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andrey2020 [161]

3 years ago

5

A mixing chamber receives a steady flow of water from different parts of a processing plant. It has two inlets and one outlet. T

he mixing chamber is NOT well insulated and DOES allow heat transfer between the device and the surroundings. Kinetic and potential energy effects are negligible. Determine the mass flow rate at inlet 1 in [kg/s]. Determine the rate of heat transfer between the mixing chamber and the surroundings in [kW]; carefully indicate direction (i.e., does energy move by heat transfer to or from the chamber

1 answer:

alex41 [277]3 years ago

4 0

Answer:

hello your question has some missing information attached below is the missing information

answer :

a) 9.17431 kg/s

b) 2598.4374 kW

Explanation:

<u>a) Calculate mass flow rate at inlet 1 </u>

Given data :

AV1 = 0.01 m^3/s

pressure ( P1 ) = 5 bar , temp ( T1 ) = 150°C

therefore : m1 = AV1 / v1 ------ ( 1 )

where : v1 = 0.00109 m^3/kg ( value gotten from compressed table )

back to equation 1

m1 = ( 0.01 m^3/s / 0.00109 m^3/kg ) = 9.17431 kg/s

<u>b) Calculate the rate of heat transfer </u>

first determine mass balance

m3 = m1 + m2

= 9.17431 + 19 = 28.17431 kg/s

apply the energy balance equation to determine the rate of heat transfer

m1*h1 + m2*h2 = m3*h3 + Q

∴ Q = m1*h1 + m2*h2 - m3*h3 ------------ ( 1 )

where :

h1 = 632.266 kJ/kg value gotten from compressed water table

h2 = 2812.45 kJ/kg value gotten from superheated steam table

h3 = 2010.2965 kJ/kg value gotten from saturated pressure table

input values into equation 1 above

Q = 2598.4374 kW . the energy moves from the chamber to the surroundings

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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

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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)

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

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3 years ago

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4 0

3 years ago

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc

wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

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and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

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v = 38.05 m/s

so radial acceleration at distance 200 m

ar = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²

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3 years ago