Draw a sinusoidal signal and illustrate how quantization and sampling is handled by

An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel castin

Maksim231197 [3]

Answer:

h = 375 KW/m^2K

Explanation:

Given:

Thermo-couple distances: L_1 = 10 mm , L_2 = 20 mm

steel thermal conductivity k = 15 W / mK

Thermo-couple temperature measurements: T_1 = 50 C , T_2 = 40 C

Air Temp T_∞ = 100 C

Assuming there are no other energy sources, energy balance equation is:

E_in = E_out

q"_cond = q"_conv

Since, its a case 1-D steady state conduction, the total heat transfer rate can be found from Fourier's Law for surfaces 1 and 2

q"_cond = k * (T_1 - T_2) / (L_2 - L_1) = 15 * (50 - 40) / (0.02 - 0.01)

=15KW/m^2

Assuming SS is solid, temperature at the surface exposed to air will be 60 C since its gradient is linear in the case of conduction, and there are two temperatures given in the problem. Convection coefficient can be found from Newton's Law of cooling:

q"_conv = h * ( T_∞ - T_s ) ----> h = q"_conv / ( T_∞ - T_s )

h = 15000 W / (100 - 60 ) C = 375 KW/m^2K

4 0

3 years ago

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

4 0

3 years ago

1. A soil core sampling tube of 4 cm diameter, 12 cm length and initial mass of 0.525 kg (sample only), was dried at 105o C and

belka [17]

Answer:

porosity = 0.07 or 7%

dry bulk density = 3.25g/cm3]

water content =

Explanation:

bulk density = dry Mass / volume of sample

dry mass = 0.490kg = 490g

volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3

density = 490/150.8 = 3.25g/cm3

porosity = $\frac{wet mass - dry mass }{wet mass}$ = $\frac{0.525 - 0.49}{0.525}$ = 0.07 or 7%

water content = $\frac{wet mass - dry mass}{wet mass}$ = 7%

8 0

3 years ago

Read 2 more answers

What is the diffrence between a small block and a big block lets see if yall know

ivann1987 [24]

Answer:

The difference in weight and size?

Explanation:

It explains itself :P

5 0

2 years ago

How to find the voltage(B Aab) in series parallel circuit?

Sindrei [870]

Answer:

Vab ≈ 3.426 V

Explanation:

First of all, it is convenient to find the equivalent parallel resistance of R5 and R6. That will be ...

R56 = (R5)(R6)/(R5 +R6) = (1000)(1500)/(1000 +1500) = 600

Then we can call V1 the voltage at the top of R2. The voltage at Va is a divider from V1:

Va = V1·(R4/(R3+R4)) = V1(560/1030) ≈ 0.543689V1

The voltage at Vb is also a divider from V1:

Vb = V1·(R7+R8)/(R2 +R56 +R7 +R8) = V1(780/1710) ≈ 0.456140V1

The parallel branches containing Va and Vb have an effective resistance of ...

(1030)(1710)/(1030+1710) = 642.81

That forms a divider with R1 to give V1:

V1 = (100 V)642.81/(1000 +642.81) ≈ 39.1287 V

The difference Va-Vb is ...

Vab = (39.1287 V)(0.543689 -0.456140) ≈ 3.426 V

_____

We have done this using parallel resistance and voltage divider calculations. You can also do it using node voltage equations. Using the same definition for V1 as above, we have ...

(Vs -V1)/R1 +(Vb -V1)/(R56+R2) +(Va-V1)/R3 = 0

(V1 -Vb)/(R56 +R2) -Vb/(R7+R8) = 0

(V1 -Va)/R3 -Va/R4 = 0

The solution of interest is the value of Vab, shown in the attachment. It computes as 154200/45013 V ≈ 3.42568 V.

4 0

3 years ago