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:
There is not going to be pressure build up in the system,that is isobaric process.
Explanation:
Assumptions to be made
1. No mass is gained or lost during the heating process.
2. There are no friction losses,so work is transmitted efficiently.
3. It was started the water in the drum and its surrounding have same temperature.
4. This system is closed,so there is no mass transfer across its boundaries.
Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Answer:
Hello your question is incomplete attached below is the complete question
Answer : Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Explanation:
Factor of safety for point A :
i) using MSS
(Fos)MSS = 3.22
ii) using DE
(Fos)DE = 3.27
Factor of safety for point B
i) using MSS
(Fos)MSS = 3.04
ii) using DE
(Fos)DE = 3.604
Attached below is the detailed solution
