Answer:
flow ( m ) = 4.852 kg/s
Explanation:
Given:
- Inlet of Turbine
P_1 = 10 MPa
T_1 = 500 C
- Outlet of Turbine
P_2 = 10 KPa
x = 0.9
- Power output of Turbine W_out = 5 MW
Find:
Determine the mass ow rate required
Solution:
- Use steam Table A.4 to determine specific enthalpy for inlet conditions:
P_1 = 10 MPa
T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg
- Use steam Table A.6 to determine specific enthalpy for outlet conditions:
P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg
x = 0.9 -------------> h_fg = 2392.1 KJ/kg
h_2 = h_f + x*h_fg
h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg
- The work produced by the turbine W_out is given by first Law of thermodynamics:
W_out = flow(m) * ( h_1 - h_2 )
flow ( m ) = W_out / ( h_1 - h_2 )
- Plug in values:
flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )
flow ( m ) = 4.852 kg/s
I believe i got this correct when i had an assignment with this question. It should be 7 1/2. B!
Answer:
The average loss for this device is $10.90
Explanation:
Given data
Specification = 24 +/- 0.4 Amps
average quality cost = $32.00
average value y = 23.9
standard deviation s = 0.211 Amps
32 = k(24.4 - 24)²
32 = k(0.4)² = k(0.16)
k = 32/0.16 = 200
To evaluate average loss, use
L = k{s² + (y - T)²}
T = 24A
L = 200{0.211² + (23.9 - 24)²}
L = $10.90
The average loss for this device is $10.90
Answer:
The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.
Explanation:
The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.
