Mechanical Engineering - Thermodynamics - Discussion

1. If we want without any loss (heat) process then it will only reversible process which is a very slow process.

2. We can understand it by taking the example of piston-cylinder arrangement. In this according to irreversible process piston move in a cylinder with some speed, if speed I'd high than heat is a loss, therefore, it is an irreversible process and due to irreversible heat is lost.

On the other hand, heat is not lost when the process is very slow means one stroke of the piston is done in like one hour, one day, one month which is useless for any mechanism.

An irreversible process is a non-quasistatic process which is a fast process in comparison to quasistatic process and the fast process has affected the surrounding around them means loses some energy in the form of friction or heat or in any other type.

For ex- wheel of automobile release some energy on the road (surrounding) in the form of friction and heat which is useless after reaching in the surrounding.

Except Carnot cycle, every process is losses heat or energy without any loss no real process exists in nature.

It is not always true. Ex- let us consider an insulated container (system) having two gases with a finite pressure difference separated by a partition. Removing the partition will result in an irreversible process due to finite pressure difference.

But in this case there is no heat transfer as the system is insulated. So, an irreversible process not necessarily result in heat loss.

(1) For example heat lose b/w body a 500 k and body b 1000 k.

(2) Heat lose b/w body c 300 k and body d 500 k. So process 1 more irreversible then process 2.

(3) For ex heat lose b/w body h 500 k and body I 499.999 k temperature.

Here process 3 is reversible because process is very slow.

So in this question both answer is possible.

Hello friends.

The correct answer is FALSE,

CHANGE IN ENTROPY OF SYSTEM = INT(DQ/T)+ENTROPY GENERATION;

For irreversible process change in entropy will be greater than 0, and as we know that entropy generation is always +,

And for no heat loss the quantity int (dq/T) will be 0 but due to entropy generation change will be +.

In this if temperature difference is very negligible between two bodies then heat lose very slow and process become reversible. But in these bodies having considerable temperature difference then heat lose become fast and process become irreversible. So in these case both answer is possible.

An irreversible process is always accompanied by an increase in entropy. This will result in heat loss.

Even in the case of the adiabatic irreversible process though there is no heat loss due to external irreversibility,

But internal irreversibility will lead to heat loss.

It's a contradictory question. Both answer is possible.

Let's take an example. A gas chamber is subdivided by partition and suddenly partition is removed and it is insulated. It's example of irreversibly but more closet answer is option A.

In reversible process heat can change in this process apple thermodynamic first law. Heat can not created or not destroyed when any process go reversible this heat chang due to some friction and some surrounding loss heat.

In thermodynamics, reversible process is know as quize static process. Irreversible process means we can't get the same amount of energy when we reverse the process. So amount of heat loses through the surroundings.