Ask question

Ask question

All categories

3 years ago

6

. In one stroke of a reciprocating compressor, helium is isothermally and reversibly compressed in a piston + cylinder from 298

K and 20 bars to 200 bars. Compute the heat removal and work required.

1 answer:

andriy [413]3 years ago

5 0

Answer:

5.7058kj/mole

Explanation:

Please see attachment for step by step guide

You might be interested in

What are practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectio

Mrrafil [7]

Answer:

The answer is below

Explanation:

The practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectional area fixed are:

1. Shapes of moment of inertia: Engineers should consider or know the different shapes of moment of inertia for different shape

2. Understanding the orientation of the beam: this will allow engineers to either increase or decrease the moment of inertia of a beam without increasing its cross sectional area.

4 0

3 years ago

What should you use to turn the water off and open the door?

galina1969 [7]

youn need to use your hands

7 0

3 years ago

Read 2 more answers

A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

$E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}$

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet $P_1$ = 96 kPa

the pressure at the exit $P_2$ = ???

the pressure $\rho$ = 1000 kg/m³

∴

$0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}$

$0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$\dfrac{800000}{2} = P_2-96000$

400000 = P₂ - 96000

400000 + 96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0

2 years ago

A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2

Svetllana [295]

$79 f t^{3}$ is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

$V_{1}=100\ \mathrm{ft}^{3}$

First has a natural water content of 25% = $\frac{25}{100}$ = 0.25

Shrinkage limit, $w_{1}=12 \%=\frac{12}{100}=0.12$

$G_{s}=2.70$

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

$V \propto[1+e]$

$\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}$ ------> eq 1

$e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}$

The above equation is at $S_{r}=1$ ,

$e_{1}=w_{1} \times G_{s}$

Applying the given values, we get

$e_{1}=0.25 \times 2.70=0.675$

Shrinkage limit is lowest water content

$e_{2}=w_{2} \times G_{s}$

Applying the given values, we get

$e_{2}=0.12 \times 2.70=0.324$

Applying the found values in eq 1, we get

$\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904$

$V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}$

7 0

3 years ago

This is it dont anwser this is for my other account

Nezavi [6.7K]

Answer:

thanks for the poiunts

Explanation:

6 0

3 years ago