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lions [1.4K]
2 years ago
13

8.28 Water is the working fluid in an ideal Rankine cycle with superheat and reheat. Steam enters the first-stage turbine at 140

0 lbf/in.2 and 10008F, expands to a pressure of 350 lbf/in.2, and is reheated to 9008F before entering the secondstage turbine. The condenser pressure is 2 lbf/in.2 The net power output of the cycle is 1 3 109 Btu/h. Determine for the cycle (a) the mass flow rate of steam, in lb/h. (b) the rate of heat transfer, in Btu/h, to the working fluid passing through the steam generator. (c) the rate of heat transfer, in Btu/h, to the working fluid passing through the reheater. (d) the thermal efficiency.
Engineering
1 answer:
zubka84 [21]2 years ago
3 0

Answer:

Betbtybrbytntrnyrnrynunjhjhnthnnhtnnthnhtnnhnhrnntnthhnhnhtnthn

Explanation:

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svlad2 [7]

Answer:

multitenancy is the term.

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2 years ago
Need help solving math problem using integration
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Ummm did you try to add or subtract and multiply or divide that can get your answer
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2 years ago
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical s
monitta

Answer:

a. \dfrac{D_{1}}{ D_{2}}  =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n} which is constant therefore, n = constant

b. The temperature at the end of the process is 109.6°C

c. The work done by the balloon boundaries = 10.81 MJ

The work done on the surrounding atmospheric air = 10.6 MJ

Explanation:

p₁ = 100 kPa

T₁ = 27°C

D₁ = 10 m

v₂ = 1.2 × v₁

p ∝ α·D

α = Constant

v_1 = \dfrac{4}{3} \times  \pi \times r^3

\therefore v_1 = \dfrac{4}{3} \times  \pi \times  \left (\dfrac{10}{2}  \right )^3 = 523.6 \ m^3

v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³

Therefore, D₂ = 10.63 m

We check the following relation for a polytropic process;

\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} = \left (\dfrac{T_{1}}{T_{2}}   \right )^{\dfrac{n}{n-1}}

We have;

\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_2}{2}  \right )^3}{\dfrac{4}{3} \times  \pi \times  \left (\dfrac{D_1}{2}  \right )^3}   \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{   \left{D_2}  }{ {D_1}}   \right )^{3\times n} =  \left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )^{-3\times n}

\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2  \right )^{n} = \left (\dfrac{   \left{D_2}  ^3}{ {D_1}^3}   \right )^{n}

log  \left (\dfrac{D_{1}}{ D_{2}}\right )  =  -3\times n \times log\left (\dfrac{   \left{D_1}  }{ {D_2}}   \right )

n = -1/3

Therefore, the relation, pVⁿ = Constant

b. The temperature T₂ is found as follows;

\left (\dfrac{628.32 }{523.6}   \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}}   \right )^{\dfrac{1}{4}}

T₂ = 300.15/0.784 = 382.75 K = 109.6°C

c. W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}

p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}}   \right )^{n} } =  \dfrac{100\times 10^3}{ \left (1.2) \right  ^{-\dfrac{1}{3} } }

p₂ =  100000/0.941 = 106.265 kPa

W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3  \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J

The work done by the balloon boundaries = 10.81 MJ

Work done against atmospheric pressure, Pₐ, is given by the relation;

Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J

The work done on the surrounding atmospheric air = 10.6 MJ

4 0
3 years ago
Write multiple if statements. If car_year is 1969 or earlier, print "Few safety features." If 1970 or later, print "Probably has
Slav-nsk [51]

Answer:

Explanation along with code and output results is provided below.

C++ Code:

#include <iostream>

using namespace std;

int main()

{

   int year;

   cout<<"Enter the car model year."<<endl;

   cin>>year;    

  if (year<=1969)

  {

cout<<"Few safety features."<<endl;

  }

else if (year>=1970 && year<1989)

{

cout<<"Probably has seat belts."<<endl;

}

else if (year>=1990 && year<1999)

{

cout<<"Probably has antilock brakes."<<endl;

}

else if (year>=2000)

{

cout<<"Probably has airbags."<<endl;

   }

   return 0;

}

Explanation:

The problem was to print feature messages of a car given its model year.

If else conditions are being used incorporate the logic. The code has been tested with several inputs and got correct output results.

Output:

Enter the car model year.

1961

Few safety features.

Enter the car model year.

1975

Probably has seat belts.

Enter the car model year.

1994

Probably has antilock brakes.

Enter the car model year.

2005

Probably has airbags.

5 0
3 years ago
Helpppp EDO segundo orden
jeka94
Can’t really see it it’s blurry

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3 years ago
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