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nalin [4]
3 years ago
6

A carbon resistor has a resistance of 976 ohms at 0 degrees C. Determine its resistance at 89 degrees C​

Engineering
1 answer:
nignag [31]3 years ago
5 0

Answer:

1028.1184 Ohms

Explanation:

<u>Given the following data;</u>

  • Initial resistance, Ro = 976 Ohms
  • Initial temperature, T1 = 0°C
  • Final temperature, T2 = 89°C

Assuming the temperature coefficient of resistance for carbon at 0°C is equal to 0.0006 per degree Celsius.

To find determine its new resistance, we would use the mathematical expression for linear resistivity;

R_{89} = R_{0} + R_{0}(\alpha T)

Substituting into the equation, we have;

R_{89} = 976 + 976*(0.0006*89)

R_{89} = 976 + 976*(0.0534)

R_{89} = 976 + 52.1184

R_{89} = 1028.1184 \ Ohms

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Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore\theta angle = 0

and radial component of given velocity is zero

we haveh = r_o v_r_o = 6378+600 =6.97*10^6 m

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)

GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s

so

\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)

solvingt for \epsilon)

\epsilon = 0.22)

therefore eccentrcity of orbit is 0.22

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3 years ago
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Explanation:

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What camera battery does the canon 250d use?
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Answer:

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Read 2 more answers
Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.
Rufina [12.5K]

Answer:

a) \dot m = 16.168\,\frac{kg}{s}, b) v_{out} = 680.590\,\frac{m}{s}, c) \dot W_{out} = 18276.307\,kW

Explanation:

A turbine is a steady-state devices which transforms fluid energy into mechanical energy and is modelled after the Principle of Mass Conservation and First Law of Thermodynamics, whose expressions are described hereafter:

Mass Balance

\frac{v_{in}\cdot A_{in}}{\nu_{in}} - \frac{v_{out}\cdot A_{out}}{\nu_{out}} = 0

Energy Balance

-q_{loss} - w_{out} + h_{in} - h_{out} = 0

Specific volumes and enthalpies are obtained from property tables for steam:

Inlet (Superheated Steam)

\nu_{in} = 0.055665\,\frac{m^{3}}{kg}

h_{in} = 3650.6\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mix)

\nu_{out} = 5.89328\,\frac{m^{3}}{kg}

h_{out} = 2500.2\,\frac{kJ}{kg}

a) The mass flow rate of the steam is:

\dot m = \frac{v_{in}\cdot A_{in}}{\nu_{in}}

\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }

\dot m = 16.168\,\frac{kg}{s}

b) The exit velocity of steam is:

\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}

v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}

v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}

v_{out} = 680.590\,\frac{m}{s}

c) The power output of the steam turbine is:

\dot W_{out} = \dot m \cdot (-q_{loss} + h_{in}-h_{out})

\dot W_{out} = \left(16.168\,\frac{kg}{s} \right)\cdot \left(-20\,\frac{kJ}{kg} + 3650.6\,\frac{kJ}{kg} - 2500.2\,\frac{kJ}{kg}\right)

\dot W_{out} = 18276.307\,kW

6 0
3 years ago
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
3 years ago
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