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Answer:
Q = 63,827.5 W
Explanation:
Given:-
- The dimensions of plate A = ( 10 mm x 1 m )
- The fluid comes at T_sat , 1 atm.
- The surface temperature, T_s = 75°C
Find:-
Determine the total condensation rate of water vapor onto the front surface of a vertical plate
Solution:-
- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.
h = 255,310 W /m^2.K
- The rate of condensation (Q) is given by Newton's cooling law:
Q = h*As*( T_sat - Ts )
Q = (255,310)*( 0.01*1)*( 100 - 75 )
Q = 63,827.5 W
Answer:
The current drawn from the outlet is 0.2 A
The number of turns on the input side is 350 turns
Explanation:
Given;
number of turns of the secondary coil, Ns = 35 turns
the output current,
= 2 A
power supplied,
= 24 W
the standard wall outlet in most homes = 120 V = input voltage
For an ideal transformer; output power = input power
the current drawn from the outlet is calculated;
![I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A](https://tex.z-dn.net/?f=I_pV_p%20%3D%20P_s%5C%5C%5C%5CI_p%20%3D%20%5Cfrac%7BP_s%7D%7BV_p%7D%20%3D%20%5Cfrac%7B24%7D%7B120%7D%20%3D%200.2%20%5C%20A)
The number of turns on the input side is calculated as;
![\frac{N_p}{N_s} = \frac{I_s}{I_p} \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns](https://tex.z-dn.net/?f=%5Cfrac%7BN_p%7D%7BN_s%7D%20%3D%20%5Cfrac%7BI_s%7D%7BI_p%7D%20%20%5C%5C%5C%5CN_p%20%3D%20%5Cfrac%7BN_sI_s%7D%7BI_p%7D%20%5C%5C%5C%5CN_p%20%3D%20%5Cfrac%7B35%20%5Ctimes%202%7D%7B0.2%7D%20%5C%5C%5C%5CN_p%20%3D%20350%20%5C%20turns)