course Phy232
Assignment 9Ch. 20
Problems: 38, 44, 45, 48, 51, 52
38.
(A): Qc/Qh=-Tc/Th e=1-[Qc]/[Qh]
0.40=-183/Th
Th= 73.2
the - sign on the right-hand side doesn't belong there; the solution to the equation as you have written it would be Th = -73.2 K.
You haven't included the details of the given information.
(B): Qc/Qh=-Tc/Th -90+273=183
0.40=Tc/183K
Tc= 73.2
44.
(A): What is the Maximum theoretical efficiency of this powerplant?
273.15+27= 300.15K
273.15+6= 279.15K
E=1-(Tc/Th)=Th-Tc/Th
E=300.15-279.15/300.14= 0.070 or 7.0%
(B): Produce 210kW of power at what rat must heat be extracted?
Work Done/e
210kw/0.070= 3000kW
Cold Water absorbed = 3000-210= 2790kW
45:
1=>2: T1T1^y-1=T2T2^y-1 T1V1^y-1=T2V2^y-1 T1=T2(V2/V1)^y-1
2=>3: T2T2^y-1=T3T3^y-1 T2V2^y-1=T3V3^y-1 T2=T3(V3/V2)^y-1
3=>4: T3T3^y-1=T4T4^y-1 T3V3^y-1=T4V4^y-1 T3=T4(V4/V3)^y-1
4=>1: T4T4^y-1=T1T1^y-1 T4V4^y-1=T1V1^y-1 T4=T1(V1/V4)^y-1
E=1-(1)/(V1/V2)^y-1
E=1-(1)/(V2/V3)^y-1
E=1-(1)/(V3 /V4)^y-1
E=1-(1)/(V4/V1)^y-1
I need to know what information was given and what the conditions of the problem were.
48: Monatomic ideal gas taken around the cycle. Path c=>a is a straight line.
(a):
Va=nRTh/Pa Vb=nRth/Pa
Not sure what to do with this problem
© efficiency=W/Qh
the answer depends on which paths are vertical, which are horizontal, etc..
On a vertical path `dV = 0 so no work is done. All energy going into the system therefore increases its internal energy. For a monatomic gas the change in interval energy is 3/2 n R `dT.
On a horizontal path the increase in interval energy is also 3/2 n R `dT, but the amount of energy going into the system is greater than this by n R `dT, which is the energy required for the expansion.
Hopefully this will help; insert additional details into a copy of this document and I'll be glad to provide additional feedback.
51:
(A):
If you do not have P then you will not have electricity.
(B):
P=Kd^2V^3
3.2=(0.5)(155)v^3
3.2=77.5v^3
0.041=v^3
This looks like it relates power, diameter and wind velocity. Assuming I'm right, these quantities should all have units. Assuming P is power then P might have units of watts, d units of meters, v units of meters / second. So k will have units of P / (d^2 v^3).
Without the units these results cannot be interpreted.
(C ):
Because mountain passes are the high winded areas. Wind turbines convert wind power to mechanical rotation velocity which produces renewable energy.
you probably need to specify more information about this problem
52:
(A) At 105km/h what is the rate of gas combustion in L/h?
Gc=740/104= 7.05
you don't say what 740 and 104 represent; if units were included it might be clearer, but you might also need to simply state the given quantities
(B) What is the theoretical efficiency of the engine? Y=1.40
E=1-(1/r^y-1)
E=1-(1/(8.5)^1.40-1
E=1-(1/2.4)
E=0.58
(C) How much power is the engine producing at 105km/h?
Cop=[Qh]/W
236=[Qh]/105
=2.48x10^4
I can tell about half of what you're doing with these problems, so my notes might be helfpul.
However I you to include more information about the problems, per my notes on your earlier submissions.
See my notes.