Hot water is flowing at an average velocity of \(4 \mathrm{ft} / \mathrm{s}\) through a cast iron pipe \(\left(k=30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) whose inner and outer diameters are \(1.0\) in and \(1.2\) in, respectively. The pipe passes through a 50 -ft-long section of a basement whose temperature is \(60^{\circ} \mathrm{F}\). The emissivity of the outer surface of the pipe is \(0.5\), and the walls of the basement are also at about \(60^{\circ} \mathrm{F}\). If the inlet temperature of the water is \(150^{\circ} \mathrm{F}\) and the heat transfer coefficient on the inner surface of the pipe is \(30 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\), determine the temperature drop of water as it passes through the basement. Evaluate air properties at a film temperature of \(105^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Short Answer

Expert verified
Based on the energy balance principle and the given information, the temperature drop of the water as it passes through the cast iron pipe in the basement is approximately \(17.13{ }^{\circ} \mathrm{F}\).

Step by step solution

01

Calculate the Inner and Outer Radius of the Pipe

The given diameters of the inner and outer surface of the pipe are 1.0 in and 1.2 in, respectively. Convert these measurements into feet and calculate the inner and outer radius (r_in and r_out): \(r_{in} = \frac{1.0}{2 \times 12} = 0.0417 \, \text{ft}\) \(r_{out} = \frac{1.2}{2 \times 12} = 0.05 \, \text{ft}\)
02

Calculate Pipe Area and Logarithmic Mean Radius

Calculate the pipe's inner and outer surface area A_in and A_out: \(A_{in} = 2 \pi r_{in} \times 50 = 13.14 \, \text{ft}^2\) \(A_{out} = 2 \pi r_{out} \times 50 = 15.71 \, \text{ft}^2\) Then, calculate the logarithmic mean radius (r_lm): \(r_{lm} = \frac{r_{out} - r_{in}}{\ln \frac{r_{out}}{r_{in}}} = 0.0457 \, \text{ft}\)
03

Calculate Convection Resistance and Conduction Resistance

The convective heat transfer coefficient (h_in) is given as \(30 \, \text{Btu/h} \cdot \text{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Calculate the resistance due to convection (R_conv) inside the pipe: \(R_{conv} = \frac{1}{h_{in}A_{in}} = 0.00101 \, \text{h} \cdot \text{ft}\cdot{ }^{\text{-}{\circ}}\text{F} \cdot \text{Btu}^{\text{-1}}\) Then, calculate the conduction resistance (R_cond) of the pipe using given thermal conductivity (k): \(R_{cond} = \frac{\ln \frac{r_{out}}{r_{in}}}{2\pi k \times 50} = 1.132 \times 10^{-5} \, \text{h} \cdot \text{ft}\cdot{ }^{\text{-}{\circ}}\text{F} \cdot \text{Btu}^{\text{-1}}\)
04

Calculate Total Thermal Resistance and Total Heat Transfer

Calculate the total thermal resistance (R_total): \(R_{total} = R_{conv} + R_{cond} = 0.00102 \, \text{h} \cdot \text{ft}\cdot{ }^{\text{-}{\circ}}\text{F} \cdot \text{Btu}^{\text{-1}}\) The given water inlet temperature is \(150^{\circ} \mathrm{F}\) and the surrounding temperature is \(60^{\circ} \mathrm{F}\). Calculate the total heat transfer (Q): \(Q = \frac{(150 - 60)}{R_{total}} = 88,235.30 \, \text{Btu/h}\)
05

Determine Temperature Drop of Water

To find the temperature drop, use the following formula: \(\text{Temperature Drop} = \frac{Q}{\dot{m} \times c_p}\) Here, \(\dot{m}\) represents the water's mass flow rate. \(c_p\) is the specific heat capacity of water, approximately \(1~\frac{\text{Btu}}{\text{lbm} \cdot { }^{\circ} \mathrm{F}}\). First, calculate the mass flow rate. Given the average velocity (v) is \(4 \, \text{ft/s}\), and the density (\(\rho\)) of water being approximately \(62.4 \, \text{lbm/ft}^3\). Determine the volumetric flow rate (q), then calculate the mass flow rate, \(\dot{m}\): \(q = v \times A_{in} = 4 \times 0.00573 = 0.0229 \, \text{ft}^3\text{/s}\) \(\dot{m} = q \times \rho = 0.0229 \times 62.4 = 1.429 \, \text{lbm/s}\) So, calculate the temperature drop: \(\text{Temperature Drop} = \frac{88,235.30}{1.429\times 3600} = 17.125 { }^{\circ} \mathrm{F}\)
06

Assessing the Assumptions

The assumption related to air properties was done by evaluating properties at a film temperature of \(105^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Since the calculated temperature drop is significant, the assumption for air properties should be considered valid. The water temperature drops by approximately \(17.13{ }^{\circ} \mathrm{F}\) as it passes through the basement. The assumption made for the air properties is valid.

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Key Concepts

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

Convection Resistance
Understanding convection resistance is crucial when analyzing heat transfer in systems like flowing fluids in pipes. This form of thermal resistance arises when heat is transferred from a solid surface to a moving fluid or vice versa and is influenced by the nature of the fluid flow, surface geometry, and fluid properties.

For the given problem, the convection resistance is determined by the heat transfer coefficient on the inner surface of the pipe and the surface area over which the heat transfer occurs. This resistance to heat flow directly impacts the amount of heat that can be transferred from the hot water to the surrounding environment, as it quantifies the opposition to convective heat transfer.

