In a heat exchanger there are 200 tubes each \(3.65 \mathrm{~m}\) long and \(30 \mathrm{~mm}\) outside diameter and \(25 \mathrm{~mm}\) bore. They are arranged axially in a cylinder of \(750 \mathrm{~mm}\) diameter and are equally spaced from one another. A liquid of relative density \(0.9\) flows at a mean velocity of \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) through the tubes and water flows at \(2.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) between the tubes in the opposite direction. For all surfaces \(f\) may be taken as \(0.01\). Neglecting entry and exit losses, calculate (a) the total power required to overcome fluid friction in the exchanger and \((b)\) the saving in power if the two liquids exchanged places but the system remained otherwise unaltered.

Short Answer

Expert verified
Calculate Reynold's number, find friction factor, compute head loss in tubes, determine power, swap the liquid, and calculate savings.

Step by step solution

01

- Calculate the Reynold's Number

First, calculate the Reynold's number for the liquid flowing through the tubes. Since the liquid is flowing at a mean velocity of \(2.5 \, \mathrm{m/s}\), with a tube bore diameter of \(25 \, \mathrm{mm} = 0.025 \, \mathrm{m}\), and using the relative density of \(0.9\), the formula for Reynold's Number \(Re\) is \[ Re = \frac{\rho vd}{\mu} \] where \(\rho\) is the density of the fluid, \(v\) is the flow velocity, \(d\) is the diameter, and \(\mu\) is the dynamic viscosity. Relative density \(0.9\) implies \(\rho = 0.9 \times 1000 \, \mathrm{kg/m^3}} = 900 \, \mathrm{kg/m^3}\). The water's density is \(\rho_w = 1000 \, \text{kg/m}^3\).
02

- Calculate the Friction Factor

The friction factor \(f\) for the flow inside the tubes is given as \(0.01\). This value also applies to the water flowing through the gaps between the tubes.
03

- Calculate Head Loss in Tubes

The head loss \(h_f\) per tube can be calculated using Darcy's equation \[ h_f = f \frac{L}{d} \frac{v^2}{2g} \] where \(L = 3.65 \, \mathrm{m}\) is the length of the tubes. Plug in the given values to find \( h_f \).
04

- Calculate Power Required for Tubes

The total power \(P = \rho Q g h_f\), where \(\rho\) is the fluid density, \(Q = Av\) is the volumetric flow rate for one tube, and \(A = \frac{\pi d^2}{4}\). Multiply by 200 tubes to get the total power required.
05

- Calculate Head Loss Between Tubes

The head loss for water flowing between the tubes is calculated the same way. Assume a cross-sectional area available for flow between tubes and calculate the effective diameter. Then use Darcy's equation to find the head loss for water.
06

- Swap the Liquids and Recalculate

When the liquids are swapped, repeat the calculations above but with the liquid now outside the tubes and water in the tubes.
07

- Calculate the Power Savings

Compare the total power requirements for both scenarios to determine the power savings if the liquids are swapped.

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Key Concepts

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

heat exchanger
A heat exchanger is a device used to transfer heat between two or more fluids without mixing them. In this exercise, we have two fluids flowing in opposite directions within the heat exchanger.
The liquid flows inside the tubes, while water flows in the space between the tubes.
The main challenge is to minimize the power required to overcome fluid friction within this setup.
To achieve this, it’s crucial to understand the flow characteristics of each fluid, calculated using various parameters such as Reynold's Number, friction factor, and head loss.
Reynold's Number
Reynold's Number \(\text{Re}\) is a dimensionless quantity used to predict flow patterns in different fluid flow situations. It helps determine whether the flow is laminar (smooth) or turbulent (chaotic).
For this exercise:
    \(\text{Re} = \frac{\rho v d}{\mu}\)

Where:
    \(\rho\) = density of the fluid
    \(v\) = flow velocity
    \(d\) = diameter of the tube
    \(\mu\) = dynamic viscosity of the fluid

Given the liquid’s density, the bore diameter of the tubes, and mean velocity, we can calculate \(\text{Re}\).
This determines if the flow inside the tubes is laminar or turbulent.
friction factor
The friction factor (f) quantifies the resistance due to friction in fluid flow through a pipe. It depends on the flow regime and surface roughness of the pipe.
For smooth pipes and turbulent flow, a common approximation of the friction factor is used, such as ===0.01=== in this exercise.
The friction factor directly influences the head loss in the tubes and gaps between the tubes.
head loss
Head loss \((h_f)\) refers to the loss of mechanical energy of the fluid due to friction as it flows through a pipe.
It is calculated using Darcy's equation:
    \( h_f = f \frac{L}{d} \frac{v^2}{2g} \)

