Anyone help me please ?

Air enters a compressor operating at steady state with pressure of 90 kPa, at a temperature of 350 K, and a volumetric flow rate

natita [175]

Answer:

$T_{out} = 457.921\,K$

Explanation:

Before determining the exit temperature of air, it is required to find the specific enthalpy at outlet by using the First Law of Thermodynamics:

$-\dot Q_{out} + \dot W_{un} + \dot m \cdot (h_{in}-h_{out})=0$

$h_{out} = \frac{\dot W_{in}}{\dot m}- q_{out} +h_{in}$

An ideal gas observes the following mathematical model:

$P\cdot V = n\cdot R_{u}\cdot T$

Where:

$P$ - Absolute pressure, in kilopascals.

$V$ - Volume, in cubic meters.

$n$ - Quantity of moles, in kilomole.

$R_{u}$ - Ideal gas universal constant, in $\frac{kPa\cdot m^{3}}{kmole\cdot K}$ .

$T$ - Absolute temperature, in kelvin.

The previous equation is re-arranged in order to calculate specific volume at inlet:

$P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T$

$\nu = \frac{R_{u}\cdot T}{P\cdot M}$

$\nu_{in} = \frac{(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} )\cdot (350\,K)}{(90\,kPa)\cdot (28.97\,\frac{kg}{kmol} )}$

$\nu_{in} = 1.116\,\frac{m^{3}}{kg}$

The mass flow is:

$\dot m = \frac{\dot V}{\nu_{in}}$

$\dot m = \frac{0.6\,\frac{m^{3}}{s} }{1.116\,\frac{m^{3}}{kg} }$

$\dot m = 0.538\,\frac{kg}{s}$

The specific enthalpy in ideal gases depends on temperature exclusively. Then:

$h_{in} = 350.49\,\frac{kJ}{kg}$

The specific enthalpy at outlet is:

$h_{out} = \frac{75\,kW}{0.538\,\frac{kg}{s} }-30\,\frac{kJ}{kg} + 350.49\,\frac{kJ}{kg}$

$h_{out} = 459.895\,\frac{kJ}{kg}$

The exit temperature of air is:

$T_{out} = 457.921\,K$

5 0

3 years ago

Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downward

andrew-mc [135]

Answer:

We choose PTFE

Explanation:

Attached are the modulus density and modulus strength chart.

Due to its young modulus, the density is near 0.5 GPa, as seen in the chart and support water gliding. The PTFE density is between 1 and 10 Mg / cubic meter (see module and chart of density), and the resistance is between 10 and 100 Mpa (see module and chart of strength). Therefore, the finest ploymer will be PTFE that meets the requirements.

8 0

4 years ago

What have you learned about designing solutions? How does this apply to engineering? Think of some engineering solutions that st

Andrew [12]

Answer:

In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.

Explanation:

4 0

3 years ago

(d) Suppose two students are memorizing a list according to the same model dL dt = 0.5(1 − L) where L represents the fraction of

Neporo4naja [7]

Answer:

Rate of learning =0

Explanation:

Please see attachment

8 0

3 years ago

A hollow pipe is submerged in a stream of water so that the length of the pipe is parallel to the velocity of the water. If the

Arlecino [84]

Answer:

increases by a factor of 6.

Explanation:

Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:

Initial flow rate = area * velocity = A * V = AV m³/s

The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:

Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate

Hence, the volume flow rate of the water passing through it increases by a factor of 6.

8 0

3 years ago