What happens to the odometer reading when a car drives beyond its maximum reading?

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$

Then,

$T_{2}=1040K(\frac{120kPa}{278kPa})^{0.287/1.004}=817.96K$

So, now it is possible to calculate $h_{2s}-h_{1}$ :

$h_{2s}-h_{1}}=Cp(T_{2s}-T_{1}})=1.004\frac{kJ}{kgK}*(817.96K-1040K)=-222.92\frac{kJ}{kg}$

Finally, the efficiency can be calculated:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}=\frac{-203.64\frac{kJ}{kg}}{-222.92\frac{kJ}{kg}}\\e=0.913=91.3$ %

4 0

3 years ago

The author states that chemical engineering is one of the most difficult and complex aspects of engineering. Why do you think th

marishachu [46]

Answer:

hope this helps

Explanation:

answers:

1. Chemical engineering is most difficult because it's a mix of physics, chemistry and math

2. Stoichiometry is so important because it shows how materials react, interact and play off each other

3. Yes I think consumers would notice if process control standards were not met. for example medicines, when people take Tylenol or cold pills, if the amount of time it took to kick it becomes longer, people will become aware that the product is not consistent and reliable.

4. i have no idea sorry :(

5. This is explaining how there are rules and regulations to make the workplace safe. it can be accomplished by following those rules and regulations

8 0

3 years ago

Read 2 more answers

Getting the bottom of your feet burned when walking on hot sand is due to what form of energy transmission? group of answer choi

Scilla [17]

Getting the bottom of your feet burned when walking on hot sand is due to a form of energy transmission known as conduction.

<h3>The types of energy transmission.</h3>

In Science, there are three (3) main types of energy transmission and these include the following:

Convection

Radiation

Conduction

In this scenario, we can infer and logically conclude that burning the bottom of your feet when walking on hot sand is primarily due to a form of energy transmission known as conduction because it involves the transfer of thermal energy (heat) due to the movement of particles.

Read more on heat conduction here: brainly.com/question/12072129

#SPJ12

6 0

1 year ago

A system consists of a disk rotating on a frictionless axle

kakasveta [241]

The system includes a disk rotating on a frictionless axle and a bit of clay transferring towards it, as proven withinside the determine above.

<h3>What is the angular momentum?</h3>

The angular momentum of the device earlier than and after the clay sticks can be the same.

Conservation of angular momentum the precept of conservation of angular momentum states that the whole angular momentum is usually conserved.

Li = Lf where;
li is the preliminary second of inertia
If is the very last second of inertia
wi is the preliminary angular velocity
wf is the very last angular velocity
Li is the preliminary angular momentum
Lf is the very last angular momentum

Thus, the angular momentum of the device earlier than and after the clay sticks can be the same.