Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer:
The heat transfer q = 18.32W
Explanation:
In this question, we are asked to calculate the heat entering the tube and also evaluate properties at T =400K
Please check attachment for complete solution and step by step explanation
Answer:
.
Explanation:
Given that
L= 50 m
Pressure drop = 130 KPa
copper tube is 3/4 standard type K drawn tube.
From standard chart ,the dimension of 3/4 standard type K copper tube given as
Outside diameter=22.22 mm
Inside diameter=18.92 mm
Dynamic viscosity for kerosene
![\mu =0.00164\ Pa.s](https://tex.z-dn.net/?f=%5Cmu%20%3D0.00164%5C%20Pa.s)
We know that
![\Delta P=\dfrac{128\mu QL}{\pi d_i^4}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cdfrac%7B128%5Cmu%20QL%7D%7B%5Cpi%20d_i%5E4%7D)
Where Q is volume flow rate
L is length of tube
is inner diameter of tube
ΔP is pressure drop
μ is dynamic viscosity
Now by putting the values
![\Delta P=\dfrac{128\mu QL}{\pi d_i^4}](https://tex.z-dn.net/?f=%5CDelta%20P%3D%5Cdfrac%7B128%5Cmu%20QL%7D%7B%5Cpi%20d_i%5E4%7D)
![130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}](https://tex.z-dn.net/?f=130%5Ctimes%201000%3D%5Cdfrac%7B128%5Ctimes%200.00164%5Ctimes%2050Q%7D%7B%5Cpi%20%5Ctimes%200.01892%5E4%7D)
![Q=4.98\times 10^{-3}\ m^3/s](https://tex.z-dn.net/?f=Q%3D4.98%5Ctimes%2010%5E%7B-3%7D%5C%20m%5E3%2Fs)
So flow rate is
.
The answer is D I’m 90% sure