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Mariulka [41]
3 years ago
6

Create a function (prob3_5) that will take inputs of vectors x and y in feet, scalar N, scalars L and W in feet and scalars T1 a

nd T2 in degrees Fahrenheit. It will output a matrix T which is the temperature of each x and y locations. T will have the number of columns equal to the number of elements in x and rows equal to the number of elements in y. Though this can be done without loops (perhaps more efficiently), your program must use a nested loop.
Engineering
1 answer:
Shalnov [3]3 years ago
3 0

Answer:

clear, clc

prob3_5([1,2,3],[6,5,7],12,11,22,55,76)

function T=prob3_5(x,y,N,L,W,T1,T2)

w=zeros(1,length(x));

for n=1:2:N

for i=1:length(x)

w(i)=w(i)+(2/pi)*(2/n)*sin(n*pi*x(i)/L).*sinh(n*pi*y(i)/L)/sinh(n*pi*W/L);

end

end

T=(T2-T1)*w+T1;

end

Explanation:

Please input the commands into MATLAB

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A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp
Furkat [3]

Answer:

1700kJ/h.K

944.4kJ/h.R

944.4kJ/h.°F

Explanation:

Conversions for different temperature units are below:

1K = 1°C + 273K

1R = T(K) * 1.8

= (1°C + 273) * 1.8

1°F = (1°C * 1.8) + 32

Q/delta T = 1700kJ/h.°C

T (K) = 1700kJ/h.°C

= 1700kJ/K

T (R) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8R

= 944.4kJ/h.R

T (°F) = 1700kJ/h.°C

= 1700kJ/h.°C * 1°C/1.8°F

= 944.4kJ/h.°F

Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C

= 2°C

(273+5) - (273+3)

= 2 K

5 0
4 years ago
When is a handrail required for stairs?
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Answer:

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3 years ago
A vertical piston-cylinder assembly contains water. The piston has a mass of 2 kg and a diameter of 7 cm . Assume atmospheric pr
Step2247 [10]

Answer:

for a) F= 744.97 N

for a) F= 167.85 N

for a) F= 764.57 N

Explanation:

the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.

Then from steam tables

T= 120°C → P required=Pr= 198.67 kPa

then the pressure developed by the piston is

P = (m*g + F)/A

where m= mass of the piston ,g= gravity F= force required and A= area of the piston

then

Pr = P = (m*g + F)/A

F = Pr*A-m*g

since A= π/4*D²

F =π/4* Pr*D²-m*g

replacing values

F =π/4* Pr*D²-m*g = π/4*198.67  *10³Pa*(0.07m)² -2kg* 9.8m/s²

F= 744.97 N

b) for T₂=80°C → Pr₂=47.41 kPa

F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²

F₂= 167.85 N

c) for m=0 (mass of the piston neglected) ,the force required is

F₃ =π/4*Pr*D² =  π/4*198.67 *10³Pa*(0.07m)²= 764.57 N

F₃ =764.57 N

4 0
3 years ago
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