The amount of work done by steady flow devices varies with the particular gas volume. The kinetic energy of gas particles decreases during cooling.
When the gas is subjected to intermediate cooling during compression, the gas specific volume is reduced, which lowers the compressor's power consumption. Compression is less adiabatic and more isothermal because the compressed gas must be cooled between stages since compression produces heat. The system's thermodynamic cycle's cold sink temperature is lowered by cooling the compressor coils. By increasing the temperature difference between the heat source and the cold sink, this improves efficiency.
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brainly.com/question/1368306
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Answer:
![\vec F_{A} = -67500\,N\cdot (i + j)](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20-67500%5C%2CN%5Ccdot%20%28i%20%2B%20j%29)
Explanation:
The position of each point are the following:
![A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)](https://tex.z-dn.net/?f=A%20%3D%20%280%5C%2Cm%2C0%5C%2Cm%2C0%5C%2Cm%29%2C%20B%20%3D%20%280.02%5C%2Cm%2C0%5C%2Cm%2C0%5C%2Cm%29%2C%20C%20%3D%20%280%5C%2Cm%2C0.02%5C%2Cm%2C0%5C%2Cm%29)
Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:
![\vec F_{A} = \vec F_{AB} + \vec F_{AC}](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20%5Cvec%20F_%7BAB%7D%20%2B%20%5Cvec%20F_%7BAC%7D)
![\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20-%5Cfrac%7Bk%5Ccdot%20q%5E%7B2%7D%7D%7Br_%7BAB%7D%5E%7B2%7D%7D%5Ccdot%20i%20-%20%5Cfrac%7Bk%5Ccdot%20q%5E%7B2%7D%7D%7Br_%7BAC%7D%5E%7B2%7D%7D%5Ccdot%20j)
![\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20-%20%5Cfrac%7Bk%5Ccdot%20q%5E%7B2%7D%7D%7Br%5E%7B2%7D%7D%20%5Ccdot%20%28i%20%2B%20j%29)
![\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20-%5Cfrac%7B%289%20%5Ctimes%2010%5E%7B9%7D%5C%2C%5Cfrac%7BN%5Ccdot%20m%5E%7B2%7D%7D%7BC%5E%7B2%7D%7D%20%29%5Ccdot%20%283%5Ctimes%2010%5E%7B-9%7D%5C%2CC%29%7D%7B%280.02%5C%2Cm%29%5E%7B2%7D%7D%5Ccdot%20%28i%20%2B%20j%29)
![\vec F_{A} = -67500\,N\cdot (i + j)](https://tex.z-dn.net/?f=%5Cvec%20F_%7BA%7D%20%3D%20-67500%5C%2CN%5Ccdot%20%28i%20%2B%20j%29)
Answer:
![L=107.6m](https://tex.z-dn.net/?f=L%3D107.6m)
Explanation:
Cold water in: ![m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C](https://tex.z-dn.net/?f=m_%7Bc%7D%3D1.2kg%2Fs%2C%20C_%7Bc%7D%3D4.18kJ%2Fkg%5C%C2%B0C%2C%20T_%7Bc%2Cin%7D%3D20%5C%C2%B0C%2C%20T_%7Bc%2Cout%7D%3D80%5C%C2%B0C)
Hot water in: ![m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C](https://tex.z-dn.net/?f=m_%7Bh%7D%3D2kg%2Fs%2C%20C_%7Bh%7D%3D4.18kJ%2Fkg%5C%C2%B0C%2C%20T_%7Bh%2Cin%7D%3D160%5C%C2%B0C%2C%20T_%7Bh%2Cout%7D%3D%3F%5C%C2%B0C)
![D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m](https://tex.z-dn.net/?f=D%3D1.5cm%3D0.015m%2C%20U%3D649W%2Fm%5E%7B2%7DK%2C%20LMTD%3D%3F%5C%C2%B0C%2C%20A_%7Bs%7D%3D%3Fm%5E%7B2%7D%2CL%3D%3Fm)
Step 1: Determine the rate of heat transfer in the heat exchanger
![Q=m_{c}C_{c}(T_{c,out}-T_{c,in})](https://tex.z-dn.net/?f=Q%3Dm_%7Bc%7DC_%7Bc%7D%28T_%7Bc%2Cout%7D-T_%7Bc%2Cin%7D%29)
![Q=1.2*4.18*(80-20)](https://tex.z-dn.net/?f=Q%3D1.2%2A4.18%2A%2880-20%29)
![Q=1.2*4.18*(80-20)](https://tex.z-dn.net/?f=Q%3D1.2%2A4.18%2A%2880-20%29)
![Q=300.96kW](https://tex.z-dn.net/?f=Q%3D300.96kW)
Step 2: Determine outlet temperature of hot water
![Q=m_{h}C_{h}(T_{h,in}-T_{h,out})](https://tex.z-dn.net/?f=Q%3Dm_%7Bh%7DC_%7Bh%7D%28T_%7Bh%2Cin%7D-T_%7Bh%2Cout%7D%29)
![300.96=2*4.18*(160-T_{h,out})](https://tex.z-dn.net/?f=300.96%3D2%2A4.18%2A%28160-T_%7Bh%2Cout%7D%29)
![T_{h,out}=124\°C](https://tex.z-dn.net/?f=T_%7Bh%2Cout%7D%3D124%5C%C2%B0C)
Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)
![dT_{1}=T_{h,in}-T_{c,out}](https://tex.z-dn.net/?f=dT_%7B1%7D%3DT_%7Bh%2Cin%7D-T_%7Bc%2Cout%7D)
![dT_{1}=160-80](https://tex.z-dn.net/?f=dT_%7B1%7D%3D160-80)
![dT_{1}=80\°C](https://tex.z-dn.net/?f=dT_%7B1%7D%3D80%5C%C2%B0C)
![dT_{2}=T_{h,out}-T_{c,in}](https://tex.z-dn.net/?f=dT_%7B2%7D%3DT_%7Bh%2Cout%7D-T_%7Bc%2Cin%7D)
![dT_{2}=124-20](https://tex.z-dn.net/?f=dT_%7B2%7D%3D124-20)
![dT_{2}=104\°C](https://tex.z-dn.net/?f=dT_%7B2%7D%3D104%5C%C2%B0C)
![LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cfrac%7BdT_%7B2%7D-dT_%7B1%7D%7D%7Bln%28%5Cfrac%7BdT_%7B2%7D%7D%7BdT_%7B1%7D%7D%29%7D)
![LMTD = \frac{104-80}{ln(\frac{104}{80})}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cfrac%7B104-80%7D%7Bln%28%5Cfrac%7B104%7D%7B80%7D%29%7D)
![LMTD = \frac{24}{ln(1.3)}](https://tex.z-dn.net/?f=LMTD%20%3D%20%5Cfrac%7B24%7D%7Bln%281.3%29%7D)
![LMTD = 91.48\°C](https://tex.z-dn.net/?f=LMTD%20%3D%2091.48%5C%C2%B0C)
Step 4: Determine required surface area of heat exchanger
![Q=UA_{s}LMTD](https://tex.z-dn.net/?f=Q%3DUA_%7Bs%7DLMTD)
![300.96*10^{3}=649*A_{s}*91.48](https://tex.z-dn.net/?f=300.96%2A10%5E%7B3%7D%3D649%2AA_%7Bs%7D%2A91.48)
![A_{s}=5.07m^{2}](https://tex.z-dn.net/?f=A_%7Bs%7D%3D5.07m%5E%7B2%7D)
Step 5: Determine length of heat exchanger
![A_{s}=piDL](https://tex.z-dn.net/?f=A_%7Bs%7D%3DpiDL)
![5.07=pi*0.015*L](https://tex.z-dn.net/?f=5.07%3Dpi%2A0.015%2AL)
![L=107.57m](https://tex.z-dn.net/?f=L%3D107.57m)
Answer:
Angle grinders are used mostly for copper, iron, steel, lead, and other metals.
Explanation:
I hope it helps! Have a great day!
Lilac~
Answer:
8 for dual-op-amp package, and 14 for quad-op-amp
Explanation;
This is because every op-amp has 2 input terminal 4 pns
So one output terminal that is 2 pins which are required for power
and the same for a minumum number of pins required by quad op amp which is 14