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faltersainse [42]
3 years ago
7

prove that the heat transfer at the constant pressure is given by the enthalpy change during the process​

Engineering
1 answer:
mihalych1998 [28]3 years ago
5 0

Answer:

at constant pressure, the heat flow for any process is equal to the change in the internal energy of the system plus the PV work done. Comparing the previous two equations shows that at constant pressure, the change in the enthalpy of a system is equal to the heat flow: ΔH=qp.

Explanation:

Give brainliest

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suter [353]

Answer:

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Explanation:

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6 0
2 years ago
If you are trying to land the plane, which part(s) would you move? How would you move it/them?
KiRa [710]

Answer: hope it helps

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7 0
3 years ago
At steady state, a reversible refrigeration cycle discharges energy at the rate QH to a hot reservoir at temperature TH, while r
ludmilkaskok [199]

Answer:

a) COP_{R} = 25.014, b) T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

Explanation:

a) The coefficient of performance of a reversible refrigeration cycle is:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

Temperatures must be written on absolute scales (Kelvin for SI units, Rankine for Imperial units)

COP_{R} = \frac{275.15\,K}{286.15\,K-275.15\,K}

COP_{R} = 25.014

b) The respective coefficient of performance is determined:

COP_{R} = \frac{Q_{L}}{Q_{H}-Q_{L}}

COP_{R} = \frac{8.75\,kW}{10.5\,kW-8.75\,kW}

COP_{R} = 5

But:

COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}

The temperature at hot reservoir is found with some algebraic help:

COP_{R} \cdot (T_{H}-T_{L})=T_{L}

T_{H}-T_{L} = \frac{T_{L}}{COP_{R}}

T_{H} = T_{L}\cdot \left(1+\frac{1}{COP_{R}}  \right)

T_{H} = 273.15\,K \cdot \left(1+\frac{1}{5}  \right)

T_{H} = 327.78\,K\,(54.63\,^{\textdegree}C)

8 0
3 years ago
Read 2 more answers
A packet weighs 40kg in air but when it is totally submerged into a 1mx1m square tank the weight of the packet is only 18kg. How
Irina18 [472]

Answer:

water  rise = 22 mm

Explanation:

weight of packet IN AIR = 40 *9.81 =392.4 N

weight of packet  IN WATER= 18 *9.81 =176.58 N

by Archimedi's principle

difference in weight = weight of displaced water

w_a - w_w = \rho_w v_d g

392.4 - 176.58 = 1000* v_d* 9.81

v_d = 0.022 m^3

v_d = A*H_rise

0.022 =1*H_rise

H_rise = 0.022 m = 22 mm

water  rise = 22 mm

5 0
3 years ago
Steam enters an adiabatic turbine at 10MPa and 500 C and leaves at 10 kPa with a quality of 90%. Neglecting the changes in kinet
Amanda [17]

Answer:

flow ( m ) = 4.852 kg/s

Explanation:

Given:

- Inlet of Turbine

        P_1 = 10 MPa

        T_1 = 500 C

- Outlet of Turbine

        P_2 = 10 KPa

        x = 0.9

- Power output of Turbine W_out = 5 MW

Find:

Determine the mass ow rate required

Solution:

- Use steam Table A.4 to determine specific enthalpy for inlet conditions:

          P_1 = 10 MPa

          T_1 = 500 C            ---------- > h_1 = 3375.1 KJ/kg

- Use steam Table A.6 to determine specific enthalpy for outlet conditions:

          P_2 = 10 KPa       -------------> h_f = 191.81 KJ/kg

          x = 0.9                  -------------> h_fg = 2392.1 KJ/kg

          h_2 = h_f + x*h_fg

          h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg

- The work produced by the turbine W_out is given by first Law of thermodynamics:

          W_out = flow(m) * ( h_1 - h_2 )

          flow ( m ) = W_out / ( h_1 - h_2 )

- Plug in values:

          flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )

          flow ( m ) = 4.852 kg/s

3 0
3 years ago
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