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2 years ago

15

Refer to Figure 9-18. A #_____ electrode lead and workpiece lead should be used to carry 150 amperes of electricity 75 feet to t

he workpiece and 75 feet back to the welding machine.

1 answer:

Anni [7]2 years ago

8 0

Answer:

AC or

Explanation:

BC

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A counter-flow double pipe heat exchanger is heat heat water from 20 degrees Celsius to 80 degrees Celsius at the rate of 1.2 kg

lakkis [162]

Answer:

$L=107.6m$

Explanation:

Cold water in: $m_{c}=1.2kg/s, C_{c}=4.18kJ/kg\°C, T_{c,in}=20\°C, T_{c,out}=80\°C$

Hot water in: $m_{h}=2kg/s, C_{h}=4.18kJ/kg\°C, T_{h,in}=160\°C, T_{h,out}=?\°C$

$D=1.5cm=0.015m, U=649W/m^{2}K, LMTD=?\°C, A_{s}=?m^{2},L=?m$

Step 1: Determine the rate of heat transfer in the heat exchanger

$Q=m_{c}C_{c}(T_{c,out}-T_{c,in})$

$Q=1.2*4.18*(80-20)$

$Q=1.2*4.18*(80-20)$

$Q=300.96kW$

Step 2: Determine outlet temperature of hot water

$Q=m_{h}C_{h}(T_{h,in}-T_{h,out})$

$300.96=2*4.18*(160-T_{h,out})$

$T_{h,out}=124\°C$

Step 3: Determine the Logarithmic Mean Temperature Difference (LMTD)

$dT_{1}=T_{h,in}-T_{c,out}$

$dT_{1}=160-80$

$dT_{1}=80\°C$

$dT_{2}=T_{h,out}-T_{c,in}$

$dT_{2}=124-20$

$dT_{2}=104\°C$

$LMTD = \frac{dT_{2}-dT_{1}}{ln(\frac{dT_{2}}{dT_{1}})}$

$LMTD = \frac{104-80}{ln(\frac{104}{80})}$

$LMTD = \frac{24}{ln(1.3)}$

$LMTD = 91.48\°C$

Step 4: Determine required surface area of heat exchanger

$Q=UA_{s}LMTD$

$300.96*10^{3}=649*A_{s}*91.48$

$A_{s}=5.07m^{2}$

Step 5: Determine length of heat exchanger

$A_{s}=piDL$

$5.07=pi*0.015*L$

$L=107.57m$

7 0

2 years ago

Hęłłõ hõw årę ÿõū dõìńg tõdāÿ

kirill [66]

Answer:

All good :)

What about you??

8 0

2 years ago

Read 2 more answers

What two elements make up fuel oil

nordsb [41]

Carbon and hydrogen!!

4 0

2 years ago

An automotive fuel cell consumes fuel at a rate of 28m3/h and delivers 80kW of power to the wheels. If the hydrogen fuel has a h

EastWind [94]

Answer:

The efficiency of this fuel cell is 80.69 percent.

Explanation:

From Physics we define the efficiency of the automotive fuel cell ( $\eta$ ), dimensionless, as:

$\eta = \frac{\dot W_{out}}{\dot W_{in}}$ (Eq. 1)

Where:

$\dot W_{in}$ - Maximum power possible from hydrogen flow, measured in kilowatts.

$\dot W_{out}$ - Output power of the automotive fuel cell, measured in kilowatts.

The maximum power possible from hydrogen flow is:

$\dot W_{in} = \dot V\cdot \rho \cdot L_{c}$ (Eq. 2)

Where:

$\dot V$ - Volume flow rate, measured in cubic meters per second.

$\rho$ - Density of hydrogen, measured in kilograms per cubic meter.

$L_{c}$ - Heating value of hydrogen, measured in kilojoules per kilogram.

If we know that $\dot V = \frac{28}{3600}\,\frac{m^{3}}{s}$ , $\rho = 0.0899\,\frac{kg}{m^{3}}$ , $L_{c} = 141790\,\frac{kJ}{kg}$ and $\dot W_{out} = 80\,kW$ , then the efficiency of this fuel cell is:

(Eq. 1)

$\dot W_{in} = \left(\frac{28}{3600}\,\frac{m^{3}}{s}\right)\cdot \left(0.0899\,\frac{kg}{m^{3}} \right)\cdot \left(141790\,\frac{kJ}{kg} \right)$

$\dot W_{in} = 99.143\,kW$

(Eq. 2)

$\eta = \frac{80\,kW}{99.143\,kW}$

$\eta = 0.807$

The efficiency of this fuel cell is 80.69 percent.

3 0

2 years ago

The constant-pressure specific heat of air at 25°C is 1.005 kJ/kg·°C. Express this value in kJ/kg·K, J/g·°C, kcal/ kg·°C, and Bt

Dmitry [639]

Answer:

a) C_v = 1.005 KJ/kgK

b) C_v = 1005.000 J/kgC

c) C_v = 0.240 kcal/kgC

d) C_v = 0.240 Btu/lbmF

Explanation:

Given:

- constant-pressure specific heat C_v = 1.005 KJ/kgC

Find C_v in units of:

a) kJ/kg·K

b) J/g·°C

c) kcal/ kg·°C

d) Btu/lbm·°F

Solution:

a) C_v is Specific heat capacity is the quantity of heat needed to raise the temperature per unit mass. Usually, it's the heat in Joules needed to raise the temperature of 1 gram of sample 1 Kelvin or 1 degree Celsius. Hence,

C_v = 1.005 KJ/kgK

b)

C_v = 1.005 KJ/kgC * ( 1000 J / KJ)

C_v = 1005.000 J/kgC

c)

C_v = 1.005 KJ/kgC * ( 0.239006 kcal / KJ)

C_v = 0.240 kcal/kgC

d)

C_v = 1.005 KJ/kgC * ( 0.947817 Btu / KJ) * ( kg / 2.205 lbm)*(Δ1 C / Δ1.8 F)

C_v = 0.240 Btu/lbmF

6 0

2 years ago