Answer:
The power produced by the turbine is 23309.1856 kW
Explanation:
h₁ = 3755.39
s₁ = 7.0955
s₂ = sf + x₂sfg =
Interpolating fot the pressure at 3.25 bar gives;
570.935 +(3.25 - 3.2)/(3.3 - 3.2)*(575.500 - 570.935) = 573.2175
2156.92 +(3.25 - 3.2)/(3.3 - 3.2)*(2153.77- 2156.92) = 2155.345
h₂ = 573.2175 + 0.94*2155.345 = 2599.2418 kJ/kg
Power output of the turbine formula =
![Q - \dot{W } = \dot{m}\left [ \left (h_{2}-h_{1} \right )+\dfrac{v_{2}^{2}- v_{1}^{2}}{2} + g(z_{2}-z_{1})\right ]](https://tex.z-dn.net/?f=Q%20-%20%5Cdot%7BW%20%7D%20%3D%20%5Cdot%7Bm%7D%5Cleft%20%5B%20%5Cleft%20%28h_%7B2%7D-h_%7B1%7D%20%20%5Cright%20%29%2B%5Cdfrac%7Bv_%7B2%7D%5E%7B2%7D-%20v_%7B1%7D%5E%7B2%7D%7D%7B2%7D%20%2B%20g%28z_%7B2%7D-z_%7B1%7D%29%5Cright%20%5D)
Which gives;
![560 - \dot{W } = 8\left [ \left (2599.2418-3755.39 \right )+\dfrac{15^{2}- 60^{2}}{2} \right ]](https://tex.z-dn.net/?f=560%20-%20%5Cdot%7BW%20%7D%20%3D%208%5Cleft%20%5B%20%5Cleft%20%282599.2418-3755.39%20%20%5Cright%20%29%2B%5Cdfrac%7B15%5E%7B2%7D-%2060%5E%7B2%7D%7D%7B2%7D%20%5Cright%20%5D)
= -8*((2599.2418 - 3755.39)+(15^2 - 60^2)/2 ) = -22749.1856
= -22749.1856 - 560 = -23309.1856 kJ
= 23309.1856 kJ
Power produced by the turbine = Work done per second = 23309.1856 kW.
Answer:
Code is given below:
Explanation:
.data
str1: .space 20
str2: .space 20
msg1:.asciiz "Please enter string (max 20 characters): "
msg2: .asciiz "\n Please enter string (max 20 chars): "
msg3:.asciiz "\nSAME"
msg4:.asciiz "\nNOT SAME"
.text
.globl main
main:
li $v0,4 #loads msg1
la $a0,msg1
syscall
li $v0,8
la $a0,str1
addi $a1,$zero,20
syscall #got string to manipulate
li $v0,4 #loads msg2
la $a0,msg2
syscall
li $v0,8
la $a0,str2
addi $a1,$zero,20
syscall #got string
la $a0,str1 #pass address of str1
la $a1,str2 #pass address of str2
jal methodComp #call methodComp
beq $v0,$zero,ok #check result
li $v0,4
la $a0,msg4
syscall
j exit
ok:
li $v0,4
la $a0,msg3
syscall
exit:
li $v0,10
syscall
methodComp:
add $t0,$zero,$zero
add $t1,$zero,$a0
add $t2,$zero,$a1
loop:
lb $t3($t1) #load a byte from each string
lb $t4($t2)
beqz $t3,checkt2 #str1 end
beqz $t4,missmatch
slt $t5,$t3,$t4 #compare two bytes
bnez $t5,missmatch
addi $t1,$t1,1 #t1 points to the next byte of str1
addi $t2,$t2,1
j loop
missmatch:
addi $v0,$zero,1
j endfunction
checkt2:
bnez $t4,missmatch
add $v0,$zero,$zero
endfunction:
jr $ra
Answer:
0.0659 A
Explanation:
Given that :
( saturation current )
at 25°c = 300 k ( room temperature )
n = 2 for silicon diode
Determine the saturation current at 100 degrees = 373 k
Diode equation at room temperature = I = Io 
next we have to determine the value of V at 373 k
q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1
Given that I is constant
Io = 
= 0.0659 A
Answer:
a) Ql=33120000 kJ
b) COP = 5.6
c) COPreversible= 29.3
Explanation:
a) of the attached figure we have:
HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:
W=Qh-Ql
Ql=Qh-W
where W=2000 kWh
Qh=120000 kJ/h
b) The coefficient of performance is:
c) The coefficient of performance of a reversible heat pump is:
Th=20+273=293 K
Tl=10+273=283K
Replacing:
