Engineers are problem blank<br> who use critical thinking to create new solutions.

A 1000 kg turbine has a rotating unbalance of 0.1 kg.m. The turbine operates at a speed between 500 to 750 rpm. What is the maxi

raketka [301]

Answer:

maximum isolator stiffness k =1764 kN-m

Explanation:

mean speed of rotation $=\frac{N_1 +N_2}{2}$

$Nm = \frac{500+750}{2} = 625 rpm$

$w =\frac{2\pi Nm}{60}$

=65.44 rad/sec

$F_T = mw^2 e$

$F_T = mew^2$

= 0.1*(65.44)^2

F_T =428.36 N

Transmission ratio $=\frac{300}{428.36} = 0.7$

also

transmission ratio $= \frac{1}{[\frac{w}{w_n}]^{2} -1}$

$0.7 =\frac{1}{[\frac{65.44}{w_n}]^2 -1}$

SOLVING FOR Wn

Wn = 42 rad/sec

$Wn = \sqrt {\frac{k}{m}$

k = m*W^2_n

k = 1000*42^2 = 1764 kN-m

k =1764 kN-m

3 0

3 years ago

A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th

inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = $\int\limits^a_b P \, dV$ = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 0

3 years ago

In the final stages of production, a pharmaceutical is sterilized by heating it from 25 to 75°C as it moves at 0.2 m/s through a

stepan [7]

Answer:

The required heat flux = 12682.268 W/m²

Explanation:

From the given information:

The initial = 25°C

The final = 75°C

The volume of the fluid = 0.2 m/s

The diameter of the steel tube = 12.7 mm = 0.0127 m

The fluid properties for density $\rho$ = 1000 kg/m³

The mass flow rate of the fluid can be calculated as:

$m = pAV$

$m = \rho \dfrac{\pi}{4}D^2V$

$m = 1000 \times \dfrac{\pi}{4} \times ( 0.0127)^2 \times 0.2$

$m = 0.0253 \ kg/s$

To estimate the amount of the heat by using the expression:

$q = mc_p(T_{final}-T_{initial})$

q = 0.0253 × 4000(75-25)

q = 101.2 (50)

q = 5060 W

Finally, the required heat of the flux is determined by using the formula:

$q" = \dfrac{q}{A_s}$

$q" = \dfrac{q}{\pi D L}$

$q" = \dfrac{5060}{\pi \times 0.0127 \times 10}$

q" = 12682.268 W/m²

The required heat flux = 12682.268 W/m²

3 0

3 years ago

A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio

natima [27]

Answer:

1) A=282.6 mm

2) $a_{max}=60.35\ m/s^2$

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm ------------1

V=2.5 m/s at x=225 mm ------------2

Where x measured from mid position.

We know that velocity in simple harmonic given as

$V=\omega \sqrt{A^2-x^2}$

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

$3.5=\omega \sqrt{A^2-0.15^2}$ ------3

$2.5=\omega \sqrt{A^2-0.225^2}$ --------4

Now by dividing equation 3 by 4

$\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}$

$1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}$

So A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

$3.5=\omega \sqrt{A^2-0.15^2}$

$3.5=\omega \sqrt{0.2826^2-0.15^2}$

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

$a_{max}=\omega ^2A$

$a_{max}=14.61 ^2\times 0.2826\ m/s^2$

$a_{max}=60.35\ m/s^2$

Time period T

$T=\dfrac{2\pi}{\omega}$

$T=\dfrac{2\pi}{14.609}$

T=0.42 sec

8 0

4 years ago

A pump is used to extract water from a reservoir and deliver it to another reservoir whose free surface elevation is 200 ft abov

babunello [35]

Answer:

a) the expected flow rate is 31.4 ft³/s

b) the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation is -8.4 ft

Explanation:

Given the data in the question;

free surface elevation = 200 ft

total length of pipe required = 1000 ft

diameter = 12 inch

Iron with relative roughness ( k/D ) = 0.0005

H $_{pump$ = 665-0.051Q² [Qinft ]

a) the expected flow rate

given that;

k/D = 0.0005

k/2R = 0.0005

R/k = 1000

now, we determine the friction factor;

1/√f = 2log₁₀( R/k ) + 1.74

we substitute

1/√f = 2log₁₀( 1000 ) + 1.74

1/√f = 6 + 1.74

1/√f = 7.74

√f = 1/7.74

√f = 0.1291989

f = (0.1291989)²

f = 0.01669

Now, Using Bernoulli theorem between two reservoirs;

(p/ρq)₁ + (v²/2g)₁ + z₁ + H $_p$ = (p/ρq)₂ + (v²/2g)₂ + z₂ + h $_L$

so

0 + 0 + 0 + 665-0.051Q² = 0 + 0 + 200 + flQ²/2gdA²

665-0.051Q² = 200 + flQ²/2gdA²

665-0.051Q² = 200 +[ ( 0.01669 × 1000 × Q² ) / (2 × 32.2 × (π/4)² × 1⁵ )

665 - 0.051Q² = 200 + [ 16.69Q² / 39.725 ]

665 - 200 - 0.051Q² = 0.420138Q²

665 - 200 = 0.420138Q² + 0.051Q²

465 = 0.471138Q²

Q² = 465 / 0.471138

Q² = 986.97196

Q = √986.97196

Q = 31.4 ft³/s

Therefore, the expected flow rate is 31.4 ft³/s

b) the brake horsepower required to drive the pump (assume an efficiency of 78%).

we know that;

P = ρgH $_p$ Q / η

where; H $_p$ = 665 - 0.051(986.97196) = 614.7

we substitute;

P = ( 62.42 × 614.7 × 31.4 ) / ( 0.78 × 550 )

P = 1204804.6236 / 429

P = 2808.4 bhp

Therefore, the required brake horsepower is 2808.4 bhp

c) the location of pump inlet to avoid cavitation (assume the required NPSH=25 ft).

NPSH = ( $P_{atom$ / ρg) - h $_s$ - ( P $_v$ / ρg )

we substitute

25 = ( 2116 / 62.42 ) - h $_s$ - ( 30 / 62.42 )

h $_s$ = 8.4 ft

Therefore, the location of pump inlet to avoid cavitation is -8.4 ft

6 0

3 years ago

Engineers are problem blank who use critical thinking to create new solutions.