Answer:
D) 1.04 Btu/s from the liquid to the surroundings.
Explanation:
Given that:
flow rate (m) = 2 lb/s
liquid specific enthalpy at the inlet (
Btu/lb)
liquid specific enthalpy at the exit (
Btu/lb)
initial elevation (
)
final elevation (
)
acceleration due to gravity (g) = 32.174 ft/s²
= 3 Btu/s
The energy balance equation is given as:
![Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2B%28%5Cfrac%7BV_1%5E2-V_2%5E2%7D%7B2%7D%29%2Bg%28z_1-z_2%29%5D%3D0)
Since kinetic energy effects are negligible, the equation becomes:
![Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0](https://tex.z-dn.net/?f=Q_%7Bcv%7D-W%7Bcv%7D%2Bm%5B%28h_1-h_2%29%2Bg%28z_1-z_2%29%5D%3D0)
Substituting values:
![Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\](https://tex.z-dn.net/?f=Q_%7Bcv%7D-%28-3%29%2B2%5B%2840.09-40.94%29%2B%5Cfrac%7B32.174%280-100%29%7D%7B778%2A32.174%7D%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%5B-0.85-0.1285%20%5D%3D0%5C%5CQ_%7Bcv%7D%2B3%2B2%28-0.9785%29%3D0%5C%5CQ_%7Bcv%7D%2B3-1.957%3D0%5C%5CQ_%7Bcv%7D%2B1.04%3D0%5C%5CQ_%7Bcv%7D%3D-1.04%5C%5C)
The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.
Answer: heat flux into the fun is 21.714 mW/m^2
Explanation:
Heat flux Q = q/A
q = heat transfer rate W
A = area m^2
q = area * conductivity * temperature gradient
Temperature gradient = difference in temperature of the metal faces divided by the thickness.
Therefore Q = k * ( temp. gradient)
Q = 200 * ((400-20)/3.5*10^-2)
Q = 21714285.71 = 21.714 mW/m^2
Answer 2: convective heat transfer flux between fin and air
is 3800W/m^2
Explanation :
q = hA*(Ts-Ta)
h = convective heat transfer coefficient
Ts = temperature of fin
Ta = temperature of air
Q = q/A
Q = h(Ts-Ta)
Q = 10(400 - 20)
Q = 3800 W/m^2
answer:
Energy' has the potential to:<u>do work</u>
b.do work
Answer:
a) 600°F days
b) 103°F days
Explanation:
Given data :
overall heat loss coefficient = 400 Btu/h.oF
Furnace efficiency = 0.78
applying the degree- day method the heating energy requirement per day can be obtained by subtracting the average temperature for a particular day from the acceptable mid/balance point temperature
lets assume the mid/balance point temperature = 65°F
a) heating energy requirement in Denver Colorado
we will have 600°F days ; considering the the average temperature - balance point temperature
= 665 - 65 = 600°F
b) heating energy requirement in Little Rock Arkansas
we will have 103° days
