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Chapter 7: Problem 113

Oil at \(60^{\circ} \mathrm{C}\) flows at a velocity of \(20 \mathrm{~cm} / \mathrm{s}\) over a \(5.0\)-m-long and \(1.0-\mathrm{m}\)-wide flat plate maintained at a constant temperature of \(20^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the oil to the plate if the average oil properties are: \(\rho=880 \mathrm{~kg} / \mathrm{m}^{3}, \mu=0.005 \mathrm{~kg} / \mathrm{m} \cdot \mathrm{s}, k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\).

Short Answer

Expert verified
Answer: The rate of heat transfer from the oil to the plate is 25403.9 W.

Step by step solution

01

Calculate the Reynolds number

The Reynolds number (Re) is important in determining the flow regime (laminar or turbulent) of the fluid and can be calculated using the formula: \(Re = \frac{\rho vL}{\mu}\) where \(蟻\) is the fluid density, \(v\) is the flow velocity, \(L\) is the characteristic length, and \(渭\) is the dynamic viscosity of the fluid. Here, the characteristic length would be the length of the flat plate. Using the given values, \(Re = \frac{880 \mathrm{~kg/m^{3}} \times 0.2 \mathrm{~m/s} \times 5.0 \mathrm{~m}}{0.005 \mathrm{~kg/m \cdot s}} = 176000\)
02

Calculate the Prandtl number

The Prandtl number (Pr) is calculated using the following formula: \(Pr = \frac{c_{p} \mu}{k}\) where \(c_{p}\) is the specific heat at constant pressure, and \(k\) is the thermal conductivity of the fluid. Using the given values, \(Pr = \frac{2000 \mathrm{~J/kg \cdot K} \times 0.005 \mathrm{~kg/m \cdot s}}{0.15 \mathrm{~W/m \cdot K}} = 66.67\)
03

Determine the Nusselt number

Since the flow is turbulent (Re > 10000), we will use the Dittus-Boelter equation for the Nusselt number (Nu) for turbulent flow over a flat plate: \(Nu = 0.0296 Re^{4 / 5} Pr^{n}\) For cooling cases, \(n = 1 / 3\). So, \(Nu = 0.0296 \times (176000)^{4 / 5} \times (66.67)^{1 / 3} = 4250.91\)
04

Calculate the convective heat transfer coefficient

The convective heat transfer coefficient (h) can be calculated using the Nusselt number and the thermal conductivity: \(h = \frac{Nu \times k}{L}\) \(h = \frac{4250.91 \times 0.15 \mathrm{~W/m \cdot K}}{5.0 \mathrm{~m}} = 127.53 \mathrm{~W/m^{2} \cdot K}\)
05

Determine the rate of heat transfer

Now that we have the convective heat transfer coefficient, we can use it to find the rate of heat transfer from the oil to the plate using the following formula: \(q = h \times A \times (T_{\mathrm{oil}} - T_{\mathrm{plate}})\) where \(A\) is the surface area of the flat plate, and \(T_{\mathrm{oil}}\) and \(T_{\mathrm{plate}}\) are the temperatures of the oil and the plate, respectively. Since the plate is 5.0 m long and 1.0 m wide, its surface area is: \(A = 5.0 \mathrm{~m} \times 1.0 \mathrm{~m} = 5.0 \mathrm{~m^{2}}\) Given temperatures are in Celsius, we have: \(T_{\mathrm{oil}} = 60^{\circ} \mathrm{C} = 333.15 \mathrm{~K}\) \(T_{\mathrm{plate}} = 20^{\circ} \mathrm{C} = 293.15 \mathrm{~K}\) So, the rate of heat transfer is: \(q = 127.53 \mathrm{~W/m^{2} \cdot K} \times 5.0 \mathrm{~m^{2}} \times (333.15 \mathrm{~K} - 293.15 \mathrm{~K}) = 25403.9 \mathrm{~W}\)

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Key Concepts

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

Reynolds Number
When discussing fluid dynamics and heat transfer, the Reynolds number (Re) is a crucial dimensionless quantity that helps identify the type of flow regime. It's essentially a measure of the relationship between inertial forces and viscous forces within a fluid flow. This tells us if the flow will be laminar, turbulent, or somewhere in between.
In simple terms, when an object moves through a fluid or a fluid flows over a surface, the motion can be smooth (laminar) or mixed and chaotic (turbulent).
If Re is less than about 2000, the flow is generally considered to be laminar, which is smooth and orderly. If it's greater than 4000, it's turbulent, characterized by chaotic changes in pressure and flow velocity.

To calculate Re, we use:
  • \(Re = \frac{\rho vL}{\mu}\)
  • \(\rho\) is the fluid density.
  • \(v\) is the velocity of the fluid.
  • \(L\) is a characteristic length (like the length of a flat surface over which the fluid flows).
  • \(\mu\) is the dynamic viscosity of the fluid.
In our example, a calculated Re of 176,000 indicates turbulent flow, implying that as oil flows over the flat plate, it's in a chaotic and mixed state, which impacts the heat transfer rate significantly.
Nusselt Number
The Nusselt number is another dimensionless number used in heat transfer problems. It provides a relationship between conductive and convective heat transfer modes.
By calculating the Nusselt number, engineers can predict how efficiently heat is being transferred from the fluid to the surface, or vice versa.

