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2 years ago

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A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

ir at 6 bar, 295 K. The valve is opened only as long as required to fill the tank with air to a pressure of 6 bar and a temperature of 350 K. Assuming the ideal gas model for the air, determine the heat transfer between the tank contents and the surroundings, in kJ

1 answer:

bazaltina [42]2 years ago

8 0

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

You might be interested in

A Rankine steam power plant is considered. Saturated water vapor enters a turbine at 8 MPa and exits at condenser at 10 kPa. The

Ray Of Light [21]

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

s1 = s-P1 = 0.6492 KJ/kg.K

v1 = v-P1 = 0.001010 m^3 / kg

P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

s2 = s1 = 0.6492 KJ/kg.K .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

$w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}$

- From the following relation we can determine ( h2 ) as follows:

h2 = h1 + wp

h2 = 191.81 + 8.0699

h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

P3 = 8 MPa

T3 = ? ( assume fluid exist in the saturated vapor phase )

h3 = hg-P3 = 2758.7 KJ/kg

s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

$q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}$

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

P4 = 10 KPa

s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

$x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947$

- Determine the isentropic ( h4s ) at this state as follows:

$h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}$

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

$h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\$

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

$w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}$

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

$W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}$

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

$n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31$

Answer: The thermal efficiency of the cycle is 0.31

7 0

3 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using

kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0

3 years ago

The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet

alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

$T_{h_i}=200^{\circ}C$

$T_{h_o}=120^{\circ}C$

$T_{c_i}=100^{\circ}C$

$T_{c_o}=125^{\circ}C$

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

$m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]$

$\frac{m_hc_{ph}}{m_cc_{pc}}$ = $\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )$

we can see that heat capacity of hot fluid is minimum

Also from energy balance

$Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )$

$NTU=\frac{UA}{\left ( mc_p\right )_{h}}$ = $\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}$

$T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}$

$T_m=41.63^{\circ}C$

NTU=1.921

$And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}$

$\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}$

$\epsilon =\frac{1-0.2669}{1-0.0834}$

$\epsilon =0.799$

5 0

3 years ago

(a) For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary e

aksik [14]

Answer:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

Explanation:

(a) Surface energy is greater than grain boundary energy due to the fact that the bonds of the atoms on the surface are lower than those of the atoms at the grain boundary. The energy is also directly proportional to the number of bonds created.

(b) The energy of a high-angle grain boundary is higher than that of a small-angle grain boundary because the high-angle grain boundary has a higher misalignment and smaller number of bonds than a small-angle grain boundary.

5 0

2 years ago

Three point charges, each with q = 3 nC, are located at the corners of a triangle in the x-y plane, with one corner at the origi

lawyer [7]

Answer:

$\vec F_{A} = -67500\,N\cdot (i + j)$

Explanation:

The position of each point are the following:

$A = (0\,m,0\,m,0\,m), B = (0.02\,m,0\,m,0\,m), C = (0\,m,0.02\,m,0\,m)$

Since the three objects report charges with same sign, then, net force has a repulsive nature. The net force experimented by point charge A is:

$\vec F_{A} = \vec F_{AB} + \vec F_{AC}$

$\vec F_{A} = -\frac{k\cdot q^{2}}{r_{AB}^{2}}\cdot i - \frac{k\cdot q^{2}}{r_{AC}^{2}}\cdot j$

$\vec F_{A} = - \frac{k\cdot q^{2}}{r^{2}} \cdot (i + j)$

$\vec F_{A} = -\frac{(9 \times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} )\cdot (3\times 10^{-9}\,C)}{(0.02\,m)^{2}}\cdot (i + j)$

$\vec F_{A} = -67500\,N\cdot (i + j)$

6 0

3 years ago