哪位英语加流体力学的大神帮我解几道题目谢谢了.说不清楚加我QQ1192649066就这么多财富值了帮帮忙~ Answer

哪位英语加流体力学的大神帮我解几道题目谢谢了.说不清楚加我QQ1192649066就这么多财富值了帮帮忙~ Answer all parts of this question
a) There are two groups of thermodynamic systems. With the aid of sketches discuss these two systems. [30%]
b) Air enters a gas turbine system with a velocity of 105 m/s and has a specific volume of 0.8m3/kg. The inlet area of the gas turbine is 0.05m2. At exit the air has a velocity of 135 m/s and has a specific volume of 1.5m3/kg. In its passage through the turbine system, the specific enthalpy of the air is reduced by 145kJ/kg and the air also has a heat transfer loss of 27 kJ/kg. Determine:
i) The mass flow rate of the air through the turbine.
ii) The exit area of the turbine
iii) The power developed by the turbine system in kW.
[70%]
2. Answer all parts of this question
a) A turbine operating on a steady flow rate of nitrogen is to produce 0.8kW of power by expanding nitrogen from 300kPa, 350K (inlet specific volume of 0.346 m3/kg) to 120kPa. For preliminary design, the inlet velocity is assumed to be 50m/s and the expansion will be considered as a quasi-equilibrium process such that PV1.40 = constant.
Determine the required flow rate. All assumptions must be clearly stated. [40%]
b) A mass of 1.32 kg of fluid is contained in cylinder at an initial pressure of 12bar. The fluid expands reversibly and moves a piston according to the law PV2= constant, until the volume is increased by 200%. The fluid is then cooled reversibly at constant pressure until the piston moves back to its original position. Heat is then supplied reversibly with the piston firmly locked in position until the pressure rises to the original value of 12 bar. Sketch the process and calculate the net work done by the fluid, for an initial volume of 0.11 m3. [60%]
EE1TF1
3
EE1TF1 Turn Over
3. Answer all parts of this question
a) An unknown mass of ideal gas occupies 370ml at 16oC and 150kPa. What is its volume at -21oC and 420kPa? [10%]
b) A mixture is made up of 3kg of O2, 5kg of N2 and 12kg of CH4. Determine the following: [40%]
删 The mass fraction of each component
删 The mole fraction of each component
删 The average molar mass of the mixture
删 The gas constant of the mixture.
[Molar mass of: O2= 32kg/kmol, N2 = 28kg/kmol, CH4 = 16kg/kmol; Gas constant R = 8.314kJ/kmol. K]
c) A room of volume 75m3 contains air at 100kPa at a relative humidity of 75%. Determine: [35%]
删 The partial pressure of the dry air
删 The specific humidity
删 The enthalpy per unit mass of dry air
[Specific heat of air at room temperature, cp = 1.005kJ/kg oC; Saturated pressure = 3.1698kPa and enthalpy = 2546.5kJ/kg for water at 25oC]
Assume the dry air and water vapour in the room are ideal gases.
d) Outline two (2) limitations of the ideal gas law and explain how they have been overcome. [15%]
女魔头 1年前 已收到1个回答 举报

原本的名字 幼苗

共回答了14个问题采纳率:100% 举报

这不是流体力学,是热力学,解答如下,如有问题,呼我.
a)
1)open system 2) closed system
b)
i)mass flow rate=v*A*rou=105*0.05*(1/0.8)=6.5625kg/s
ii)conseveration of mass:Exit area=mass flow rate/(v*rou)=6.5625/135*1.5=0.073m^2
iii)steady state:
o=q-w+delta entropy+(v1^2-v2^2)/2----->w=-27+145-3.6=114.4KJ/Kg
power=mass flow rate *w=114.4*6.5625=750KW.
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a)Assume:I)NO HEAT LOSS.II)STEADY STATE
since pv^1.4=constant=300000*0.346^1.4=67893.p2=120000pa
power=mass flow rate*w=mass flow rate*integal-vdp=mass flow rate*67893^(1/1.4)*(1-1.4^-1)^-1*-1*(120000^0.2857-300000^0.2857)=800w
solve for mass flow rate=0.0096kg/s
b)1-->2.
v1=0.11/1.32=0.00833.v2=3*v1=0.25 since pv^2=c=p1v1^2=12*10^5*0.0833^2=8333.3,thus,p2=133333pa
w12=integal pdv=-8333.3*(0.25^-1-0.0833^-1)=66.71kJ/kg
2--->3
w23=integal pdv=p2*(v3-v2)=133333*(0.0833-0.25)=-22.23kj/kg
3--->1 no work
therefore, net work=m(w12+w23)=1.32(66.7-22.23)=1.32*44.48=58.7kj
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3
a)
idea gas law:pv=RT,v2=115mL
b)
simple
c)
1)pv=psat*relative humidity=0.75*3.1698kpa=2.377kpa
pa=100-2.377=97.62kpa
2)w=0.6219*(2.377/97.62)=0.015g/g
3)h=(ha*ma+hv*mv)/(mv+mw)
ha=cpT=1.005*(25+273)=299.5KJ/KG(dry air)
h=332.7kj/kg(dry air +vapor)
4) i) The vapor can be condensed when temperature goes down, therefore, the mass of the vapor varies as the temperature.
ii) It is the same that when the temperature rises up, more water maybe absorbed by the air.

1年前

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