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olya-2409 [2.1K]
4 years ago
10

3. Given the Taylor series expansion for sin x = x − x3/ 3! + x5/5! − x7/7! . Write a program that takes in an arbitrary integer

for an angle in degrees, converted the degrees to radians and performs the calculation above. Remember you will need a loop to calculate the ! term and float variables to hold the term(s) in the expansion and the converted degrees to radian.
Engineering
1 answer:
katen-ka-za [31]4 years ago
8 0

Answer:

clc

clear

x = input('type value of angle in degrees:\n');

x = x*pi/180; %convverting fron degree to radian

sin_x = x; %as first term of taylor series is x

E = 1; %just giving a value of error greater than desired error

n = 0;

while E > 0.000001

previous = sin_x;

n = n+1;

sin_x = sin_x + ((-1)^n)*(x^(2*n+1))/factorial(2*n+1);

E = abs(sin_x - previous); %calculating error

end

a = sprintf('sin(x) = %1.6f',sin_x);

disp(a)

Explanation:

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On a summer day in New Orleans, Louisiana, the pressure is 1 atm: the temperature is 32°C; and the relative humidity is 95 perce
FinnZ [79.3K]

Answer:

Δw =7.95 kg/1000m^3

q = 62362.3 kg/1000m^3

Explanation:

To solve this problem we first need to use the psycrometric chart to determine the enthalpy h1, specific volume vi and absolute humidity col by using the given temperature T1 = 32°C and the relative humidity Ф1 = 95%.  

h_1 =  106.5 kJ/kg

v_1 = 0.91 m^3/kg

w_1 = 0.02905

We will also need the enthalpy h2 and the absolute humidity w_2 at the exit point. We will again use the pyscrometric chart and the given temperature T_2 = 24°C. From the problem we also know that the exit relative humidity is = 60%.  

h2 = 52.6 kJ/kg

w_2 = 0.01119

We need to express the final results in units per 1000 m^3. To do that we will need the mass m of this volume of air V and to calculate that we will use the given pressure p = 1 atm = 101.3 kPa.  

m = R_a*T_1/V.p

m = 1000*101.3/0.287*305K

m = 1157 kg

Because it is a closed system, the amount of water removed Δw can be calculated as:  

Δw =w_1 - w_2

Δw =0.02905- 0.01119

Δw =0.00687 kg/kg* 1157kg/1000m^3

Δw =7.95 kg/1000m^3

From the energy balance equation we can calculate the specific heat q removed from the air.  

q = h_1 - h_2

q = 106.5 kJ/kg - 52.6 kJ/kg

q = 53.9 kJ/kg * 1157kg/1000m^3

q = 62362.3 kg/1000m^3

3 0
3 years ago
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