AUTO TECH/TASK SHEET
AUTO TECH/TASK SHEET
6th Edition
ISBN: 9780135257630
Author: Halderman
Publisher: PEARSON
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Textbook Question
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Chapter 27, Problem 1RQ

How can a thermostat fail?

Expert Solution & Answer
Check Mark
To determine

Explain how a thermostat can fail.

Explanation of Solution

A thermostat can fail in the following four ways.

  • Stuck open.
  • Stuck closed.
  • Stuck partially open.
  • Skewed.

Stuck open:

A thermostat can fail when open or partially open because the normal temperature of the engine will be higher than the operating temperature.

Stuck closed:

The thermostat can fail when closed or almost closed because the engine will start to overheat.

Stuck partially open:

The thermostat can fail when partially open because the engine will start to warm up gradually indicating that the coolant temperature of the engine will not attain the desired temperature.

Skewed:

The thermostat can fail when skewed because the engine could overheat or function cooler than regular, or even perform both.

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Students have asked these similar questions
A short brass cyclinder (denisty=8530 kg/m^3, cp=0.389 kJ/kgK, k=110 W/mK, and alpha=3.39*10^-5 m^2/s) of diameter 4 cm and height 20 cm is initially at uniform temperature of 150 degrees C. The cylinder is now placed in atmospheric air at 20 degrees C, where heat transfer takes place by convection with a heat transfer coefficent of 40 W/m^2K. Calculate (a) the center temp of the cylinder, (b) the center temp of the top surface of the cylinder, and (c) the total heat transfer from the cylinder 15 min after the start of the cooling. Solve this problem using the analytical one term approximation method. (Answer: (a) 45.7C, (b)45.3C, (c)87.2 kJ)
A short brass cyclinder (denisty=8530 kg/m^3, cp=0.389 kJ/kgK, k=110 W/mK, and alpha=3.39*10^-5 m^2/s) of diameter 4 cm and height 20 cm is initially at uniform temperature of 150 degrees C. The cylinder is now placed in atmospheric air at 20 degrees C, where heat transfer takes place by convection with a heat transfer coefficent of 40 W/m^2K. Calculate (a) the center temp of the cylinder, (b) the center temp of the top surface of the cylinder, and (c) the total heat transfer from the cylinder 15 min after the start of the cooling. Solve this problem using the analytical one term approximation method.
A 6 cm high rectangular ice block (k=2.22 W/mK, and alpha=0.124*10^-7 m^2/s) initially at -18 degrees C is placed on a table on its square base 4 cm by 4cm in size in a room at 18 degrees C. The heat transfer coefficent on the exposed surfaces of the ice block is 12 W/m^2K. Disregarding any heat transfer from the base to the table, determine how long it will be before the ice block starts melting. Where on the ice block will the first liquid droplets appear? Solve this problem using the analytical one-term approximation method.

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