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faltersainse [42]

3 years ago

12

2. One of the many methods used for drying air is to cool the air below the dew point so that condensation or freezing of the mo

isture takes place. To what temperature must atmospheric air be cooled in order to have humidity ratio of 0.005 lb/lb?

1 answer:

REY [17]3 years ago

6 0

Answer:

0.5°c

Explanation:

Humidity ratio by mass can be expressed as

the ratio between the actual mass of water vapor present in moist air - to the mass of the dry air

Humidity ratio is normally expressed in kilograms (or pounds) of water vapor per kilogram (or pound) of dry air.

Humidity ratio expressed by mass:

x = mw / ma (1)

where

x = humidity ratio (kgwater/kgdry_air, lbwater/lbdry_air)

mw = mass of water vapor (kg, lb)

ma = mass of dry air (kg, lb)

It can be as:

x = 0.005 (100) / [(100 - 100)]

x = 0.005 x 100 / (100 - 100)

x = 0.005 x 100 / 0

x = 0.5°c

So the temperature to which atmospheric air must be cooled in order to have humidity ratio of 0.005 lb/lb is 0.5°c

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Answer:

it is wave inversion

Explanation:

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Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each repl

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Answer:

The answers to the question are

(1) Process 1 to 2

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

(2) Process 2 to 3

W = 0

Q = 1135.376 kJ/kg

(3) Process 3 to 4

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

(4) Process 4 to 3

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency = 49.9 %

(c) The mean effective pressure is 9.44 bar

Explanation:

(a) Volume compression ratio $\frac{v_1}{v_2} = 10$

Initial pressure p₁ = 1 bar

Initial temperature, T₁ = 310 K

cp = 1.005 kJ/kg⋅K

Temperature T₃ = 2200 K from the isentropic chart of the Otto cycle

For a polytropic process we have

$\frac{p_1}{p_2} = (\frac{v_2}{v_1} )^n$ Therefore p₂ = p₁ ÷ $(\frac{v_2}{v_1} )^n$ = (1 bar) ÷ $(\frac{1}{10} )^{1.3}$ = 19.953 bar

Similarly for a polytropic process we have

$\frac{T_1}{T_2} = (\frac{v_2}{v_1} )^{n-1}$ or T₂ = T₁ ÷ $(\frac{v_2}{v_1} )^{n-1}$ = $\frac{310}{0.1^{0.3}}$ = 618.531 K

The molar mass of air is 28.9628 g/mol.

Therefore R = $\frac{8.3145}{28.9628}$ = 0.287 kJ/kg⋅K

cp = 1.005 kJ/kg⋅K Therefore cv = cp - R = 1.005- 0.287 = 0.718 kJ/kg⋅K

1). For process 1 to 2 which is polytropic process we have

W = $\frac{R(T_2-T_1)}{n-1}$ = $\frac{0.287(618.531-310)}{1.3 - 1}$ = 295.16 kJ/kg

Q = $(\frac{n-\gamma}{\gamma - 1} )W$ = $(\frac{1.3-1.4}{1.4-1} ) 295.16 kJ/kg$ = -73.79 kJ/kg

W = 295.16 kJ/kg

Q = -73.79 kJ/kg

2). For process 2 to 3 which is reversible constant volume heating we have

W = 0 and Q = cv×(T₃ - T₂) = 0.718× (2200-618.531) = 1135.376 kJ/kg

W = 0

Q = 1135.376 kJ/kg

3). For process 3 to 4 which is polytropic process we have

W = $\frac{R(T_4-T_3)}{n-1}$ = Where T₄ is given by $\frac{T_4}{T_3} = (\frac{v_3}{v_4} )^{n-1}$ or T₄ = T₃ × $0.1^{0.3}$

= 2200 × $0.1^{0.3}$ T₄ = 1102.611 K

W = $\frac{0.287(1102.611-2200)}{1.3 - 1}$ = -1049.835 kJ/kg

and Q = 262.459 kJ/kg

W = -1049.835 kJ/kg

Q = 262.459 kJ/kg

4). For process 4 to 1 which is reversible constant volume cooling we have

W = 0 and Q = cv×(T₁ - T₄) = 0.718×(310 - 1102.611) = -569.09 kJ/kg

W=0

Q = -569.09 kJ/kg

(b) The thermal efficiency is given by

$\eta = 1-\frac{T_4-T_1}{T_3-T_2}$ = $1-\frac{1102.611-310}{2200-618.531}$ = 0.499 or 49.9 % Efficient

(c) The mean effective pressure is given by

$p_{m} = \frac{p_1r[(r^{n-1}-1)(r_p-1)]}{ (n-1)(r-1)}$ where r = compression ratio and $r_p$ = $\frac{p_3}{p_2}$

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$r_p$ = $\frac{p_3}{p_2}$ = $\frac{70.97}{19.953} = 3.56$

Therefore $p_m =\frac{1*10*[(10^{0.3}-1)(3.56-1)]}{0.3*9}$ = 9.44 bar

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Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

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% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

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if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

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error('The number of output variables does not equal the number of fields');

end

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3 years ago

A hypothetical metal alloy has a grain diameter of 2.4 × 10−2 mm. After a heat treatment at 575°C for 500 min, the grain diamete

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Answer:

The time required is 10.078 hours or 605 min

Explanation:

The formula to apply here is ;

K=(d²-d²₀ )/t

where t is time in hours

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d₀ is the grain diameter before heating in mm

Given

d=5.5 × 10^-2 mm

d₀=2.4 × 10^-2 mm

t₁= 500 min = 500/60 =25/3 hrs

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First find K

K=(d²-d²₀ )/t₁

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Re-arrange equation for K ,to get the equation for d as;

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4 0

3 years ago

A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?

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Answer:

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Explanation:

Given;

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The magnetic moment is given by;

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A is area of the coil = πr² = π(0.098)² = 0.03018 m²

The current through the coil is given by;

I = P / A

I = (6.2 x 10⁻² ) / (0.03018)

I = 2.05 A

Therefore, the current through the coil is 2.05 A

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3 years ago