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Triss [41]
4 years ago
11

Question 1/5

Engineering
1 answer:
erastova [34]4 years ago
6 0
It’s between D and C
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Air enters the compressor of a cold air-standard Brayton cycle with regeneration and reheat at 100 kPa, 300 K, with a mass flow
yanalaym [24]

Answer:

a. 47.48%

b. 35.58%

c. 2957.715 KW

Explanation:

T_2 =T_1 + \dfrac{T_{2s} - T_1}{\eta _c}

T₁ = 300 K

\dfrac{T_{2s}}{T_1} = \left( \dfrac{P_{2}}{P_1} \right)^{\dfrac{k-1}{k} }

T_{2s} = 300 \times (10) ^{\dfrac{0.4}{1.4} }

T_{2s} = 579.21 K

T₂ = 300+ (579.21 - 300)/0.8 = 649.01 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₄ = 1400 K

Given that the pressure ratios across each turbine stage are equal, we have;

\dfrac{T_{5s}}{T_4} = \left( \dfrac{P_{5}}{P_4} \right)^{\dfrac{k-1}{k} }

T_{5s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }  = 1007.6 K

T₅ = T₄ + (T_{5s} - T₄)/\eta _t = 1400 + (1007.6- 1400)/0.8 = 909.5 K

T₃ = T₂ + \epsilon _{regen}(T₅ - T₂)

T₃ = 649.01 + 0.8*(909.5 - 649.01 ) = 857.402 K

T₆ = 1400 K

\dfrac{T_{7s}}{T_6} = \left( \dfrac{P_{7}}{P_6} \right)^{\dfrac{k-1}{k} }

T_{7s} = 1400×\left( 1/\sqrt{10}  \right)^{\dfrac{0.4}{1.4} }   = 1007.6 K

T₇ = T₆ + (T_{7s} - T₆)/\eta _t = 1400 + (1007.6 - 1400)/0.8 = 909.5 K

a. W_{net \ out} = 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 = \dfrac{W_{c,in}}{W_{t,out}}

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

3 0
3 years ago
The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c
marissa [1.9K]

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

3 0
3 years ago
Select the statement that is false.
ra1l [238]

Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

8 0
3 years ago
Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus
Alina [70]

Answer:

Explanation:

find attached the solution to the question

4 0
3 years ago
Which is not an affect of alcohol
Gre4nikov [31]
Pooping problems is not an affect
6 0
3 years ago
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