Answer:
Viscosity is notated using the common classification “XW-XX”. The number preceding the “W” (winter) rates the oil's flow (viscosity) at zero degrees Fahrenheit (-17.8 degrees Celsius). The lower the number, the less the oil thickens in cold weather.
Answer: heat loss through wall is 16.58034kW
Temperature of inside wall surface is 47°c
Temperature of outside wall surface is -2.7°c
Explanation:detailed calculation and explanation is shown in the image below.
Below is the program to separate odd and even numbers
<u>Explanation</u>:
<u>L1:</u>
mov ah,00
mov al,[BX]
mov dl,al
div dh
cmp ah,00
je EVEN1
mov [DI],dl
add OddAdd,dl
INC DI
INC BX
Loop L1
jmp CAL
<u>EVEN1:</u>
mov [SI],dl
add Even Add,dl
INC SI
INC BX
Loop L1
<u>CAL: </u>
mov ax,0000
mov bx,0000
mov al,OddAdd
mov bl,EvenAdd
MOV ax,4C00h
int 21h
end
The above program separates odd and even numbers from the array using 8086 microprocessor. It has odd numbers in 2000h and even numbers in 3000h.
Explanation:
Superheater has two types of parts which are:
- The primary super-heater
- The secondary super-heater
Primary super-heater is first heater which is passed by the steam after steam comes out of steam drum.
After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.
Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.
Answer:
The temperature of the first exit (feed to water heater) is at 330.15ºC. The second exit (exit of the turbine) is at 141ºC. The turbine Power output (if efficiency is %100) is 3165.46 KW
Explanation:
If we are talking of a steam turbine, the work done by the steam is done in an adiabatic process. To determine the temperature of the 2 exits, we have to find at which temperature of the steam with 1000KPa and 200KPa we have the same entropy of the steam entrance.
In this case for steam at 3000 kPa, 500°C, s= 7.2345Kj/kg K. i=3456.18 KJ/Kg
For steam at 1000 kPa and s= 7.2345Kj/kg K → T= 330.15ºC i=3116.48KJ/Kg
For steam at 200 kPa and s= 7.2345Kj/kg K → T= 141ºC i=2749.74KJ/Kg
For the power output, we have to multiply the steam flow with the enthalpic jump.
The addition of the 2 jumps is the total power output.