Where:
    \( f \) = friction factor
    \( L \) = length of the pipe
    \( d \) = diameter of the pipe
    \( v \) = velocity of the fluid
    \( g \) = acceleration due to gravity

Calculating head loss allows us to determine the energy required to overcome the frictional resistance in the tubes and gaps.
Darcy's equation
Darcy's equation is fundamental for calculating head loss in fluid flow through pipes. The equation is given by:
    \( h_f = f \frac{L}{d} \frac{v^2}{2g} \)

Understanding this equation is critical for evaluating the power required to maintain fluid flow within the heat exchanger.
This applies to both situations when either liquid is flowing through the tubes and water in the gaps, or vice versa.
Using Darcy's equation, we can optimize the flow and achieve power savings by swapping the fluids as demonstrated in the problem.

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Most popular questions from this chapter

A pump delivers water through two pipes laid in parallel. One pipe is \(100 \mathrm{~mm}\) diameter and \(45 \mathrm{~m}\) long and discharges to atmosphere at a level \(6 \mathrm{~m}\) above the pump outlet. The other pipe, \(150 \mathrm{~mm}\) diameter and \(60 \mathrm{~m}\) long, discharges to atmosphere at a level \(8 \mathrm{~m}\) above the pump outlet. The two pipes are connected to a junction immediately adjacent to the pump and both have \(f=0.008 .\) The inlet to the pump is \(600 \mathrm{~mm}\) below the level of the outlet. Taking the datum level as that of the pump inlet, determine the total head at the pump outlet if the flow rate through it is \(0.037 \mathrm{~m}^{3} \cdot \mathrm{s}^{-1}\). Losses at the pipe junction may be neglected.

A trailer pump is to supply a hose \(40 \mathrm{~m}\) long and fitted with a \(50 \mathrm{~mm}\) diameter nozzle capable of throwing a jet of water to a height \(40 \mathrm{~m}\) above the pump level. If the power lost in friction in the hose is not to exceed \(15 \%\) of the available hydraulic power, determine the diameter of hose required. Friction in the nozzle may be neglected and \(f\) for the hose assumed to be in the range \(0.007-0.01\).

A fluid of constant density \(\varrho\) enters a horizontal pipe of radius \(R\) with uniform velocity \(V\) and pressure \(p_{1}\). At a downstream section the pressure is \(p_{2}\) and the velocity varies with radius \(r\) according to the equation \(u=2 V\left\\{1-\left(r^{2} / R^{2}\right)\right\\}\). Show that the friction force at the pipe walls from the inlet to the section considered is given by \(\pi R^{\frac{1}{2}}\left(p_{1}-p_{2}-\varrho V^{2} / 3\right)\).

A pipe \(900 \mathrm{~m}\) long and \(200 \mathrm{~mm}\) diameter discharges water to atmosphere at a point \(10 \mathrm{~m}\) below the level of the inlet. With a pressure at inlet of \(40 \mathrm{kPa}\) above atmospheric the steady discharge from the end of the pipe is \(49 \mathrm{~L} \cdot \mathrm{s}^{-1}\). At a point half way along the pipe a tapping is then made from which water is to be drawn off at a rate of \(18 \mathrm{~L} \cdot \mathrm{s}^{-1}\). If conditions are such that the pipe is always full, to what value must the inlet pressure be raised so as to provide an unaltered discharge from the end of the pipe? (The friction factor may be assumed unaltered.)

Kerosene of relative density \(0.82\) and kinematic viscosity \(2.3 \mathrm{~mm}^{2} \cdot \mathrm{s}^{-1}\) is to be pumped through \(185 \mathrm{~m}\) of galvanized iron pipe \((k=0.15 \mathrm{~mm})\) at \(40 \mathrm{~L} \cdot \mathrm{s}^{-1}\) into a storage tank. The pressure at the inlet end of the pipe is \(370 \mathrm{kPa}\) and the liquid level in the storage tank is \(20 \mathrm{~m}\) above that of the pump. Neglecting losses other than those due to pipe friction determine the size of pipe necessary.

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