Ask question

Ask question

All categories

Gelneren [198K]

3 years ago

6

Water at 70 kPa and 1008C is compressed isentropically in a closed system to 4 MPa. Determine the final temperature of the water

and the work required, in kJ/kg, for this compression.

1 answer:

Nikitich [7]3 years ago

3 0

Answer:

The answer is "909.3928 KJ".

Explanation:

$70 \ kPa \ \ and \ \ 100^{\circ}C \\\\s_i= 7.56162\ \frac{kJ}{kgK}\\\\u_i= 2509.39 \ \frac{kJ}{kg}\\\\$

The method is isentropic since the cylinders are shielded.

Calculating the work:

$w= u_2-u_i \\\\$

$= 3418.7728-2509.38 \\\\=909.3928 \ KJ$

You might be interested in

Which option identifies the concept represented in the following scenario?

dlinn [17]

Answer:

project object

Explanation:

3 0

3 years ago

A PMMA plate with a 25 mm (width) x 6.5 mm (thickness) cross-section has a contained crack of length 2c = 0.5 mm in the center o

victus00 [196]

Answer:

LAOD = 6669.86 N

Explanation:

Given data:

width $= 25 mm = 25\times 10^{-3} m$

thickness $= 6.5 mm = 6.5\times 10^{-3} m$

crack length 2c = 0.5 mm at centre of specimen

$\sigma _{applied} = 1000 N/cross sectional area$

stress intensity factor = k will be

$\sigma_{applied} = \frac{1000}{25\times 10^{-3}\times 6.5\times 10^{-3}}$

$= 6.154\times 10^{6} Pa$

we know that

$k =\sigma_{applied} (\sqrt{\pi C})$

$=6.154\sqrt{\pi (2.5\times 10^{-04})}$ [c =0.5/2 = 2.5*10^{-4}]

K = 0.1724 Mpa m^{1/2} for 1000 load

if $K_C = 1.15 Mpa m^{1/2}$ then load will be

$Kc = \sigma _{frac}(\sqrt{\pi C})$

$1.15 MPa = \sigma _{frac}\times \sqrt{\pi (2.5\times 10^{-04})}$

$\sigma _{frac} = 41.04 MPa$

$load = \sigma _{frac}\times Area$

$load = 41.04 \times 10^6 \times 25\times 10^{-3}\times 6.5\times 10^{-3} N$

LAOD = 6669.86 N

3 0

3 years ago

A coal fired power plant geneartes 2.4 lbs. of CO2 per kWh. A lighting system consumes 300,000kWh per year. A corporation is con

Serjik [45]

Answer:

The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.

Explanation:

The CO2 requirement for the plant is:

Amount of CO2 per year = (2.4 lb / KWh)(300,000 KWh)

Amount of CO2 per year = (720000 lb)(1 ton/ 2204.62 lb)

Amount of CO2 per year = 326.59 ton

The perceived economic impact of CO2 generated per year will then be:

Economic Impact = ($25 / ton)(326.59 ton)

<u>Economic Impact = $8164.67</u>

7 0

3 years ago

Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.

mezya [45]

I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!

4 0

3 years ago

A traffic flow has density 61 veh/km when the speed is 59 veh/hr. If a flow has a jam density of 122 veh/km, what is the maximum

antoniya [11.8K]

Since this traffic flow has a jam density of 122 veh/km, the maximum flow is equal to 3,599 veh/hr.

<u>Given the following data:</u>

Density = 61 veh/km.

Speed = 59 km/hr.

Jam density = 122 veh/km.

<h3>How to calculate the maximum flow.</h3>

According to Greenshield Model, maximum flow is given by this formula:

$q_{max}=\frac{V_f \times K_i}{4}$

<u>Where:</u>

$V_f$ is the free flow speed.

$K_i$ is the Jam density.

In order to calculate the free flow speed, we would use this formula:

$V_f =2 V\\\\V_f =2\times 59\\\\V_f=118\;km/hr$

Substituting the parameters into the model, we have:

$q_{max}=\frac{118 \times 122}{4}\\\\q_{max}=\frac{14396}{4}$

Max flow = 3,599 veh/hr.

Read more on traffic flow here: brainly.com/question/15236911

6 0

2 years ago