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Chapter 8: Problem 59

A heating contractor must heat \(0.2 \mathrm{~kg} / \mathrm{s}\) of water from \(15^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) using hot gases in cross flow over a thinwalled tube. Your assignment is to develop a series of design graphs that can be used to demonstrate acceptable combinations of tube dimensions \((D\) and \(L)\) and of hot gas conditions \(\left(T_{\text {e }}\right.\) and \(V\) ) that satisfy this requirement. In your analysis, consider the following parameter ranges: \(D=20,30\), or \(40 \mathrm{~mm} ; L=3,4\), or \(6 \mathrm{~m}\); \(T_{=}=250,375\), or \(500^{\circ} \mathrm{C}\) : and \(20 \leq V \leq 40 \mathrm{~m} / \mathrm{s}\),

Short Answer

Expert verified
Develop graphs of \(U\) versus \(T_e\) and \(V\) for different \(D\) and \(L\) values.

Step by step solution

01

Determine Heating Requirements

To calculate the power required to heat the water, we use the formula \(Q = \dot{m} \cdot c_p \cdot (T_{out} - T_{in})\), where \(\dot{m}\) is the mass flow rate of the water, \(c_p\) is the specific heat capacity of water (approximately \(4.18 \text{ } \mathrm{kJ/kg} \degree\text{C}\)), and \(T_{out}\) and \(T_{in}\) are the outlet and inlet temperatures, respectively. Substituting the given values, \(Q = 0.2 \cdot 4.18 \cdot (35 - 15) = 16.72 \text{ } \, \mathrm{kW}\).
02

Calculate Required Heat Transfer Coefficient

To achieve the required heat transfer rate, we need to consider the heat transfer coefficient \(U\). The formula relating heat transfer rate \(Q\) to \(U\), surface area \(A\), and the logarithmic mean temperature difference \(\Delta T_{lm}\) is \(Q = U \cdot A \cdot \Delta T_{lm}\). Calculate \(\Delta T_{lm}\) using the formula: \(\Delta T_{lm} = \frac{(T_{hot,in} - T_{cold,out}) - (T_{hot,out} - T_{cold,in})}{\ln\left(\frac{T_{hot,in} - T_{cold,out}}{T_{hot,out} - T_{cold,in}}\right)}\).
03

Determine Tube Surface Area

The surface area \(A\) of the tube is given by \(A = \pi \cdot D \cdot L\), where \(D\) is the diameter and \(L\) is the length of the tube. Evaluate surface areas for all combinations of \(D\) and \(L\) using the given parameter ranges.
04

Construct Design Graphs

Create graphs to represent combinations of \(D\), \(L\), \(T_{e}\), and \(V\) that satisfy the required heating. Plot graphs of \(U\) versus other parameters like \(T_{e}\) and \(V\) for each combination of \(D\) and \(L\), by solving iteratively or using design software. These graphs will help visualize acceptable combinations that achieve the required heat transfer.

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Key Concepts

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

Heat Transfer Coefficient
In heat transfer design, the heat transfer coefficient, often denoted as \( U \), plays a critical role. It represents how well heat is transferred between two mediums, such as water and hot gases in this case. This coefficient can be affected by several factors, including the materials involved and the types of fluids interacting.

To calculate the heat transfer coefficient, you can use the formula:
  • \( Q = U \cdot A \cdot \Delta T_{lm} \)
Where:
  • \( Q \) is the heat transfer rate in watts (W)
  • \( A \) is the surface area available for heat exchange
  • \( \Delta T_{lm} \) is the logarithmic mean temperature difference
This formula highlights the direct relationship between \( U \) and effective heating, meaning a higher \( U \) is generally better for efficient heat transfer.

In practical applications, \( U \) may require experimental data to determine accurately, especially in complex systems where theoretical predictions might not suffice. This involves understanding fluid properties and flow dynamics over the tube's surface.
Logarithmic Mean Temperature Difference
The concept of logarithmic mean temperature difference (LMTD) is essential in heat exchanger design. It quantifies an average temperature difference between the two fluids involved in the heat exchange process. This average temperature helps determine the efficiency of the heat exchanger.

The formula for LMTD is:
  • \( \Delta T_{lm} = \frac{(T_{hot,in} - T_{cold,out}) - (T_{hot,out} - T_{cold,in})}{\ln\left(\frac{T_{hot,in} - T_{cold,out}}{T_{hot,out} - T_{cold,in}}\right)} \)
This equation considers temperature gradients at both ends of the heat exchanger, giving you a more precise measure than simple arithmetic averages.

Understanding LMTD is crucial because it directly affects the calculation of \( U \) and the required surface area \( A \). A higher LMTD typically means more efficient heat transfer, requiring less surface area for the same amount of thermal exchange.
Tube Surface Area
When it comes to designing heat exchangers, the tube surface area \( A \) is crucial. It's the actual area through which heat is transferred between the flowing mediums. The surface area of a tube is calculated using the formula:
  • \( A = \pi \cdot D \cdot L \)
Where:
  • \( D \) is the diameter of the tube
  • \( L \) is the length of the tube
By varying \( D \) and \( L \), you can influence \( A \), thus changing your system's ability to transfer heat effectively. A larger surface area generally allows for a higher heat transfer rate, given the same conditions.

