____ is used in windshield washer fluid to prevent freezing and
1 answer:
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Answer:
See explanation
Explanation:
//Include the
//required header files.
#include <stdio.h>
//Define the
//main() function.
int main(void) {
//Declare the
//required variables.
char input_char;
int input_int;
double input_double;
char input_string[100];
//Prompt the user
//to enter an integer.
printf("Enter integer: ");
//Read and store
//the integer.
scanf("%d", &input_int);
//Prompt the user
//to enter a double value.
printf("Enter double: ");
//Read and store
//the double value.
scanf("%lf", &input_double);
//Prompt the user
//to enter a character.
printf("Enter character: ");
//Read and store
//the character.
scanf(" %c", &input_char);
//Prompt user to
//enter the string
printf("Enter string: ");
//Read and
//store the string.
scanf("%s", input_string);
//(1)
//Display the values.
printf("%d %lf %c %s\n",
input_int, input_double,
input_char, input_string);
//(2)
//Display the values
//in reverse order.
printf("%s %c %lf %d\n",
input_string, input_char,
input_double, input_int);
//(3)
//Cast the double to
//an integer and display it.
printf("%lf cast to an integer is %d",
input_double, (int)(input_double));
//Return from the
//main() function.
return 0;
}
Question
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.
If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?
Answer:
a. 4.52W/m
b. 13mm
Explanation:
Given
Diameter of electrical wire = 2mm
Wire Thickness = 2-mm
Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)
Thermal contact resistance = 3E-4m².K/W
Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,
Temperature of the ambient air = 20°C.
Maximum Allowable Sheet Temperature = 50°C.
From the thermal circuit (See attachment), we my write
E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)
= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))
Where r in,i = D/2
= 2mm/2
= 1 mm
= 0.001m
r in,o = r in,i + t = 0.003m
T in, i = Tmax = 50°C
Hence
q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]
= 30/[(Ln3/0.26π) + 1/0.06π)]
= 30/[(1.34) + 5.30)]
= 30/6.64
= 4.52W/m
The critical radius is unaffected by the constant resistance.
Hence
Critical Radius = k/h
= 0.13/10
= 0.013m
= 13mm
Answer:
Explanation:
Given that:
diameter = 100 mm
initial temperature = 500 ° C
Conventional coefficient = 500 W/m^2 K
length = 1 m
We obtain the following data from the tables A-1;
For the stainless steel of the rod
Here, we can't apply the lumped capacitance method, since Bi > 0.1
However, on a single rod, the energy extracted is:
Hence, for centerline temperature at 50 °C;
The surface temperature is: