Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 35

Consider a hot automotive engine, which can be approximated as a \(0.5-\mathrm{m}\)-high, \(0.40\)-m-wide, and \(0.8-\mathrm{m}\)-long rectangular block. The bottom surface of the block is at a temperature of \(100^{\circ} \mathrm{C}\) and has an emissivity of \(0.95\). The ambient air is at \(20^{\circ} \mathrm{C}\), and the road surface is at \(25^{\circ} \mathrm{C}\). Determine the rate of heat transfer from the bottom surface of the engine block by convection and radiation as the car travels at a velocity of \(80 \mathrm{~km} / \mathrm{h}\). Assume the flow to be turbulent over the entire surface because of the constant agitation of the engine block.

Short Answer

Expert verified
Answer: The rates of heat transfer from the bottom surface of an engine block are 3571.04 W by convection and 341.74 W by radiation.

Step by step solution

01

Calculate the surface area of the bottom surface of the engine block

First, we need to find the area of the bottom surface of the engine block, which is a rectangle with a width of 0.4 m and a length of 0.8 m. We can calculate the area as: Area = width × length Area = 0.4 m × 0.8 m Area = 0.32 m²
02

Convert the car's velocity to m/s

We are given the car's velocity in km/h, but we need it to be in m/s. To do this, we can use the following conversion: 1 km/h = 1000 m / 3600 s So, the car's velocity in m/s would be: 80 km/h × (1000 m / 3600 s) = 22.22 m/s
03

Calculate the Reynolds number

We'll use the Reynolds number to determine the proper correlation for the heat transfer coefficient. The Reynolds number can be calculated as: Re = (ρ × V × L) / µ For air at 20°C, we will use the following properties: ρ = 1.204 kg/m³ (density) µ = 1.825 × 10⁻⁵ kg/(m.s) (dynamic viscosity) The Reynolds number for this case is: Re = (1.204 kg/m³ × 22.22 m/s × 0.8 m) / (1.825 × 10⁻⁵ kg/(m.s)) Re = 106657 Since Re > 4000, the flow is turbulent over the bottom surface of the engine block.
04

Calculate the heat transfer coefficient by convection

Using the Dittus-Boelter equation, we can find the heat transfer coefficient for turbulent flow as: h = 0.023 × Re^0.8 × Pr^1/3 × k/L For air at 20°C, we will use the following properties: k = 0.02624 W/(m.K) (thermal conductivity) Pr = 0.707 (Prandtl number) So the heat transfer coefficient can be calculated as: h = 0.023 × 106657^0.8 × 0.707^1/3 × (0.02624 W/(m.K)) / (0.8 m) h = 138.95 W/(m².K)
05

Calculate the heat transfer by convection

Now we can find the heat transfer by convection using the formula: q_conv = h × A × (T_s - T_inf) Where: T_s = 100°C (surface temperature of the engine block) T_inf = 20°C (ambient air temperature) q_conv = 138.95 W/(m².K) × 0.32 m² × (100 - 20) K q_conv = 3571.04 W
06

Calculate the heat transfer by radiation

To find the heat transfer by radiation, we will use the Stefan-Boltzmann law: q_rad = ε × σ × A × (T_s^4 - T_sur^4) Where: ε = 0.95 (emissivity of the surface) σ = 5.67 × 10⁻⁸ W/(m².K⁴) (Stefan-Boltzmann constant) T_s = 100°C + 273.15 K (converted to Kelvin) T_sur = 25°C + 273.15 K (temperature of the road surface converted to Kelvin) q_rad = 0.95 × 5.67 × 10⁻⁸ W/(m².K⁴) × 0.32 m² × ((373.15 K)^4 - (298.15 K)^4) q_rad = 341.74 W So the rate of heat transfer from the bottom surface of the engine block is: By convection: 3571.04 W By radiation: 341.74 W

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the difference between skin friction drag and pressure drag? Which is usually more significant for slender bodies such as airfoils?

Air at \(15^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) flows over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) at a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Compute the following quantities at \(x=0.3 \mathrm{~m}\) : (a) Hydrodynamic boundary layer thickness, \(\mathrm{m}\) (b) Local friction coefficient (c) Average friction coefficient (d) Total drag force due to friction, \(\mathrm{N}\) (e) Local convection heat transfer coefficient, W/m² \(\mathbf{K}\) (f) Average convection heat transfer coefficient, W/m² \(\mathrm{K}\) (g) Rate of convective heat transfer, W

Engine oil at \(105^{\circ} \mathrm{C}\) flows over the surface of a flat plate whose temperature is \(15^{\circ} \mathrm{C}\) with a velocity of \(1.5 \mathrm{~m} / \mathrm{s}\). The local drag force per unit surface area \(0.8 \mathrm{~m}\) from the leading edge of the plate is (a) \(21.8 \mathrm{~N} / \mathrm{m}^{2}\) (b) \(14.3 \mathrm{~N} / \mathrm{m}^{2}\) (c) \(10.9 \mathrm{~N} / \mathrm{m}^{2}\) (d) \(8.5 \mathrm{~N} / \mathrm{m}^{2}\) (e) \(5.5 \mathrm{~N} / \mathrm{m}^{2}\) (For oil, use \(\nu=8.565 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}, \rho=864 \mathrm{~kg} / \mathrm{m}^{3}\) )

A thin, square flat plate has \(1.2 \mathrm{~m}\) on each side. Air at \(10^{\circ} \mathrm{C}\) flows over the top and bottom surfaces of a very rough plate in a direction parallel to one edge, with a velocity of \(48 \mathrm{~m} / \mathrm{s}\). The surface of the plate is maintained at a constant temperature of \(54^{\circ} \mathrm{C}\). The plate is mounted on a scale that measures a drag force of \(1.5 \mathrm{~N}\). Determine the total heat transfer rate from the plate to the air.

During flow over a given body, the drag force, the upstream velocity, and the fluid density are measured. Explain how you would determine the drag coefficient. What area would you use in calculations?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Radiation

Read Explanation

Electromagnetism

Read Explanation

Turning Points in Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Physical Quantities And Units

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free