Designating a standard part is an example of a use for ______

A 2 in. diameter pipe supplying steam at 300°F is enclosed in a 1 ft square duct at 70°F. The outside of the duct is perfectly i

Shkiper50 [21]

Answer:

The value of heat transferred watt per foot length Q = 54.78 Watt per foot length.

Explanation:

Diameter of pipe = 2 in = 0.0508 m

Steam temperature $T_{1}$ = 300 F = 422.04 K

Duct temperature $T_{2}$ = 70 F = 294.26 K

Emmisivity of surface 1 = 0.79

Emmisivity of surface 2 = 0.276

Net emmisivity of both surfaces ∈ = 0.25

Stefan volazman constant $\sigma$ = 5.67 × $10^{-8}$ $\frac{W}{m^{2} K^{4} }$

Heat transfer per foot length is given by

Q = ∈ $\sigma$ A ( $T_{1}^{4} - T_{2} ^{4}$ ) ------ (1)

Put all the values in equation (1) , we get

Q = 0.25 × 5.67 × $10^{-8}$ × 3.14 × 0.0508 × 1 × ( $422.04^{4} - 294.26^{4}$ )

Q = 54.78 Watt per foot.

This is the value of heat transferred watt per foot length.

4 0

3 years ago

Who is the inventor that created the light bulb

uranmaximum [27]

Answer:

Thomas Elva Edison

He invented light.

4 0

3 years ago

An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir

Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

$\frac{A}{F}$ = $\frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}$

Formula Used:

$A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}$

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

$\omega = \frac{2 \pi\times N}{60}$

so,

$\omega = \frac{2 \pi\times 720}{60}$ = 75.39 rad/s

Using the given formula:

Damping is negligible, so, $\zeta = 0$

$\frac{A}{F}$ will give the tranfer function

Therefore,

$\frac{A}{F}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

$0.1\times 10^{-3}$ = $\frac{1}{\sqrt{(4K - 120\ ^{2})}}$

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

8 0

3 years ago

Calculate the volume of a hydraulic accumulator capable of delivering 5 liters of oil between 180 and 80 bar, using as a preload

Vinil7 [7]

Answer:

1) $V_o = 10 liters$

2) $V_o = 12.26 liters$

Explanation:

For isothermal process n =1

$V_o =\frac{\Delta V}{(\frac{p_o}{p_1})^{1/n} -(\frac{p_o}{p_2})^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1} -[\frac{72}{180}]^{1/1}}$

$V_o = 10 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.03

$actual \ volume = c1\times 10 = 10.3 liters$

b) for adiabatic process

n =1.4

volume of hydraulic accumulator is given as

$V_o =\frac{\Delta V}{[\frac{p_o}{p_1}]^{1/n} -[\frac{p_o}{p_2}]^{1/n}}$

$V_o = \frac{5}{[\frac{72}{80}]^{1/1.4} -[\frac{72}{180}]^{1/1.4}}$

$V_o = 12.26 liters$

calculate pressure ratio to determine correction factor

$\frac{p_2}{p_1} =\frac{180}{80} = 2.25$

correction factor for calculate dpressure ration for isothermal process is

c1 = 1.15

$actual \volume = c1\times 10 = 11.5 liters$

8 0

3 years ago

An engineer lives in Hawaii at a location where the annual rain fall is 300 inches. She decides to use the rain to generate elec

Alex777 [14]

Answer:

80.7lbft/hr

Explanation:

Flow rate of water in the system = 3.6x10^-6

The height h = 100

1s = 1/3600h

This implies that

Q = 3.6x10^-6/[1/3600]

Q = 0.0000036/0.000278

Q = 0.01295

Then the power is given as

P = rQh

The specific weight of water = 62.3 lb/ft³

P = 62.3 x 0.01295 x 100

P = 80.675lbft/h

When approximated

P = 80.7 lbft/h

This is the average power that could be generated in a year.

This answers the question and also corresponds with the answer in the question.

4 0

3 years ago

Designating a standard part is an example of a use for _______ notes.