Ball Throws

A heat pump operates on a Carnot heat pump cycle with a COP of 12.5. It keeps a space at 24°C by consuming 2.15 kW of power. Det

Vinil7 [7]

Answer:

a) $T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$ , b) $\dot Q_{H} = 26.875\,kW$

Explanation:

a) The Coefficient of Performance of the Carnot Heat Pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

After some algebraic handling, the temperature of the cold reservoir is determined:

$T_{H}-T_{L} = \frac{T_{H}}{COP_{HP}}$

$T_{L} = T_{H}\cdot \left(1-\frac{1}{COP_{HP}} \right)$

$T_{L} = (297.15\,K)\cdot \left(1-\frac{1}{12.5}\right)$

$T_{L} = 273.378\,K\,(0.228\,^{\textdegree}C)$

b) The heating load provided by the heat pump is:

$\dot Q_{H} = COP_{HP}\cdot \dot W$

$\dot Q_{H} = (12.5)\cdot (2.15\,kW)$

$\dot Q_{H} = 26.875\,kW$

4 0

3 years ago

An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no

Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus $s_{1} =s_{2}$

From the refrigerant table A11-A13

$P_{1} =0.8MPa$ $\left \{ {{ {{v_{1}=v_{g} @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g} @0.8MPa =246.82 kJ/kg } - also {{s_{1}=s_{g} @0.8MPa =0.91853 kJ/kgK } } \right.$

sat vapor

m= $\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339} = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) = 38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}$

$T_{2} =T_{sat @ 0.2MPa} = -10.09^{o} C$

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

$E_{in} - E_{out} =$ ΔE

$w_{b, out} =$ ΔU

$w_{b, out} =m([tex]u_{1} -u_{2$ )

$w_{b, out} =$ workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0

3 years ago

A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma

MArishka [77]

Answer:

5320.6 Pascal

Explanation:

Manometer is a pressure measuring device use to measure gas pressure .

Pressure difference in Manometer is a function of density,gravity and the height difference of the liquid.

Pressure difference = density x acceleration due to gravity x difference in height of liquid

Density of liquid = specific gravity of object x density of water.

Density of water = 997 kg/m^3

Specific gravity of liquid = 1.7

Density of liquid = 997 x 1.7 =1694.9kg/m^3

g= 9.81 m/s^2

h =320mm = 0.320m

Pressure difference = 1694.9 x 9.81 x 0.320 = 5320.6 Pascal

3 0

3 years ago

Read 2 more answers

Gaining unauthorized access to a computer's data is called (5 points)

yanalaym [24]

Hacking is correcttttttttt

6 0

3 years ago

The importance of reading a circuit diagram to interpret a wiring diagram?

Nataly [62]

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation:

7 0

2 years ago