Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. <u>Assume that the air conditioner operates steadily.</u>
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.
Answer:
88750 N
Explanation:
given data:
plastic deformation σy=266 MPa=266*10^6 N/m^2
cross-sectional area Ao=333 mm^2=333*10^-6 m^2
solution:
To determine the maximum load that can be applied without
plastic deformation (Fy).
Fy=σy*Ao
=88750 N
the function is to provide sealed combustion so that the loss of gas is minimized
Answer:
![\Delta m = 102.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20102.25%5C%2Ckg)
Explanation:
The mass inside the rigid tank before the high pressure stream enters is:
![m_{o} = \rho_{air}\cdot V_{tank}](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%5Crho_%7Bair%7D%5Ccdot%20V_%7Btank%7D)
![m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%20%281.25%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%2825%5C%2Cm%5E%7B3%7D%29)
![m_{o} = 31.25\,kg](https://tex.z-dn.net/?f=m_%7Bo%7D%20%3D%2031.25%5C%2Ckg)
The final mass inside the rigid tank is:
![m_{f} = \rho \cdot V_{tank}](https://tex.z-dn.net/?f=m_%7Bf%7D%20%3D%20%5Crho%20%5Ccdot%20V_%7Btank%7D)
![m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})](https://tex.z-dn.net/?f=m_%7Bf%7D%20%3D%20%285.34%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%2825%5C%2Cm%5E%7B3%7D%29)
![m_{f}= 133.5\,kg](https://tex.z-dn.net/?f=m_%7Bf%7D%3D%20133.5%5C%2Ckg)
The supplied air mass is:
![\Delta m = m_{f}-m_{o}](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20m_%7Bf%7D-m_%7Bo%7D)
![\Delta m = 133.5\,kg-31.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20133.5%5C%2Ckg-31.25%5C%2Ckg)
![\Delta m = 102.25\,kg](https://tex.z-dn.net/?f=%5CDelta%20m%20%3D%20102.25%5C%2Ckg)
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:
![v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2} }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2} }{4}) } =0.729m/s](https://tex.z-dn.net/?f=v_%7Bw%7D%20%3D%5Cfrac%7Bm%7D%7B%5Crho%20%5Cpi%20%28%5Cfrac%7Bd_%7B2%7D%5E%7B2%7D-d_%7B1%7D%5E%7B2%7D%20%20%7D%7B4%7D%20%29%7D%20%3D%5Cfrac%7B0.3%7D%7B998%2A%5Cpi%20%28%5Cfrac%7B0.025%5E%7B2%7D-0.01%5E%7B2%7D%20%20%7D%7B4%7D%29%20%7D%20%3D0.729m%2Fs)
It is necessary to get the Reynold's number:
![Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343](https://tex.z-dn.net/?f=Re%3D%5Cfrac%7Bv_%7Bw%7D%28d_%7B2%7D-d_%7B1%7D%29%20%7D%7Bv%7D%20%3D%5Cfrac%7B0.729%2A%280.025-0.01%29%7D%7B1.004x10%5E%7B-6%7D%20%7D%20%3D10891.4343)
Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:
![Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517](https://tex.z-dn.net/?f=Nu%3D0.023Re%5E%7B0.8%7D%20Pr%5E%7B0.4%7D%20%3D0.023%2A%2810891.4343%29%5E%7B0.8%7D%20%2A%287.01%29%5E%7B0.4%7D%20%3D85.0517)
The overall heat transfer coefficient:
![Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} } }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B1%7D%7Bh_%7B1%7D%20%7D%2B%5Cfrac%7B1%7D%7Bh_%7B2%7D%20%7D%20%20%7D)
Here
![h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C](https://tex.z-dn.net/?f=h_%7B2%7D%20%3D%5Cfrac%7BkNu%7D%7Bd_%7B2%7D-d_%7B1%7D%7D%20%3D%5Cfrac%7B0.598%2A85.0517%7D%7B0.025-0.01%7D%20%3D3390.7278W%2Fm%5E%7B2%7DC)
Substituting values:
![Q=\frac{1}{\frac{1}{4100}+\frac{1}{3390.7278} } =1855.8923W/m^{2} C](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B1%7D%7B%5Cfrac%7B1%7D%7B4100%7D%2B%5Cfrac%7B1%7D%7B3390.7278%7D%20%20%7D%20%3D1855.8923W%2Fm%5E%7B2%7D%20C)