a) The initial moisture content of a food product is 77% (wet basis), and the critical moisture content is 30% (wet basis). If the constant drying rate in a fluidized bed dryer is 0.1 kg water removed/m2-s, determine the time required for the product to begin the falling-rate drying period. The product has a cube shape with 5-cm sides; the initial product density is 950 kg/m3.

Answer:

  (√6)/3 ≈ 0.8165

Explanation:

The RMS value is the square root of the mean of the square of the waveform over one period. It will be ...

  [tex]\displaystyle\sqrt{\frac{1}{T}\left(\int_{\frac{T}{4}}^{\frac{3T}{4}}{1^2}\,dt+\int_{\frac{3t}{4}}^{\frac{5t}{4}}{(\frac{-4}{T}}(t-T))^2\,dt\right)}=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\left.\frac{16}{T^2}\cdot\frac{1}{3}(t-T)^3\right|_{\frac{3t}{4}}^{\frac{5t}{4}}\right)}\\\\=\sqrt{\frac{1}{T}\left(\frac{T}{2}+\frac{T}{6}\right)}=\sqrt{\frac{2}{3}}=\boxed{\frac{\sqrt{6}}{3}\approx0.8165}[/tex]

Answer:

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Explanation:

From the information given:

Life requirement = 40 kh = 40 [tex]40 \times 10^{3} \ h[/tex]

Speed (N) = 520 rev/min

Reliability goal [tex](R_D)[/tex] = 0.9

Radial load [tex](F_D)[/tex] = 2600 lbf

To find C10 value by using the formula:

[tex]C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}[/tex]

where;

[tex]x_D = \text{bearing life in million revolution} \\ \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}[/tex]

[tex]\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}[/tex]

The Weibull parameters include:

[tex]x_o = 0.02[/tex]

[tex](\theta - x_o) = 4.439[/tex]

[tex]b= 1.483[/tex]

∴

Using the above formula:

[tex]C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}[/tex]

[tex]C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}[/tex]

[tex]C_{10} = 30962.449 \ lbf[/tex]

Recall that:

1 kN = 225 lbf

∴

[tex]C_{10} = \dfrac{30962.449}{225}[/tex]

[tex]\mathbf{C_{10} = 137.611 \ kN}[/tex]

Answer:

a. 41

b. i. 15 mA ii. 625 mA

c. 192 Ω

Explanation:

Here is the complete question

A high-voltage discharge tube is often used to study atomic spectra. The tubes require a large voltage across their terminals to operate. To get the large voltage, a step-up transformer is connected to a line voltage (120 V rms) and is designed to provide 5000 V (rms) to the discharge tube and to dissipate 75.0 W. (a) What is the ratio of the number of turns in the secondary to the number of turns in the primary? (b) What are the rms currents in the primary and secondary coils of the transformer? (c) What is the effective resistance that the 120-V source is subjected to?

Solution

(a) What is the ratio of the number of turns in the secondary to the number of turns in the primary?

For a transformer N₂/N₁ = V₂/V₁

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V and V₂ = voltage of secondary coil = 5000 V

So,  N₂/N₁ = V₂/V₁

N₂/N₁ = 5000 V/120 V = 41.6 ≅ 41 (rounded down because we cannot have a decimal number of turns)

(b) What are the rms currents in the primary and secondary coils of the transformer?

i. The rms current in the secondary

We need to find the current in the secondary from

P = IV where P = power dissipated in secondary coil = 75.0 W, I =rms current in secondary coil and V = rms voltage in secondary coil = 5000 V

P = IV

I = P/V = 75.0 W/5000 V = 15 × 10⁻³ A = 15 mA

ii. The rms current in the primary

Since N₂/N₁ = V₂/V₁ = I₁/I₂

where N₁ = number of turns of primary coil, N₂ =number of coil of secondary, V₁ = voltage of primary coil = 120 V, V₂ = voltage of secondary coil = 5000 V, I₁ = current in primary coil and I₂ = current in secondary coil = 15 mA

So, V₂/V₁ = I₁/I₂

V₂I₂/V₁ = I₁

I₁ = V₂I₂/V₁

= P/V₁

= 75.0 W/120 V

= 0.625 A

= 625 mA

(c) What is the effective resistance that the 120-V source is subjected to?

Using V = IR where V =  voltage = 120 V, I = current in primary = 0.625 A and R = resistance of primary coil

R = V/I

= 120 V/0.625 A

= 192 V/A

= 192 Ω

Here are the eight stages of the recruitment process and what Garth might expect to occur or carry out at each stage in the explanation part.

The process of identifying, attracting, and selecting qualified candidates for a job opening in an organisation is known as recruitment.

Here are the eight stages of the recruitment process, as well as what Garth might expect to happen or do at each stage:

Thus, these are the stages of recruitment.

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