Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality = 
exit quality = 
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Answer:You are correct, no need to change.
Explanation:
Answer:
809.98°C
Explanation:
STEP ONE: The first step to take in order to solve this particular Question or problem is to find or determine the Biot value.
Biot value = (heat transfer coefficient × length) ÷ thermal conductivity.
Biot value = (220 × 0.1)÷ 110 = 0.2.
Biot value = 0.2.
STEP TWO: Determine the Fourier number. Since the Biot value is greater than 0.1. Tis can be done by making use of the formula below;
Fourier number = thermal diffusivity × time ÷ (length)^2.
Fourier number = (3 × 60 × 33.9 × 10^-6)/( 0.1)^2 = 0.6102.
STEP THREE: This is the last step for the question, here we will be calculating the temperature of the center plane of the brass plate after 3 minutes.
Thus, the temperature of the center plane of the brass plane after 3 minutes = (1.00705) (0.89199) (900- 15) + 15.
= > the temperature of the center plane of the brass plane after 3 minutes = 809.98°C.
Answer:
7.615 kW
Explanation:
Solution in pen paper form in the attachment section
Answer:
Rate of heat transfer is 0.56592 kg/hour
Explanation:
Q = kA(T2 - T1)/t
Q is rate of heat transfer in Watts or Joules per second
k is thermal conductivity of the styrofoam = 0.035 W/(mK)
A is area of the cubical picnic chest = 6L^2 = 6(0.5)^2 = 6×0.25 = 1.5 m^2
T1 is initial temperature of ice = 0 °C = 0+273 = 273 K
T2 is temperature of the styrofoam = 25 °C = 25+273 = 298 K
t is thickness of styrofoam = 0.025 m
Q = 0.035×1.5(298-273)/0.025 = 1.3125/0.025 = 52.5 W = 52.5 J/s
Mass flow rate = rate of heat transfer ÷ latent heat of melting of ice = 52.5 J/s ÷ 3.34×10^ 5 J/kg = 1.572×10^-4 kg/s = 1.572×10^-4 kg/s × 3600 s/1 hr = 0.56592 kg/hr
