A) chilled water from evaporator
Answer:
1. ![\dot Q=19600\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D19600%5C%20W)
2. ![\dot Q=120\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D120%5C%20W)
Explanation:
1.
Given:
- height of the window pane,
![h=2\ m](https://tex.z-dn.net/?f=h%3D2%5C%20m)
- width of the window pane,
![w=1\ m](https://tex.z-dn.net/?f=w%3D1%5C%20m)
- thickness of the pane,
![t=5\ mm= 0.005\ m](https://tex.z-dn.net/?f=t%3D5%5C%20mm%3D%200.005%5C%20m)
- thermal conductivity of the glass pane,
![k_g=1.4\ W.m^{-1}.K^{-1}](https://tex.z-dn.net/?f=k_g%3D1.4%5C%20W.m%5E%7B-1%7D.K%5E%7B-1%7D)
- temperature of the inner surface,
![T_i=15^{\circ}C](https://tex.z-dn.net/?f=T_i%3D15%5E%7B%5Ccirc%7DC)
- temperature of the outer surface,
![T_o=-20^{\circ}C](https://tex.z-dn.net/?f=T_o%3D-20%5E%7B%5Ccirc%7DC)
<u>According to the Fourier's law the rate of heat transfer is given as:</u>
![\dot Q=k_g.A.\frac{dT}{dx}](https://tex.z-dn.net/?f=%5Cdot%20Q%3Dk_g.A.%5Cfrac%7BdT%7D%7Bdx%7D)
here:
A = area through which the heat transfer occurs = ![2\times 1=2\ m^2](https://tex.z-dn.net/?f=2%5Ctimes%201%3D2%5C%20m%5E2)
dT = temperature difference across the thickness of the surface = ![35^{\circ}C](https://tex.z-dn.net/?f=35%5E%7B%5Ccirc%7DC)
dx = t = thickness normal to the surface = ![0.005\ m](https://tex.z-dn.net/?f=0.005%5C%20m)
![\dot Q=1.4\times 2\times \frac{35}{0.005}](https://tex.z-dn.net/?f=%5Cdot%20Q%3D1.4%5Ctimes%202%5Ctimes%20%5Cfrac%7B35%7D%7B0.005%7D)
![\dot Q=19600\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D19600%5C%20W)
2.
- air spacing between two glass panes,
![dx=0.01\ m](https://tex.z-dn.net/?f=dx%3D0.01%5C%20m)
- area of each glass pane,
![A=2\times 1=2\ m^2](https://tex.z-dn.net/?f=A%3D2%5Ctimes%201%3D2%5C%20m%5E2)
- thermal conductivity of air,
![k_a=0.024\ W.m^{-1}.K^{-1}](https://tex.z-dn.net/?f=k_a%3D0.024%5C%20W.m%5E%7B-1%7D.K%5E%7B-1%7D)
- temperature difference between the surfaces,
![dT=25^{\circ}C](https://tex.z-dn.net/?f=dT%3D25%5E%7B%5Ccirc%7DC)
<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>
![\dot Q=k_a.A.\frac{dT}{dx}](https://tex.z-dn.net/?f=%5Cdot%20Q%3Dk_a.A.%5Cfrac%7BdT%7D%7Bdx%7D)
![\dot Q=0.024\times 2\times \frac{25}{0.01}](https://tex.z-dn.net/?f=%5Cdot%20Q%3D0.024%5Ctimes%202%5Ctimes%20%5Cfrac%7B25%7D%7B0.01%7D)
![\dot Q=120\ W](https://tex.z-dn.net/?f=%5Cdot%20Q%3D120%5C%20W)
Answer:
import java.util.Scanner;
public class InputExample {
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int birthMonth;
int birthYear;
birthMonth = scnr.nextInt();
birthYear = scnr.nextInt();
System.out.println(birthMonth+"/"+birthYear);
}
}
Answer:
![10.8\ \text{lb/ft^2}](https://tex.z-dn.net/?f=10.8%5C%20%5Ctext%7Blb%2Fft%5E2%7D)
![101.96\ \text{lb/ft}^2](https://tex.z-dn.net/?f=101.96%5C%20%5Ctext%7Blb%2Fft%7D%5E2)
Explanation:
= Velocity of car = 65 mph = ![65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}](https://tex.z-dn.net/?f=65%5Ctimes%20%5Cdfrac%7B5280%7D%7B3600%7D%3D95.33%5C%20%5Ctext%7Bft%2Fs%7D)
= Density of air = ![0.00237\ \text{slug/ft}^3](https://tex.z-dn.net/?f=0.00237%5C%20%5Ctext%7Bslug%2Fft%7D%5E3)
![v_2=0](https://tex.z-dn.net/?f=v_2%3D0)
![P_1=0](https://tex.z-dn.net/?f=P_1%3D0)
![h_1=h_2](https://tex.z-dn.net/?f=h_1%3Dh_2)
From Bernoulli's law we have
![P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}](https://tex.z-dn.net/?f=P_1%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%2Bh_1%3DP_2%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2%2Bh_2%5C%5C%5CRightarrow%20P_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%5C%5C%5CRightarrow%20P_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%200.00237%5Ctimes%2095.33%5E2%5C%5C%5CRightarrow%20P_2%3D10.8%5C%20%5Ctext%7Blb%2Fft%5E2%7D)
The maximum pressure on the girl's hand is ![10.8\ \text{lb/ft^2}](https://tex.z-dn.net/?f=10.8%5C%20%5Ctext%7Blb%2Fft%5E2%7D)
Now
= 200 mph = ![200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}](https://tex.z-dn.net/?f=200%5Ctimes%20%5Cdfrac%7B5280%7D%7B3600%7D%3D293.33%5C%20%5Ctext%7Bft%2Fs%7D)
![P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2](https://tex.z-dn.net/?f=P_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%5C%5C%5CRightarrow%20P_2%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%200.00237%5Ctimes%20293.33%5E2%5C%5C%5CRightarrow%20P_2%3D101.96%5C%20%5Ctext%7Blb%2Fft%7D%5E2)
The maximum pressure on the girl's hand is ![101.96\ \text{lb/ft}^2](https://tex.z-dn.net/?f=101.96%5C%20%5Ctext%7Blb%2Fft%7D%5E2)