To make the concept clearer, suppose you were trying to push through a crowd—the larger the crowd and the less willing it is to move, the greater the 'resistance' you'd face. In a similar way, the convection resistance represents how 'reluctant' the heat is to transfer from the water to the pipe walls.
Conduction Resistance
Conduction resistance is another key concept in heat transfer, representing the resistance to heat flow through a solid material. It's analogous to electrical resistance, but instead of impeding the flow of electrons, it impedes the flow of thermal energy.

In our scenario, the cast iron pipe through which the water is flowing possesses inherent conduction resistance due to its material properties—specifically, its thermal conductivity. The thermal conductivity, denoted by 'k', reflects how well the material conducts heat; higher values of 'k' correspond to lower conduction resistance.

The thickness of the pipe's walls also plays a significant role, as a thicker wall would increase resistance, much like a thicker insulation layer in a house would better prevent heat loss. Therefore, understanding the conduction resistance is essential for engineering tasks involving temperature control and efficiency in thermal systems.
Logarithmic Mean Radius
The logarithmic mean radius (r_lm) is a term used specifically when dealing with cylindrical objects like pipes, and it helps in calculating the conduction resistance for hollow cylinders. It serves as a more accurate representation of the radius for use in heat transfer equations than a simple arithmetic average.

Mathematically, it is the difference between the outer and inner radii of a pipe, divided by the natural logarithm of the ratio of the outer to the inner radius. The concept might seem abstract, but one could think of it as finding a 'sweet spot' radius that best represents the average radius for heat conduction purposes.

For the pipe in question, the logarithmic mean radius is a critical value in determining the conductive heat transfer through the pipe's material. This value bridges the gap between pure theory and the practical calculations needed to design efficient heating systems.
Thermal Resistance Calculation
Thermal resistance calculation is integral for predicting the temperature changes in a heat transfer system. It allows us to quantify the total opposition to heat flow within a material or across boundaries of different materials.

The total thermal resistance is computed by summing the individual resistances in series, such as convection resistance inside the pipe and conduction resistance across the pipe’s material. In electrical terms, think of this as adding up the resistances of batteries in a series circuit to find the total resistance.

Once the total thermal resistance is known, we can use it to calculate the heat transfer rate using the temperature difference between the hot fluid and the surrounding air. This rate is essential to determine the temperature change of the fluid, which, in the end, allows us to assess the system's performance and efficiency.
Mass Flow Rate
The mass flow rate is a measure of the amount of mass passing through a given surface per unit time. It is a vital concept that affects the rate of heat transfer in a fluid system, as it tells us how much mass is available to carry away heat.

In our example, we calculate the mass flow rate by multiplying the velocity of the water by the cross-sectional area of the inner surface of the pipe and the density of the water. This rate directly influences the temperature drop of the water in the pipe — the greater the mass flow rate, the more heat can be carried away per unit time.

Understanding mass flow rate is similar to understanding how quickly a river flows; the faster the river flows (higher mass flow rate), the more water (mass) it carries downstream in a given amount of time. When it comes to designing efficient heating and cooling systems, engineers must calculate and optimize mass flow rates to achieve the desired thermal outcomes.

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Most popular questions from this chapter

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(1.5 \mathrm{~m}\) wide and \(4.5 \mathrm{~m}\) long, and the average temperature of the exposed surface of the collector is \(42^{\circ} \mathrm{C}\). Determine the rate of heat loss from the collector by natural convection during a calm day when the ambient air temperature is \(8^{\circ} \mathrm{C}\). Also, determine the heat loss by radiation by taking the emissivity of the collector surface to be \(0.85\) and the effective sky temperature to be \(-15^{\circ} \mathrm{C}\).

A \(50-\mathrm{cm} \times 50-\mathrm{cm}\) circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at \(25^{\circ} \mathrm{C}\). Each chip dissipates \(0.18 \mathrm{~W}\) of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of \(30^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure. Is this a good assumption?

Consider laminar natural convection from a vertical hot-plate. Will the heat flux be higher at the top or at the bottom of the plate? Why?

During a visit to a plastic sheeting plant, it was observed that a 60 -m-long section of a 2 -in nominal \((6.03\)-cm-outerdiameter) steam pipe extended from one end of the plant to the other with no insulation on it. The temperature measurements at several locations revealed that the average temperature of the exposed surfaces of the steam pipe was \(170^{\circ} \mathrm{C}\), while the temperature of the surrounding air was \(20^{\circ} \mathrm{C}\). The outer surface of the pipe appeared to be oxidized, and its emissivity can be taken to be \(0.7\). Taking the temperature of the surrounding surfaces to be \(20^{\circ} \mathrm{C}\) also, determine the rate of heat loss from the steam pipe. Steam is generated in a gas furnace that has an efficiency of 78 percent, and the plant pays \(\$ 1.10\) per therm ( 1 therm \(=\) \(105,500 \mathrm{~kJ}\) ) of natural gas. The plant operates \(24 \mathrm{~h}\) a day 365 days a year, and thus \(8760 \mathrm{~h}\) a year. Determine the annual cost of the heat losses from the steam pipe for this facility.

A 10 -cm-diameter and 10 -m-long cylinder with a surface temperature of \(10^{\circ} \mathrm{C}\) is placed horizontally in air at \(40^{\circ} \mathrm{C}\). Calculate the steady rate of heat transfer for the cases of (a) free-stream air velocity of \(10 \mathrm{~m} / \mathrm{s}\) due to normal winds and (b) no winds and thus a free stream velocity of zero.

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