Choosing the correct dimensions for \( D \) and \( L \) is a vital part of the design process and is often iteratively adjusted via graphs or computational models to meet specific heating requirements. These adjustments aim to balance efficiency with material costs and spatial constraints in the system's layout.

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Most popular questions from this chapter

Consider a concentric tube annulus for which the inner and outer diameters are 25 and \(50 \mathrm{~mm}\). Water enters the annular region at \(0.04 \mathrm{~kg} / \mathrm{s}\) and \(25^{\circ} \mathrm{C}\). If the inner tube wall is heated electrically at a rate (per unit length) of \(q^{\prime}=4000 \mathrm{~W} / \mathrm{m}\), while the outer tube wall is insulated, how long must the tubes be for the water to acticue \(\boldsymbol{z}\) outlet temperature of \(85^{\circ} \mathrm{C}\) ? What is the inner nube serface temperature at the outlet, where fully devekped conditions may be assumed?

Consider a horizontal, thin-walled circular tube of diameler \(D=0.025 \mathrm{~m}\) submerged in a container of \(n\)-octadecane (paraffin), which is used to store thermal energy. As hot water flows through the tube, heat is transferred to the paraffin, converting it from the solid to liq: uid state at the phase change temperature of \(T_{\mathrm{s}}=\) \(27.4^{\circ} \mathrm{C}\). The latent heat of fusion and density of paraffin are \(h_{t j}=244 \mathrm{~kJ} / \mathrm{kg}\) and \(\rho=770 \mathrm{~kg} / \mathrm{m}^{3}\), respectively, and thermophysical properties of the water may be taken as \(c_{\mathrm{r}}=4.185 \mathrm{~kJ} / \mathrm{kg}+\mathrm{K}, k=0.653 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \mu=467 \times\) \(10^{-6} \mathrm{~kg} / \mathrm{s} \cdot \mathrm{m}\), and \(\operatorname{Pr}=2.99\). (a) Assuming the tube surface to have a uniform temperature corresponding to that of the phase change, determine the water outlet temperature and total heat transfer rate for a water flow rate of \(0,1 \mathrm{~kg} / \mathrm{s}\) and an inlet temperature of \(60^{\circ} \mathrm{C}\). If \(H=W=\) \(0.25 \mathrm{~m}\), how long would it take to completely liquefy the paraffin, from an initial state for which all the paraffin is solid and at \(27.4^{\circ} \mathrm{C}\) ? (b) The liquefaction process can be accelerated by increasing the flow rate of the water. Compute and plot the heat rate and outlet temperature as a function of flow rate for \(0.1 \leq m \leq 0.5 \mathrm{~kg} / \mathrm{s}\). How long would it take to melt the paraffin for in \(=0.5 \mathrm{~kg} / \mathrm{s}\) ?

Compare Nusselt predictions based on the Colburn, Dittus-Boelter, Sieder and Tate, and Gnielinski conels tions for the fully developed turbulent flow of water in a smooth circular tube at Reynolds numbers of \(4000,16^{2}\). and \(10^{5}\) when the average mean temperature is \(295 \mathrm{~K}\) and the surface temperature is \(305 \mathrm{~K}\).

To cool electronic components that are mounted to a printed circuit board and hermetically sealed from their surroundings, two boards may be joined to form an intermediate channel through which the coolant is passed. Termed a hollow- core PCB, all of the heat generated by the components may be assumed to be transferred to the coolant. Consider a hollow-core PCB of length \(L=300 \mathrm{~mm}\) and equivalent width \(W\) (normial to the page). Under normal operating conditions, \(40 \mathrm{~W}\) of power are dissipated on each side of the PCB and a uniform distribution of the attendant heat transfer may be assumed for each of the hollow-core surfaces. If air enters a core of height \(H=4 \mathrm{~mm}\) at a temperature of \(T_{m e}=20^{\circ} \mathrm{C}\) and a flow rate of \(i m=0.002 \mathrm{~kg} / \mathrm{s}\), what is its outlet temperature \(T_{\text {mnn }}\) ? What are the surface temperatures at the inlet and outlet of the core? What are the foregoing temperatures if the flow rate is increased by a factor of five?

Heated air required for a food-drying process is generated by passing ambient air at \(20^{\circ} \mathrm{C}\) through long, circular tubes \((D=50 \mathrm{~mm}, L=5 \mathrm{~m})\) housed in a steam condenser. Saturated steam at atmospheric pressure condenses on the outer surface of the tubes, maintaining a uniform surface temperature of \(100^{\circ} \mathrm{C}\). (a) If an air flow rate of \(0.01 \mathrm{~kg} / \mathrm{s}\) is maintained in each tube, determine the air outlet tempersture \(T_{\text {me }}\) and the total heat rate \(q\) for the tube. (b) The air outlet temperature may be controlled by adjusting the tube mass flow rate. Compute and plot \(T_{m}\) as a function of \(m\) for \(0.005 \leq m \leq 0.050 \mathrm{~kg} / \mathrm{s}\). If a particular drying process requires approximately \(1 \mathrm{~kg} / \mathrm{s}\) of air at \(75^{\circ} \mathrm{C}\), what design and operating conditions, should be prescribed for the air heater, subject to the constraint that the tube diameter and length be fixed at \(50 \mathrm{~mm}\) and \(5 \mathrm{~m}\), respectively?
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