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lord [1]
3 years ago
13

Someone please helpppp!

Physics
1 answer:
Vlad1618 [11]3 years ago
8 0
I believe it’s A, i could be wrong tho 3
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Air is compressed from 14.7 psia and 60°F to a pressure of 150 psia while being cooled at a rate of 10 Btu/lbm by circulating wa
iren2701 [21]

Answer:

A. 6.36 lbm/s

b. T_2=341\textdegree F

Explanation:

a. Given the following information;

#Compressor inlet:

Air pressure,P_1=14.7psia,T_1=60\textdegree F

#Compressor outlet:

Air \ Pressure, P_2=150psia\\\\Air \ volume \ flow \ rate, \dot V_1=5000ft^3/min

#Cooling rate,

q_{out}=10Btu/lbm, \dot W=700hp

# From table A-1E

Gas constant of air R=0.3704\ psia.ft^3/lbm.R

Specific enthalpy at P_1=520R-h_1=124.27Btu/lbm

Using the mass balance:

\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{0.3704\times520}{14.7}\\\\\\v_1=13.1026ft^3/lbm\\\\\dot m=\frac{5000/60}{13.1025}\\\\\dot m=6.36\ lbm/s

Hence, the mass flow rate of the air is 6.36lbm/s

b The specific enthalpy at the exit is defined as the energy balance on the system:

\dot m_{in}-\dot m_{out}=\bigtriangleup \dot m_{sys}=0.0\\\\\dot m_{in}=\dot m_{out}\\\\\dot W_{in}+ \dot mh_1=\dot Q_{out}+\dot m_2\\\\h_2=\frac{\dot W_{in}-\dot Q_{out}}{\dot m}+h_1\\\\h_2=\frac{700\times 0.7068-10\times 6.36}{6.36}+124.27\\\\h_2=192.06\ Btu/lbm\\\\#at \ h_2=182.06Btu/lbm, T_2=801R=341\textdegree F

Hence, the temperature at the compressor exit T_2=341\textdegree F

5 0
3 years ago
What is the acceleration of a rocket that speeds up from 50 m/s to 1000 m/s in 3 seconds?
tresset_1 [31]

Answer:

95 m/s^2

Explanation:

Givens

vi = 50 m/s

vf = 1000 m/s

t = 10 sec

a = ?

Equation

a = (vf - vi)/t

Solution

a = (1000 - 50)/10

a = 950/10

a = 95 m/s^2

6 0
3 years ago
Convert 0.0779 kg to g
Mariana [72]
I think it would be 77.9 grams
5 0
3 years ago
If the solar system shrank so that the sun were located just one centimeter from Earth, about how far away could you find the ne
DedPeter [7]
E                  S                                                               *

The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).

The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years

Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.

To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.

As a result, you get a mere 268335.7 centimeters.

To put that into perspective, that's only about 1.7 miles

A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
8 0
3 years ago
An eagle flying at 35 m/s emits a cry whose frequency is 440 Hz. A blackbird is moving in the same direction as the eagle at 10
Akimi4 [234]

Answer:

a)  F=475.7Hz

b)  F'=410.899Hz

Explanation:

From the question we are told that:

Velocity of eagle V_1=35m/s

Frequency of eagle F_1=440Hz

Velocity of Black bird V_2=10m/s

Speed of sound s=343m/s

a)

Generally the equation for Frequency is mathematically given by

 F=f_0(\frac{v-v_2}{v-v_1})

 F=440(\frac{343-10}{343-35})

 F=475.7Hz

b)

Generally the equation for Frequency is mathematically given by

 F'=f_0(\frac{v+v_2}{v+v_1})

 F'=440(\frac{343+10}{343+35})

 F'=410.899Hz

7 0
3 years ago
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