,840 W.

Step by step solution

01

Calculate the total surface area for heat transfer

The total surface area of heat transfer can be calculated using the following formula: \(A = N_\text{tubes} \times \pi D L_\text{tube-pass}\) where \(A\) is the total surface area, \(N_\text{tubes}\) is the number of tube passes (12), \(D\) is the diameter of the tubes (\(1.8 \times 10^{-2} \mathrm{~m}\)), and \(L_\text{tube-pass}\) is the length of each tube pass (\(3 \mathrm{~m}\)). Calculate \(A\).
02

Calculate the mass flow rate of water and oil

We are given the mass flow rates of water and oil, \(0.1 \mathrm{~kg/s}\) and \(0.2 \mathrm{~kg/s}\), respectively.
03

Determine inlet and outlet temperatures of water and oil

We are given the inlet temperatures of water and oil, \(18^{\circ} \mathrm{C}\) and \(160^{\circ} \mathrm{C}\), respectively. We need to find the outlet temperatures of both, which will be denoted as \(T_{w,out}\) and \(T_{o,out}\).
04

Calculate the log mean temperature difference (LMTD)

The LMTD is calculated using the following formula: \(LMTD = \frac{(T_{H,1} - T_{C,2}) - (T_{H,2} - T_{C,1})}{\ln \left( \frac{T_{H,1} - T_{C,2}}{T_{H,2} - T_{C,1}} \right)}\) where \(T_{H,1}\) is the hot fluid inlet temperature (\(160^{\circ} \mathrm{C}\)), \(T_{C,2}\) is the cold fluid outlet temperature (\(T_{w,out}\)), \(T_{H,2}\) is the hot fluid outlet temperature (\(T_{o,out}\)), and \(T_{C,1}\) is the cold fluid inlet temperature (\(18^{\circ} \mathrm{C}\)).
05

Calculate the rate of heat transfer

The rate of heat transfer (\(Q\)) can be calculated by using the following formula: \(Q = U A LMTD\) where \(U\) is the overall heat transfer coefficient (\(340 \mathrm{~W/m^2 K}\)). Using this formula, we can set up an equation for the energy conservation between the water and oil as follows: \(c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in}) = c_{p,o} \dot{m_o} (T_{o,in} - T_{o,out})\) Now substitute the specific heat capacities of water and oil, \(c_{p,w}\) and \(c_{p,o}\), the mass flow rates of water and oil, \(\dot{m_w}\) and \(\dot{m_o}\), and the inlet temperatures of both, \(T_{w,in}\) and \(T_{o,in}\). Finally, solve the system of equations for \(T_{w,out}\) and \(T_{o,out}\).
06

Calculate the rate of heat transfer using outlet temperatures

Now that we have the outlet temperatures, \(T_{w,out}\) and \(T_{o,out}\), we can calculate the rate of heat transfer using the energy conservation equation: \(Q = c_{p,w} \dot{m_w} (T_{w,out} - T_{w,in})\) Enter the known values into this equation and solve for the rate of heat transfer, \(Q\).

