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Tasya [4]
3 years ago
9

Identifying Mechanical Energy

Physics
2 answers:
sergij07 [2.7K]3 years ago
6 0
Ummmm ok so I don’t really know
arsen [322]3 years ago
3 0
Bruh ima be honest with you idek
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A student wants to draw a model of an atom. Which statement describes how to find the number of neutrons to include in the model
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1 year ago
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

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