Answer:
Babies come from heaven didn't you know?
Answer:
(i) 169.68 volt
(ii) 16.90 volt
(iii) 16.90 volt
(iv) 108.07 volt
(v) 2.161 A
Explanation:
Turn ratio is given as 10:1
We have given that input voltage 
(i) We know that peak voltage is give by 
(ii) We know that for transformer 
So 

So peak voltage in secondary will be 16.90 volt
(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary
So peak voltage of the rectifier will be 16.90 volt
(iv) Dc voltage of the rectifier is given by 
(v) Now dc current is given by 
Answer:
The split is given by including spaces in both tabs
Explanation:
The bracket notation can be used to indicate the split. Here is an example:
String [ ] parts = s. split ( "[/]")
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
Answer:
the rate of heat loss by convection across the air space = 82.53 W
Explanation:
The film temperature

to kelvin = (5 + 273)K = 278 K
From the " thermophysical properties of gases at atmospheric pressure" table; At
= 278 K ; by interpolation; we have the following
→ v 13.93 (10⁻⁶) m²/s
→ k = 0.0245 W/m.K
→ ∝ = 19.6(10⁻⁶)m²/s
→ Pr = 0.713

The Rayleigh number for vertical cavity

= 
= 

For the rectangular cavity enclosure , the Nusselt number empirical correlation:





h = 1.99 W/m².K
Finally; the rate of heat loss by convection across the air space;
q = hA(T₁ - T₂)
q = 1.99(1.4*0.96)(20-(-10))
q = 82.53 W