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2 years ago

The drag force, Fd, imposed by the surrounding air on a

Engineering

1 answer:

Ad libitum [116K]2 years ago

3 0

Answer:

a) 23.551 hp

b) 516.89 hp

Explanation:

given:

$F_{d} =\frac{1}{2} C_{d} A_{p} V^{2} \\V_{a}=25 m/hr-->25*\frac{5280}{3600} =36.67ft/s\\V_{b}=70 m/hr-->70*\frac{5280}{3600} =102.67ft/s\\\\C_{d}=.28\\A=25 ft^2\\p=.075lb/ft^2$

required:

the power in hp

solution:

$(F_{d})_{a} =\frac{1}{2} C_{d} A_{p} V_{a} ^{2}$ .............(1)

by substituting in the equation (1)

=353.27 lbf

$(F_{d})_{b} =\frac{1}{2} C_{d} A_{p} V_{b} ^{2}$ ..........(2)

by substituting in the equation (2)

= 2769.29 lbf

power is defined by

P=F.V

$P_{a}=$ 353.27*36.67

=12954.411 lbf.ft/s

=12954.411*.001818

=23.551 hp

$P_{a}=$ 2769.29*102.67

= 284323 lbf.ft/s

= 284323*.001818

= 516.89 hp

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Water vapor at 10bar, 360°C enters a turbine operatingat steady state with a volumetric flow rate of 0.8m3/s and expandsadiabati

Artyom0805 [142]

Answer:

A) W' = 178.568 KW

B) ΔS = 2.6367 KW/k

C) η = 0.3

Explanation:

We are given;

Temperature at state 1;T1 = 360 °C

Temperature at state 2;T2 = 160 °C

Pressure at state 1;P1 = 10 bar

Pressure at State 2;P2 = 1 bar

Volumetric flow rate;V' = 0.8 m³/s

A) From table A-6 attached and by interpolation at temperature of 360°C and Pressure of 10 bar, we have;

Specific volume;v1 = 0.287322 m³/kg

Mass flow rate of water vapour at turbine is defined by the formula;

m' = V'/v1

So; m' = 0.8/0.287322

m' = 2.784 kg/s

Now, From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific enthalpy;h1 = 3179.46 KJ/kg

Now, From table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific enthalpy;h2 = 3115.32 KJ/kg

Now, since stray heat transfer is neglected at turbine, we have;

-W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2 - h1)

Plugging in relevant values, the work of the turbine is;

W' = -2.784(3115.32 - 3179.46)

W' = 178.568 KW

B) Still From table A-6 attached and by interpolation at state 1 with temperature of 360°C and Pressure of 10 bar, we have;

Specific entropy: s1 = 7.3357 KJ/Kg.k

Still from table A-6 attached and by interpolation at state 2 with temperature of 160°C and Pressure of 1 bar, we have;

Specific entropy; s2 = 8.2828 KJ/kg.k

The amount of entropy produced is defined by;

ΔS = m'(s2 - s1)

ΔS = 2.784(8.2828 - 7.3357)

ΔS = 2.6367 KW/k

C) Still from table A-6 attached and by interpolation at state 2 with s2 = s2s = 8.2828 KJ/kg.k and Pressure of 1 bar, we have;

h2s = 2966.14 KJ/Kg

Energy equation for turbine at ideal process is defined as;

Q' - W' = m'[(h2 - h1) + ((V2)² - (V1)²)/2 + g(z2 - z1)]

Again, Potential and kinetic energy can be neglected and so we have;

-W' = m'(h2s - h1)

W' = -2.784(2966.14 - 3179.46)

W' = 593.88 KW

the isentropic turbine efficiency is defined as;

η = W_actual/W_ideal

η = 178.568/593.88 = 0.3

8 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

Technician A says that weld-on dent removal attachments may be used on a steel panel.

spin [16.1K]

Answer:

sjasfjajfhajshfdff

Explanation:

7 0

2 years ago

A single-degree-of-freedom mass-spring-damper system is observed during its free vibration and the displacement amplitude decays

AleksandrR [38]

Answer:

Logarithmic decrement is equal to 0.182

Explanation:

given,

amplitude decay = 9 dB

number of cycles = 12 cycles

mass of the system = 7 kg

spring stiffness = 3000 N/m

logarithmic decrement = ?

now,

logarithmic decreament = $ln\ D^{\frac{1}{n}}$

= $ln\ 9^{\frac{1}{12}}$

=ln (1.2)

= 0.182

Hence, Logarithmic decrement is equal to 0.182

5 0

3 years ago

(a) The reverse-saturation current of a pn junction diode is IS = 10−11 A. Determine the diode voltage to produce currents of (i

kirill115 [55]

Answer:

The equation used to solve a diode is

$i_d = I_se^\frac{V_d}{V_T}-1$

$i_d$ is the current going through the diode
$I_s$ is your saturation current
$V_D$ is the voltage across your diode
$V_T$ is the voltage of the diode at a certain room temperature. by default, you always use $V_T=25.9mV$ for room temperature.

If you look at the equation, $i_d = I_se^\frac{V_d}{V_T}-1$ , you'd notice that the $e^\frac{V_d}{V_T}$ grow exponentially fast, so we can ignore the -1 in the equation because it's so small compared to the exponential.