Answer:
a. 47.48%
b. 35.58%
c. 2957.715 KW
Explanation:
T₁ = 300 K
= 579.21 K
T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K
T₃ = T₂ + 
(T₅ - T₂)
T₄ = 1400 K
Given that the pressure ratios across each turbine stage are equal, we have;
= 1400×
= 1007.6 K
T₅ = T₄ + ( - T₄)/
= 1400 + (1007.6- 1400)/0.8 = 909.5 K
T₃ = T₂ + (T₅ - T₂)
T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K
T₆ = 1400 K
= 1400× = 1007.6 K
T₇ = T₆ + ( - T₆)/ = 1400 + (1007.6 - 1400)/0.8 = 909.5 K
a. 
= cp(T₆ -T₇) = 1.005 * (1400 - 909.5) = 492.9525 KJ/kg
Heat supplied is given by the relation
cp(T₄ - T₃) + cp(T₆ - T₅) = 1.005*((1400 - 857.402) + (1400 - 909.5)) = 1038.26349 kJ/kg
Thermal efficiency of the cycle = (Net work output)/(Heat supplied)
Thermal efficiency of the cycle = (492.9525 )/(1038.26349 ) =0.4748 = 47.48%
b. 
bwr = (T₂ -T₁)/[(T₄ - T₅) +(T₆ -T₇)] = (649.01 - 300)/((1400 - 909.5) + (1400 - 909.5)) = 35.58%
c. Power = 6 kg *492.9525 KJ/kg = 2957.715 KW
Answer:
Answered below.
Explanation:
A) Both spheroidite & tempered martensite possess sphere - like cementite particles within their microstructure known as a ferrite matrix. However, the difference is that these particles are much larger for spheroidite than tempered.
B) Tempered martensite is much harder and stronger than spheroidite primarily because there is much more ferrite - cementite phase boundary area for its sphere - like cementite particles.
This is because the greater the boundary area, the more the hardness.
The answer is windmill it was used for electricity
Answer:
learn from their mistakes
Explanation:
and so u could do better next time
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
