Thermodynamics Review Problems for Mechanical Engineering Students | Series 7

Hi folks!

This is the 7^th series of my "Thermodynamics Review Problems for Mechanical Engineering Students". If you've missed the previous series you may try scrolling this blog and head over to the "Curriculum". This series features two problems for which the first review problem requires us to obtain the change in entropy of a steel rivet whereas the second review problem requires us to obtain the volume that a certain hydraulic hoist needs to pumped out as it lifts a car with a given mass. Without much ado, let's get started in solving these problems.

Review Problem 1

A 0.5 kilogram steel (c = 0.5 kilojoule per kilogram per degree Kelvin) rivet cools from 800 degree Kelvin to 300 degree Kelvin upon being installed in a riveted structure. The entropy change in kilojoule per degree Kelvin of this rivet is:

Solution

For this review problem, we were not sure if the steel rivets are undergoing either a constant pressure or constant volume process so with that we will just leave the given specific heat capacity as is. And since the formula for computation of the change in entropy for both isobaric and isometric process follows this one,
, for which x can either be p for constant pressure or v for constant volume. We will just use that formula and with that we’ve found out that the change in entropy being experienced by the steel rivets when its temperature had dropped is equal to - 0.245 kilojoule per degree Kelvin, for which the calculation is shown below.

---

Review Problem 2

A 1750 kilogram car is raised to a height of 1.8 meters by a hydraulic hoist. The hydraulic pump has a constant pressure of 800 kilo-pascal on its piston. How much volume in cubic meters (m3) should the pump displace to deliver the required work for lifting the car?

Solution

For the second review problem, the very first thing to obtain is the work that is done by the hydraulic hoist as it lifts that 1750 kilogram car. And we know that work can be expressed as a product of force and displacement, wherein it has a formula of
, wherein F is force and d is the displacement. And since we are provided with the mass of the car, we need to obtain its weight which is its force with respect to gravity, or in layman’s terms we simply multiply it with 9.81 meters per square seconds, and with that we’ve obtained the weight of the car which is equal to 17.2 kilonewtons. Thus, we now have work that is equal to 30.96 kilojoule, as shown in the photo below.

For the computation for the volume of the fluid that is being displaced by the hydraulic hoist, we will be using the formula, W = PV. And with that we’ve found out that the required volume of the fluid that the hydraulic hoist needs to displaced in order to lift that 1750 kilogram car, is equal to 0.0387 cubic meters, for which the computation is shown in the photo below.

---

Curriculum

• Series 1
• Series 2
• Series 3
• Series 4
• Series 5
• Series 6

---

_{Computations and screenshots are made by the author.
Special thanks to @jbeguna04 for designing the GIF photos.}