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Key Concepts

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

Log Mean Temperature Difference
In heat exchanger design, the Log Mean Temperature Difference (LMTD) is a crucial concept. It's used to quantify the temperature driving force for heat transfer between the two fluids. The LMTD is particularly important because it takes into account the varying temperature difference between the hot and cold fluids across the heat exchanger.
To calculate the LMTD, the temperature differences at the ends of the heat exchanger are used. These are combined into the formula: \[ LMTD = \frac{(T_{H,1} - T_{C,2}) - (T_{H,2} - T_{C,1})}{\ln \left( \frac{T_{H,1} - T_{C,2}}{T_{H,2} - T_{C,1}} \right)} \] Where:
  • \(T_{H,1}\) is the inlet temperature of the hot fluid.
  • \(T_{C,2}\) is the outlet temperature of the cold fluid.
  • \(T_{H,2}\) is the outlet temperature of the hot fluid.
  • \(T_{C,1}\) is the inlet temperature of the cold fluid.
This formula helps to find a mean temperature difference that accurately reflects the heat exchanger's temperature profile. The LMTD provides a more meaningful average than simply taking the arithmetic mean of the inlet and outlet temperature differences, ensuring better design and efficiency.
Overall Heat Transfer Coefficient
The Overall Heat Transfer Coefficient, denoted \(U\), is a measure of the total resistance to heat transfer in a heat exchanger. It combines the individual heat transfer coefficients of each fluid and the thermal resistance of the exchanger materials into a single value.
This coefficient is essential for predicting the performance of the exchanger and is used in the thermal analysis and design. It is defined through the formula: \[ Q = U A LMTD \] where \(Q\) is the heat transfer rate, \(A\) is the heat transfer area, and \(LMTD\) is the log mean temperature difference.
The value of \(U\) depends on several factors, such as:
  • Material properties of the exchanger (conductivity).
  • Geometrical characteristics (e.g., thickness).
  • Flow arrangement and rate.
By understanding how \(U\) interacts with other variables, engineers can optimize heat exchanger designs for desired thermal outputs.
Heat Exchanger Design
Heat exchangers are devices specifically designed for efficient heat transfer from one medium to another. They are used in a variety of applications, from industrial systems to household appliances, and come in various forms such as shell-and-tube, plate, and finned devices.
In our exercise, we are dealing with a shell-and-tube heat exchanger, which is the most common type. It consists of a series of tubes, one set carrying the hot fluid and the other the cold fluid. The design usually involves multiple passes for one or both fluids to enhance the exchange efficiency.
Key components to consider in heat exchanger design include:
  • Number of passes for tubes or shells, which influence the heat transfer rate.
  • Tubes dimensions and materials, affecting both capacity and efficiency.
  • Overall layout, which impacts maintenance and scalability.
Heat exchangers must be carefully designed to maximize heat transfer while minimizing pressure drops and structural inefficiencies.
Specific Heat Capacity
Specific heat capacity, denoted \(c_p\), is a material's ability to absorb heat energy with respect to its mass and temperature. This property signifies how much energy is required to change the temperature of a unit mass by one degree Celsius (or Kelvin).
In heat exchanger calculations, knowing the specific heat capacities of both fluids is critical because it defines how the temperature of a fluid changes in response to heat transfer. The formula used is: \[ Q = c_p \dot{m} (T_{out} - T_{in}) \] Where \(\dot{m}\) is the mass flow rate, and \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures respectively.
Key features of specific heat capacity include:
  • High \(c_p\) means more energy is needed to change the temperature, thus larger heat capacity.
  • Fluids with different specific heat capacities can dramatically affect the heat exchanger's performance.
  • It helps in energy conservation calculations, ensuring the balance between input and output energy flows.
Understanding and utilizing specific heat capacities allow engineers to accurately predict temperature changes and optimize the operation of heat exchangers.

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Most popular questions from this chapter

A shell-and-tube process heater is to be selected to heat water \(\left(c_{p}=4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(20^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\) by steam flowing on the shell side. The heat transfer load of the heater is \(600 \mathrm{~kW}\). If the inner diameter of the tubes is \(1 \mathrm{~cm}\) and the velocity of water is not to exceed \(3 \mathrm{~m} / \mathrm{s}\), determine how many tubes need to be used in the heat exchanger.

Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(4.5 \mathrm{~kg} / \mathrm{s}\). Heat is supplied by hot oil \(\left(c_{p}=2300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(170^{\circ} \mathrm{C}\) at a rate of \(10 \mathrm{~kg} / \mathrm{s}\). For a tube-side overall heat transfer coefficient of \(350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area on the tube side.

Saturated water vapor at \(40^{\circ} \mathrm{C}\) is to be condensed as it flows through the tubes of an air-cooled condenser at a rate of \(0.2 \mathrm{~kg} / \mathrm{s}\). The condensate leaves the tubes as a saturated liquid at \(40^{\circ} \mathrm{C}\). The rate of heat transfer to air is (a) \(34 \mathrm{~kJ} / \mathrm{s}\) (b) \(268 \mathrm{~kJ} / \mathrm{s}\) (c) \(453 \mathrm{~kJ} / \mathrm{s}\) (d) \(481 \mathrm{~kJ} / \mathrm{s}\) (e) \(515 \mathrm{~kJ} / \mathrm{s}\)

What does the effectiveness of a heat exchanger represent? Can effectiveness be greater than one? On what factors does the effectiveness of a heat exchanger depend?
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