Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N
Answer:
V = 6.33 m/s
Explanation:
Given:
- The length of the wire L = 0.02 m
- The diameter of the wire D = 0.0005 m
- The calibration expression V = 0.0000625*h^2
- Environment temperature T_inf = 298 K
- Surface temperature T_s = 348 K
- The voltage drop dV = 5 V
- The electric current I = 0.1 A
Find:
- the velocity of Air
Solution:
- Calculate the surface area of the wire:
A = pi*D*L
A = pi*(0.0005)*(0.02) = 0.00003142 m^2
- The rate of energy in the wire P:
P = I*dV = 0.1*5 = 0.5 W
- Apply Newton's Law of Cooling:
P = h*A*(T_s - T_inf)
h = P /A*(T_s - T_inf)
Plug in the values:
h= 0.5/ 0.00003142*(348 - 298)
h = 318.27 W /m^2K
- Using the calibration relationship given, compute the velocity of air:
V = 6.25*10^-5 * h^2
V = 6.25*10^-5 * (318.27)^2
V = 6.33 m/s
