Describe the greatest power in design according to Aravena?

Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surro

loris [4]

Answer:

(a)The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) The final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) The volume flow rate at exit is V₃ =V₁ + V₂

Explanation:

Solution

Now

The system comprises of two inlets and on exit.

Mass flow rate enthalpy of fluid from inlet -1 be m₁ and h₁

Mass flow rate enthalpy of fluid from inlet -2 be m₂ and h₂

Mass flow rate enthalpy of fluid from exit be m₃ and h₃

Mixing chambers do not include any kind of work (w = 0)

So, both the kinetic and potential energies of the fluid streams are usually negligible (ke =0, pe =0)

(a) Applying the mass balance of mixing chamber, min = mout

Applying the energy balance of mixing chamber,

Ein = Eout

min hin =mout hout

miCpT₁ + m₂CpT₂ +Qin =m₃CpT₃

T₃ = miCpT₁/m₃CpT₃ + m₂CpT₂/m₃CpT₃ + Qin/m₃CpT₃ +

T₃ =m₁T₁/m₃+ m₂T/m₃ + Qin/m₃Cp

The final temperature of mixture is T₃ =m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp

(b) From the ideal gas equation,

v =RT/PT

v₃ = RT₃/P₃

The volume flow rate at the exit, V₃ =m₃v₃

V₃ = m₃ RT₃/P₃

Substituting the value of T₃, we have

V₃=m₃ R/P₃ (=m₁T₁/m₃+ m₂T₂/m₃ + Qin/m₃Cp)

V₃ = R/P₃ (m₁T₁+ m₂T₂ + Qin/Cp)

Now

The mixing process occurs at constant pressure P₃=P₂=P₁.

Hence V₃ becomes:

V₃=m₁RT₁/P₁ +m₂RT₂/P₂ + RQin/P₃Cp

V₃ =V₁ + V₂ + RQin/P₃Cp

Therefore, the final volume is V₃ =V₁ + V₂ + RQin/P₃Cp

(c) Now for an adiabatic mixing, Qin =0

Hence V₃ becomes:

V₃ =V₁ + V₂ + r * 0/P₃Cp

V₃ =V₁ + V₂ + 0

V₃ =V₁ + V₂

Therefore the volume flow rate at exit is V₃ =V₁ + V₂

8 0

3 years ago

All of the following are examples of capital intensive industries EXCEPT: *

OverLord2011 [107]

i believe it is soft drink production

8 0

2 years ago

Read 2 more answers

All circuits need three basic parts: an energy source, wires, and the object that is going to change the electrical energy into

Radda [10]

load every electric circuit,regardless of where it is or how large or small, has four basic parts: an energy source (ac or dc),a conductor (wire), an electrical load (device), and at least one controller(switch)

7 0

3 years ago

Read 2 more answers

A 1020 CD steel shaft is to transmit 15 kW while rotating at 1750 rpm. Determine the minimum diameter for the shaft to provide a

vladimir2022 [97]

Answer:

diameter is 14 mm

Explanation:

given data

power = 15 kW

rotation N = 1750 rpm

factor of safety = 3

to find out

minimum diameter

solution

we will apply here power formula to find T that is

power = 2π×N×T / 60 .................1

put here value

15 × $10^{3}$ = 2π×1750×T / 60

so

T = 81.84 Nm

and

torsion = T / Z ..........2

here Z is section modulus i.e = πd³/ 16

so from equation 2

torsion = 81.84 / πd³/ 16

so torsion = 416.75 / / d³ .................3

so from shear stress theory

torsion = σy / factor of safety

so here σy = 530 for 1020 steel

so

torsion = σy / factor of safety

416.75 / d³ = 530 × $10^{6}$ / 3

so d = 0.0133 m

so diameter is 14 mm

3 0

3 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

3 years ago