The Nusselt number (Nu) is calculated using formulas specific to the flow type, such as the Dittus-Boelter equation for turbulent flow:
  • \( Nu = 0.0296 Re^{4/5} Pr^{n} \)
  • For cooling cases, \(n = 1/3\).
Here, Nu (4250.91) suggests effective heat transfer between the oil and the plate.
The higher the Nu, the more effective the convective heat transfer compared to conduction. This means as oil flows over the surface, it transfers heat efficiently, critical when determining the performance of systems involving fluid flow.
Convective Heat Transfer Coefficient
The convective heat transfer coefficient (h) quantifies how well convection is carrying heat between a surface and a fluid flowing over it.
This coefficient is particularly useful in assessing the rate at which heat is added to or removed from the surface by the moving fluid.

To find this, we use the formula:
  • \( h = \frac{Nu \times k}{L} \)
  • \(Nu\) is the Nusselt number.
  • \(k\) is the thermal conductivity of the fluid.
  • \(L\) is the characteristic length.
In the given scenario, h was computed to be 127.53 W/m虏路K.
This indicates a moderately high rate of heat transfer efficiency from the oil to the plate surface, which is significant in applications that require precise thermal management, like cooling systems.

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Most popular questions from this chapter

Solar radiation is incident on the glass cover of a solar collector at a rate of \(700 \mathrm{~W} / \mathrm{m}^{2}\). The glass transmits 88 percent of the incident radiation and has an emissivity of \(0.90\). The entire hot water needs of a family in summer can be met by two collectors \(1.2 \mathrm{~m}\) high and \(1 \mathrm{~m}\) wide. The two collectors are attached to each other on one side so that they appear like a single collector \(1.2 \mathrm{~m} \times 2 \mathrm{~m}\) in size. The temperature of the glass cover is measured to be \(35^{\circ} \mathrm{C}\) on a day when the surrounding air temperature is \(25^{\circ} \mathrm{C}\) and the wind is blowing at \(30 \mathrm{~km} / \mathrm{h}\). The effective sky temperature for radiation exchange between the glass cover and the open sky is \(-40^{\circ} \mathrm{C}\). Water enters the tubes attached to the absorber plate at a rate of \(1 \mathrm{~kg} / \mathrm{min}\). Assuming the back surface of the absorber plate to be heavily insulated and the only heat loss to occur through the glass cover, determine \((a)\) the total rate of heat loss from the collector, \((b)\) the collector efficiency, which is the ratio of the amount of heat transferred to the water to the solar energy incident on the collector, and \((c)\) the temperature rise of water as it flows through the collector.

The local atmospheric pressure in Denver, Colorado (elevation \(1610 \mathrm{~m}\) ), is \(83.4 \mathrm{kPa}\). Air at this pressure and \(20^{\circ} \mathrm{C}\) flows with a velocity of \(8 \mathrm{~m} / \mathrm{s}\) over a \(1.5 \mathrm{~m} \times 6 \mathrm{~m}\) flat plate whose temperature is \(140^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the plate if the air flows parallel to the \((a)\)-m-long side and \((b)\) the \(1.5 \mathrm{~m}\) side.

Water at \(43.3^{\circ} \mathrm{C}\) flows over a large plate at a velocity of \(30.0 \mathrm{~cm} / \mathrm{s}\). The plate is \(1.0 \mathrm{~m}\) long (in the flow direction), and its surface is maintained at a uniform temperature of \(10.0^{\circ} \mathrm{C}\). Calculate the steady rate of heat transfer per unit width of the plate. 7-24 The forming section of a plastics plant puts out a continuous sheet of plastic that is \(1.2 \mathrm{~m}\) wide and \(2 \mathrm{~mm}\) thick at a rate of \(15 \mathrm{~m} / \mathrm{min}\). The temperature of the plastic sheet is \(90^{\circ} \mathrm{C}\) when it is exposed to the surrounding air, and the sheet is subjected to air flow at \(30^{\circ} \mathrm{C}\) at a velocity of \(3 \mathrm{~m} / \mathrm{s}\) on both sides along its surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in \(2 \mathrm{~s}\). Determine the rate of heat transfer from the plastic sheet to the air.

Conduct this experiment to determine the heat loss coefficient of your house or apartment in \(\mathrm{W} /{ }^{\circ} \mathrm{C}\) or \(\mathrm{Btu} / \mathrm{h} \cdot{ }^{\circ} \mathrm{F}\). First make sure that the conditions in the house are steady and the house is at the set temperature of the thermostat. Use an outdoor thermometer to monitor outdoor temperature. One evening, using a watch or timer, determine how long the heater was on during a 3 -h period and the average outdoor temperature during that period. Then using the heat output rating of your heater, determine the amount of heat supplied. Also, estimate the amount of heat generation in the house during that period by noting the number of people, the total wattage of lights that were on, and the heat generated by the appliances and equipment. Using that information, calculate the average rate of heat loss from the house and the heat loss coefficient.

A heated long cylindrical rod is placed in a cross flow of air at \(20^{\circ} \mathrm{C}(1 \mathrm{~atm})\) with velocity of \(10 \mathrm{~m} / \mathrm{s}\). The rod has a diameter of \(5 \mathrm{~mm}\) and its surface has an emissivity of \(0.95\). If the surrounding temperature is \(20^{\circ} \mathrm{C}\) and the heat flux dissipated from the rod is \(16000 \mathrm{~W} / \mathrm{m}^{2}\), determine the surface temperature of the rod. Evaluate the air properties at \(70^{\circ} \mathrm{C}\).
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